Question

The sides of a triangle are 21, 20 and 13cm respectively, find the area of the larger triangle into which it is divided by the perpendicular upon the longest side from the opposite vertex.

Answer 1

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Answer 2

Concept

The area of a triangle is up to Base*Height/2. The perpendicular could be a line that creates a 90 degree angle on the opposite side. The pythagoras theorem states that in right angled triangle the square of hypotenuse is up to the squares of base and height.

Given

sides of the triangle=21,20,13cm

To find

The area of the larger triangle?

Explanation

In triangle ABD

$y^{2} +(21-x)^{2}$

In triangle BCD

$x^{2} +y^{2} =13^{2}$

$(21-x)^{2} +x^{2} =20^{2} -13^{2}$

x=5

AD=21-5=16

$x^{2} +y^{2} =13^{2}$

y=12

Area of triangle ABD=1/2*16*12

=96 units square.

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