Question

$<b>{Find the area of the triangle formed by the points (8, -5), (-2, -7) and (5, 1) by using Heron’s formula}$

Answer 1

Let point ( 8 , - 5 ) = A
Point ( - 2 , - 7 ) = B
Point ( 5 , 1 ) = C



By Distance Formula :


$\text{Distance between A and B }=\sqrt{(8+2)^2 + (7-5)^2}\\\quad = \sqrt{(10)^2+(2)^2}\\\quad=\sqrt{100+4}\\\quad=\sqrt{104}units$


Hence, distance between A and B is √104 units.



Then,
$\text{Distance between A and C}=\sqrt{(8-5)^2+(-5-1)^2}\\\quad=\sqrt{(3)^2+(-6)^2}\\\quad=\sqrt{9+36}\\\quad=\sqrt{45}units$


Hence, distance between A and C is √45 units



Now,
$\text{Distance between B and C}=\sqrt{(5+2)^2+(1+7)^2}\\\quad=\sqrt{(7)^2+(8)^2}\\\quad=\sqrt{49+64}\\\quad=\sqrt{113} units$


Hence distance between B and C is √113 units



Now,
$\text{Semi perimeter of triangle}=\dfrac{\sqrt{104}+\sqrt{45}+\sqrt{113}}{2}$



By Heron's Formula,
$\implies Area=\sqrt{\bigg\{\dfrac{\sqrt{104}+\sqrt{45}+\sqrt{113}}{2}\bigg\}\bigg\{\dfrac{\sqrt{104}+\sqrt{45}+\sqrt{113}}{2}-\sqrt{104}\bigg\}\bigg\{\dfrac{\sqrt{104}+\sqrt{45}+\sqrt{113}}{2}-\sqrt{45}\bigg\}\bigg\{\dfrac{\sqrt{104}+\sqrt{45}+\sqrt{113}}{2}-\sqrt{113}\bigg\}}$


$\\\implies Area=\sqrt{\bigg\{\dfrac{\sqrt{104}+\sqrt{45}+\sqrt{113}}{2}\bigg\}\bigg\{\dfrac{\sqrt{113}+\sqrt{45}-\sqrt{104}}{2}\bigg\}\bigg\{\dfrac{\sqrt{104}-\sqrt{45}+\sqrt{113}}{2}\bigg\}\bigg\{\dfrac{\sqrt{104}+\sqrt{45}-\sqrt{113}}{2}\bigg\}}$


$\\\implies Area=\sqrt{\bigg\{\dfrac{\sqrt{104}+\sqrt{45}+\sqrt{113}}{2}\bigg\}\bigg\{\dfrac{\sqrt{104}+\sqrt{45}-\sqrt{113}}{2}\bigg\}\bigg\{\dfrac{\sqrt{113}+\sqrt{45}-\sqrt{104}}{2}\bigg\}\bigg\{\dfrac{\sqrt{104}+\sqrt{113}-\sqrt{45}}{2}\bigg\}}$


$\\\implies Area=\sqrt{\bigg\{\dfrac{\sqrt{104}+\sqrt{45}+\sqrt{113}}{2}\bigg\}\bigg\{\dfrac{\sqrt{104}+\sqrt{45}-\sqrt{113}}{2}\bigg\}\bigg\{\dfrac{\sqrt{113}+(\sqrt{45}-\sqrt{104})}{2}\bigg\}\bigg\{\dfrac{\sqrt{113}-(\sqrt{45}-\sqrt{104}}{2}\bigg\}}$



From factorization, we know,
( a + b )( a - b ) = a² - b²


$\implies Area = \dfrac{1}{4}\times\sqrt{\{(\sqrt{104}+\sqrt{45}\}^2-\{\sqrt{113}\}^2 )(\{\sqrt{113}\}^2-\{\sqrt{45}-\sqrt{104}\}^2)}$


$\implies Area = \dfrac{1}{4}\times \sqrt{(12\sqrt{130}+36)(12\sqrt{130}-36)}$


$\implies Area = \dfrac{1}{4}\times\sqrt{(12\sqrt{130})^2-(36)^2}$


$\implies Area=\dfrac{1}{4}\times\sqrt{132^2}$


$\implies Area = \dfrac{132}{4} unit^2$


$\implies Area = 33 unit^2$



Therefore the area of the triangle is 33 unit².