Question

Find the area of an isosceles triangle whose perimeter is 32 cm and its base is 12 cm.

Answer 1

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Answer 2

$\huge\rm \underline \red{QUESTION}$

Find the area of an isosceles triangle whose perimeter is 32 cm and its base is 12 cm.

$\rm\underline \red {Given :}$

• Base of isosceles triangle =12 cm
• Perimeter of isosceles triangle = 32cm

$\rm \underline \red{To \: Find :}$

• Area of isosceles triangle

$\huge \rm \underline \red{SOLUTION}$

In isosceles triangle 2 sides are equal.

Perimeter = 32cm

Base = 12 cm

Let other sides be x.

$x + x + 12 = 32$

$2x = 32 - 12$

$2x = 20$

$x = \frac{20}{2} = 10$

Therefore the other 2 sides are 10cm.

Area of isosceles triangle

$= \frac{1}{2} \times b \times \sqrt{a ^{2} - \frac{b ^{2} }{4} }$

$= 6 \sqrt{10 ^{2} - \frac{12 ^{2} }{4} }$

$= 6 \sqrt{100 - \frac{144}{4} }$

$= 6 \sqrt{\frac{400 - 144}{2} }$

$= 6 \sqrt{ \frac{256}{4} }$

$= 6\sqrt{64} →6 \sqrt{8 \times 8}$

$6 \times 8 = 48cm ^{2}$

Therefore area of isosceles triangle is 48cm².

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