Table of Contents
Exercise 7.3 Solutions – Class X Mathematics
1. Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
Using the area formula: Area = \( \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right| \)
Here, \( (x_1, y_1) = (2, 3) \), \( (x_2, y_2) = (-1, 0) \), \( (x_3, y_3) = (2, -4) \).
Area = \( \frac{1}{2} \left| 2(0 – (-4)) + (-1)(-4 – 3) + 2(3 – 0) \right| = \frac{1}{2} \left| 2(4) + (-1)(-7) + 2(3) \right| = \frac{1}{2} \left| 8 + 7 + 6 \right| = \frac{1}{2} \times 21 = \frac{21}{2} \).
Area = \( \frac{21}{2} \) square units.
Using the area formula: Area = \( \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right| \)
Here, \( (x_1, y_1) = (2, 3) \), \( (x_2, y_2) = (-1, 0) \), \( (x_3, y_3) = (2, -4) \).
Area = \( \frac{1}{2} \left| 2(0 – (-4)) + (-1)(-4 – 3) + 2(3 – 0) \right| = \frac{1}{2} \left| 2(4) + (-1)(-7) + 2(3) \right| = \frac{1}{2} \left| 8 + 7 + 6 \right| = \frac{1}{2} \times 21 = \frac{21}{2} \).
Area = \( \frac{21}{2} \) square units.
(ii) (-5, -1), (3, -5), (5, 2)
\( (x_1, y_1) = (-5, -1) \), \( (x_2, y_2) = (3, -5) \), \( (x_3, y_3) = (5, 2) \).
Area = \( \frac{1}{2} \left| (-5)(-5 – 2) + 3(2 – (-1)) + 5(-1 – (-5)) \right| = \frac{1}{2} \left| (-5)(-7) + 3(3) + 5(4) \right| = \frac{1}{2} \left| 35 + 9 + 20 \right| = \frac{1}{2} \times 64 = 32 \).
Area = 32 square units.
\( (x_1, y_1) = (-5, -1) \), \( (x_2, y_2) = (3, -5) \), \( (x_3, y_3) = (5, 2) \).
Area = \( \frac{1}{2} \left| (-5)(-5 – 2) + 3(2 – (-1)) + 5(-1 – (-5)) \right| = \frac{1}{2} \left| (-5)(-7) + 3(3) + 5(4) \right| = \frac{1}{2} \left| 35 + 9 + 20 \right| = \frac{1}{2} \times 64 = 32 \).
Area = 32 square units.
(iii) (0, 0), (3, 0), and (0, 2)
\( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (3, 0) \), \( (x_3, y_3) = (0, 2) \).
Area = \( \frac{1}{2} \left| 0(0 – 2) + 3(2 – 0) + 0(0 – 0) \right| = \frac{1}{2} \left| 0 + 3(2) + 0 \right| = \frac{1}{2} \times 6 = 3 \).
Area = 3 square units.
\( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (3, 0) \), \( (x_3, y_3) = (0, 2) \).
Area = \( \frac{1}{2} \left| 0(0 – 2) + 3(2 – 0) + 0(0 – 0) \right| = \frac{1}{2} \left| 0 + 3(2) + 0 \right| = \frac{1}{2} \times 6 = 3 \).
Area = 3 square units.
2. Find the value of ‘K’ for which the points are collinear:
Points are collinear if the area of the triangle formed is zero.
(i) (7, -2), (5, 1), (3, K)
Area = \( \frac{1}{2} \left| 7(1 – K) + 5(K – (-2)) + 3(-2 – 1) \right| = 0 \).
\( \frac{1}{2} \left| 7(1 – K) + 5(K + 2) + 3(-3) \right| = 0 \),
\( 7 – 7K + 5K + 10 – 9 = 0 \),
\( -2K + 8 = 0 \), \( 2K = 8 \), \( K = 4 \).
(i) (7, -2), (5, 1), (3, K)
Area = \( \frac{1}{2} \left| 7(1 – K) + 5(K – (-2)) + 3(-2 – 1) \right| = 0 \).
\( \frac{1}{2} \left| 7(1 – K) + 5(K + 2) + 3(-3) \right| = 0 \),
\( 7 – 7K + 5K + 10 – 9 = 0 \),
\( -2K + 8 = 0 \), \( 2K = 8 \), \( K = 4 \).
(ii) (K, K), (2, 3), and (4, -1)
Area = \( \frac{1}{2} \left| K(3 – (-1)) + 2(-1 – K) + 4(K – 3) \right| = 0 \).
\( \frac{1}{2} \left| K(4) + 2(-1 – K) + 4(K – 3) \right| = 0 \),
\( 4K – 2 – 2K + 4K – 12 = 0 \),
\( 6K – 14 = 0 \), \( 6K = 14 \), \( K = \frac{14}{6} = \frac{7}{3} \).
Area = \( \frac{1}{2} \left| K(3 – (-1)) + 2(-1 – K) + 4(K – 3) \right| = 0 \).
\( \frac{1}{2} \left| K(4) + 2(-1 – K) + 4(K – 3) \right| = 0 \),
\( 4K – 2 – 2K + 4K – 12 = 0 \),
\( 6K – 14 = 0 \), \( 6K = 14 \), \( K = \frac{14}{6} = \frac{7}{3} \).
(iii) (8, 1), (K, -4), (2, -5)
Area = \( \frac{1}{2} \left| 8(-4 – (-5)) + K(-5 – 1) + 2(1 – (-4)) \right| = 0 \).
\( \frac{1}{2} \left| 8(1) + K(-6) + 2(5) \right| = 0 \),
\( 8 – 6K + 10 = 0 \),
\( 18 – 6K = 0 \), \( 6K = 18 \), \( K = 3 \).
Area = \( \frac{1}{2} \left| 8(-4 – (-5)) + K(-5 – 1) + 2(1 – (-4)) \right| = 0 \).
\( \frac{1}{2} \left| 8(1) + K(-6) + 2(5) \right| = 0 \),
\( 8 – 6K + 10 = 0 \),
\( 18 – 6K = 0 \), \( 6K = 18 \), \( K = 3 \).
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.
Step 1: Find the midpoints of the sides.
\( A(0, -1) \), \( B(2, 1) \), \( C(0, 3) \).
Midpoint of \( AB \): \( D = \left( \frac{0 + 2}{2}, \frac{-1 + 1}{2} \right) = (1, 0) \).
Midpoint of \( BC \): \( E = \left( \frac{2 + 0}{2}, \frac{1 + 3}{2} \right) = (1, 2) \).
Midpoint of \( CA \): \( F = \left( \frac{0 + 0}{2}, \frac{3 + (-1)}{2} \right) = (0, 1) \).
Step 2: Area of triangle \( DEF \).
\( D(1, 0) \), \( E(1, 2) \), \( F(0, 1) \).
Area = \( \frac{1}{2} \left| 1(2 – 1) + 1(1 – 0) + 0(0 – 2) \right| = \frac{1}{2} \left| 1(1) + 1(1) + 0 \right| = \frac{1}{2} \times 2 = 1 \).
Area of \( \triangle DEF = 1 \) square unit.
Step 3: Area of triangle \( ABC \).
Area = \( \frac{1}{2} \left| 0(1 – 3) + 2(3 – (-1)) + 0(-1 – 1) \right| = \frac{1}{2} \left| 0 + 2(4) + 0 \right| = \frac{1}{2} \times 8 = 4 \).
Area of \( \triangle ABC = 4 \) square units.
Step 4: Ratio of areas.
Ratio = \( \frac{\text{Area of } \triangle DEF}{\text{Area of } \triangle ABC} = \frac{1}{4} \).
The ratio is 1 : 4.
\( A(0, -1) \), \( B(2, 1) \), \( C(0, 3) \).
Midpoint of \( AB \): \( D = \left( \frac{0 + 2}{2}, \frac{-1 + 1}{2} \right) = (1, 0) \).
Midpoint of \( BC \): \( E = \left( \frac{2 + 0}{2}, \frac{1 + 3}{2} \right) = (1, 2) \).
Midpoint of \( CA \): \( F = \left( \frac{0 + 0}{2}, \frac{3 + (-1)}{2} \right) = (0, 1) \).
Step 2: Area of triangle \( DEF \).
\( D(1, 0) \), \( E(1, 2) \), \( F(0, 1) \).
Area = \( \frac{1}{2} \left| 1(2 – 1) + 1(1 – 0) + 0(0 – 2) \right| = \frac{1}{2} \left| 1(1) + 1(1) + 0 \right| = \frac{1}{2} \times 2 = 1 \).
Area of \( \triangle DEF = 1 \) square unit.
Step 3: Area of triangle \( ABC \).
Area = \( \frac{1}{2} \left| 0(1 – 3) + 2(3 – (-1)) + 0(-1 – 1) \right| = \frac{1}{2} \left| 0 + 2(4) + 0 \right| = \frac{1}{2} \times 8 = 4 \).
Area of \( \triangle ABC = 4 \) square units.
Step 4: Ratio of areas.
Ratio = \( \frac{\text{Area of } \triangle DEF}{\text{Area of } \triangle ABC} = \frac{1}{4} \).
The 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).
Divide the quadrilateral into two triangles by drawing a diagonal, say from \( (-4, -2) \) to \( (3, -2) \).
Triangle 1: (-4, -2), (-3, -5), (3, -2)
Area = \( \frac{1}{2} \left| (-4)(-5 – (-2)) + (-3)(-2 – (-2)) + 3(-2 – (-5)) \right| = \frac{1}{2} \left| (-4)(-3) + (-3)(0) + 3(3) \right| = \frac{1}{2} \left| 12 + 0 + 9 \right| = \frac{1}{2} \times 21 = \frac{21}{2} \).
Triangle 2: (-4, -2), (3, -2), (2, 3)
Area = \( \frac{1}{2} \left| (-4)(-2 – 3) + 3(3 – (-2)) + 2(-2 – (-2)) \right| = \frac{1}{2} \left| (-4)(-5) + 3(5) + 2(0) \right| = \frac{1}{2} \left| 20 + 15 + 0 \right| = \frac{1}{2} \times 35 = \frac{35}{2} \).
Total area = \( \frac{21}{2} + \frac{35}{2} = \frac{56}{2} = 28 \).
Area of the quadrilateral = 28 square units.
Triangle 1: (-4, -2), (-3, -5), (3, -2)
Area = \( \frac{1}{2} \left| (-4)(-5 – (-2)) + (-3)(-2 – (-2)) + 3(-2 – (-5)) \right| = \frac{1}{2} \left| (-4)(-3) + (-3)(0) + 3(3) \right| = \frac{1}{2} \left| 12 + 0 + 9 \right| = \frac{1}{2} \times 21 = \frac{21}{2} \).
Triangle 2: (-4, -2), (3, -2), (2, 3)
Area = \( \frac{1}{2} \left| (-4)(-2 – 3) + 3(3 – (-2)) + 2(-2 – (-2)) \right| = \frac{1}{2} \left| (-4)(-5) + 3(5) + 2(0) \right| = \frac{1}{2} \left| 20 + 15 + 0 \right| = \frac{1}{2} \times 35 = \frac{35}{2} \).
Total area = \( \frac{21}{2} + \frac{35}{2} = \frac{56}{2} = 28 \).
Area of the quadrilateral = 28 square units.
5. Find the area of the triangle formed by the points (2, 3), (6, 3), and (2, 6) by using Heron’s formula.
Step 1: Find the lengths of the sides.
\( A(2, 3) \), \( B(6, 3) \), \( C(2, 6) \).
\( AB = \sqrt{(6 – 2)^2 + (3 – 3)^2} = 4 \),
\( BC = \sqrt{(6 – 2)^2 + (3 – 6)^2} = \sqrt{16 + 9} = 5 \),
\( CA = \sqrt{(2 – 2)^2 + (6 – 3)^2} = 3 \).
Step 2: Apply Heron’s formula.
Semi-perimeter \( s = \frac{4 + 5 + 3}{2} = 6 \).
Area = \( \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{6(6 – 4)(6 – 5)(6 – 3)} = \sqrt{6 \times 2 \times 1 \times 3} = \sqrt{36} = 6 \).
Area = 6 square units.
\( A(2, 3) \), \( B(6, 3) \), \( C(2, 6) \).
\( AB = \sqrt{(6 – 2)^2 + (3 – 3)^2} = 4 \),
\( BC = \sqrt{(6 – 2)^2 + (3 – 6)^2} = \sqrt{16 + 9} = 5 \),
\( CA = \sqrt{(2 – 2)^2 + (6 – 3)^2} = 3 \).
Step 2: Apply Heron’s formula.
Semi-perimeter \( s = \frac{4 + 5 + 3}{2} = 6 \).
Area = \( \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{6(6 – 4)(6 – 5)(6 – 3)} = \sqrt{6 \times 2 \times 1 \times 3} = \sqrt{36} = 6 \).
Area = 6 square units.
