Question

Solve this Question this isssss so difficult koi kar ke to dikhaye dum hai to losers. ​

Answer 1

(i) If we take A as origin then AD is X-axis and AB is the Y-axis Then the coordinates of the vertices of △PQR is

P=(4,6),Q=(3,2),R=(6,5)

(ii) If we take C as origin then CB is X-axis and CD is the Y-axis Then the coordinates of the vertices of △PQR is

P=(12,2),Q=(13,6),R=(10,3)

Area of the triangle =

$\small\bold{\frac{1}{2} [x_1(y_2−y_3)+x_2(y_3−y_1)+x_3(y_1−y_2)]}$

Area of △PQR=

$= \frac{1}{2} [4(2−5)+3(5−6)+6(6−2)]$

$\frac{1}{2} (4×−3+3×−1+6×4)$

$\small \bold{ = \frac{1}{2} (−12−3+24) = \frac{9}{2} \: sq.units}$

Second case-

Area of △PQR=

$\small \bold{\frac{1}{2}[12(6−3)+13(3−2)+10(2−6)]}$

$= \frac{1}{2} (12×3+13×1+10×−4)$

$\frac{1}{2} (36+13−40) = \frac{9}{2} sq.units$

Hence we observed that the area of the triangle in both case is equal.