# NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3

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## Chapter 7 Ex.7.3 Question 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

Reasoning:

Let $$ABC$$  be any triangle whose vertices are $$A(x_1, y_1)$$, $$B(x_2, y_2)$$ and $$C(x_3, y_3)$$

Area of a triangle

\begin{align} \! = \! \frac{1}{2} \begin{bmatrix} x_1 \left( y_2 \! - \! y_3 \right) \! + \! x_2 \left( y_3 \! - \! y_1 \right) + \\ x_3 \left( y_1 \! - \! y_2 \right) \end{bmatrix} \end{align}

What is the known:

The $$x$$ and $$y$$ co-ordinates of the vertices of the triangle.

What is the Unknown:

The area of the triangle.

Solution:

(i) Given,

• Let  $$A\,(x_1, y_1) = (2, 3)$$
• Let  $$B\,(x_2, y_2) = (-1 , 0)$$
• Let  $$C\,(x_3, y_3) = (2, -4)$$

Area of a triangle

\begin{align} \! = \! \frac{1}{2} \begin{bmatrix} x_1 \left( y_2 \! - \! y_3 \right) \! + \! x_2 \left( y_3 \! - \! y_1 \right) \! + \! \\ x_3 \left( y_1 \! - \! y_2 \right) \end{bmatrix} \cdots (1) \end{align}

By substituting the values of vertices, $$A, B, C$$ in the Equation $$(1)$$,

Area of the given triangle

\begin{align}&= \! \frac{1}{2} \! \begin{bmatrix}2 ( 0 \! - \! ( - 4)) \! + \! ( - 1)(( - 4) \! - \! (3)) \\ + 2(3 \! - \! 0) \end{bmatrix} \\&= \! \frac{1}{2}(8 + 7 + 6)\\ &= \! \frac{{{\text{21}}}}{{\text{2}}}{\text{ Square units }}\end{align}

(ii) Given,

• Let $$A(x_1, y_1) = (-5, -1)$$
• Let $$B(x_2, y_2) = (3, -5)$$
• Let $$C(x_3, y_3) = (5, 2)$$

Area of a triangle

\begin{align}= \! \frac{1}{2} \! \begin{bmatrix} x_1 \left( y_2 \! - \! y_3 \right) \! + \! x_2 \left( y_3 \! - \! y_1 \right) + \\x_3 \left( y_1 \! - \! y_2 \right) \end{bmatrix} \;\; \cdots (1) \end{align}

By substituting the values of vertices, $$A, B, C$$ in the Equation $$(1)$$,

Area of the given triangle

\begin{align} &= \! \frac{1}{2} \begin{bmatrix} ( - 5) (- 5) \! - \! ( - 2) \! + \! 3(2 \! - \! ( - 1)) + \\5 \! - \! 1 \! - \! ( - 5) \end{bmatrix} \\&= \frac{1}{2}(35 \! + \! 9 \! + \! 20) \\ &= \text{32 Square units } \end{align}

## Chapter 7 Ex.7.3 Question 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

Reasoning:

Three or more points are said to be collinear if they lie on a single straight line .

What is Known:

The $$x$$ and $$y$$ co-ordinates of the points.

What is  Unknown:

The value of ‘$$k$$’, for which the points are collinear.

Solution:

(i) Given,

• Let A $$(x_1, y_1) = (7, -2)$$
• Let B $$(x_2, y_2) = (5 , 1)$$
• Let C $$(x_3, y_3) = (3, -k)$$

\begin{align} & =\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) \\ + x_2 \left( y_3 - y_1 \right) \\ + x_3 \left( y_1 - y_2 \right) \end{bmatrix} \\ &= 0\end{align}

(For Collinear Points)

By substituting the values of vertices, $$A, B, C$$ in the Equation $$(1)$$,

\begin{align}\frac{1}{2} \begin{bmatrix} 7(1 \! - \! k) \! + \! 5(k \! - \! ( - 2)) \! + \! \\ 3(( - 2) \! - \! 1) \end{bmatrix} \! = \! 0 \\ \begin{bmatrix} 7 \! - \! 7k \! + \! 5k \! + \! 10 \! - \! 9 \end{bmatrix} \! = \! 0 \\ - 2k \! + \! 8 \! = \! 0 \\ k \! = \! 4 \end{align}

Hence the given points are collinear for \begin{align}k = 4\end{align}

(ii) Given,

• Let A $$(x_1, y_1) = (8, 1)$$
• Let B $$(x_2, y_2) = (k , -4)$$
• Let C $$(x_3, y_3) = (2, -5)$$

\begin{align} & = \! \frac{1}{2} \! \begin{bmatrix} x_1 \left( y_2 \! - \! y_3 \right) \! + \! x_2 \left( y_3 \! - \! y_1 \right) \! + \! \\ x_3 \left( y_1 \! - \! y_2 \right) \end{bmatrix} \\ &= \! 0 \end{align}

(For Collinear Points)

By substituting the values of vertices, $$A, B, C$$ in the Equation $$(1)$$,

\begin{align} {\frac{1}{2}} \!\begin{bmatrix} 8(- 4 \! - \! ( - 5)) \! + \! k(( - 5) \! - \! (1)) \\ + 2(1 \! - \! ( - 4)) \end{bmatrix} & \! = \! 0\\ 8 \! - \! 6k \! + \! 10 & \! = \! 0 \\ \left( {\text{By Transposing}} \right)\\ 6k & \! = \! 18\\ k & \! = \! 3\end{align}

Hence, the given points are collinear for \begin{align}k = 3 \end{align}

## Chapter 7 Ex.7.3 Question 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

Reasoning:

Let ABC be any triangle whose vertices are  $$A(x_1, y_1),$$  $$B(x_2, y_2)$$ and $$C(x_3, y_3).$$

Area of a triangle

\begin{align}=\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) \\ + x_2 \left( y_3 - y_1 \right) \\ + x_3 \left( y_1 - y_2 \right) \end{bmatrix} \end{align}

What is Known:

The $$x$$ and $$y$$ co-ordinates of the vertices of the triangle.

What is Unknown?

The ratio of this area to the area of the given triangle.

Solution:

From the given figure,

Given,

• Let A $$(x_1, y_1) = (0, -1)$$
• Let B $$(x_2, y_2) = (2 , 1)$$
• Let C $$(x_3, y_3) = (0, 3)$$

Area of a triangle

\begin{align}=\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) \\ + x_2 \left( y_3 - y_1 \right) \\ + x_3 \left( y_1 - y_2 \right) \end{bmatrix}\;\; \dots (1) \end{align}

By substituting the values of vertices, $$A, B, C$$ in $$(1)$$,

Let $$P, Q, R$$ be the mid-points of the sides of this triangle.

Coordinates of $$P, Q,$$ and $$R$$ are given by

\begin{align}{P = \left[ {\frac{{0 + 2}}{2},\frac{{ - 1 + 1}}{2}} \right] = (1,0)}\\{Q = \left[ {\frac{{0 + 0}}{2},\frac{{3 - 1}}{2}} \right] = (0,1)}\\{R = \left[ {\frac{{2 + 0}}{2},\frac{{1 + 3}}{2}} \right] = (1,2)}\end{align}

By substituting the values of Points $$P, Q, R$$

Area of $$\Delta PQR$$,

\begin{align}&= \frac{1}{2}\begin{bmatrix} (2 - 1) + 1(1 - 0) \\ + 0(0 - 2)\end{bmatrix} \\ &= \frac{1}{2}(1 + 1)\\ &= 1{\text{ Square units }}\end{align}

By substituting the values of Points $$A, B, C$$

Area of $$\Delta ABC$$,

\begin{align}&= \frac{1}{2} \begin{bmatrix} 0(1 - 3) + 2( 3 - ( - 1)) \\ + 0( - 1 - 1) \end{bmatrix} \\ &= \frac{1}{2}(8) \\ &= 4{\text{ Square units }}\end{align}

Therefore, Ratio of this area $$\Delta \, PQR$$ to the area of the triangle \begin{align}\Delta {\text{ABC = 1: 4}}\end{align}

## Chapter 7 Ex.7.3 Question 4

Find the area of the quadrilateral whose vertices, taken in order, are $$(-4, -2), (-3, -5), (3, -2)$$ and $$(2, 3)$$

### Solution

Reasoning:

Let $$ABC$$ be any triangle whose vertices are $$A$$$$(x_1, y_1),$$ $$B$$$$(x_2, y_2)$$ and $$C$$$$(x_3, y_3).$$

Area of a triangle

\begin{align}=\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) \\ + x_2 \left( y_3 - y_1 \right) \\ + x_3 \left( y_1 - y_2 \right) \end{bmatrix} \end{align}

What is known?

The $$x$$ and $$y$$ co-ordinates of the vertices of the quadrilateral.

What is  unknown?

Solution:

From the figure,

Given,

• Let the vertices of the quadrilateral be $$A$$$$(-4, -2),$$ $$B$$$$(-3, -5),$$ $$C$$$$(3, -2),$$ and $$D$$$$(2, 3).$$
• Join $$AC$$ to form two triangles $$\Delta \, ABC$$ and $$\Delta \, ACD$$.

We know that,

Area of a triangle

\begin{align}=\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) \\ + x_2 \left( y_3 - y_1 \right) \\ + x_3 \left( y_1 - y_2 \right) \end{bmatrix}\;\; \dots(1) \end{align}

By substituting the values of vertices, $$A, B, C$$ in the Equation ($$1$$),

Area of $$\Delta {ABC}$$

\begin{align}&= \! \frac{1}{2} \begin{bmatrix} ( - 4)(( - 5) \! - \! ( - 2)) \! + \!\\ ( - 3)(( - 2) \! - \! ( - 2)) +\\ 3(( - 2) \! - \! ( - 2)) \end{bmatrix} \\& = \! \frac{1}{2}(12 \! + \! 0 \! + \! 9)\\& = \! \frac{{21}}{2}{\text{ Square units }}\end{align}

By substituting the values of vertices, $$A, C, D$$ in the Equation ($$1$$),

Area of $$\Delta ACD$$

\begin{align}\\ &= \frac{1}{2}\begin{bmatrix} ( - 4)(( - 2) - ( - 3)) + \\ 3((3) - ( - 2)) + \\ 2(( - 2)) - ( - 2)) \end{bmatrix} \\ &= \frac{1}{2}(20 + 15 + 0) \\ &= \frac{{35}}{2}{\text{ Square units}}\end{align}

\begin{align} \text{Area of }ABCD=\begin{bmatrix}\text{Area of }\Delta {{ABC}}+\\\text{Area of }\Delta {{ACD}} \end{bmatrix} \end{align}

\begin{align} \\ &= \left[{\frac{{21}}{2} + \frac{{35}}{2}} \right]{\text{ Square units }}\\ &= 28{\text{ Square units }}\end{align}

## Chapter 7 Ex.7.3 Question 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 $$\Delta ABC$$ whose vertices are $$A(4, -6),$$ $$B(3, -2)$$ and $$C(5, 2)$$

### Solution

Reasoning:

Let $$ABC$$ be any triangle whose vertices are  $$A(x_1, y_1)$$, $$B(x_2, y_2)$$ and $$C(x_3, y_3).$$

Area of a triangle

\begin{align}=\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) + \\ x_2 \left( y_3 - y_1 \right) + \\ x_3 \left( y_1 - y_2 \right) \end{bmatrix} \end{align}

What is known?

The $$x$$ and $$y$$ co-ordinates of the vertices of the triangle.

What is unknown?

To verify that a median of a triangle divides it into two triangles of equal areas.

Steps:

From the figure,

Given,

• Let the vertices of the triangle be $$A$$$$(4, -6),$$ $$B$$$$(3, -2),$$ and $$C$$$$(5, 2).$$
• Let $$M$$ be the mid-point of side $$BC$$ of $$\Delta ABC.$$

Therefore, $$AM$$ is the median in $$\Delta ABC.$$

Coordinates of point $$M$$

\begin{align}{\text{}}\;{{}} &= \left[ {\frac{{3 + 5}}{2} + \frac{{ - 2 + 2}}{2}} \right] \\ &= [4,0]\end{align}

Area of a triangle

\begin{align}=\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) + \\ x_2 \left( y_3 - y_1 \right) + \\ x_3 \left( y_1 - y_2 \right) \end{bmatrix}\;\;\dots(1) \end{align}

By substituting the values of vertices, $$A, B, M$$ in the Equation $$(1)$$

Area of $$\Delta ABM$$

\begin{align} &= \frac{1}{2} \begin{bmatrix} (4)\{ ( - 2) - (0)\} \\ + (3)\{ (0) - ( - 6)\} \\ + (4)\{ ( - 6)\} - ( - 2)\} \end{bmatrix} \\ &= \frac{1}{2}[ - 8 + 18 - 16]\\& = 3\;\text{ Square units }\end{align}

By substituting the values of vertics, $$A,D,C$$ in the Equation $$(1)$$

Area of $$\Delta ABD$$

\begin{align} &= \frac{1}{2} \begin{bmatrix} (4)(( - 2) - (0)) + \3)((0) - ( - 6))+\\((4){ ( - 6)) - ( - 2)} \end{bmatrix} \\&= \frac{1}{2}[- 8 + 18 - 16]\\&= 3{\text{ Square units }}\end{align} However, area cannot be negative. Therefore, area of \(\Delta AMC is $$3\, \rm{square \,units}.$$

Hence, clearly, median $$AM$$ has divided $$\Delta ABC$$ in two triangles of equal areas.