The base of an isoceles of …
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Manoj Kumar 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Here, base, b = 24 cm, and let each equal side be a cm. Then,
area = {tex}\frac { 1 } { 4 } b \sqrt { 4 a ^ { 2 } - b ^ { 2 } } \text { sq units } = \frac { 1 } { 4 } \times 24 \times \sqrt { 4 a ^ { 2 } - 576 } \mathrm { cm } ^ { 2 }{/tex}
={tex}12 \times \sqrt { a ^ { 2 } - 144 } \mathrm { cm } ^ { 2 }{/tex}
But, area = 192 cm2 [given].
{tex}\therefore \quad 12 \times \sqrt { a ^ { 2 } - 144 } = 192 \Rightarrow \sqrt { a ^ { 2 } - 144 } = 16{/tex}
{tex}\Rightarrow a ^ { 2 } - 144 = 256 \Rightarrow a ^ { 2 } = 400 \Rightarrow a = 20{/tex}
{tex}\therefore\quad {/tex}perimeter of the triangle = (2a + b) cm
= (2 {tex}\times{/tex} 20 + 24) cm = 64 cm.
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