Prasanna

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 1 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 1 Solutions Maharashtra Board

Std 7 Maths Practice Set 1 Solutions Answers

Question 1.
Draw line segments of the lengths given below and draw their perpendicular bisectors:
i. 5.3 cm
ii. 6.7 cm
iii. 3.8 cm
Solution:
i.

Line AB is the perpendicular bisector of seg PQ.

ii.

Line UV is the perpendicular bisector of seg ST.

iii.

Line ST is the perpendicular bisector of seg LM.

Question 2.
Draw angles of the measures given below and draw their bisectors:
i. 105°
ii. 55°
iii. 90°
Solution:
i. 105°

ii. 55°

iii. 90°

Question 3.
Draw, an obtuse-angled triangle and a right-angled triangle. Find the points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
Solution:

The points of concurrence of the angle bisectors of both the triangles lie in the interior of the triangles.

Question 4.
Draw a right-angled triangle. Draw the perpendicular bisectors of its sides. Where does the point of concurrence lie?
Solution:

The point of concurrence of the perpendicular bisectors of the sides of the right angled triangle lies on the hypotenuse.

Question 5.
Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.
Solution:
Since, Maithili, Shaila and Ajay live in three different places, lines joining their houses will form a triangle.
The position of the toy shop which is equidistant from three houses can be found out by drawing the perpendicular bisector of the sides of the triangle joining the three houses.
The shop will be at the point of concurrence of the perpendicular bisectors.

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 1 Intext Questions and Activities

Question 1.
Draw a line segment PS of length 4cm and draw its perpendicular bisector. (Textbook pg. no. 1)

1. How will your verify that CD is the perpendicular bisector? m∠CMS = __°
2. Is l(PM) = l(SM)?

Solution:

1. Here, m∠CMS = 90°
2. Also, l(PM) = l(SM) = 2cm
   ∴ line CD is the perpendicular bisector of seg PS.

Std 7 Maths Digest

• Practice Set 1 Class 7 Answers
• Practice Set 2 Class 7 Answers
• Practice Set 3 Class 7 Answers
• Practice Set 4 Class 7 Answers
• Practice Set 5 Class 7 Answers
• Practice Set 6 Class 7 Answers
• Practice Set 7 Class 7 Answers

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 24 Answers Solutions Chapter 5 Operations on Rational Numbers.

Operations on Rational Numbers Class 7 Maths Chapter 5 Practice Set 24 Solutions Maharashtra Board

Std 7 Maths Practice Set 24 Solutions Answers

Question 1.
Write the following rational numbers in decimal form.
i. \(\frac { 13 }{ 4 }\)
ii. \(\frac { -7 }{ 8 }\)
iii. \(7\frac { 3 }{ 5 }\)
iv. \(\frac { 5 }{ 12 }\)
v. \(\frac { 22 }{ 7 }\)
vi. \(\frac { 4 }{ 3 }\)
vii. \(\frac { 7 }{ 9 }\)
Solution:
i. \(\frac { 13 }{ 4 }\)

ii. \(\frac { -7 }{ 8 }\)

iii. \(7\frac { 3 }{ 5 }\)

iv. \(\frac { 5 }{ 12 }\)

v. \(\frac { 22 }{ 7 }\)

vi. \(\frac { 4 }{ 3 }\)

vii. \(\frac { 7 }{ 9 }\)

Maharashtra Board Class 7 Maths Chapter 5 Operations on Rational Numbers Practice Set 24 Intext Questions and Activities

Question 1.
Without using division, can we tell from the denominator of a fraction, whether the decimal form of the fraction will be a terminating decimal? Find out. (Textbook pg. no. 40)
Solution:
If the prime factorization of the denominator of a fraction has only factors as 2 or 5 or a combination of 2 and 5 then the decimal form of that fractional will be a terminating decimal form.
Consider the fractions \(\frac { 17 }{ 20 }\) and \(\frac { 19 }{ 6 }\)
Now, 20 = 2 x 2 x 5, and 6 = 2 x 3
∴ \(\frac { 17 }{ 20 }\) is terminating decimal form while \(\frac { 19 }{ 6 }\) is recurring decimal form.

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 4 Answers Solutions.

Integers Class 6 Maths Chapter 3 Practice Set 4 Solutions Maharashtra Board

Std 6 Maths Practice Set 4 Solutions Answers

Question 1.
Classify the following numbers as positive numbers and negative numbers.
-5, +4, -2, 7, +26, -49, -37, 19, -25, +8, 5, -4, -12, 27
Solution:

Question 2.
Given below are the temperatures in some cities. Write them using the proper signs.

Solution:

Question 3.
Write the numbers in the following examples using the proper signs.

1. A submarine is at a depth of 512 meters below sea level.
2. The height of Mt Everest, the highest peak in the Himalayas, is 8848 meters.
3. A kite is flying at a distance of 120 meters from the ground.
4. The tunnel is at a depth of 2 meters under the ground.

Solution:

1. A submarine is at a depth of -512 meters from sea level.
2. The height of Mt Everest, the highest peak in the Himalayas is +8848 meters.
3. A kite is flying at a distance of +120 meters from the ground.
4. The tunnel is at a depth of -2 meters from the ground.

Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 4 Intext Questions and Activities

Question 1.
Take warm water in one beaker, some crushed ice in another and a mixture of salt and crushed ice in a third beaker. Ask your teacher for help in measuring the temperature of the substance in each of the beakers using a thermometer. Note the temperatures. (Textbook pg. no. 13)

Solution:
( Students should attempt this activity on their own)

Question 2.
Look at the picture of the kulfi man. Why do you think he keeps the kulfi moulds in a mixture of salt and ice? (Textbook pg. no. 14)

Solution:
Kulfi man keeps the kulfi moulds in a mixture of salt and ice because such a mixture helps in keeping the kulfi cool for a longer period of time. The kulfi kept in the said mixture relatively takes more time to melt. This mixture is Considered ideal as it has the temperature of -4°C as against the temperature of ice i.e. 0°C.

Question 3.
My class, i.e. Std. VI, is a part of my school. My school is in my town. My town is a part of a taluka. In the same way, the taluka is a part of a district, and the district is a part of Maharashtra State. In the same way, what can you say about these groups of numbers? Textbook pg. no. 15)

Solution:
By observing the above given groups of numbers, we can infer that natural numbers are a part of whole numbers. In turn, whole numbers are a part of integers.

Std 6 Maths Digest

• Practice Set 1 Class 6 Answers
• Practice Set 2 Class 6 Answers
• Practice Set 3 Class 6 Answers
• Practice Set 4 Class 6 Answers
• Practice Set 5 Class 6 Answers
• Practice Set 6 Class 6 Answers
• Practice Set 7 Class 6 Answers
• Practice Set 8 Class 6 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

9th Standard Maths 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 1.
Find the distances with the help of the number line given below.

i. d(B, E)
ii. d (J, J)
iii. d(P, C)
iv. d(J, H)
v. d(K, O)
vi. d(O, E)
vii. d(P, J)
viii. d(Q, B)
Solution:
i. Co-ordinate of the point B is 2.
Co-ordinate of the point E is 5.
Since, 5 > 2
∴ d(B, E) = 5 – 2
∴ d(B, E) = 3

ii. Co-ordinate of the point J is -2.
Co-ordinate of the point A is 1.
Since, 1 > -2
∴ d(J, A) = 1 – (-2)
= 1 + 2
∴ d(J, A) = 3

iii. Co-ordinate of the point P is -4.
Co-ordinate of the point C is 3.
Since, 3 > -4
∴ d(P,C) = 3 – (-4)
= 3 + 4
∴ d(P,C) = 7

iv. Co-ordinate of the point J is -2.
Co-ordinate of the point H is -1.
Since, -1 > -2
∴ d(J,H) = – 1 – (-2)
= -1 + 2
∴ d(J,H) = 1

v. Co-ordinate of the point K is -3.
Co-ordinate of the point O is 0.
Since,0 > -3
∴ d(K, O) = 0 – (-3)
= 0 + 3
∴ d(K, O) = 3

vi. Co-ordinate of the point O is 0.
∴ Co-ordinate of the point E is 5.
Since, 5 > 0
∴ d(O, E) = 5 – 0
∴ d(O, E) = 5

vii. Co-ordinate of the point P is -4.
Co-ordinate of the point J is -2.
Since -2 > -4
∴ d(P, J) = -2 – (-4)
= – 2+ 4
∴ d(P, J) = 2

viii. Co-ordinate of the point Q is -5.
Co-ordinate of the point B is 2.
Since,2 > -5
∴ d(Q,B) = 2 – (-5)
= 2 + 5
∴ d(Q, B) = 7

Question 2.
If the co-ordinate of A is x and that of B is . y, find d(A, B).
i. x = 1, y = 7
ii. x = 6, y = -2
iii. x = -3, y = 7
iv. x = -4, y = -5
v. x = -3, y = -6
vi. x = 4, y = -8
Solution:
i. Co-ordinate of point A is x = 1.
Co-ordinate of point B is y = 7
Since, 7 > 1
∴ d(A, B) = 7 – 1
∴ d(A, B) = 6

ii. Co-ordinate of point A is x = 6.
Co-ordinate of point B is y = -2.
Since, 6 > -2
∴ d(A, B) = 6 – ( -2) = 6 + 2
∴ d(A, B) = 8

iii. Co-ordinate of point A is x = -3.
Co-ordinate of point B is y = 7.
Since, 7 > -3
∴ d(A, B) = 7 – (-3) = 7 + 3
∴ d(A, B) = 10

iv. Co-ordinate of point A is x = -4.
Co-ordinate of point B is y = -5.
Since, -4 > -5
∴ d(A, B) = -4 – (-5)
= -4 + 5
∴ d(A, B) = 1

v. Co-ordinate of point A is x =-3.
Co-ordinate of point B is y = -6.
Since, -3 > -6
∴ d(A, B) = -3 – (-6)
= -3 + 6
∴ d(A, B) = 3

vi. Co-ordinate of point A is x = 4.
Co-ordinate of point B is y = -8.
Since, 4 > -8
∴ d(A, B) = 4 – (-8)
= 4 + 8
∴d(A, B) = 12

Question 3.
From the information given below, find which of the point is between the other two. If the points are not collinear, state so.
i. d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
ii. d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
iii. d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
iv. d(L, M) =11, d(M, N) = 12, d(N, L) = 8
v. d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
vi. d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Solution:
i. Given, d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
d(P, Q) = 10 …(i)
d(P, R) + d(Q, R) = 7 + 3 = 10 .. .(ii)
∴ d(P, Q) = d(P, R) + d(Q, R) …[From (i) and (ii)]
∴ Point R is between the points P and Q
i. e., P – R – Q or Q – R – P.
∴ Points P, R, Q are collinear.

ii. Given, d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
d(R, S) = 8 …(i)
d(S, T) + d(R, T) = 6 + 4 = 10 …(h)
∴ d(R, S) ≠ d(S, T) + d(R, T) … [From (i) and (ii)]
∴ The given points are not collinear.

iii. Given, d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
d(A, B) = 16 …(i)
d(C, A) + d(B, C) = 9 + 7 = 16 …(ii)
∴ d(A, B) = d(C, A) + d(B, C) …[From(i) and (ii)]
∴ Point C is between the points A and B.
i. e., A – C – B or B – C – A.
∴ Points A, C, B are collinear

iv. Given, d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
d(M, N) = 12 …(i)
d(L, M) + d(N, L) = 11 + 8 = 19 …(ii)
∴d(M, N) + d(L, M) + d(N, L) … [From (i) and (ii)]
∴ The given points are not collinear.

v. Given, d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
d(X, Y) = 15 …(i)
d(X,Z) + d(Y, Z) = 8 + 7= 15 …(ii)
∴ d(X, Y) = d(X, Z) + d(Y, Z) …[From (i) and (ii)]
∴ Point Z is between the points X and Y
i. e.,X – Z – Y or Y – Z – X.
∴ Points X, Z, Y are collinear.

vi. Given, d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
d(E, F) = 8 …(i)
d(D, E) + d(D, F) = 5 + 6 = 11 …(ii)
∴ d(E, F) ≠ d(D, E) + d(D, F) … [From (i) and (ii)]
∴ The given points are not collinear.

Question 4.
On a number line, points A, B and C are such that d(A, C) = 10, d(C, B) = 8. Find d(A, B) considering all possibilities.
Solution:
Given, d(A, C) = 10, d(C, B) = 8.

Case I: Points A, B, C are such that, A – B – C.

∴ d(A, C) = d(A, B) + d(B, C)
∴ 10 = d(A, B) + 8
∴ d(A, B) = 10 – 8
∴ d(A, B) = 2

Case II: Points A, B, C are such that, A – C – B.

∴ d(A, B) = d(A, C) + d(C, B)
= 10 + 8
∴ d(A, B) = 18

Case III: Points A, B, C are such that, B – A – C.

From the diagram,
d (A, C) > d(B, C)
Which is not possible
∴ Point A is not between B and C.
∴ d(A, B) = 2 or d(A, B) = 18.

Question 5.
Points X, Y, Z are collinear such that d(X, Y) = 17, d(Y, Z) = 8, find d(X, Z).
Solution:
Given,d(X, Y) = 17, d(Y, Z) = 8
Case I: Points X, Y, Z are such that, X – Y – Z.

∴ d(X, Z) = d(X, Y) + d(Y, Z)
= 17 + 8
∴ d(X, Z) = 25

Case II: Points X, Y, Z are such that, X – Z – Y.

∴ d(X,Y) = d(X,Z) + d(Z,Y)
∴ 17 = d(X, Z) + 8
∴ d(X, Z) = 17 – 8
∴ d(X, Z) = 9

Case III: Points X, Y, Z are such that, Z – X – Y.

From the diagram,
d(X, Y) > d (Y, Z)
Which is not possible
∴ Point X is not between Z and Y.
∴ d(X, Z) = 25 or d(X, Z) = 9.

Question 6.
Sketch proper figure and write the answers of the following questions. [2 Marks each]
i. If A – B – C and l(AC) = 11,
l(BC) = 6.5, then l(AB) = ?
ii. If R – S – T and l(ST) = 3.7,
l(RS) = 2.5, then l(RT) = ?
iii. If X – Y – Z and l(XZ) = 3√7,
l(XY) = √7, then l(YZ) = ?
Solution:
i. Given, l(AC) =11, l(BC) = 6.5

l(AC) = l(AB) + l(BC) … [A – B – C]
∴ 11= l(AB) + 6.5
∴ l(AB) = 11 – 6.5
∴ l(AB) = 4.5

ii. Given, l(ST) = 3.7, l(RS) = 2.5

l(RT) = l(RS) + l(ST) … [R – S – T]
= 2.5 + 3.7
∴ (RT) = 6.2

iii. l(XZ) = 3√7 , l(XY) = √7,

l(XZ) = l(X Y) + l(YZ) … [X – Y – Z]
∴ 3 √7 ⇒ √7 + l(YZ)
∴ l(YZ)= 3√7 – √7
∴ l(YZ) = 2 √7

Question 7.
Which figure is formed by three non-collinear points?
Solution:
Three non-collinear points form a triangle.

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Problem Set 2 Class 9 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.3 Geometry Class 10 Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.
i. 45°
ii. 60°
iii. 90°
Solution:
i. Angle made with the positive direction of
X-axis (θ) = 45°
Slope of the line (m) = tan θ
∴ m = tan 45° = 1
∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis (θ) = 60°
Slope of the line (m) = tan θ
∴ m = tan 60° = \(\sqrt { 3 }\)
∴ The slope of the line is \(\sqrt { 3 }\).

iii. Angle made with the positive direction of
X-axis (θ) = 90°
Slope of the line (m) = tan θ
∴ m = tan 90°
But, the value of tan 90° is not defined.
∴ The slope of the line cannot be determined.

Practice Set 5.3 Geometry Question 2. Find the slopes of the lines passing through the given points.
i. A (2, 3), B (4, 7)
ii. P(-3, 1), Q (5, -2)
iii. C (5, -2), D (7, 3)
iv. L (-2, -3), M (-6, -8)
v. E (-4, -2), F (6, 3)
vi. T (0, -3), s (0,4)
Solution:
i. A (x₁, y₁) = A (2, 3) and B (x₂, y₂) = B (4, 7)
Here, x₁ = 2, x₂ = 4, y₁ = 3, y₂ = 7

∴ The slope of line AB is 2.

ii. P (x₁, y₁) = P (-3, 1) and Q (x₂, y₂) = Q (5, -2)
Here, x₁ = -3, x₂ = 5, y₁ = 1, y₂ = -2

∴ The slope of line PQ is \(\frac { -3 }{ 8 } \)

iii. C (x₁, y₁) = C (5, -2) and D (x₂, y₂) = D (7, 3)
Here, x₁ = 5, x₂ = 7, y₁ = -2, y₂ = 3

∴ The slope of line CD is \(\frac { 5 }{ 2 } \)

iv. L (x₁, y₁) = L (-2, -3) and M (x₂,y₂) = M (-6, -8)
Here, x₁ = -2, x₂ = – 6, y₁ = – 3, y₂ = – 8

∴ The slope of line LM is \(\frac { 5 }{ 4 } \)

v. E (x₁, y₁) = E (-4, -2) and F (x₂, y₂) = F (6, 3)
Here,x₁ = -4, x₂ = 6, y₁ = -2, y₂ = 3

∴ The slope of line EF is \(\frac { 1 }{ 2 } \).

vi. T (x₁, y₁) = T (0, -3) and S (x₂, y₂) = S (0, 4)
Here, x₁ = 0, x₂ = 0, y₁ = -3, y₂ = 4

∴ The slope of line TS cannot be determined.

5.3.5 Practice Question 3. Determine whether the following points are collinear.
i. A (-1, -1), B (0, 1), C (1, 3)
ii. D (- 2, -3), E (1, 0), F (2, 1)
iii. L (2, 5), M (3, 3), N (5, 1)
iv. P (2, -5), Q (1, -3), R (-2, 3)
v. R (1, -4), S (-2, 2), T (-3,4)
vi. A(-4,4),K[-2,\(\frac { 5 }{ 2 } \)], N (4,-2)
Solution:


∴ slope of line AB = slope of line BC
∴ line AB || line BC
Also, point B is common to both the lines.
∴ Both lines are the same.
∴ Points A, B and C are collinear.

∴ slope of line DE = slope of line EF
∴ line DE || line EF
Also, point E is common to both the lines.
∴ Both lines are the same.
∴ Points D, E and F are collinear.

∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.

∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.

∴ slope of line RS = slope of line ST
∴ line RS || line ST
Also, point S is common to both the lines.
∴ Both lines are the same.
∴ Points R, S and T are collinear.

∴ slope of line AK = slope of line KN
∴ line AK || line KN
Also, point K is common to both the lines.
∴ Both lines are the same.
∴ Points A, K and N are collinear.

Practice Set 5.3 Geometry 9th Standard Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.
Solution:

∴ The slopes of the sides AB, BC and AC are -5, \(\frac { 1 }{ 5 } \) and \(\frac { -2 }{ 3 } \) respectively.

Geometry 5.3 Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.
Proof:


∴ Slope of side AB = Slope of side CD … [From (i) and (iii)]
∴ side AB || side CD
Slope of side BC = Slope of side AD … [From (ii) and (iv)]
∴ side BC || side AD
Both the pairs of opposite sides of ꠸ABCD are parallel.
꠸ABCD is a parallelogram.
Points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram.

Question 6.
Find k, if R (1, -1), S (-2, k) and slope of line RS is -2.
Solution:
R(x₁, y₁) = R (1, -1), S (x₂, y₂) = S (-2, k)
Here, x₁ = 1, x₂ = -2, y₁ = -1, y₂ = k

But, slope of line RS is -2. … [Given]
∴ -2 = \(\frac { k+1 }{ -3 } \)
∴ k + 1 = 6
∴ k = 6 – 1
∴ k = 5

5.3 Class 10 Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.
Solution:
B(x₁, y₁) = B (k, -5), C (x₂, y₂) = C (1, 2)
Here, x₁ = k, x₂ = 1, y₁ = -5, y₂ = 2

But, slope of line BC is 7. …[Given]
∴ 7 = \(\frac { 7 }{ 1-k } \)
∴ 7(1 – k) = 7
∴ 1 – k = \(\frac { 7 }{ 7 } \)
∴ 1 – k = 1
∴ k = 0

Question 8.
Find k, if PQ || RS and P (2, 4), Q (3, 6), R (3,1), S (5, k).
Solution:

But, line PQ || line RS … [Given]
∴ Slope of line PQ = Slope of line RS
∴ 2 = \(\frac { k-1 }{ 2 } \)
∴ 4 = k – 1
∴ k = 4 + 1
∴ k = 5

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Find the co-ordinates of point P if P divides the line segment joining the points A (-1, 7) and B (4, -3) in the ratio 2:3.
Solution:
Let the co-ordinates of point P be (x, y) and A (x₁, y₁) B (x₂, y₂) be the given points.
Here, x₁ = -1, y₁ = 7, x₂ = 4, y₂ = -3, m = 2, n = 3
∴ By section formula,

∴ The co-ordinates of point P are (1,3).

Question 2.
In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
i. P (-3, 7), Q (1, -4), a : b = 2 : 1
ii. P (-2, -5), Q (4, 3), a : b = 3 : 4
iii. P (2, 6), Q (-4, 1), a : b = 1 : 2
Solution:
Let the co-ordinates of point A be (x, y).
i. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -3, y₁ = 7, x₂ = 1, y₂ = -4, a = 2, b = 1
∴ By section formula,

∴ The co-ordinates of point A are (\(\frac { -1 }{ 3 } \),\(\frac { -1 }{ 3 } \)).

ii. Let P (x₁,y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -2, y₁ = -5, x₂ = 4, y₂ = 3, a = 3, b = 4
By section formula,

∴ The co-ordinates of point A are (\(\frac { 4 }{ 7 } \),\(\frac { -11 }{ 7 } \))

iii. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here,x₁ = 2,y₁ = 6, x₂ = -4, y₂ = 1, a = 1,b = 2
∴ By section formula,

∴ The co-ordinates of point A are (0,\(\frac { 13 }{ 3 } \))

Question 3.
Find the ratio in which point T (-1, 6) divides the line segment joining the points P (-3,10) and Q (6, -8).
Solution:
Let P (x₁, y₁), Q (x₂, y₂) and T (x, y) be the given points.
Here, x₁ = -3, y₁ = 10, x₂ = 6, y₂ = -8, x = -1, y = 6
∴ By section formula,

∴ Point T divides seg PQ in the ratio 2 : 7.

Question 4.
Point P is the centre of the circle and AB is a diameter. Find the co-ordinates of point B if co-ordinates of point A and P are (2, -3) and (-2,0) respectively.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 2, y₁ =-3,
x = -2, y = 0

Point P is the midpoint of seg AB.
∴ By midpoint formula,

∴ The co-ordinates of point B are (-6,3).

Question 5.
Find the ratio in which point P (k, 7) divides the segment joining A (8, 9) and B (1,2). Also find k.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 8, y₁ = 9, x₂ = 1, y₂ = 2, x = k, y = 7
∴ By section formula,

∴ Point P divides seg AB in the ratio 2 : 5, and the value of k is 6.

Question 6.
Find the co-ordinates of midpoint of the segment joining the points (22, 20) and (0,16).
Solution:
Let A (x₁, y₁) = A (22, 20),
B (x₂,y₂) = B (0, 16)
Let the co-ordinates of the midpoint be P (x,y).
∴ By midpoint formula,

The co-ordinates of the midpoint of the segment joining (22, 20) and (0, 16) are (11,18).

Question 7.
Find the centroids of the triangles whose vertices are given below.
i. (-7, 6), (2,-2), (8, 5)
ii. (3, -5), (4, 3), (11,-4)
iii. (4, 7), (8, 4), (7, 11)
Solution:
i. Let A (x₁, y₁) = A (-7, 6),
B (x₂, y₂) = B (2, -2),
C (x₃, y₃) = C(8, 5)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (1,3).

ii. Let A (x₁ y₁) = A (3, -5),
B (x₂, y₂) = B (4, 3),
C(x₃, y₃) = C(11,-4)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (6, -2).

iii. Let A (x₁, y₁) = A (4, 7),
B (x₂, y₂) = B (8,4),
C (x₃, y₃) = C(7,11)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (\(\frac { 19 }{ 3 } \),\(\frac { 22 }{ 3 } \))

Question 8.
In ∆ABC, G (-4, -7) is the centroid. If A (-14, -19) and B (3, 5), then find the co-ordinates of C.
Solution:
G (x, y) = G (-4, -7),
A (x₁, y₁) = A (-14, -19),
B(x₂, y₂) = B(3,5)
Let the co-ordinates of point C be (x₃, y₃).
G is the centroid.
By centroid formula,

∴ The co-ordinates of point C are (-1, – 7).

Question 9.
A (h, -6), B (2, 3) and C (-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1,5). Find h and k.
Solution:
A(x₁,y₁) = A(h, -6),
B (x₂, y₂) = B(2, 3),
C (x₃, y₃) = C (-6, k)
∴ centroid G (x, y) = G (1, 5)
G is the centroid.
By centroid formula,

∴ 3 = h – 4
∴ h = 3 + 4
∴ h = 7

∴ 15 = -3 + k
∴ k = 15 + 3
∴ k = 18
∴ h = 7 and k = 18

Question 10.
Find the co-ordinates of the points of trisection of the line segment AB with A (2,7) and B (-4, -8).
Solution:
A (2, 7), B H,-8)
Suppose the points P and Q trisect seg AB.
∴ AP = PQ = QB

∴ Point P divides seg AB in the ratio 1:2.
∴ By section formula,

Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,

Co-ordinates of Q are (-2, -3).
∴ The co-ordinates of the points of trisection seg AB are (0,2) and (-2, -3).

Question 11.
If A (-14, -10), B (6, -2) are given, find the co-ordinates of the points which divide segment AB into four equal parts.
Solution:
Let the points C, D and E divide seg AB in four equal parts.

Point D is the midpoint of seg AB.
∴ By midpoint formula,

∴ Co-ordinates of D are (-4, -6).
Point C is the midpoint of seg AD.
∴ By midpoint formula,

∴ Co-ordinates of C are (-9, -8).
Point E is the midpoint of seg DB.
∴ By midpoint formula,


∴ Co-ordinates of E are (1,-4).
∴ The co-ordinates of the points dividing seg AB in four equal parts are C(-9, -8), D(-4, -6) and E(1, – 4).

Question 12.
If A (20, 10), B (0, 20) are given, find the co-ordinates of the points which divide segment AB into five congruent parts.
Solution:
Suppose the points C, D, E and F divide seg AB in five congruent parts.
∴ AC = CD = DE = EF = FB

∴ co-ordinates of C are (16, 12).
E is the midpoint of seg CB.
By midpoint formula,

∴ co-ordinates of E are (8, 16).
D is the midpoint of seg CE.

∴ co-ordinates of F are (4, 18).
∴ The co-ordinates of the points dividing seg AB in five congruent parts are C (16, 12), D (12, 14), E (8, 16) and F (4, 18).

Maharashtra Board Class 10 Maths Chapter 5 Co-ordinate Geometry Intext Questions and Activities

Question 1.
A (15, 5), B (9, 20) and A-P-B. Find the ratio in which point P (11, 15) divides segment AB. Find the ratio using x and y co-ordinates. Write the conclusion. (Textbook pg. no. 113)
Solution:
Suppose point P (11,15) divides segment AB in the ratio m : n.
By section formula,

∴ Point P divides seg AB in the ratio 2 : 1.
The ratio obtained by using x and y co-ordinates is the same.

Question 2.
External division: (Textbook pg. no. 115)
Suppose point R divides seg PQ externally in the ratio 3:1.

Let the common multiple be k.
Let PR = 3k and QR = k
Now, PR = PQ + QR … [P – Q – R]
∴ 3k = PQ + k
∴ \(\frac { PQ }{ QR } \) = \(\frac { 2k }{ k } \) = \(\frac { 2 }{ 1 } \)
∴ Point Q divides seg PR in the ratio 2 : 1 internally.
Thus, we can find the co-ordinates of point R, when co-ordinates of points P and Q are given.

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.1 Geometry Class 10 Question 1. Find the distance between each of the following pairs of points.
i. A (2, 3), B (4,1)
ii. P (-5, 7), Q (-1, 3)
iii. R (0, -3), S (0,\(\frac { 5 }{ 2 } \))
iv. L (5, -8), M (-7, -3)
v. T (-3, 6), R (9, -10)
vi. W(\(\frac { -7 }{ 2 } \),4), X(11, 4)
Solution:
i. Let A (x₁, y₁) and B (x₂, y₂) be the given points.
∴ x₁ = 2, y₁ = 3, x₂ = 4, y₂ = 1
By distance formula,

∴ d(A, B) = 2\(\sqrt { 2 }\) units
∴ The distance between the points A and B is 2\(\sqrt { 2 }\) units.

ii. Let P (x₁, y₁ ) and Q (x₂, y₂) be the given points.
∴ x₁ = -5, y₁ = 7, x₂ = -1, y₂ = 3
By distance formula,

∴ d(P, Q) = 4\(\sqrt { 2 }\) units
∴ The distance between the points P and Q is 4\(\sqrt { 2 }\) units.

iii. Let R (x₁, y₁) and S (x₂, y₂) be the given points.
∴ x₁ = 0, y₁ = -3, x₂ = 0, y₂ = \(\frac { 5 }{ 2 } \)
By distance formula,

∴ d(R, S) = \(\frac { 11 }{ 2 } \) units
∴ The distance between the points R and S is \(\frac { 11 }{ 2 } \) units.

iv. Let L (x₁, y₁) and M (x₂, y₂) be the given points.
∴ x₁ = 5, y₁ = -8, x₂ = -7, y₂ = -3
By distance formula,

∴ d(L, M) = 13 units
∴ The distance between the points L and M is 13 units.

v. Let T (x₁,y₁) and R (x₂, y₂) be the given points.
∴ x₁ = -3, y₁ = 6,x₂ = 9,y₂ = -10
By distance formula,

∴ d(T, R) = 20 units
∴ The distance between the points T and R 20 units.

vi. Let W (x₁, y₁) and X (x₂, y₂) be the given points.

∴ d(W, X) = \(\frac { 29 }{ 2 } \) units
∴ The distance between the points W and X is \(\frac { 29 }{ 2 } \) units.

Practice Set 5.1 Geometry 10th Question 2. Determine whether the points are collinear.
i. A (1, -3), B (2, -5), C (-4, 7)
ii. L (-2, 3), M (1, -3), N (5, 4)
iii. R (0, 3), D (2, 1), S (3, -1)
iv. P (-2, 3), Q (1, 2), R (4, 1)
Solution:
i. By distance formula,


∴ d(A, B) = \(\sqrt { 5 }\) …(i)
On adding (i) and (iii),
d(A, B) + d(A, C)= \(\sqrt { 5 }\) + 5\(\sqrt { 5 }\) = 6\(\sqrt { 5 }\)
∴ d(A, B) + d(A, C) = d(B, C) … [From (ii)]
∴ Points A, B and C are collinear.

ii. By distance formula,

On adding (i) and (iii),
d(L, M) + d(L, N) = 3\(\sqrt { 5 }\) + 5\(\sqrt { 2 }\) ≠ \(\sqrt { 65 }\)
∴ d(L, M) + d(L, N) ≠ d(M, N) … [From (ii)]
∴ Points L, M and N are not collinear.

iii. By distance formula,

On adding (i) and (ii),
∴ d(R, D) + d(D, S) = \(\sqrt { 8 }\) + \(\sqrt { 5 }\) ≠ 5
∴ d(R, D) + d(D, S) ≠ d(R, S) … [From (iii)]
∴ Points R, D and S are not collinear.

iv. By distance formula,

On adding (i) and (ii),
d(P, Q) + d(Q, R) = \(\sqrt { 10 }\) + \(\sqrt { 10 }\) = 2\(\sqrt { 10 }\)
∴ d(P, Q) + d(Q, R) = d(P, R) … [From (iii)]
∴ Points P, Q and R are collinear.

Coordinate Geometry Class 10 Practice Set 5.1 Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Solution:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
∴ its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
∴ AC = BC

∴ (x + 3)² + (-4)² = (x- 1)² + 4²
∴ x² + 6x + 9 + 16 = x² – 2x + 1 + 16
∴ 8x = – 8
∴ x = – \(\frac { 8 }{ 8 } \) = -1
∴ The point on X-axis which is equidistant from points A and B is (-1,0).

10th Geometry Practice Set 5.1 Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Solution:
Distance between two points

Consider, PQ² + QR² = 42 + 52 = 16 + 25 = 41 … [From (i) and (ii)]
∴ PR² = PQ² + QR² … [From (iii)]
∴ ∆PQR is a right angled triangle. … [Converse of Pythagoras theorem]
∴ Points P, Q and R are the vertices of a right angled triangle.

Question 5.
Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Proof:
Distance between two points

PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
∴ □ PQRS is a parallelogram.
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 6.
Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Proof:
Distance between two points


∴ AB = BC = CD = AD …[From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
∴ □ ABCD is a rhombus.
∴ Points A, B, C and D are the vertices of rhombus ABCD.

Practice Set 5.1 Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Solution:
X₁ = x, y₁ = 7, x₂ = 1, y₂ = 15
By distance formula,

∴ 1 – x = ± 6
∴ 1 – x = 6 or l – x = -6
∴ x = – 5 or x = 7
∴ The value of x is – 5 or 7.

Geometry 5.1 Question 8. Show that the points A (1, 2), B (1, 6), C (1 + 2\(\sqrt { 3 }\), 4) are vertices of an equilateral triangle.
Proof:
Distance between two points

∴ AB = BC = AC … [From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.

Maharashtra Board Class 10 Maths Chapter 5 Coordinate Geometry Intext Questions and Activities

Question 1.
In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)

Solution:
In ∆ABC, ∠B = 900
∴ (AB)² + (BC)² = [(Ac)² …(i) … [Pythagoras theorem]
seg CB || X-axis
∴ y co-ordinate of B = 2
seg BA || Y-axis
∴ x co-ordinate of B = 2
∴ co-ordinate of B is (2, 2) = (x₁,y₁)
co-ordinate of A is (2, 3) = (x₂, Y₂)
Since, AB || to Y-axis,
d(A, B) = Y₂ – Y₁
d(A,B) = 3 – 2 = 1
co-ordinate of C is (-2,2) = (x₁,y₁)
co-ordinate of B is (2, 2) = (x₂, y₂)
Since, BC || to X-axis,
d(B, C) = x₂ – x₁
d(B,C) = 2 – -2 = 4
∴ AC² = 12 + 42 …[From (i)]
= 1 + 16 = 17
∴ AC = \(\sqrt { 17 }\) units …[Taking square root of both sides]

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

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.

10th Standard Maths 2 Practice Set 1.2 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.2 Chapter 1 Similarity Questions With Answers Maharashtra Board

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:

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