Class 10

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.

10th Standard Maths 2 Problem Set 2 Chapter 2 Pythagoras Theorem Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Problem Set 2 Chapter 2 Pythagoras Theorem Questions With Answers Maharashtra Board

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

Class 10 Maths Digest

• Practice Set 2.1 Class 10 Answers
• Practice Set 2.2 Class 10 Answers
• Problem Set 2 Class 10 Answers
• Practice Set 3.1 Class 10 Answers
• Practice Set 3.2 Class 10 Answers
• Practice Set 3.3 Class 10 Answers
• Practice Set 3.4 Class 10 Answers
• Practice Set 3.5 Class 10 Answers
• Problem Set 3 Class 10 Answers

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.

10th Standard Maths 2 Practice Set 2.2 Chapter 2 Pythagoras Theorem Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 2.2 Chapter 2 Pythagoras Theorem Questions With Answers Maharashtra Board

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

Class 10 Maths Digest

• Practice Set 2.1 Class 10 Answers
• Practice Set 2.2 Class 10 Answers
• Problem Set 2 Class 10 Answers
• Practice Set 3.1 Class 10 Answers
• Practice Set 3.2 Class 10 Answers
• Practice Set 3.3 Class 10 Answers
• Practice Set 3.4 Class 10 Answers
• Practice Set 3.5 Class 10 Answers
• Problem Set 3 Class 10 Answers

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.

10th Standard Maths 2 Practice Set 2.1 Chapter 2 Pythagoras Theorem Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 2.1 Chapter 2 Pythagoras Theorem Questions With Answers Maharashtra Board

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.

Class 10 Maths Digest

• Practice Set 2.1 Class 10 Answers
• Practice Set 2.2 Class 10 Answers
• Problem Set 2 Class 10 Answers
• Practice Set 3.1 Class 10 Answers
• Practice Set 3.2 Class 10 Answers
• Practice Set 3.3 Class 10 Answers
• Practice Set 3.4 Class 10 Answers
• Practice Set 3.5 Class 10 Answers
• Problem Set 3 Class 10 Answers

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

10th Standard Maths 2 Problem Set 1 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Problem Set 1 Chapter 1 Similarity Questions With Answers Maharashtra Board

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

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.4 Algebra 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.4 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.4 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.
Solution:
Let the corresponding sides of similar triangles be S₁ and S₂.
Let A₁ and A₂ be their corresponding areas.

∴ Ratio of areas of similar triangles = 9 : 25

Question 2.
If ∆ABC ~ ∆PQR and AB : PQ = 2:3, then fill in the blanks.
Solution:

Question 3.
If ∆ABC ~ ∆PQR, A(∆ABC) = 80, A(∆PQR) = 125, then fill in the blanks.
Solution:

Question 4.
∆LMN ~ ∆PQR, 9 × A(∆PQR) = 16 × A(∆LMN). If QR = 20, then find MN.
Solution:
9 × A(∆PQR) = 16 × A(∆LMN) [Given]
∴ \(\frac{\mathrm{A}(\Delta \mathrm{LMN})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{9}{16}\) (i)
Now, ∆LMN ~ ∆PQR [Given]
∴ \(\frac{\mathrm{A}(\Delta \mathrm{LMN})}{\mathrm{A}(\Delta \mathrm{PQR})}=\frac{\mathrm{MN}^{2}}{\mathrm{QR}^{2}}\) (ii) [Theorem of areas of similar triangles]
∴ \(\frac{\mathrm{MN}^{2}}{\mathrm{QR}^{2}}=\frac{9}{16}\) [From (i) and (ii)]
∴ \(\frac{M N}{Q R}=\frac{3}{4}\) [Taking square root of both sides]
∴ \(\frac{\mathrm{MN}}{20}=\frac{3}{4}\)
∴ MN = \(\frac{20 \times 3}{4}\)
∴ MN = 15 units

Question 5.
Areas of two similar triangles are 225 sq. cm. and 81 sq. cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle.
Solution:
Let the areas of two similar triangles be A₁ and A₂.
A₁ = 225 sq. cm. A₂ = 81 sq. cm.
Let the corresponding sides of triangles be S₁ and S₂ respectively.
S₁ = 12 cm

∴ The length of the corresponding side of the bigger triangle is 20 cm.

Question 6.
∆ABC and ∆DEF are equilateral triangles. If A(∆ABC): A(∆DEF) = 1:2 and AB = 4, find DE.
Solution:
In ∆ABC and ∆DEF,

Question 7.
In the adjoining figure, seg PQ || seg DE, A(∆PQF) = 20 sq. units, PF = 2 DP, then find A (꠸ DPQE) by completing the following activity.
Solution:
A(∆PQF) = 20 sq.units, PF = 2 DP, [Given]
Let us assume DP = x.
∴ PF = 2x

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

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

10th Standard Maths 1 Problem Set 6.1 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Problem Set 6.1 Chapter 6 Statistics Questions With Answers Maharashtra Board

10th Geometry Problem Set 6 Question 1.
Find the correct answer from the alternatives given.

i. The persons of O – blood group are 40%. The classification of persons based on blood groups is to be shown by a pie diagram. What should be the measures of angle for the persons of O – blood group?
(A) 114°
(B) 140°
(C) 104°
(D) 144°
Answer:
Measure of the central angle = \(\frac { 40 }{ 100 } \) × 360° = 144°
(D)

ii. Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure of ₹ 45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction?
(A) 2,16,000
(B) 3,60,000
(C) 4,50,000
(D) 7,50,000
Answer:
Measure of the central angle = \(\frac{\text { Expenditure of cement }}{\text { Total expenditure }} \times 360^{\circ}\)

(A)

iii. Cumulative frequencies in a grouped frequency table are useful to find.
(A) Mean
(B) Median
(C) Mode
(D) All of these
Answer:
(B)

iv. The formula to find mean from a grouped frequency table is \(\overline{\mathrm{X}}=\mathrm{A}+\frac{\sum f_{i} u_{i}}{\sum f_{i}} \times g\)
in the formula u_i = _________.

Answer:

(C)

v.

The median of the distances covered per litre shown in the above data is in the group
(A) 12 – 14
(B) 14 – 16
(C) 16 – 18
(D) 18 – 20
Answer:
(C)

vi.

The above data is to be shown by a frequency polygon. The coordinates of the points to show number of students in the class 4 – 6 are.
(A) (4, 8)
(B) (3,5)
(C) (5,8)
(D) (8,4)
Answer:
Class mark = 5
Frequency = 8
∴ Co-ordinates of the point = (5, 8)
(C)

Statistics Problem Set 6 Question 2.
The following table shows the income of farmers in a grape season. Find the mean of their income.

Solution:

∴ The mean of the income of the farmers is ₹ 52,500.

Statistics Problem Set Question 3.
The loans sanctioned by a bank for construction of farm ponds are shown in the following table. Find the mean of the loans.

Solution:

∴ The mean of the loans given by the bank is ₹ 65,400.

Question 4.
The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.

Solution:

∴ The mean of the weekly wages of the workers is ₹ 4250.

Problem Set 6 Algebra Class 9 Question 5.
The following frequency distribution table shows the amount of aid given to 50 flood affected families. Find the mean of the amount of aid.

Solution:

∴ The mean of the amount of aid given to families is ₹ 72,400.
[Note: The above problems are solved using direct method. Students can solve these problems by using other method.]

Problem Set 6 Algebra Class 10 Question 6.
The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distances.

Solution:

Cumulative frequency which is just greater than (or equal) to 125 is 180.
∴ The median class is 220 – 230.
Now, L = 220, f = 80, cf = 100, h = 10

∴ The median of the distances is 223.13 km (approx.).

Algebra 10th Class Problem Set 6 Question 7.
The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.

Solution:

Cumulative frequency which is just greater than (or equal) to 200 is 240.
∴ The median class is 20 – 40.
Now,L = 20, f = 100,cf = 140, h = 20

∴ The median of the prices of different articles is ₹ 32.

10th Algebra Problem Set 6 Question 8.
The following frequency table shows the demand for a sweet and the number of customers. Find the mode of demand of sweet.

Solution:


∴ The mode of the demand of sweet is 397.06 grams.

Question 9.
Draw a histogram for the following frequency distribution.

Solution:

Question 10.
In a handloom factory different workers take different periods of time to weave a saree. The number of workers and their required periods are given below. Present the information by a frequency polygon.

Solution:

Problem Set 6 Question 11.
The time required for students to do a science experiment and the number of students is shown in the following grouped frequency distribution table. Show the information by a histogram and also by a frequency polygon.

Solution:

Question 12.
Draw a frequency polygon for the following grouped frequency distribution table.

Solution:

Question 13.
The following table shows the average rainfall in 150 towns. Show the information by a frequency polygon.

Solution:

Question 14.
Observe the given pie diagram. It shows the percentages of number of vehicles passing a signal in a town between 8 am and 10 am.
i. Find the central angle for each type of vehicle.
ii. If the number of two-wheelers is 1200, find the number of all vehicles.

Solution:

∴ The total number of vehicles is 3000.

Problem Set 6 Geometry Class 10 Question 15.
The following table shows causes of noise pollution. Show it by a pie diagram.

Solution:

Question 16.
A survey of students was made to know which game they like. The data obtained in the survey is
presented in the given pie diagram. If the total number of students are 1000,
i. how many students like cricket?
ii. how many students like football?
iii. how many students prefer other games?

Solution:

∴ 225 students like cricket.

ii. Central angle for football (θ) = 63°

∴ 175 students like football.

iii. Central angle for other games (θ) = 72°

∴ 200 students like other games.

Question 17.
Medical check up of 180 women was conducted in a health centre in a village. 50 of them were short of hemoglobin, 10 suffered from cataract and 25 had respiratory disorders. The remaining women were healthy. Show the information by a pie diagram.
Solution:
Total number of women = 180

Question 18.
On an environment day, students in a school planted 120 trees under plantation project. The information regarding the project is shown in the following table. Show it by a pie diagram.

Solution:
Total number of trees planted = 120

Maharashtra Board Class 10 Maths Solutions

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

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.

10th Standard Maths 1 Practice Set 6.6 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.6 Chapter 6 Statistics Questions With Answers Maharashtra Board

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.

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

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.

10th Standard Maths 1 Practice Set 6.5 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.5 Chapter 6 Statistics Questions With Answers Maharashtra Board

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:

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.4 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.4 Chapter 6 Statistics Questions With Answers Maharashtra Board

Question 1.
Draw a histogram of the following data.

Solution:

Question 2.
The table below shows the yield of jowar per acre. Show the data by histogram.

Solution:

Question 3.
In the following table, the investment made by 210 families is shown. Present it in the form of a histogram.

Solution:

Question 4.
Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.

Solution:

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.3 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.3 Chapter 6 Statistics Questions With Answers Maharashtra Board

Question 1.
The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.

Solution:

Here, the maximum frequency is 80.
∴ The modal class is 4 – 5.
L = lower class limit of the modal class = 4
h = class interval of the modal class = 1
f₁ = frequency of the modal class = 80
f₀ = frequency of the class preceding the modal class = 70
f₂ = frequency of the class succeeding the modal class = 60

∴ The mode of the fat content is 4.33%.

Question 2.
Electricity used by some families is shown in the following table. Find the mode of use of electricity.

Solution:

Here, the maximum frequency is 100.
∴ The modal class is 60 – 80.
L = lower class limit of the modal class = 60
h = class interval of the modal class = 20
f₁ = frequency of the modal class = 100
f₀ = frequency of the class preceding the modal class = 70
f₂ = frequency of the class succeeding the modal class = 80

∴ The mode of use of electricity is 72 units.

Question 3.
Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.

Solution:

Here, the maximum frequency is 35.
∴ The modal class is 9 – 11.
L = lower class limit of the modal class = 9
h = class interval of the modal class = 2
f₁ = frequency of the modal class = 35
f₀ = frequency of the class preceding the modal class = 20
f₂ = frequency of the class succeeding the modal class = 18

∴ The mode of the supply of milk is 9.94 litres (approx.).

Question 4.
The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.

Solution:

Here, the maximum frequency is 50.
The modal class is 9.5 – 14.5.
L = lower class limit of the modal class = 9.5
h = class interval of the modal class = 5
f₁ = frequency of the modal class = 50
f₀ = frequency of the class preceding the modal class = 32
f₂ = frequency of the class succeeding the modal class = 36

∴ The mode of the ages of the patients is 12.31 years (approx.).

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers