 # NCERT Solutions for Class 10 Maths Exercise 7.3 Chapter 7 Coordinates Geometry

NCERT exercise Solutions provided here ensures that you get all the vital information on calculating Areas of a triangle. Areas of a triangle are very important from the point of CBSE Class 10 first term examination. Hence, students are advised to get comprehended with the formulas and methods of calculating areas of triangles.

Great way of learning Maths is by solving the problems. To make you learn the methods of calculating areas of triangle we have provided detailed solutions to questions provided in exercise 7.3 of NCERT class 10 textbook.

## Topics covered in Exercise 7.3

1. Section formula
2. Solved examples

• Offers in-depth knowledge of solving problems on areas of triangle
• Guides you in clearing all the fundamental concepts
• Offers different types of question to practice

## Download PDF of NCERT Solutions for Class 10 Maths Chapter 7- Coordinate Geometry Exercise 7.3    ### Access Other Exercise Solutions of Class 10 Maths Chapter 7- Coordinate Geometry

Exercise 7.1– 10 Questions (8 practical based questions, 2 reasoning questions)

Exercise 7.2– 10 Questions (8 long answer questions, 2 short answer questions)

Exercise 7.4– 8 Questions (6 long answer questions, 1 practical based question, 1 reasoning question)

## Exercise 7.3 Page No: 170

1. Find the area of the triangle whose vertices are:

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

Solution:

Area of a triangle formula = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

(i) Here,

x1 = 2, x2 = -1, x3 = 2, y1 = 3, y2 = 0 and y3 = -4

Substitute all the values in the above formula, we get

Area of triangle = 1/2 [2 {0- (-4)} + (-1) {(-4) – (3)} + 2 (3 – 0)]

= 1/2 {8 + 7 + 6}

= 21/2

So, area of triangle is 21/2 square units.

(ii) Here,

x1 = -5, x2 = 3, x3 = 5, y1 = -1, y2 = -5 and y3 = 2

Area of the triangle = 1/2 [-5 { (-5)- (2)} + 3(2-(-1)) + 5{-1 – (-5)}]

= 1/2{35 + 9 + 20} = 32

Therefore, the area of the triangle is 32 square units.

2. In each of the following find the value of ‘k’, for which the points are collinear.

(i) (7, -2), (5, 1), (3, -k)

(ii) (8, 1), (k, -4), (2, -5)

Solution:

(i) For collinear points, area of triangle formed by them is always zero.

Let points (7, -2) (5, 1), and (3, k) are vertices of a triangle.

Area of triangle = 1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) – 1}] = 0

7 – 7k + 5k +10 -9 = 0

-2k + 8 = 0

k = 4

(ii) For collinear points, area of triangle formed by them is zero.

Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0

1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0

8 – 6k + 10 = 0

6k = 18

k = 3

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Solution:

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle.

Coordinates of D, E, and F are given by

D = (0+2/2, -1+1/2 ) = (1, 0)

E = ( 0+0/2, -1+3/2 ) = (0, 1)

F = ( 0+2/2, 3+1/2 ) = (1, 2) Area of a triangle = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)} = 1/2 (1+1) = 1

Area of ΔDEF is 1 square units

Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)] = 1/2 {8} = 4

Area of ΔABC is 4 square units

Therefore, the required ratio is 1:4.

4. Find the area of the quadrilateral whose vertices, taken in order, are

(-4, -2), (-3, -5), (3, -2) and (2, 3).

Solution:

Let the vertices of the quadrilateral be A (- 4, – 2), B ( – 3, – 5), C (3, – 2), and D (2, 3).

Join AC and divide the quadrilateral into two triangles. We have two triangles ΔABC and ΔACD.

Area of a triangle = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Area of ΔABC = 1/2 [(-4) {(-5) – (-2)} + (-3) {(-2) – (-2)} + 3 {(-2) – (-5)}]

= 1/2 (12 + 0 + 9)

= 21/2 square units

Area of ΔACD = 1/2 [(-4) {(-2) – (3)} + 3{(3) – (-2)} + 2 {(-2) – (-2)}]

= 1/2 (20 + 15 + 0)

= 35/2 square units

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD

= (21/2 + 35/2) square units = 28 square units

5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

Solution:

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2). Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

Coordinates of point D = Midpoint of BC = ((3+5)/2, (-2+2)/2) = (4, 0)

Formula, to find Area of a triangle = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

Now, Area of ΔABD = 1/2 [(4) {(-2) – (0)} + 3{(0) – (-6)} + (4) {(-6) – (-2)}]

= 1/2 (-8 + 18 – 16)

= -3 square units

However, area cannot be negative. Therefore, area of ΔABD is 3 square units.

Area of ΔACD = 1/2 [(4) {0 – (2)} + 4{(2) – (-6)} + (5) {(-6) – (0)}]

= 1/2 (-8 + 32 – 30) = -3 square units

However, area cannot be negative. Therefore, the area of ΔACD is 3 square units.

The area of both sides is the same. Thus, median AD has divided ΔABC in two triangles of equal areas.