Question

Find the area of the quadrilateral ABCD whose vertices are A (-1, -3), B (5, -7), C (10, -2), D (5, 17).​

Answer 1

Given :-

A(-1, -3)

B(5, -7)

C(10, -2)

D(5, 17)

To Find :-

Area of the quadrilateral

Solution :-

Here

Area of ABCD = Area ΔABC + Area ΔBCD

For ABC

$\sf Area=\dfrac{1}{2}\big|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big|$

$\sf Area = \dfrac{1}{2}\big|-1\{-7-(-2)\}+5\{-2-(-3)\}+10\{-3-(-7)\}\big|$

$\sf Area = \dfrac{1}{2}\big|-1(-7+2)+5(-2+3)+10(-3+7)\big|$

$\sf Area = \dfrac{1}{2}\big|-1(-5)+5(1)+10(4)\big|$

$\sf Area = \dfrac{1}{2}\big|5+5+40\big|$

$\sf Area = \dfrac{1}{2}\big|50\big|$

$\sf Area = \dfrac{1}{2}\times 50$

$\sf Area = 25\;cm^2$

For BCD

$\sf Area=\dfrac{1}{2}\big|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\big|$

$\sf Area = \dfrac{1}{2}\big|5(-2-17)+10\{17-(-7)\}+5(-7+2)\big|$

$\sf Area = \dfrac{1}{2}\big|5(-19)+10(17+7)+5(-5)\big|$

$\sf Area = \dfrac{1}{2}\big|-95+240-25\big|$

$\sf Area = \dfrac{1}{2}\big|240-120\big|$

$\sf Area = \dfrac{1}{2}\big|120\big|$

$\sf Area = \dfrac{1}{2}\times 120$

$\sf Area=60\;cm^2$

$\bf Total\;area=Area\;\;\triangle ABC+Area\;\; \triangle BCD$

$\sf Total \;area=25+60$

$\sf Total \;area=85\;cm^2$