Question

If the base of an isosceles triangle is 12cm its perimeter is 35 cm find the area of the triangle

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{Question:-}}}}}}$.

• If the base of an isosceles triangle is 12cm its perimeter is 35 cm find the area of the triangle

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$\large{\underline{\underline{\mathfrak{\sf{Given\:Here:-}}}}}$.

• Base of triangle b = 12cm

• Perimeter of triangle = 35cm

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$\large{\underline{\underline{\mathfrak{\sf{Find\:Here-}}}}}$.

• Area of triangle

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$\large{\underline{\underline{\mathfrak{\sf{Explanation:-}}}}}$.

$\bold{\:we\:know\:that}$

$\large{\boxed{\:Perimeter\:of\:isosceles\:Triangle\\ \:(P)\:=\:(2a+b)}}$

Where ,

• a = two equal side of isosceles triangle ,

• b = base of isosceles triangle

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So ,

$\leadsto\:perimeter\:(P)\:=\:(2a+12)$

$\leadsto\:(35)\:=\:(2a+12)$

$\leadsto\:2a\:=\:35-12$

$\leadsto\boxed{\:a\:=\frac{23}{2}}$

Now, We know

$\large\boxed{\:Area\:of\:isosceles\:triangle\:=\frac{1}{2}(base\:×\:height)}$

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Now, Calculate height of triangle

$\leadsto\boxed{\:Height\:=\sqrt{(side)^2-(base)^2}}$

$\leadsto\:Height\:=\sqrt{(\frac{23}{2})^2-\:(\frac{12}{2})^2}$

$\leadsto\:Height\:=\sqrt{(\frac{529}{4})-\:(\:36)}$

$\leadsto\:Height\:=\sqrt{(\frac{529-144}{4})}$

$\leadsto\:Height\:=\sqrt{(\frac{385}{4})}$

$\leadsto\:Height\:=\frac{1}{2}\:×\sqrt{385}$

Or,

$\leadsto\:Height\:=\frac{1}{2}\:×\:19.62$

$\leadsto\boxed{\:Height\:=\:9.81}$

$\large{\underline{\underline{\mathfrak{\sf{Area\:Of\:Triangle:-}}}}}$.

$\leadsto\boxed{\:Area\:=\frac{1}{2}\:×\:(base\:×\:Height)}$

Keep value of Base and Height ,

$\leadsto\:Area\:=\frac{1}{2}\:×\:(12×9.81)$

$\leadsto\:Area\:=\:(6×9.81)$

$\leadsto\boxed{\:Area\:=\:58.86}$

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