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

Video 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

Video 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

Video 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

Video 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?

The area of the quadrilateral.

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

Video 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.

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