Delta ABC is an isosceles triangle in which B | vedclass

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(A) In an isosceles triangle $\Delta ABC$,let $AB = AC = 13 \, cm$ and $BC = 24 \, cm$.Draw an altitude $AD$ from vertex $A$ to the base $BC$. Since t

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) In an isosceles triangle $\Delta ABC$,let $AB = AC = 13 \, cm$ and $BC = 24 \, cm$.Draw an altitude $AD$ from vertex $A$ to the base $BC$. Since the triangle is isosceles,the altitude $AD$ bisects the base $BC$.Therefore,$BD = DC = \frac{BC}{2} = \frac{24}{2} = 12 \, cm$.In the right-angled triangle $\Delta ABD$,by the Pythagorean theorem:$AB^2 = AD^2 + BD^2$$13^2 = AD^2 + 12^2$$169 = AD^2 + 144$$AD^2 = 169 - 144 = 25$$AD = \sqrt{25} = 5 \, cm$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes AD$.Area $= \frac{1}{2} 	imes 24 	imes 5 = 12 	imes 5 = 60 \, cm^2$.

$\Delta ABC$ is an isosceles triangle in which $BC = 24 \, cm$ and $AB = AC = 13 \, cm$. Then,the area of $\Delta ABC = \dots \, cm^2$.

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