Class 10

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 3 Circle.

Practice Set 3.2 Geometry 10th Std Maths Part 2 Answers Chapter 3 Circle

Question 1.
Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.
Solution:
Let the two circles having centres P and Q touch each other internally at point R.
Here, QR = 3.5 cm, PR = 4.8 cm

The two circles touch each other internally.
∴ By theorem of touching circles,
P – Q – R
PQ = PR – QR
= 4.8 – 3.5
= 1.3 cm
[The distance between the centres of circles touching internally is equal to the difference in their radii]

Question 2.
Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.
Solution:
Let the two circles having centres P and R touch each other externally at point Q.
Here, PQ = 5.5 cm, QR = 4.2 cm

The two circles touch each other externally.
∴ By theorem of touching circles,
P – Q – R
PR = PQ + QR
= 5.5 + 4.2
= 9.7 cm
[The distance between the centres of the circles touching externally is equal to the sum of their radii]

Question 3.
If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other
i. externally
ii. internally.
Solution:
i. Circles touching externally:

ii. Circles touching internally:

Question 4.
In the adjoining figure, the circles with centres P and Q touch each other at R A line passing through R meets the circles at A and B respectively. Prove that –

i. seg AP || seg BQ,
ii. ∆APR ~ ∆RQB, and
iii. Find ∠RQB if ∠PAR = 35°.
Solution:
The circles with centres P and Q touch each other at R.
∴ By theorem of touching circles,
P – R – Q
i. In ∆PAR,
seg PA = seg PR [Radii of the same circle]
∴ ∠PRA ≅ ∠PAR (i) [Isosceles triangle theorem]
Similarly, in ∆QBR,
seg QR = seg QB [Radii of the same circle]
∴ ∠RBQ ≅ ∠QRB (ii) [Isosceles triangle theorem]
But, ∠PRA ≅ ∠QRB (iii) [Vertically opposite angles]
∴ ∠PAR ≅ ∠RBQ (iv) [From (i) and (ii)]
But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ.
∴ seg AP || seg BQ [Alternate angles test]
ii. In ∆APR and ∆RQB,
∠PAR ≅ ∠QRB [From (i) and (iii)]
∠APR ≅ ∠RQB [Alternate angles]
∴ ∆APR – ∆RQB [AA test of similarity]
iii. ∠PAR = 35° [Given]
∴ ∠RBQ = ∠PAR= 35° [From (iv)]
In ∆RQB,
∠RQB + ∠RBQ + ∠QRB = 180° [Sum of the measures of angles of a triangle is 180°]
∴ ∠RQB + ∠RBQ + ∠RBQ = 180° [From (ii)]
∴ ∠RQB + 2 ∠RBQ = 180°
∴ ∠RQB + 2 × 35° = 180°
∴ ∠RQB + 70° = 180°
∴ ∠RQB = 110°

Question 5.
In the adjoining figure, the circles with centres A and B touch each other at E. Line l is a common tangent which touches the circles at C and D respectively. Find the length of seg CD if the radii of the circles are 4 cm, 6 cm.

Construction : Draw seg AF ⊥ seg BD.
Solution:
i. The circles with centres A and B touch each other at E. [Given]
∴ By theorem of touching circles,
A – E – B

∴ ∠ACD = ∠BDC = 90° [Tangent theorem]
∠AFD = 90° [Construction]
∴ ∠CAF = 90° [Remaining angle of ꠸AFDC]
∴ ꠸AFDC is a rectangle. [Each angle is of measure 900]
∴ AC = DF = 4 cm [Opposite sides of a rectangle]
Now, BD = BF + DF [B – F – C]
∴ 6 = BF + 4 BF = 2 cm
Also, AB = AE + EB
= 4 + 6 = 10 cm
[The distance between the centres of circles touching externally is equal to the sum of their radii]

ii. Now, in ∆AFB, ∠AFB = 90° [Construction]
∴ AB² = AF² + BF² [Pythagoras theorem]
∴ 10² = AF² + 2²
∴ 100 = AF² + 4
∴ AF² = 96
∴ AF = \(\sqrt { 96 }\) [Taking square root of both sides]
= \(\sqrt{16 \times 6}\)
= 4 \(\sqrt { 6 }\) cm
But, CD = AF [Opposite sides of a rectangle]
∴ CD = 4 \(\sqrt { 6 }\) cm

Question 1.
Take three collinear points X – Y – Z as shown in figure.
Draw a circle with centre X and radius XY. Draw another circle with centre Z and radius YZ.
Note that both the circles intersect each other at the single point Y. Draw a line / through point Y and perpendicular to seg XZ. What is line l (Textbook pg. no. 56)
Solution:

Line l is a common tangent of the two circles.

Question 2.
Take points Y – X – Z as shown in the figure. Draw a circle with centre Z and radius ZY.
Also draw a circle with centre X and radius XY. Note that both the circles intersect each other at the point Y.
Draw a line l perpendicular to seg YZ through point Y. What is line l? (Textbook pg. no. 56)
Solution:

Line l is a common tangent of the two circles.

If two circles in the same plane intersect with a line in the plane in only one point, they are said to be touching circles and the line is their common tangent.

The point common to the circles and the line is called their common point of contact.

1. Circles touching externally:
For circles touching externally, the distance between their centres is equal to sum of their radii, i.e. AB = AC + BC

2. Circles touching internally:
For circles touching internally, the distance between their centres is equal to difference of their radii,
i. e. AB = AC – BC

Question 3.
The circles shown in the given figure are called externally touching circles. Why? (iexthook pg. no. 57)

Answer:
Circles with centres R and S lie in the same plane and intersect with a line l in the plane in one and only one point T [R – T – S].
Hence the given circles are externally touching circles.

Question 4.
The circles shown in the given figure are called internally touching circles, why? (Textbook pg. no. 57)

Answer:
Circles with centres N and M lie in the same plane and intersect with a line p in the plane in one and only one point T [K – N – M].
Hence, the given circles are internally touching circles.

Question 5.
In the given figure, the radii of the circles with centres A and B are 3 cm and 4 cm respectively. Find
i. d(A,B) in figure (a)
ii. d(A,B) in figure (b) (Textbook pg. no. 57)

Solution:
i. Here, circle with centres A and B touch each other externally at point C.
∴ d(A, B) = d(A, C) + d(B ,C)
= 3 + 4
∴ d(A,B) = 7 cm
[The distance between the centres of circles touching externally is equal to the sum of their radii]
ii. Here, circle with centres A and 13 touch each other internally at point C.
∴ d(A, B) = d(A, C) – d(B, C)
= 4 – 3
∴ d(A,B) = 1 cm
[The distance between the centres of circles touching internally is equal to the difference in their radii]

Maharashtra Board Class 10 Maths Solutions

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 3 Circle.

Practice Set 3.1 Geometry 10th Std Maths Part 2 Answers Chapter 3 Circle

Question 1.
In the adjoining figure, the radius of a circle with centre C is 6 cm, line AB is a tangent at A. Answer the following questions.
i. What is the measure of ∠CAB? Why?
ii. What is the distance of point C from line AB? Why?
iii. d(A, B) = 6 cm, find d(B, C).
iv. What is the measure of ∠ABC? Why?

Solution:
i. line AB is the tangent to the circle with centre C and radius AC. [Given]
∴ ∠CAB = 90° (i) [Tangent theorem]
ii. seg CA ⊥ line AB [From (i)]
radius = l(AC) = 6 cm
∴ The distance of point C from line AB is 6 cm.
iii. In ∆CAB, ∠CAB = 90° [From (i)]
∴ BC² = AB² + AC² . [Pythagoras theorem]

= 6² + 6²
= 2 × 6²
∴ BC = \(\sqrt{2 \times 6^{2}}\) [Taking square root of both sides]
= 6 \(\sqrt { 2 }\) cm
∴ d(B, C) = 6 cm
iv. In ∆ABC,
AC = AB = 6cm
∴ ∠ABC = ∠ACB [Isosceles triangle theorem]
Let ∠ABC = ∠ACB =x
In ∆ABC,
∠CAB + ∠ABC + ∠ACB = 180° [Sum of the measures of angles of a triangle is 180°]
∴ 90° + x + x = 180°
∴ 90 + 2x = 180°
∴ 2x = 180°- 90°
∴ x = \(\frac{90^{\circ}}{2}\)
∴ x = 45°
∴ ∠ABC = 45°

Question 2.
In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then
i. What is the length of each tangent segment?
ii. What is the measure of ∠MRO?
iii. What is the measure of ∠MRN?

Solution:
seg RM and seg RN are tangents to the circle with centre O. [Given]
∴ ∠OMR = ∠ONR = 90° [Tangent theorem]

i. In ∆OMR, ∠OMR = 90°
∴ OR² = OM² + RM² [Pythagoras theorem]
∴ 10² = 5² + RM²
∴ 100 = 25 + RM²
∴ RM² = 75
∴ RM = \(\sqrt { 75 }\) [Taking square root of both sides]
∴ RM = RN [Tangent segment theorem]
Length of each tangent segment is 5 \(\sqrt { 3 }\) cm.
ii. In ∆RMO,
∠OMR = 90° [Tangent theorem]
OM = 5 cm and OR = 10 cm
∴ OM = \(\frac { 1 }{ 2 } \) OR
∴ ∠MRO = 30° (i) [Converse of 30° – 60° – 90° theorem]
Similarly, ∠NRO = 30°
iii. But, ∠MRN = ∠MRO + ∠NRO [Angle addition property]
= 30° + 30° [From (i)]
∴ ∠MRN = 60°

Question 3.
Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠MRN as well as ∠MON.

Solution:
Proof:
In ∆OMR and ∆ONR,
seg RM ≅ seg RN [Tangent segment theorem]
seg OM ≅ seg ON [Radii of the same circle]
seg OR ≅ seg OR [Common side]
∴ ∆OMR ≅ ∆ONR [SSS test of congruency]
{∴ ∠MRO ≅ ∠NRO
∠MOR ≅ ∠NOR } [c.a.c.t.]
∴ seg OR bisects ∠MRN and ∠MON.

Question 4.
What is the distance between two parallel tangents of a circle having radius 4.5 cm? Justify your answer.
Solution:
Let the lines PQ and RS be the two parallel tangents to circle at M and N respectively. Through centre O, draw line AB || line RS.

OM = ON = 4.5 [Given]
line AB || line RS [Construction]
line PQ || line RS [Given]
∴ line AB || line PQ || line RS
Now, ∠OMP = ∠ONR = 90° (i) [Tangent theorem]
For line PQ || line AB,
∠OMP = ∠AON = 90° (ii) [Corresponding angles and from (i)]
For line RS || line AB,
∠ONR = ∠AOM = 90° (iii) [Corresponding angles and from (i)]
∠AON + ∠AOM = 90° + 90° [From (ii) and (iii)]
∴ ∠AON + ∠AOM = 180°
∴ ∠AON and ∠AOM form a linear pair.
∴ ray OM and ray ON are opposite rays.
∴ Points M, O, N are collinear. (iv)
∴ MN = OM + ON [M – O – N, From (iv)]
∴ MN = 4.5 + 4.5
∴ MN = 9 cm
∴ Distance between two parallel tangents PQ and RS is 9 cm.

Question 1.
In the adjoining figure, seg QR is a chord of the circle with centre O. P is the midpoint of the chord QR. If QR = 24, OP = 10, find radius of the circle. To find solution of the problem, write the theorems that are useful. Using them, solve the problem. (Textbook pg. no. 48)

Solution:
Theorems which are useful to find solution:
i. The segment joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.
ii. In a right angled triangle, sum of the squares of the perpendicular sides is equal to square of its hypotenuse.
QP = \(\frac { 1 }{ 2 } \) (QR) [P is the midpoint of chord QR]

\(\frac { 1 }{ 2 } \) × 24 = 12 units
Also, seg OP ⊥ chord QR [The segment joining centre of a circle and midpoint of a chord is perpendicular to the chord]
In ∆OPQ, ∠OPQ = 90°
∴ OQ² = OP² + QP² [Pythagoras theorem]
= 10² + 12²
= 100 + 144
= 244
∴ OQ = \(\sqrt { 244 }\) = 2\(\sqrt { 61 }\) units.
∴ The radius of the circle is 2\(\sqrt { 61 }\) units.

Question 2.
In the adjoining figure, M is the centre of the circle and seg AB is a diameter, seg MS ⊥ chord AD, seg MT ⊥ chord AC, ∠DAB ≅ ∠CAB.
i. Prove that: chord AD ≅ chord AC.
ii. To solve this problem which theorems will you use?
a. The chords which are equidistant from the centre are equal in length.
b. Congruent chords of a circle are equidistant from the centre.

iii. Which of the following tests of congruence of triangles will be useful?
a. SAS b. ASA c. SSS d. AAS e. Hypotenuse-side test.
Using appropriate test and theorem write the proof of the above example. (Textbook pg. no, 48)
Solution:
Proof:
i. ∠DAB ≅ ∠CAB [Given]
∴ ∠SAM ≅ ∠TAM (i) [A – S – D, A – M – B, A -T – C]
In ∆SAM and ∆TAM,
∠SAM ≅ ∠TAM [From (i)]
∠ASM ≅ ∠ATM [Each angle is of measure 90°]
seg AM ≅ seg AM [Common side]
∴ ∆SAM ≅ ∆TAM [AAS [SAA] test of congruency]
∴ side MS ≅ side MT [c.s.c.t]
But, seg MS ⊥ chord AD [Given]
seg MT ⊥chord AC
∴ chord AD ≅ chord AC [Chords of a circle equidistant from the centre are congruent]
ii. Theorem used for solving the problem:
The chords which are equidistant from the centre are equal in length.
iii. Test of congruency useful in solving the above problem is AAS ISAAI test of congruency.

Question 3.
i. Draw segment AB. Draw perpendicular bisector l of the segment AB. Take point P on the line l as centre,
PA as radius and draw a circle. Observe that the circle passes through point B also. Find the reason.
ii. Taking any other point Q on the line l, if a circle is drawn with centre Q and radius QA, will it pass through B? Think.
iii. How many such circles can be drawn, passing through A and B? Where will their centres lie? (Textbook pg. no. 49)
Solution:

i. Draw the circle with centre P and radius PA.
line l is the perpendicular bisector of seg AB.
Every point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.
∴ PA = PB … [Perpendicular bisector theorem]
∴ PA = PB = radius
∴ The circle with centre P and radius PA passes through point B.

ii. The circle with any other point Q and radius QA is drawn.
QA = QB = radius … [Perpendicular bisector theorem]
∴ The circle with centre Q and radius QA passes through point B.

iii. We can draw infinite number of circles passing through A and B.
All their centres will lie on the perpendicular bisector of AB (i.e., line l)

Question 4.
i. Take any three non-collinear points. What should be done to draw a circle passing through all these points? Draw a circle through these points.
ii. Is it possible to draw one more circle passing through these three points? Think of it. (Textbook pg. no. 49)
Solution:

i. Let points A, B, C be any three non collinear points.
Draw the perpendicular bisector of seg AB (line l).
∴ Points A and B are equidistant from any point of line l ….(i)[Perpendicular bisector theorem]
Draw the perpendicular bisector of seg BC (line m) to intersect line l at point P.
∴ Points B and C are equidistant from any point of line m ….(ii) [Perpendicular bisector theorem]
∴ PA = PB …[From (i)]
PB = PC … [From (ii)]
∴ PA = PB = PC = radius
∴ With PA as radius the required circle is drawn through points A, B, C.
ii. It is not possible to draw more than one circle passing through these three points.

Question 5.
Take 3 collinear points D, E, F. Try to draw a circle passing through these points. If you are not able to draw a circle, think of the reason. (Textbook pg. no. 49)
Solution:

Let D, E, F be the collinear points.
The perpendicular bisector of DE and EF drawn (i.e., line l and line m) do not intersect at a common point.
∴ There is no single common point (centre of circle) from which a circle can be drawn passing through points D, E and F.
Hence, we cannot draw a circle passing through points D, E and F.

Question 6.
Which theorem do we use in proving that hypotenuse is the longest side of a right angled triangle? (Textbook pg. no. 52)

Solution:
In ∆ABC,
∠ABC = 90°
∴ ∠BAC < 90° and ∠ACB < 90° [Given]
∴ ∠ABC > ∠BAC and ∠ABC > ∠ACB
∴ AC > BC and AC > AB [Side opposite to greater angle is greater]
∴ Hypotenuse is the longest side in right angled triangle.
We use theorem, If two angles of a triangle are not equal, then the side opposite to the greater angle is greater than the side opposite to the smaller angle.

Question 7.
Theorem: Tangent segments drawn from an external point to a circle are congruent
Draw radius AP and radius AQ and complete the following proof of the theorem.
Given: A is the centre of the circle.
Tangents through external point D touch the circle at the points P and Q.
To prove: seg DP ≅ seg DQ
Construction: Draw seg AP and seg AQ.

Solution:
Proof:
In ∆PAD and ∆QAD,
seg PA ≅ [segQA] [Radii of the same circle]
seg AD ≅ seg AD [Common side]

∠APD = ∠AQD = 90° [Tangent theorem]
∴ ∆PAD = ∆QAD [By Hypotenuse side test]
∴ seg DP = seg DQ [c.s.c.t]

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 2 Pythagoras Theorem.

Problem Set 2 Geometry 10th Std Maths Part 2 Answers Chapter 2 Pythagoras Theorem

Question 1.
Some questions and their alternative answers are given. Select the correct alternative. [1 Mark each]

i. Out of the following which is the Pythagorean triplet?
(A) (1,5,10)
(B) (3,4,5)
(C) (2,2,2)
(D) (5,5,2)
Answer: (B)
Hint: Refer Practice set 2.1 Q.1 (i)

ii. In a right angled triangle, if sum of the squares of the sides making right angle is 169, then what is the length of the hypotenuse?
(A) 15
(B) 13
(C) 5
(D) 12
Answer: (B)
Hint:

iii. Out of the dates given below which date constitutes a Pythagorean triplet?
(A) 15/08/17
(B) 16/08/16
(C) 3/5/17
(D) 4/9/15
Answer: (A)
Hint:

iv. If a, b, c are sides of a triangle and a² + b² = c², name the type of the triangle.
(A) Obtuse angled triangle
(B) Acute angled triangle
(C) Right angled triangle
(D) Equilateral triangle
Answer: (C)

v. Find perimeter of a square if its diagonal is 10\(\sqrt { 2 }\) cm.
(A) 10 cm
(B) 40\(\sqrt { 2 }\) cm
(C) 20 cm
(D) 40 cm
Answer: (D)
Hint:

vi. Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.
(A) 9 cm
(B) 4 cm
(C) 6 cm
(D) 2\(\sqrt { 6 }\)
Answer: (C)
Hint:

vii. Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse.
(A) 24 cm
(B) 30 cm
(C) 15 cm
(D) 18 cm
Answer: (B)
Hint:

viii. In ∆ABC, AB = 6\(\sqrt { 3 }\) cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A.
(A) 30°
(B) 60°
(C) 90°
(D) 45°
Answer: (A)
Hint:

Question 2.
Solve the following examples.
i. Find the height of an equilateral triangle having side 2a.
ii. Do sides 7 cm, 24 cm, 25 cm form a right angled triangle? Give reason.
iii. Find the length of a diagonal of a rectangle having sides 11 cm and 60 cm.
iv. Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
v. A side of an isosceles right angled triangle is x. Find its hypotenuse.
vi. In ∆PQR, PQ = \(\sqrt { 8 }\), QR = \(\sqrt { 5 }\), PR = \(\sqrt { 3 }\) . Is ∆PQR a right angled triangle? If yes, which angle is of 90°?
Solution:
i. Let ∆ABC be the given equilateral triangle.

∴ ∠B = 60° [Angle of an equilateral triangle]
Let AD ⊥BC, B – D – C.
In ∆ABD, ∠B = 60°, ∠ADB = 90°
∴ ∠BAD = 30° [Remaining angle of a triangle]
∴ ∆ABD is a 30° – 60° – 90° triangle.
∴ AD = \(\frac{\sqrt{3}}{2}\) AB [Side opposite to 60°]
= \(\frac{\sqrt{3}}{2}\) × 2a
= a\(\sqrt { 3 }\) units
The height of the equilateral triangle is a\(\sqrt { 3 }\) units.

ii. The sides of the triangle are 7 cm, 24 cm and 25 cm.
The longest side of the triangle is 25 cm.
∴ (25)² = 625
Now, sum of the squares of the remaining sides is,
(7)² + (24)² = 49 + 576
= 625
∴ (25)² = (7)² + (24)²
∴ Square of the longest side is equal to the sum of the squares of the remaining two sides.
∴ The given sides will form a right angled triangle. [Converse of Pythagoras theorem]

iii. Let ꠸ABCD be the given rectangle.
AB = 11 cm, BC = 60 cm

In ∆ABC, ∠B = 90° [Angle of a rectangle]
∴ AC² = AB² + BC² [Pythagoras theorem]
= 112 + 602
= 121 +3600
= 3721
∴ AC = \(\sqrt { 3721 }\) [Taking square root of both sides]
= 61 cm
The length of the diagonal of the rectangle is 61 cm.
∴ The length of the diagonal of the rectangle is 61 cm.

iv. Let ∆PQR be the given right angled triangle.
In ∆PQR, ∠Q = 90°

∴ PR² = PQ² + QR² [Pythagoras theorem]
= 9² + 12²
= 81 + 144
= 225
∴ PR = \(\sqrt { 225 }\) [Taking square root of both sides]
= 15 cm
∴ The length of the hypotenuse of the right angled triangle is 15 cm.

v. Let ∆PQR be the given right angled isosceles triangle.
PQ = QR = x.

In ∆PQR, ∠Q = 90° [Pythagoras theorem]
∴ PR² = PQ² + QR²
= x² + x²
= 2x²
∴ PR = \(\sqrt{2 x^{2}}\) [Taking square root of both sides]
= x \(\sqrt { 2 }\) units
∴ The hypotenuse of the right angled isosceles triangle is x \(\sqrt { 2 }\) units.
∴ The hypotenuse of the right angled isosceles triangle is x \(\sqrt { 2 }\) units.

vi. Longest side of ∆PQR = PQ = \(\sqrt { 8 }\)
∴ PQ² = (\(\sqrt { 8 }\))² = 8
Now, sum of the squares of the remaining sides is,

QR² + PR² = (\(\sqrt { 5 }\))² + (\(\sqrt { 3 }\))²
= 5 + 3
= 8
∴ PQ² = QR² + PR²
∴ Square of the longest side is equal to the sum of the squares of the remaining two sides.
∴ ∆PQR is a right angled triangle. [Converse of Pythagoras theorem]
Now, PQ is the hypotenuse.
∴∠PRQ = 90° [Angle opposite to hypotenuse]
∴ ∆PQR is a right angled triangle in which ∠PRQ is of 90°.

Question 3.
In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm, then find RS and ST.
Solution:
in ∆RST, ∠S = 900, ∠T = 30° [Given]
∴ ∠R = 60° [Remaining angle of a triangle]
∴ ∆RST is a 30° – 60° – 90° triangle.

∴ RS = \(\frac { 1 }{ 2 } \) RT [Side opposite to 30°]
= \(\frac { 1 }{ 2 } \) × 12 = 6cm
Also, ST = \(\frac{\sqrt{3}}{2}\) RT [Side opposite to 60°]
= \(\frac{\sqrt{3}}{2}\) × 12 = 6 \(\sqrt { 3 }\) cm
∴ RS = 6 cm and ST = 6 \(\sqrt { 3 }\) cm

Question 4.
Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq. cm.
Solution:
Let ꠸ABCD be the given rectangle.
BC = 16cm
Area of rectangle = length × breadth
Area of ꠸ABCD = BC × AB

∴ 192 = I6 × AB
∴ AB = \(\frac { 192 }{ 16 } \)
= 12cm
Now, in ∆ABC, ∠B = 90° [Angle of a rectangle]
∴ AC² = AB² + BC² [Pythagoras theorem]
= 122 + 162
= 144 + 256
=400
∴ AC = \(\sqrt { 400 }\) [Taking square root of both sides]
= 20cm
∴ The diagonal of the rectangle is 20 cm.

Question 5.
Find the length of the side and perimeter of an equilateral triangle whose height is \(\sqrt { 3 }\) cm.
Solution:
Let ∆ABC be the given equilateral triangle.
∴ ∠B = 60° [Angle of an equilateral triangle]

AD ⊥ BC, B – D – C.
In ∆ABD, ∠B =60°, ∠ADB = 90°
∴ ∠BAD = 30° [Remaining angle of a triangle]
∴ ∆ABD is a 30° – 60° – 90° triangle.
∴ AD = \(\frac{\sqrt{3}}{2}\)AB [Side opposite to 600]
∴ \(\sqrt { 3 }\) = \(\frac{\sqrt{3}}{2}\)AB
∴ AB = \(\frac{2 \sqrt{3}}{\sqrt{3}}\)
∴ AB = 2cm
∴ Side of equilateral triangle = 2cm
Perimeter of ∆ABC = 3 × side
= 3 × AB
= 3 × 2
= 6cm
∴ The length of the side and perimeter of the equilateral triangle are 2 cm and 6 cm respectively.

Question 6.
In ∆ABC, seg AP is a median. If BC = 18, AB2 + AC2 = 260, find AP.
Solution:
PC = \(\frac { 1 }{ 2 } \) BC [P is the midpoint of side BC]
= \(\frac { 1 }{ 2 } \) × 18 = 9cm

in ∆ABC, seg AP is the median,
Now, AB² + AC² = 2 A² + 2 PC² [Apollonius theorem]
∴ 260 = 2 AP² + 2 (9)²
∴ 130 = AP² + 81 [Dividing both sides by 2]
∴ AP² = 130 – 81
∴ AP² = 49
∴ AP = \(\sqrt { 49 }\) [Taking square root of both sides]
∴ AP = 7 units

Question 7.
∆ABC is an equilateral triangle. Point P is on base BC such that PC = \(\frac { 1 }{ 3 } \) BC, if AB = 6 cm find AP.

Given: ∆ABC is an equilateral triangle.
PC = \(\frac { 1 }{ 3 } \) BC, AB = 6cm.
To find: AP
Consttuction: Draw seg AD ± seg BC, B – D – C.
Solution:
∆ABC is an equilateral triangle.
∴ AB = BC = AC = 6cm [Sides of an equilateral triangle]
pc = \(\frac { 1 }{ 3 } \) BC [Given]
= \(\frac { 1 }{ 3 } \) (6)
∴ PC = 2cm
In ∆ADC,
∠D = 90° [Construction]
∠C = 60° [Angle of an equilateral triangle]
∠DAC = 30° [Remaining angle of a triangle]
∴ ∆ ADC is a 30° – 60° – 90° triangle.
∴ AD = \(\frac{\sqrt{3}}{2}\) AC [Side opposite to 60°]
∴ AD = \(\frac{\sqrt{3}}{2}\) (6)
∴ AD = 3 \(\sqrt { 3 }\)cm
CD = \(\frac { 1 }{ 2 } \) AC [Side opposite to 30°]
∴ CD = \(\frac { 1 }{ 2 } \) (6)
∴ CD = 3cm
Now DP + PC = CD [D – P – C]
∴ DP + 2 = 3
∴ DP = 1cm
In ∆ADP,
∠ADP = 900
AP² = AD² + DP² [Pythagoras theorem]
∴ AP² = (3\(\sqrt { 3 }\))² + (1)²
∴ AP² = 9 × 3 + 1 = 27 + 1
∴ AP² = 28
∴ AP = \(\sqrt { 28 }\)
∴ AP = \(\sqrt{4 \times 7}\)
∴ AP = 2 \(\sqrt { 7 }\)cm

Question 8.
From the information given in the adjoining figure, prove that
PM = PN = \(\sqrt { 3 }\) × a

Solution:
Proof:
In ∆PMR,
QM = QR = a [Given]
∴ Q is the midpoint of side MR.
∴ seg PQ is the median.
∴ PM² + PR² = 2PQ² + 2QM² [Apollonius theorem]

∴ PM² + a² = 2a² + 2a²
∴ PM² + a² = 4a²
∴ PM² = 3a²
∴ PM,= \(\sqrt { 3 }\)a (i) [Taking square root of both sides]
SimlarIy, in ∆PNQ,
R is the midpoint of side QN.
∴ seg PR is the median.
∴ PN² + PQ² = 2 PR² + 2 RN² [Apollonius theorem]

∴ PN² + a² = 2a² + 2a²
∴ PN² + a² = 4a²
∴ PN² = 3a²
∴ PN = \(\sqrt { 3 }\)a (ii) [Taking square root of both sides]
∴ PM = PN = \(\sqrt { 3 }\) a [From (i) and (ii)]

Question 9.
Prove that the sum of the squares of the diagonals of a parallelogram ¡s equal to the sum of the squares of its sides.

Given: ꠸ABCD is a parallelogram, diagonals AC and BD intersect at point M.
To prove: AC² + BD² = AB² + BC² + CD² + AD²
Solution:
Proof:
꠸ABCD is a parallelogram.
∴ AB = CD and BC = AD (i) [Opposite sides of a parallelogram]
AM = \(\frac { 1 }{ 2 } \) AC and BM = \(\frac { 1 }{ 2 } \) BD (ii) [Diagonals of a parallelogram bisect each other]
∴ M is the midpoint of diagonals AC and BD. (iii)
In ∆ABC.
seg BM is the median. [From (iii)]
AB² + BC² = 2AM² + 2BM² (iv) [Apollonius theorem]
∴ AB² + BC² = 2(\(\frac { 1 }{ 2 } \) AC)² + 2(\(\frac { 1 }{ 2 } \) BD)² [From (ii) and (iv)]
∴ AB² + BC² = 2 × \(\frac{\mathrm{BD}^{2}}{4}+2 \times \frac{\mathrm{AC}^{2}}{4}\)
∴ AB² + BC² = \(\frac{B D^{2}}{2}+\frac{A C^{2}}{2}\)
∴ 2AB² + 2BC² = BD² + AC² [Multiplying both sides by 2]
∴ AB² + AB² + BC² + BC² = BD² + AC²
∴ AB² + CD² + BC² + AD² = BD² + AC² [From(i)]
i.e. AC² + BD² = AB² + BC² + CD² + AD²

Question 10.
Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 15\(\sqrt { 2 }\) km. Find their speed per hour.
Solution:
Suppose Pranali and Prasad started walking from point A, and reached points B and C respectively after 2 hours.
Distance between them = BC = 15\(\sqrt { 2 }\) km

Since, their speed is same, both travel the same distance in the given time.
∴ AB = AC
Let AB = AC = x km (i)
Now, in ∆ ABC, ∠A = 90°
∴ BC² = AB² + AC² [Pythagoras theorem]
∴ (15\(\sqrt { 2 }\))² = x² + x² [From (i)]
∴ 225 × 2 = 2 x²
∴ x² = 225
∴ x = \(\sqrt { 225 }\) [Taking square root of both sides]
∴ x = 15 km
∴ AB = AC = 15km
\(\text { Now, speed }=\frac{\text { distance }}{\text { time }}=\frac{15}{2}\)
= 7.5 km/hr
∴ The speed of Pranali and Prasad is 7.5 km/hour.

Question 11.
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that 4 (BL² + CM²) = 5 BC².

Given : ∠BAC = 90°
seg BL and seg CM are the medians.
To prove: 4(BL² + CM²) = 5BC²
Solution:
Proof:
In ∆BAL, ∠BAL 90° [Given]
∴ BL² = AB² + AL² (i) [Pythagoras theorem]
In ∆CAM, ∠CAM = 90° [Given]
∴ CM² = AC² + AM² (ii) [Pythagoras theorem]
∴ BL² + CM² = AB² + AC² + AL² + AM² (iii) [Adding (i) and (ii)]
Now, AL = \(\frac { 1 }{ 2 } \) AC and AM = \(\frac { 1 }{ 2 } \) AB (iv) [seg BL and seg CM are the medians]
∴ BL² + CM²
= AB² + AC² + (\(\frac { 1 }{ 2 } \) AC)² + (\(\frac { 1 }{ 2 } \) AB)² [From (iii) and (iv)]
\(=A B^{2}+A C^{2}+\frac{A C^{2}}{4}+\frac{A B^{2}}{4}\)
\(=A B^{2}+\frac{A B^{2}}{4}+A C^{2}+\frac{A C^{2}}{4}\)
\(=\frac{5 \mathrm{AB}^{2}}{4}+\frac{5 \mathrm{AC}^{2}}{4}\)
∴ BL² + CM² = \(\frac { 5 }{ 4 } \) (AB² + AC²)
∴ 4(BL² + CM²) = 5(AB² + AC²) (v)
In ∆BAC, ∠BAC = 90° [Given]
∴ BC² = AB² + AC² (vi) [Pythagoras theorem]
∴ 4(BL² + CM²) = 5BC² [From (v) and (vi)]

Question 12.
Sum of the squares of adjacent sides of a parallelogram is 130 cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.
Solution:
Let ꠸ABCD be the given
parallelogram and its diagonals AC and BD intersect at point M.

∴ AB² + AD² = 130cm, BD = 14cm
MD = \(\frac { 1 }{ 2 } \) BD (i) [Diagonals of a parallelogram bisect each other]
= \(\frac { 1 }{ 2 } \) × 14 = 7 cm
In ∆ABD, seg AM is the median. [From (i)]
∴ AB² + = 2AM² + 2MD² [Apollonius theorem]
∴ 130 = 2 AM² + 2(7)²
∴ 65 = AM² +49 [Dividing both sides by 2]
∴ AM² = 65 – 49
∴ AM² = 16 [Taking square root of both sides]
∴ AM = \(\sqrt { 16 }\)
= 4cm
Now, AC =2 AM [Diagonals of a parallelogram bisect each other]
2 × 4 = 8 cm
∴ The length of the other diagonal of the parallelogram is 8 cm.

Question 13.
In ∆ABC, seg AD ⊥ seg BC and DB = 3 CD. Prove that: 2 AB2 = 2 AC2 + BC2.
Given: seg AD ⊥ seg BC
DB = 3CD
To prove: 2AB² = 2AC² + BC²

Solution:
DB = 3CD (i) [Given]
In ∆ADB, ∠ADB = 90° [Given]
∴ AB² = AD² + DB² [Pythagoras theorem]
∴ AB² = AD² + (3CD)² [From (i)]
∴ AB² = AD² + 9CD² (ii)
In ∆ADC, ∠ADC = 90° [Given]
∴ AC² = AD² + CD² [Pythagoras theorem]
∴ AD² = AC² – CD² (iii)
AB² = AC² – CD² + 9CD² [From (ii) and(iii)]
∴ AB² = AC² + 8CD² (iv)
CD + DB = BC [C – D – B]
∴ CD + 3CD = BC [From (i)]
∴ 4CD = BC
∴ CD = \(\frac { BC }{ 4 } \) (v)
AB² = AC² + 8(\(\frac { BC }{ 4 } \))² [From (iv) and (v)]
∴ AB² = AC² + 8 × \(\frac{\mathrm{BC}^{2}}{16}\)
∴ AB² = AC² + \(\frac{\mathrm{BC}^{2}}{2}\)
∴ 2AB² = 2AC² + BC² [Multiplying both sides by 2]

Question 14.
In an isosceles triangle, length of the congruent sides is 13 em and its.base is 10 cm. Find the distance between the vertex opposite to the base and the centroid.
Given: ∆ABC is an isosceles triangle.
G is the centroid.
AB = AC = 13 cm, BC = 10 cm.
To find: AG
Construction: Extend AG to intersect side BC at D, B – D – C.

Solution:
Centroid G of ∆ABC lies on AD
∴ seg AD is the median. (i)
∴ D is the midpoint of side BC.
∴ DC = \(\frac { 1 }{ 2 } \) BC
= \(\frac { 1 }{ 2 } \) × 10 = 5
In ∆ABC, seg AD is the median. [From (i)]
∴ AB² + AC² = 2 AD² + 2 DC² [Apollonius theorem]
∴ 13² + 13² = 2 AD² + 2 (5)²
∴ 2 × 13² = 2 AD² + 2 × 25
∴ 169 = AD² + 25 [Dividing both sides by 2]
∴ AD² = 169 – 25
∴ AD² = 144
∴ AD = \(\sqrt { 144 }\) [Taking square root of both sides]
= 12 cm
We know that, the centroid divides the median in the ratio 2 : 1.
∴ \(\frac { AG }{ GD } \) = \(\frac { 2 }{ 1 } \)
∴ \(\frac { GD }{ AG } \) = \(\frac { 1 }{ 2 } \) [By invertendo]
∴ \(\frac { GD+AG }{ AG } \) = \(\frac { 1+2 }{ 2 } \) [By componendo]
∴ \(\frac { AD }{ AG } \) = \(\frac { 3 }{ 2 } \) [A – G – D]
∴ \(\frac { 12 }{ AG } \) = \(\frac { 3 }{ 2 } \)
∴ AG = \(\frac{12 \times 2}{3}\)
= 8cm
∴ The distance between the vertex opposite to the base and the centroid is 8 cm.

Question 15.
In a trapezium ABCD, seg AB || seg DC, seg BD ⊥ seg AD, seg AC ⊥ seg BC. If AD = 15, BC = 15 and AB = 25, find A (꠸ABCD).

Construction: Draw seg DE ⊥ seg AB, A – E – B
and seg CF ⊥ seg AB, A – F- B.
Solution:
In ∆ ACB, ∠ACB = 90° [Given]
∴ AB2 = AC2 + BC2 [Pythagoras theorem]
∴ 252 = AC2 + 152
∴ AC2 = 625 – 225
= 400

∴ AC = \(\sqrt { 400 }\) [Taking square root of both sides]
= 20 units
Now, A(∆ABC) = \(\frac { 1 }{ 2 } \) × BC × AC
Also, A(∆ABC) = \(\frac { 1 }{ 2 } \) × AB × CF
∴ \(\frac { 1 }{ 2 } \) × BC × AC = \(\frac { 1 }{ 2 } \) × AB × CF
∴ BC × AC = AB × CF
∴ 15 × 20 = 25 × CF
∴ CF = \(\frac{15 \times 20}{25}\) = 12 units
In ∆CFB, ∠CFB 90° [Construction]
∴ BC² = CF² + FB² [Pythagoras theorem]
∴ 15² = 12² + FB²
∴ FB² = 225 – 144
∴ FB² = 81
∴ FB = \(\sqrt { 81 }\) [Taking square root of both sides]
= 9 units
Similarly, we can show that, AE = 9 units
Now, AB = AE + EF + FB [A – E – F, E – F – B]
∴ 25 = 9 + EF + 9
∴ EF = 25 – 18 = 7 units
In ꠸CDEF,
seg EF || seg DC [Given, A – E – F, E – F – B]
seg ED || seg FC [Perpendiculars to same line are parallel]
∴ ꠸CDEF is a parallelogram.
∴ DC = EF 7 units [Opposite sides of a parallelogram]
A(꠸ABCD) = \(\frac { 1 }{ 2 } \) × CF × (AB + CD)
= \(\frac { 1 }{ 2 } \) × 12 × (25 + 7)
= \(\frac { 1 }{ 2 } \) × 12 × 32
∴ A(꠸ABCD) = 192 sq. units

Question 16.
In the adjoining figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that QS = \(\frac { 1 }{ 3 } \) QR. Prove that: 9 PS² = 7 PQ².

Given: ∆PQR is an equilateral triangle.
QS = \(\frac { 1 }{ 3 } \) QR
To prove: 9PS² = 7PQ²
Solution:
Proof:
∆PQR is an equilateral triangle [Given]
∴ ∠P = ∠Q = ∠R = 60° (i) [Angles of an equilateral triangle]
PQ = QR = PR (ii) [Sides of an equilateral triangle]
In ∆PTS, ∠PTS = 90° [Given]
PS² = PT² + ST² (iii) [Pythagoras theorem]
In ∆PTQ,
∠PTQ = 90° [Given]
∠PQT = 60° [From (i)]
∴ ∠QPT = 30° [Remaining angle of a triangle]
∴ ∆PTQ is a 30° – 60° – 90° triangle
∴ PT = \(\frac{\sqrt{3}}{2}\) PQ (iv) [Side opposite to 60°]
QT = \(\frac { 1 }{ 2 } \) PQ (v) [Side opposite to 30°]
QS + ST = QT [Q – S – T]
∴ \(\frac { 1 }{ 3 } \) QR + ST = \(\frac { 1 }{ 2 } \) PQ [Given and from (v)]
∴ \(\frac { 1 }{ 3 } \) PQ + ST = \(\frac { 1 }{ 2 } \) PQ [From (ii)]
∴ ST = \(\frac { PQ }{ 2 } \) – \(\frac { PQ }{ 3 } \)
∴ ST = \(\frac { 3PQ-2PQ }{ 6 } \)
∴ ST = \(\frac { PQ }{ 6 } \) (vi)
\(\mathrm{PS}^{2}=\left(\frac{\sqrt{3}}{2} \mathrm{PQ}\right)^{2}+\left(\frac{\mathrm{PQ}}{6}\right)^{2}\) [From (iii), (iv) and (vi)]
∴ \(\mathrm{PS}^{2}=\frac{3 \mathrm{PQ}^{2}}{4}+\frac{\mathrm{PQ}^{2}}{36}\)
∴ \(\mathrm{PS}^{2}=\frac{27 \mathrm{PQ}^{2}}{36}+\frac{\mathrm{PQ}^{2}}{36}\)
∴\(\mathrm{PS}^{2}=\frac{28 \mathrm{PQ}^{2}}{36}\)
∴PS² = \(\frac { 7 }{ 3 } \) PQ²
∴ 9PS² = 7 PQ²

Question 17.
Seg PM is a median of APQR. If PQ = 40, PR = 42 and PM = 29, find QR.
Solution:
In ∆PQR, seg PM is the median. [Given]
∴ M is the midpoint of side QR.
∴ PQ² + PR² = 2 PM² + 2 MR² [Apollonius theorem]

∴ 40² + 42² = 2 (29)² + 2 MR²
∴ 1600 + 1764 = 2 (841) + 2 MR²
∴ 3364 = 2 (841) + 2 MR²
∴ 1682 = 841 +MR² [Dividing both sides by 2]
∴ MR² = 1682 – 841
∴ MR² = 841
∴ MR = \(\sqrt { 841 }\) [Taking square root of both sides]
= 29 units
Now, QR = 2 MR [M is the midpoint of QR]
= 2 × 29
∴ QR = 58 units

Question 18.
Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM.
Solution:
In ∆ABC, seg AM is the median. [Given]
∴ M is the midpoint of side BC.

∴ MC = \(\frac { 1 }{ 2 } \) BC
= \(\frac { 1 }{ 2 } \) × 24 = 12 units
Now, AB² + AC² = 2 AM² + 2 MC² [Apollonius theorem]
∴ 22² + 34² = 2 AM² + 2 (12)²
∴ 484 + 1156 = 2 AM² + 2 (144)
∴ 1640 = 2 AM² + 2 (144)
∴ 820 = AM² + 144 [Dividing both sides by 2]
∴ AM² = 820 – 144
∴ AM² = 676
∴ AM = \(\sqrt { 676 }\) [Taking square root of both sides]
∴ AM = 26 units

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 2 Pythagoras Theorem.

Practice Set 2.2 Geometry 10th Std Maths Part 2 Answers Chapter 2 Pythagoras Theorem

Question 1.
In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

Solution:
In ∆PQR, point S is the midpoint of side QR. [Given]
∴ seg PS is the median.
∴ PQ² + PR² = 2 PS² + 2 SR² [Apollonius theorem]
∴ 11² + 17² = 2 (13)² + 2 SR²
∴ 121 + 289 = 2 (169)+ 2 SR²
∴ 410 = 338+ 2 SR²
∴ 2 SR² = 410 – 338
∴ 2 SR² = 72
∴ SR² = \(\frac { 72 }{ 2 } \) = 36
∴ SR = \(\sqrt{36}\) [Taking square root of both sides]
= 6 units Now, QR = 2 SR [S is the midpoint of QR]
= 2 × 6
∴ QR = 12 units

Question 2.
In ∆ABC, AB = 10, AC = 7, BC = 9, then find the length of the median drawn from point C to side AB.
Solution:
Let CD be the median drawn from the vertex C to side AB.
BD = \(\frac { 1 }{ 2 } \) AB [D is the midpoint of AB]
= \(\frac { 1 }{ 2 } \) × 10 = 5 units

In ∆ABC, seg CD is the median. [Given]
∴ AC² + BC² = 2 CD² + 2 BD² [Apollonius theorem]
∴ 7² + 9² = 2 CD² + 2 (5)²
∴ 49 + 81 = 2 CD² + 2 (25)
∴ 130 = 2 CD² + 50
∴ 2 CD² = 130 – 50
∴ 2 CD² = 80
∴ CD² = \(\frac { 80 }{ 2 } \) = 40
∴ CD = \(\sqrt { 40 }\) [Taking square root of both sides]
= 2 \(\sqrt { 10 }\) units
∴ The length of the median drawn from point C to side AB is 2 \(\sqrt { 10 }\) units.

Question 3.
In the adjoining figure, seg PS is the median of APQR and PT ⊥ QR. Prove that,

i. PR² = PS² + QR × ST + (\(\frac { QR }{ 2 } \))²
ii. PQ² = PS² – QR × ST + (\(\frac { QR }{ 2 } \))²
Solution:
i. QS = SR = \(\frac { 1 }{ 2 } \) QR (i) [S is the midpoint of side QR]

∴ In ∆PSR, ∠PSR is an obtuse angle [Given]
and PT ⊥ SR [Given, Q-S-R]
∴ PR² = SR² +PS² + 2 SR × ST (ii) [Application of Pythagoras theorem]
∴ PR² = (\(\frac { 1 }{ 2 } \) QR)² + PS² + 2 (\(\frac { 1 }{ 2 } \) QR) × ST [From (i) and (ii)]
∴ PR² = (\(\frac { QR }{ 2 } \))² + PS² + QR × ST
∴ PR² = PS² + QR × ST + (\(\frac { QR }{ 2 } \))²

ii. In.∆PQS, ∠PSQ is an acute angle and [Given]

PT ⊥QS [Given, Q-S-R]
∴ PQ² = QS² + PS² – 2 QS × ST (iii) [Application of Pythagoras theorem]
∴ PR² = (\(\frac { 1 }{ 2 } \) QR)² + PS² – 2 (\(\frac { 1 }{ 2 } \) QR) × ST [From (i) and (iii)]
∴ PR² = (\(\frac { QR }{ 2 } \))² + PS² – QR × ST
∴ PR² = PS² – QR × ST + (\(\frac { QR }{ 2 } \))²

Question 4.
In ∆ABC, point M is the midpoint of side BC. If AB² + AC² = 290 cm, AM = 8 cm, find BC.

Solution:
In ∆ABC, point M is the midpoint of side BC. [Given]
∴ seg AM is the median.
∴ AB² + AC² = 2 AM² + 2 MC² [Apollonius theorem]
∴ 290 = 2 (8)² + 2 MC²
∴ 145 = 64 + MC² [Dividing both sides by 2]
∴ MC² = 145 – 64
∴ MC² = 81
∴ MC = \(\sqrt{81}\) [Taking square root of both sides]
MC = 9 cm
Now, BC = 2 MC [M is the midpoint of BC]
= 2 × 9
∴ BC = 18 cm

Question 5.
In the adjoining figure, point T is in the interior of rectangle PQRS. Prove that, TS² + TQ² = TP² + TR². (As shown in the figure, draw seg AB || side SR and A – T – B)

Given: ꠸PQRS is a rectangle.
Point T is in the interior of ꠸PQRS.
To prove: TS² + TQ² = TP² + TR²
Construction: Draw seg AB || side SR such that A – T – B.
Solution:
Proof:
꠸PQRS is a rectangle. [Given]
∴ PS = QR (i) [Opposite sides of a rectangle]
In ꠸ASRB,
∠S = ∠R = 90° (ii) [Angles of rectangle PQRS]
side AB || side SR [Construction]
Also ∠A = ∠S = 90° [Interior angle theorem, from (ii)]
∠B = ∠R = 90°
∴ ∠A = ∠B = ∠S = ∠R = 90° (iii)
∴ ꠸ASRB is a rectangle.
∴ AS = BR (iv) [Opposite sides of a rectanglel

In ∆PTS, ∠PST is an acute angle
and seg AT ⊥ side PS [From (iii)]
∴ TP² = PS² + TS² – 2 PS.AS (v) [Application of Pythagoras theorem]
In ∆TQR., ∠TRQ is an acute angle
and seg BT ⊥ side QR [From (iii)]
∴ TQ² = RQ² + TR² – 2 RQ.BR (vi) [Application of pythagoras theorem]

TP² – TQ² = PS² + TS² – 2PS.AS
-RQ² – TR² + 2RQ.BR [Subtracting (vi) from (v)]
∴ TP² – TQ² = TS² – TR² + PS²
– RQ² -2 PS.AS +2 RQ.BR
∴ TP² – TQ² = TS² – TR² + PS²
– PS² – 2 PS.BR + 2PS.BR [From (i) and (iv)]
∴ TP² – TQ² = TS² – TR²
∴ TS² + TQ² = TP² + TR²

Question 1.
In ∆ABC, ∠C is an acute angle, seg AD Iseg BC. Prove that: AB² = BC² + A² – 2 BC × DC. (Textbook pg. no. 44)
Given: ∠C is an acute angle, seg AD ⊥ seg BC.
To prove: AB² = BC² + AC² – 2BC × DC
Solution:
Proof:
∴ LetAB = c, AC = b, AD = p,

∴ BC = a, DC = x
BD + DC = BC [B – D – C]
∴ BD = BC – DC
∴ BD = a – x
In ∆ABD, ∠D = 90° [Given]
AB² = BD² + AD² [Pythagoras theorem]
∴ c² = (a – x)² + [P²] (i)
∴ c² = a² – 2ax + x² + [P²]
In ∆ADC, ∠D = 90° [Given]
AC² = AD² + CD² [Pythagoras theorem]
∴ b² = p² + [X²]
∴ p² = b² – [X²] (ii)
∴ c² = a² – 2ax + x² + b² – x² [Substituting (ii) in (i)]
∴ c² = a² + b² – 2ax
∴ AB² = BC² + AC² – 2 BC × DC

Question 2.
In ∆ABC, ∠ACB is an obtuse angle, seg AD ⊥ seg BC. Prove that: AB² = BC² + AC² + 2 BC × CD. (Textbook pg. no. 40 and 4.1)
Given: ∠ACB is an obtuse angle, seg AD ⊥ seg BC.
To prove: AB² = BC² + AC² + 2BC × CD
Solution:
Proof:

Let AD = p, AC = b, AB = c,
BC = a, DC = x
BD = BC + DC [B – C – D]
∴ BD = a + x
In ∆ADB, ∠D = 90° [Given]
AB² = BD² + AD² [Pythagoras theorem]
∴ c² = (a + x)² + p² (i)
∴ c² = a² + 2ax + x² + p²
Also, in ∆ADC, ∠D = 90° [Given]
AC² = CD² + AD² [Pythagoras theorem]
∴ b² = x² + p²
∴ p² = b² – x² (ii)
∴ c² = a² + 2ax + x² + b² – x² [Substituting (ii) in (i)]
∴ c² = a² + b² + 2ax
∴ AB² = BC² + AC² + 2 BC × CD

Question 3.
In ∆ABC, if M is the midpoint of side BC and seg AM ⊥seg BC, then prove that
AB² + AC² = 2 AM² + 2 BM². (Textbook pg, no. 41)
Given: In ∆ABC, M is the midpoint of side BC and seg AM ⊥ seg BC.
To prove: AB² + AC² = 2 AM² + 2 BM²
Solution:
Proof:

In ∆AMB, ∠M = 90° [segAM ⊥ segBC]
∴ AB2 = AM2 + BM2 (i) [Pythagoras theorem]
Also, in ∆AMC, ∠M = 90° [seg AM ⊥ seg BC]
∴ AC2 = AM2 + MC2 (ii) [Pythagoras theorem]
∴ AB² + AC² = AM² + BM² + AM² + MC² [Adding (i) and (ii)]
∴ AB² + AC² = 2 AM² + BM² + BM² [∵ BM = MC (M is the midpoint of BC)]
∴ AB² + AC² = 2 AM² + 2 BM²

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 2 Pythagoras Theorem.

Practice Set 2.1 Geometry 10th Std Maths Part 2 Answers Chapter 2 Pythagoras Theorem

Question 1.
Identify, with reason, which of the following are Pythagorean triplets.
i. (3,5,4)
ii. (4,9,12)
iii. (5,12,13)
iv. (24,70,74)
v. (10,24,27)
vi. (11,60,61)
Solution:
i. Here, 5² = 25
3² + 4² = 9 + 16 = 25
∴ 5² = 3² + 4²
The square of the largest number is equal to the sum of the squares of the other two numbers.
∴ (3,5,4) is a Pythagorean triplet.

ii. Here, 12² = 144
4² + 9²= 16 + 81 =97
∴ 12² ≠ 4² + 92
The square of the largest number is not equal to the sum of the squares of the other two numbers.
∴ (4,9,12) is not a Pythagorean triplet.

iii. Here, 13² = 169
5² + 12² = 25 + 144 = 169
∴ 13² = 5² + 12²
The square of the largest number is equal to the sum of the squares of the other two numbers.
∴ (5,12,13) is a Pythagorean triplet.

iv. Here, 74² = 5476
24² + 70² = 576 + 4900 = 5476
∴ 74² = 24² + 70²
The square of the largest number is equal to the sum of the squares of the other two numbers.
∴ (24, 70,74) is a Pythagorean triplet.

v. Here, 27² = 729
10² + 24² = 100 + 576 = 676
∴ 27² ≠ 10² + 24²
The square of the largest number is not equal to the sum of the squares of the other two numbers.
∴ (10,24,27) is not a Pythagorean triplet.

vi. Here, 61² = 3721
11² + 60² = 121 + 3600 = 3721
∴ 61² = 11² + 60²
The square of the largest number is equal to the sum of the squares of the other two numbers.
∴ (11,60,61) is a Pythagorean triplet.

Question 2.
In the adjoining figure, ∠MNP = 90°, seg NQ ⊥ seg MP,MQ = 9, QP = 4, find NQ.

Solution:
In ∆MNP, ∠MNP = 90° and [Given]
seg NQ ⊥ seg MP
NQ² = MQ × QP [Theorem of geometric mean]
∴ NQ = \(\sqrt { MQ\times QP }\) [Taking square root of both sides]
= \(\sqrt { 9\times 4 } \)
= 3 × 2
∴NQ = 6 units

Question 3.
In the adjoining figure, ∠QPR = 90°, seg PM ⊥ seg QR and Q – M – R, PM = 10, QM = 8, find QR.

Solution:
In ∆PQR, ∠QPR = 90° and [Given]
seg PM ⊥ seg QR
∴ PM² = OM × MR [Theorem of geometric mean]
∴ 10² = 8 × MR
∴ MR = \(\frac { 100 }{ 8 } \)
= 12.5
Now, QR = QM + MR [Q – M – R]
= 8 + 12.5
∴ QR = 20.5 units

Question 4.
See adjoining figure. Find RP and PS using the information given in ∆PSR.

Solution:
In ∆PSR, ∠S = 90°, ∠P = 30° [Given]
∴ ∠R = 60° [Remaining angle of a triangle]
∴ ∆PSR is a 30° – 60° – 90° triangle.
RS = \(\frac { 1 }{ 2 } \) RP [Side opposite to 30°]
∴6 = \(\frac { 1 }{ 2 } \) RP
∴ RP = 6 × 2 = 12 units
Also, PS = \(\frac{\sqrt{3}}{2}\) RP [Side opposite to 60°]
= \(\frac{\sqrt{3}}{2}\) × 12
= \(6 \sqrt{3}\) units
∴ RP = 12 units, PS = 6 \(\sqrt { 3 }\) units

Question 5.
For finding AB and BC with the help of information given in the adjoining figure, complete the following activity.

Solution:
AB = BC [Given]
∴ ∠BAC = ∠BCA [Isosceles triangle theorem]
Let ∠BAC = ∠BCA = x (i)
In ∆ABC, ∠A + ∠B + ∠C = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ x + 90° + x = 180° [From (i)]
∴ 2x = 90°
∴ x = \(\frac { 90° }{ 2 } \) [From (i)]
∴ x = 45°

Question 6.
Find the side and perimeter of a square whose diagonal is 10 cm.

Solution:
Let ꠸ABCD be the given square.
l(diagonal AC) = 10 cm
Let the side of the square be ‘x’ cm.
In ∆ABC,
∠B = 90° [Angle of a square]
∴ AC² = AB² + BC² [Pythagoras theorem]
∴ 10² = x² + x²
∴ 100 = 2x²
∴ x² = \(\frac { 100 }{ 2 } \)
∴x² = 50
∴ x = \(\sqrt { 50 }\) [Taking square root of both sides]
= \(=\sqrt{25 \times 2}=5 \sqrt{2}\)
∴side of square is 5\(\sqrt { 2 }\) cm.
= 4 × 5 \(\sqrt { 2 }\)
∴ Perimeter of square = 20 \(\sqrt { 2 }\) cm

Question 7.
In the adjoining figure, ∠DFE = 90°, FG ⊥ ED. If GD = 8, FG = 12, find
i. EG
ii. FD, and
iii. EF

Solution:
i. In ∆DEF, ∠DFE = 90° and FG ⊥ ED [Given]
∴ FG² = GD × EG [Theorem of geometric mean]
∴ 122 = 8 × EG .
∴ EG = \(\frac { 144 }{ 8 } \)
∴ EG = 18 units

ii. In ∆FGD, ∠FGD = 90° [Given]
∴ FD² = FG² + GD² [Pythagoras theorem]
= 122 + 82 = 144 + 64
= 208
∴ FD = \(\sqrt { 208 }\) [Taking square root of both sides]
∴ FD = 4 \(\sqrt { 13 }\) units

iii. In ∆EGF, ∠EGF = 90° [Given]
∴ EF² = EG² + FG² [Pythagoras theorem]
= 182 + 122 = 324 + 144
= 468
∴ EF = \(\sqrt { 468 }\) [Taking square root of both sides]
∴ EF = 6 \(\sqrt { 13 }\) units

Question 8.
Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.

Solution:
Let ꠸ABCD be the given rectangle.
AB = 12 cm, BC 35 cm
In ∆ABC, ∠B = 90° [Angle of a rectangle]
∴ AC² = AB² + BC² [Pythagoras theorem]
= 122 + 352
= 144 + 1225
= 1369
∴ AC = \(\sqrt { 1369 }\) [Taking square root of both sides]
= 37 cm
∴ The diagonal of the rectangle is 37 cm.

Question 9.
In the adjoining figure, M is the midpoint of QR. ∠PRQ = 90°.
Prove that, PQ² = 4 PM² – 3 PR².

Solution:
Proof:
In ∆PQR, ∠PRQ = 90° [Given]
PQ² = PR² + QR² (i) [Pythagoras theorem]
RM = \(\frac { 1 }{ 2 } \) QR [M is the midpoint of QR]
∴ 2RM = QR (ii)
∴ PQ² = PR² + (2RM)² [From (i) and (ii)]
∴ PQ² = PR² + 4RM² (iii)
Now, in ∆PRM, ∠PRM = 90° [Given]
∴ PM² = PR² + RM² [Pythagoras theorem]
∴ RM² = PM² – PR² (iv)
∴ PQ² = PR² + 4 (PM² – PR²) [From (iii) and (iv)]
∴ PQ² = PR² + 4 PM² – 4 PR²
∴ PQ² = 4 PM² – 3 PR²

Question 10.
Walls of two buildings on either side of a street are parallel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street, its top touches the window of the other building at a height 4.2 m. Find the width of the street.
Solution:
Let AC and CE represent the ladder of length 5.8 m, and A and E represent windows of the buildings on the opposite sides of the street. BD is the width of the street.

AB = 4 m and ED = 4.2 m
In ∆ABC, ∠B = 90° [Given]
AC² = AB² + BC² [Pythagoras theorem]
∴ 5.8² = 4² + BC²
∴ 5.8² – 4² = BC²
∴ (5.8 – 4) (5.8 + 4) = BC²
∴ 1.8 × 9.8 = BC²

CE² = CD² + DE² [Pythagoras theorem]
∴ 5.8² = CD² + 4.22
∴ 5.8² – 4.2² = CD²
∴ (5.8 – 4.2) (5.8 + 4.2) = CD²
∴ 1.6 × 10 = CD²
∴ CD² = 16
∴ CD = 4m (ii) [Taking square root of both sides]
Now, BD = BC + CD [B – C – D]
= 4.2 + 4 [From (i) and (ii)]
= 8.2 m
∴ The width of the street is 8.2 metres.

Question 1.
Verify that (3,4,5), (5,12,13), (8,15,17), (24,25,7) are Pythagorean triplets. (Textbook pg. no. 30)
Solution:
i. Here, 5² = 25
3² + 4² = 9 + 16 = 25
∴ 5² = 3² + 4²
The square of the largest number is equal to the sum of the squares of the other two numbers.
∴ 3,4,5 is a Pythagorean triplet.

ii. Here, 13² = 169
5² + 12² = 25 + 144 = 169
∴ 13² = 5² + 12²
The square of the largest number is equal to the sum of the squares of the other two numbers.
∴ 5,12,13 is a Pythagorean triplet.

iii. Here, 17² = 289
8² + 15² = 64 + 225 = 289
∴ 17² = 8² + 15²
The square of the largest number is equal to the sum of the squares of the other two numbers.
∴ 8,15,17 is a Pythagorean triplet.

iv. Here, 25² = 625
7² + 24² = 49 + 576 = 625
∴ 25² = 7² + 24²
The square of the largest number is equal to the sum of the squares of the other two numbers.
∴ 24,25, 7 is a Pythagorean triplet.

Question 2.
Assign different values to a and b and obtain 5 Pythagorean triplets. (Textbook pg. no. 31)
Solution:
i. Let a = 2, b = 1
a² + b² = 2² + 1² = 4 + 1 = 5
a² – b² = 2² – 1² = 4 – 1 = 3
2ab = 2 × 2 × 1 = 4
∴ (5, 3, 4) is a Pythagorean triplet.

ii. Let a = 4,b = 3
a² + b² = 4² + 3² = 16 + 9 = 25
a² – b² = 4² – 3² = 16 – 9 = 7
2ab = 2 × 4 × 3 = 24
∴ (25, 7, 24) is a Pythagorean triplet.

iii. Let a = 5, b = 2
a² + b² = 5² + 2² = 25 + 4 = 29
a² – b² = 5² – 2² = 25 – 4 = 21
2ab = 2 × 5 × 2 = 20
∴ (29, 21, 20) is a Pythagorean triplet.

iv. Let a = 4,b = 1
a² + b² = 4² + 1² = 16 + 1 = 17
a² – b² = 42 – 12 = 16 – 1 = 15
2ab = 2 × 4 × 1 = 8
∴ (17, 15, 8) is a Pythagorean triplet.

v. Let a = 9, b = 7
a² + b² = 92 + 72 = 81 + 49 = 130
a² – b² = 92 – 72 = 81 – 49 = 32
2ab = 2 × 9 × 7 = 126
∴ (130,32,126) is a Pythagorean triplet.

Note: Numbers in Pythagorean triplet can be written in any order.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 6 Statistics.

Problem Set 1 Geometry 10th Std Maths Part 2 Answers Chapter 1 Similarity

Question 1.
Select the appropriate alternative.
i. In ∆ABC and ∆PQR, in a one to one correspondence \(\frac { AB }{ QR } \) = \(\frac { BC }{ PR } \) = \(\frac { CA }{ PQ } \), then

(A) ∆PQR – ∆ABC
(B) ∆PQR – ∆CAB
(C) ∆CBA – ∆PQR
(D) ∆BCA – ∆PQR
Answer:
(B)

ii. If in ∆DEF and ∆PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E, then which of the following statements is false?

(A) \(\frac { EF }{ PR } \) = \(\frac { DF }{ PQ } \)
(B) \(\frac { DE }{ PQ } \) = \(\frac { EF }{ RP } \)
(C) \(\frac { DE }{ QR } \) = \(\frac { DF }{ PQ } \)
(D) \(\frac { EF }{ RP } \) = \(\frac { DE }{ QR } \)
Answer:
∆DEF ~ ∆QRP … [AA test of similarity]
∴ \(\frac { DE }{ QR } \) = \(\frac { EF }{ RP } \) = \(\frac { DF }{ PQ } \) …[Corresponding sides of similar triangles]
(B)

iii. In ∆ABC and ∆DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE, then which of the statements regarding the two triangles is true?

(A) The triangles are not congruent and not similar.
(B) The triangles are similar but not congruent.
(C) The triangles are congruent and similar.
(D) None of the statements above is true.
Answer:
(B)

iv. ∆ABC and ∆DEF are equilateral triangles, A(∆ABC) : A(∆DEF) = 1 : 2. If AB = 4, then what is length of DE?

(A) 2√2
(B) 4
(C) 8
(D) 4√2
Answer:
Refer Q. 6 Practice Set 1.4
(D)

v. In the adjoining figure, seg XY || seg BC, then which of the following statements is true?

(A) \(\frac { AB }{ AC } \) = \(\frac { AX }{ AY } \)
(B) \(\frac { AX }{ XB } \) = \(\frac { AY }{ AC } \)
(C) \(\frac { AX }{ YC } \) = \(\frac { AY }{ XB } \)
(D) \(\frac { AB }{ YC } \) = \(\frac { AC }{ XB } \)
Answer:
∆ABC ~ ∆AXY … [AA test of similarity]
∴ \(\frac { AB }{ AX } \) = \(\frac { AC }{ AY } \) …[Corresponding sides of similar triangles]
∴ \(\frac { AB }{ AC } \) = \(\frac { AX }{ AY } \) …[Altemendo]
(A)

Question 2.
In ∆ABC, B-D-C and BD = 7, BC = 20, then find following ratios.

Solution:

Draw AE ⊥ BC, B – E – C.
BC = BD + DC [B – D – C]
∴ 20 = 7 + DC
∴ DC = 20 – 7 = 13

i. ∆ABD and ∆ADC have same height AE.
\(\frac{\mathrm{A}(\Delta \mathrm{ABD})}{\mathrm{A}(\Delta \mathrm{ADC})}=\frac{\mathrm{BD}}{\mathrm{DC}}\) [Triangles having equal height]
∴ \(\frac{A(\Delta A B D)}{A(\Delta A D C)}=\frac{7}{13}\)

ii. ∆ABD and ∆ABC have same height AE.
\(\frac{\mathrm{A}(\Delta \mathrm{ABD})}{\mathrm{A}(\Delta \mathrm{ABC})}=\frac{\mathrm{BD}}{\mathrm{BC}}\) [Triangles having equal height]
∴ \(\frac{A(\Delta A B D)}{A(\Delta A B C)}=\frac{7}{20}\)

iii. ∆ADC and ∆ABC have same height AE.
\(\frac{A(\Delta A D C)}{A(\Delta A B C)}=\frac{D C}{B C}\) [Triangles having equal height]
∴ \(\frac{A(\Delta A D C)}{A(\Delta A B C)}=\frac{13}{20}\)

Question 3.
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm, then what is the corresponding base of the bigger triangle?
Solution:
Let A₁ and A₂ be the areas of two triangles. Let b₁ and b₂ be their corresponding bases.
A₁ : A₂ = 2 : 3

∴ The corresponding base of the bigger triangle is 9 cm.

Question 4.
In the adjoining figure, ∠ABC = ∠DCB = 90°, AB = 6, DC = 8, then \(\frac{\mathbf{A}(\Delta \mathbf{A} \mathbf{B} \mathbf{C})}{\mathbf{A}(\mathbf{\Delta D C B})}=?\)

Solution:
∆ABC and ∆DCB have same base BC.

Question 5.
In the adjoining figure, PM = 10 cm, A(∆PQS) = 100 sq. cm,
A(∆QRS) = 110 sq. cm, then find NR.

Solution:

∴ NR = 11 cm

Question 6.
∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio \(\frac{A(\Delta M N T)}{A(\Delta Q R S)}\)

Solution:
∆MNT- ∆QRS [Given]
∴ ∠M ≅ ∠Q (i) [Corresponding angles of similar triangles]
In ∆MLT and ∆QPS,
∠M ≅ ∠Q [From (i)]
∠MLT ≅ ∠QPS [Each angle is of measure 90°]
∴ ∆MLT ~ ∆QPS [AA test of similarity]

Question 7.
In the adjoining figure, A – D – C and B – E – C. seg DE || side AB. If AD = 5, DC = 3, BC = 6.4, then find BE.

Solution:
In ∆ABC,
seg DE || side AB [Given]
∴ \(\frac { DC }{ AD } \) = \(\frac { EC }{ BE } \) [Basic proportionality theorem]
∴ \(\frac { 3 }{ 4 } \) = \(\frac { 6.4-x }{ x } \)
∴ 3x = 5 (6.4 – x)
∴ 3x = 32 – 5x
∴ 8x = 32
∴ x = \(\frac { 32 }{ 8 } \) =4
∴ BE = 4 units

Question 8.
In the adjoining figure, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280, then find PQ, QR and RS.

Solution:
seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. [Given]
∴ seg PA || seg QB || seg RC || seg SD (i) [Lines perpendicular to the same line are parallel to each other]
Let the value of PQ be x and that of QR be y.
PS = PQ + QS [P – Q – S]
∴ 280 – x + QS
∴ QS = 280 – x (ii)
Now, seg PA || seg QB || seg SD [From (i)]
∴ \(\frac { AB }{ BD } \) = \(\frac { PQ }{ QS } \) [Property of three parallel lines and their transversals]
∴\(\frac { AB }{ BC+CD } \) = \(\frac { PQ }{ QS } \) [B – C – D]
∴ \(\frac { 60 }{ 70+80 } \) = \(\frac { x }{ 280-x } \)
∴ \(\frac { 60 }{ 150 } \) = \(\frac { x }{ 280-x } \)
∴ \(\frac { 2 }{ 5 } \) = \(\frac { x }{ 280-x } \)
∴ 5x = 2 (280 – x)
∴ 5x = 560 – 2x
∴ 7x = 560
∴ x = \(\frac { 560 }{ 7 } \) = 80
∴ PQ = 80 units
QS = 280 – x [From (ii)]
= 280 – 80
= 200 units
But, QS = QR + RS [Q – R – S]
∴ 200 = y + RS
∴ RS = 200 – y (ii)
Now, seg QB || seg RC || seg SD [From (i)]
∴\(\frac { BC }{ CD } \) = \(\frac { QR }{ RS } \) [Property of three parallel lines and their transversals]
∴ \(\frac { 70 }{ 80 } \) = \(\frac { y }{ 200-y } \)
∴ \(\frac { 7 }{ 8 } \) = \(\frac { y }{ 200-y } \)
∴ 8y = 7(200 – y)
∴ 8y = 1400 – 7y
∴ 15y = 1400
∴ y = \(\frac { 1400 }{ 15 } \) = \(\frac { 280 }{ 3 } \)
∴ QR = \(\frac { 280 }{ 3 } \) units
RS = 200 – 7 [From (iii)]
= 200 – \(\frac { 280 }{ 3 } \)
= \(\frac{200 \times 3-280}{3}\)
= \(\frac { 600-280 }{ 3 } \)
∴ RS = \(\frac { 320 }{ 3 } \) units

Question 9.
In ∆PQR, seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR
Complete the proof by filling in the boxes.

Solution:
Proof:
In ∆PMQ, ray MX is bisector of ∠PMQ.

Question 10.
In the adjoining figure, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y.
AB = 5, AC = 4, BC = 6, then find \(\frac { AX }{ XY } \).

Solution:
Let the value of BY be x.
BC = BY + YC [B – Y – C]
∴ 6 = x + YC
∴ YC = 6 – x
in ∆BAY, ray BX bisects ∠B. [Given]
∴ \(\frac { AB }{ BY } \) = \(\frac { AX }{ XY } \) (i) [Property of angle bisector of a triangle]
Also, in ∆CAY, ray CX bisects ∠C. [Given]

Question 11.
In ꠸ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that \(\frac { AP }{ PD } \) = \(\frac { PC }{ BP } \)

Solution:
proof:
seg AD || seg BC and BD is their transversal. [Given]
∴ ∠DBC ≅ ∠BDA [Alternate angles]
∴ ∠PBC ≅ ∠PDA (i) [D – P – B]
In ∆PBC and ∆PDA,
∠PBC ≅ ∠PDA [From (i)]
∠BPC ≅ ∠DPA [Vertically opposite angles]
∴ ∆PBC ~ ∆PDA [AA test of similarity]
∴ \(\frac { BP }{ PD } \) = \(\frac { PC }{ AP } \) [Corresponding sides of similar triangles]
∴ \(\frac { AP }{ PD } \) = \(\frac { PC }{ BP } \) [By altemendo]

Question 12.
In the adjoining figure, XY || seg AC. If 2 AX = 3 BX and XY = 9, complete the activity to find the value of AC.

Solution:
2 AX = 3 BX [Given]

Question 13.
In the adjoining figure, the vertices of square DEFG are on the sides of ∆ABC. If ∠A = 90°, then prove that DE² = BD × EC.
(Hint: Show that ∆GBD is similar to ∆ CFE. Use GD = FE = DE.)

Solution:
proof:
꠸DEFG is a square.
∴ DE = EF = GF = GD (i) [Sides of a square]
∠GDE = ∠DEF = 90° [Angles of a square]
∴ seg GD ⊥ side BC, seg FE ⊥ side BC (ii)
In ∆BAC and ∆BDG,
∠BAC ≅ ∠BDG [From (ii), each angle is of measure 90°]
∠ABC ≅ ∠DBG [Common angle]
∴ ∆BAC – ∆BDG (iii) [AA test of similarity]
In ∆BAC and ∆FEC,
∠BAC ≅ ∠FEC [From (ii), each angle is measure 90°]
∠ACB ≅ ∠ECF [Common angle]
∴ ∆BAC – ∆FEC (iv) [AA test of similarity]
∴ ∆BDG – ∆FEC [From (iii) and (iv)]
∴ \(\frac { BD }{ EF } \) = \(\frac { GD }{ EC } \) (v) [Corresponding sides of similar triangles]
∴ \(\frac { BD }{ DE } \) = \(\frac { DE }{ EC } \) [From (i) and (v)]
∴ DE² = BD × EC

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

Practice Set 1.2 Geometry10th Std Maths Part 2 Answers Chapter 1 Similarity

Question 1.
Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.

Solution:
In ∆ PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 7 }{ 3 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.5 }{ 1.5 } \) = \(\frac { 35 }{ 15 } \) = \(\frac { 7 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR. [Converse of angle bisector theorem]

ii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 10 }{ 7 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 8 }{ 6 } \) = \(\frac { 4 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) ≠ \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is not the bisector of ∠QPR

iii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 9 }{ 10 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.6 }{ 4 } \) = \(\frac { 36 }{ 40 } \) = \(\frac { 9 }{ 10 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR [Converse of angle bisector theorem]

Question 2.
In ∆PQR PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Solution:
PN + NR = PR [P – N – R]
∴ PN + 8 = 20
∴ PN = 20 – 8 = 12
Also, PM + MQ = PQ [P – M – Q]
∴ 15 + MQ = 25

∴ line NM || side RQ [Converse of basic proportionality theorem]

Question 3.
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7, MQ = 2.5, then find QP.

Solution:
In ∆MNP, NQ is the bisector of ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 7 }{ 5 } \) = \(\frac { QP }{ 2.5 } \)
∴ QP = \(\frac { 7\times 2.5 }{ 5 } \)
∴ QP = 3.5 units

Question 4.
Measures of some angles in the figure are given. Prove that \(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \)

Solution:
Proof
∠APQ = ∠ABC = 60° [Given]
∴ ∠APQ ≅ ∠ABC
∴ side PQ || side BC (i) [Corresponding angles test]
In ∆ABC,
sidePQ || sideBC [From (i)]
∴\(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \) [Basic proportionality theorem]

Question 5.
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.

Solution:
side AB || side PQ || side DC [Given]
∴\(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \) [Property of three parallel lines and their transversals]
∴\(\frac { 15 }{ 12 } \) = \(\frac { BQ }{ 14 } \)
∴ BQ = \(\frac { 15\times 14 }{ 12 } \)
∴ BQ = 17.5 units

Question 6.
Find QP using given information in the figure.

Solution:
In ∆MNP, seg NQ bisects ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 40 }{ 25 } \) = \(\frac { QP }{ 14 } \)
∴ QP = \(\frac { 40\times 14 }{ 25 } \)
∴ QP = 22.4 units

Question 7.
In the adjoining figure, if AB || CD || FE, then find x and AE.

Solution:
line AB || line CD || line FE [Given]
∴\(\frac { BD }{ DF } \) = \(\frac { AC }{ CE } \) [Property of three parallel lines and their transversals]
∴\(\frac { 8 }{ 4 } \) = \(\frac { 12 }{ X } \)
∴ X = \(\frac { 12\times 4 }{ 8 } \)
∴ X = 6 units
Now, AE AC + CE [A – C – E]
= 12 + x
= 12 + 6
= 18 units
∴ x = 6 units and AE = 18 units

Question 8.
In ∆LMN, ray MT bisects ∠LMN. If LM = 6, MN = 10, TN = 8, then find LT.

Solution:
In ∆LMN, ray MT bisects ∠LMN. [Given]
∴\(\frac { LM }{ MN } \) = \(\frac { LT }{ TN } \) [Property of angle bisector of a triangle]
∴\(\frac { 6 }{ 10 } \) = \(\frac { LT }{ 8 } \)
∴ LT = \(\frac { 6\times 8 }{ 10 } \)
∴ LT = 4.8 units

Question 9.
In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.

Solution:
In ∆ABC, seg BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AD }{ CD } \) [Property of angle bisector of a triangle]
∴\(\frac { x }{ x+5 } \) = \(\frac { x-2 }{ x+2 } \)
∴ x(x + 2) = (x – 2)(x + 5)
∴ x2 + 2x = x2 + 5x – 2x – 10
∴ 2x = 3x – 10
∴ 10 = 3x – 2x
∴ x = 10

Question 10.
In the adjoining figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.

Solution:

Question 11.
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.

Solution:
In ∆ABC, ray BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (i) [Property of angle bisector of a triangle]
Also, in ∆ABC, ray CE bisects ∠ACB. [Given]
∴\(\frac { AC }{ BC } \) = \(\frac { AE }{ EB } \) (ii) [Property of angle bisector of a triangle]
But, seg AB = seg AC (iii) [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (iv) [From (ii) and (iii)]
∴\(\frac { AD }{ DC } \) = \(\frac { AE }{ EB } \) [From (i) and (iv)]
∴ ED || BC [Converse of basic proportionality theorem]

Question 1.
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.

iv. Find ratios \(\frac { AB }{ BC } \) and \(\frac { AD }{ DC } \)
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above. Verify that the ratios are equal. (Textbook pg. no. 8)
Solution:

Note: Students should bisect the remaining angles and verify that the ratios are equal.

Question 2.
Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof.
i. The areas of two triangles of equal height are proportional to their bases.
ii. Every point on the bisector of an angle is equidistant from the sides of the angle. (Textbook pg. no. 9)

Given: In ∆CAB, ray AD bisects ∠A.
To prove: \(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \)
Construction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.
Solution:
Proof:
In ∆ABC,
Point D is on angle bisector of ∠A. [Given]
∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle]
\(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B \times D M}{A C \times D N}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B}{A C}\) (ii) [From (i)]
Also, ∆ABD and ∆ACD have equal height.
∴ \(\frac{\mathrm{A}(\Delta \mathrm{ABD})}{\mathrm{A}(\Delta \mathrm{ACD})}=\frac{\mathrm{BD}}{\mathrm{CD}}\) (iii) [Triangles having equal height]
∴\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}\) [From (ii) and (iii)]

Question 3.
i. Draw three parallel lines.
ii. Label them as l, m, n.
iii. Draw transversals t₁ and t₂.
iv. AB and BC are intercepts on transversal t₁.
v. PQ and QR are intercepts on transversal t₂.
vi. Find ratios \(\frac { AB }{ BC } \) and \(\frac { PQ }{ QR } \). You will find that they are almost equal. Verify that they are equal.(Textbook pg, no. 10)
Solution:

(Students should draw figures similar to the ones given and verify the properties.)

Question 4.
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.(Textbook pg, no. 12)

Solution:

Question 5.
In ∆ABC, ray BD bisects ∠ABC. A – D – C, side DE || side BC, A – E – B, then prove that \(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (Textbook pg, no. 13)

Solution:

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

Practice Set 1.1 Geometry 10th Std Maths Part 2 Answers Chapter 1 Similarity

Question 1.
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Solution:
Let the base, height and area of the first triangle be b₁, h₁, and A₁ respectively.
Let the base, height and area of the second triangle be b₂, h₂ and A₂ respectively.

[Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ The ratio of areas of the triangles is 3:4.

Question 2.
In the adjoining figure, BC ± AB, AD _L AB, BC = 4, AD = 8, then find \(\frac{A(\Delta A B C)}{A(\Delta A D B)}\)

Solution:
∆ABC and ∆ADB have same base AB.

[Since Triangles having equal base]

Question 3.
In the adjoining figure, seg PS ± seg RQ, seg QT ± seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.

Solution:
In ∆PQR, PR is the base and QT is the corresponding height.
Also, RQ is the base and PS is the corresponding height.
\(\frac{A(\Delta P Q R)}{A(\Delta P Q R)}=\frac{P R \times Q T}{R Q \times P S}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{1}{1}=\frac{P R \times Q T}{R Q \times P S}\)
∴ PR × QT = RQ × PS
∴ 12 × QT = 6 × 6
∴ QT = \(\frac { 36 }{ 12 } \)
∴ QT = 3 units

Question 4.
In the adjoining figure, AP ⊥ BC, AD || BC, then find A(∆ABC) : A(∆BCD).

Solution:
Draw DQ ⊥ BC, B-C-Q.

AD || BC [Given]
∴ AP = DQ   (i)  [Perpendicular distance between two parallel lines is the same]
∆ABC and ∆BCD have same base BC.

Question 5.
In the adjoining figure, PQ ⊥ BC, AD ⊥ BC, then find following ratios.

Solution:
i. ∆PQB and tPBC have same height PQ.

ii. ∆PBC and ∆ABC have same base BC.

iii. ∆ABC and ∆ADC have same height AD.

Question 1.
Find \(\frac{A(\Delta A B C)}{A(\Delta A P Q)}\)

Solution:
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.6 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

Practice Set 6.6 Algebra 10th Std Maths Part 1 Answers Chapter 6 Statistics

Question 1.
The age group and number of persons, who donated blood in a blood donation camp is given below.
Draw a pie diagram from it.

Solution:
Total number of persons = 80 + 60 + 35 + 25 = 200

Question 2.
The marks obtained by a student in different subjects are shown. Draw a pie diagram showing the information.

Solution:
Total marks obtained = 50 + 70 + 80 + 90 + 60 + 50 = 400

Question 3.
In a tree plantation programme, the number of trees planted by students of different classes is given in the following table. Draw a pie diagram showing the information.

Solution:
Total number of trees planted = 40 + 50 + 75 + 50 + 70 + 75 = 360

Question 4.
The following table shows the percentages of demands for different fruits registered with a fruit vendor. Show the information by a pie diagram.

Solution:
Total percentage = 30 + 15 + 25 + 20 + 10 = 100%

Question 5.
The pie diagram in the given figure shows the proportions of different workers in a town. Answer the following questions with its help.
i. If the total workers is 10,000, how many of them are in the field of construction?
ii. How many workers are working in the administration?
iii. What is the percentage of workers in production?

Solution:

∴ There are 2000 workers working in the field of construction.

∴ There are 1000 workers working in the administration.

∴ 25% of workers are working in the production field.

Question 6.
The annual investments of a family are shown in the given pie diagram. Answer the following questions based on it.
i. If the investment in shares is ? 2000, find the total investment.
ii. How much amount is deposited in bank?
iii. How much more money is invested in immovable property than in mutual fund?
iv. How much amount is invested in post?

Solution:

The total investment is ₹ 12000.

ii. Central angle for deposit in bank (θ) = 90°

∴ The amount deposited in bank is ₹ 3000.

iii. Difference in central angle for immovable property and mutual fund (θ) = 120° – 60° = 60°

∴ ₹ 2000 more is invested in immovable property than in mutual fund.

iv. Central angle for post (θ) = 30°

∴ The amount invested in post is ₹ 1000.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

Practice Set 6.5 Algebra 10th Std Maths Part 1 Answers Chapter 6 Statistics

Question 1.
Observe the following frequency polygon and write the answers of the questions below it.
i. Which class has the maximum number of students?
ii. Write the classes having zero frequency.
iii. What is the class mark of the class, having frequency of 50 students?
iv. Write the lower and upper class limits of the class whose class mark is 85.
v. How many students are in the class 80 – 90?

Solution:
i. The class 60 – 70 has the maximum number of students.
ii. The classes 20 – 30 and 90 – 100 have frequency zero.
iii. The class mark of the class having 50 students is 55.
iv. The lower and upper class limits of the class having class mark 85 are 80 and 90 respectively.
v. There are 15 students in the class 80 – 90.

Question 2.
Show the following data by a frequency polygon.

Solution:

Question 3.
The following table shows the classification of percentages of marks of students and the number of students. Draw a frequency polygon from the table.

Solution: