In Delta ABC,AB = 20 , cm,BC = 48 , cm,and AC | vedclass

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(B) First,check if the triangle is a right-angled triangle by verifying the Pythagorean theorem: $AB^2 + BC^2 = 20^2 + 48^2 = 400 + 2304 = 2704$. Sinc

Vedclass · Accepted Answer

$480$

Vedclass · Answer

(B) First,check if the triangle is a right-angled triangle by verifying the Pythagorean theorem: $AB^2 + BC^2 = 20^2 + 48^2 = 400 + 2304 = 2704$. Since $AC^2 = 52^2 = 2704$,we have $AB^2 + BC^2 = AC^2$. Thus,$\Delta ABC$ is a right-angled triangle with base $BC = 48 \, cm$ and height $AB = 20 \, cm$. The area of the triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 48 	imes 20 = 24 	imes 20 = 480 \, cm^2$.

In $\Delta ABC$,$AB = 20 \, cm$,$BC = 48 \, cm$,and $AC = 52 \, cm$,find the area of $\Delta ABC$. (in $, cm^2$)

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