# Ex.7.4 Q6 Coordinate Geometry Solution - NCERT Maths Class 10

## Question

The vertices of a \(\Delta ABC\) are \(A \;(4, 6)\), \(B \;(1, 5)\) and \(C\; (7, 2)\). \(A\) line is drawn to intersect sides \(AB\) and \(AC\) at \(D\) and \(E\) respectively, such that \(\begin{align}\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}} = \frac{1}{4}\,\,.\end{align}\)

Calculate the area of the \(\Delta ADE\) and compare it with the area of \(\Delta ABC\).

(Recall Converse of basic proportionality theorem and Theorem 6.6 related to Ratio of areas of two similar triangles)

## Text Solution

**Reasoning:**

Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Theorem 6.6 : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

**What is known?**

The \(x\) and \(y\) coordinates of the vertices of the triangle.

\(\begin{align}\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}} = \frac{1}{4}\,\,.\end{align}\)

where \(AB\) and \(AC\) are sides of the triangle intersected at \(D\) and \(E\).

**What is unknown?**

The area of the \(\Delta ADE\) is to be calculated and compared with the area of \(\Delta ABC\).

**Steps:**

From the figure,

Given

\(\begin{align}\frac{{{\text{AD}}}}{{{\text{AB}}}} = \frac{{{\text{AE}}}}{{{\text{AC}}}} = \frac{1}{4}\end{align}\)

Therefore, \(D\) and \(E\) are two points on side \(AB\) and \(AC\) respectively such that they divide side \(AB\) and \(AC\) in a ratio of \(1:3\).

By Section formula

\(\begin{align}P(x, y)=\left[\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right] \quad \dots \dots(1)\end{align}\)

Coordintaes of Point D \(\begin{align} &= \left( {\frac{{1 \times 1 + 3 \times 4}}{{1 + 3}},\frac{{1 \times 5 + 3 \times 6}}{{1 + 3}}} \right)\end{align}\)

\(\begin{align} &= \left( {\frac{{13}}{4},\frac{{23}}{4}} \right){\text{ }}\\ \end{align}\)

Coordintaes of Point E* *\(\begin{align} &= \left( {\frac{{1 \times 7 + 3 \times 4}}{{1 + 3}},\frac{{1 \times 2 + 3 \times 6}}{{1 + 3}}} \right) \end{align}\)

\(\begin{align} &= \left( {\frac{{19}}{4},\frac{{20}}{4}} \right)\end{align}\)

Area of triangle\(\begin{align} = \frac{1}{2}\left[ {{{\text{x}}_1}\left( {{{\text{y}}_2} - {{\text{y}}_3}} \right) + {{\text{x}}_2}\left( {{{\text{y}}_3} - {{\text{y}}_1}} \right) + {{\text{x}}_3}\left( {{{\text{y}}_1} - {{\text{y}}_2}} \right)} \right]\quad \ldots \left( 2 \right)\end{align}\)

By substituting the vertices \(A\), \(D\), \(E\) in (2),

Area of \(\Delta ADE\) \(\begin{align} &= \frac{1}{2}\end{align}\)\(\begin{align}\left[ {4\left( {\frac{{23}}{4} - \frac{{20}}{4}} \right) + \frac{{13}}{4}\left( {\frac{{20}}{4} - 6} \right) + \frac{{19}}{4}\left( {6 - \frac{{23}}{4}} \right)} \right]\end{align}\)

\(\begin{align} &= \frac{1}{2}\left[ {3 - \frac{{13}}{4} + \frac{{19}}{{16}}} \right] \end{align}\)

\(\begin{align} &= \frac{1}{2}\left[ {\frac{{48 - 52 + 19}}{{16}}} \right] \end{align}\)

\(\begin{align} &= \frac{{15}}{{32}}\;\rm {} Square\; units\end{align}\)

By substituting the vertices \(A\), \(B\), \(C\) in (2)

Area of \(\Delta ADE\) \(\begin{align} &= \frac{1}{2}\end{align}\)\(\begin{align}\left[ {4\left( {\frac{{23}}{4} - \frac{{20}}{4}} \right) + \frac{{13}}{4}\left( {\frac{{20}}{4} - 6} \right) + \frac{{19}}{4}\left( {6 - \frac{{23}}{4}} \right)} \right] \end{align}\)

\(\begin{align} &= \frac{1}{2}\left[ {3 - \frac{{13}}{4} + \frac{{19}}{{16}}} \right] \end{align}\)

\(\begin{align} &= \frac{1}{2}\left[ {\frac{{48 - 52 + 19}}{{16}}} \right] \end{align}\)

\(\begin{align} &= \frac{{15}}{{32}}{\text{ Square units}}\end{align}\)

Clearly, the ratio between the areas of \(\Delta ADE\) and \(\Delta ABC\) is \(1:16\).

Alternatively, we know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle[Converse of Basic Proportionality Theorem]. These two triangles so formed (here \(\Delta ADE\) and \(\Delta ABC\)) will be similar to each other.

Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles.[Theorem 6.6]

Therefore, ratio between the areas of \(\Delta ADE\) and

\(\begin{align}\Delta {\text{ABC}}\,{\text{ = }}\,{\left( {\frac{1}{2}} \right)^2} = \frac{1}{{16}}\end{align}\)