Question

the base of an isosceles triangle is 48cm and one of its equal sides is 30cm find the area of the triangle

Answer 1

HOPE IT HELPS............

Answer 2

Answer:

$Area \:of \:an \\ isosceles\: triangle (A)\\=432 \:cm^{2}$

Step-by-step explanation:

$Dimensions \:of \:an\\isosceles\: triangle\\base(b)=48\:cm\\and\:one\:of \:its\: equal \:side(a)=30\:cm$

$Area \:of \:an \\ isosceles\: triangle (A)=\frac{b}{2} \times \sqrt{a^{2}-\big(\frac{b}{2}\big)^{2}}$

$=\frac{48}{2} \times \sqrt{(30)^{2}-\big(\frac{48}{2}\big)^{2}}$

$=24\times \sqrt{900-(24)^{2}}$

$=24\times \sqrt{900-576}$

$=24\times \sqrt{324}\\=24\times \sqrt{18^{2}}\\=24\times 18\\=432 \:cm^{2}$

Therefore,

$Area \:of \:an \\ isosceles\: triangle (A)\\=432 \:cm^{2}$

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