# If D ( − 1 5 , 5 2 ) , E ( 7 , 3 ) and F ( 7 2 , 7 2 ) Are the Mid-points of Sides of δ a B C , Find the Area of δ a B C . - Mathematics

If  $D\left( - \frac{1}{5}, \frac{5}{2} \right), E(7, 3) \text{ and } F\left( \frac{7}{2}, \frac{7}{2} \right)$  are the mid-points of sides of  $∆ ABC$ ,  find the area of  $∆ ABC$ .

#### Solution

The midpoint of BC is $D\left( - \frac{1}{5}, \frac{5}{2} \right)$,

The midpoint of AB is $F\left( \frac{7}{2}, \frac{7}{2} \right)$ ,

The midpoint of AC is $E\left( 7, 3 \right)$ Consider the line segment BC,

$\Rightarrow \frac{p + r}{2} = - \frac{1}{2} ; \frac{q + s}{2} = \frac{5}{2}$
$\Rightarrow p + r = - 1 ; q + s = 5 . . . . . (i)$
$\text{ Consider the line segment AB, }$
$\Rightarrow \frac{p + x}{2} = \frac{7}{2} ; \frac{q + y}{2} = \frac{7}{2}$
$\Rightarrow p + x = 7 ; q + y = 7 . . . . . (ii)$

$\text{ Consider the line segment AC, }$

$\Rightarrow \frac{r + x}{2} = 7 ; \frac{s + y}{2} = 3$

$\Rightarrow r + x = 14 ; s + y = 6 . . . . . (iii)$

Solve (i), (ii) and (iii) to get

$A\left( x, y \right) = A\left( 11, 4 \right), B\left( p, q \right) = B\left( - 4, 3 \right), C\left( r, s \right) = C\left( 3, 2 \right)$
Let us assume that BC is base of the triangle,

$BC = \sqrt{\left( - 4 - 3 \right)^2 + \left( 3 - 2 \right)^2} = \sqrt{50}$

$\text{ Equation of the line BC is }$

$\frac{x + 4}{- 4 - 3} = \frac{y - 3}{3 - 2}$

$\Rightarrow x + 7y - 17 = 0$

$\text{ The perpendicular distance from a point } P\left( x_1 , y_1 \right)is$

$P = \left| \frac{1\left( 11 \right) + 7\left( 4 \right) - 17}{\sqrt{50}} \right| = \frac{22}{\sqrt{50}}$

The area of the triangle is $A = \frac{1}{2} \times \sqrt{50} \times \frac{22}{\sqrt{50}} = 11 \text{ sq . units }$

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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.5 | Q 35 | Page 55