The base of an isosceles triangle is $12$ cm and it's perimeter is $32$ cm then the area of triangle is

30 {cm}^{2}

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48 {cm}^{2}

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40 {cm}^{2}

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20 {cm}^{2}

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Solution

The correct option is B $48 {cm}^{2}$
Perimeter $= 32$
$2 x + 12 = 32$
or, $2 x = 20$
or, $x = 10 c m$
Altitude $= \sqrt{x^{2} - 6^{2}}$
$= \sqrt{10^{2} - 6^{2}}$
$= 8 c m$
Area $= \frac{1}{2} \times 12 \times 8 = 48 c m^{2}$

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The base of an isosceles triangle is 12 cm and it's perimeter is 32 cm then the area of triangle is

The correct option is B 48 cm2Perimeter =322x+12=32or, 2x=20or, x=10cmAltitude =√x2−62=√102−62=8cmArea =12×12×8=48cm2

The base of an isosceles triangle is $12$ cm and it's perimeter is $32$ cm then the area of triangle is

The correct option is B $48 {cm}^{2}$
Perimeter $= 32$
$2 x + 12 = 32$
or, $2 x = 20$
or, $x = 10 c m$
Altitude $= \sqrt{x^{2} - 6^{2}}$
$= \sqrt{10^{2} - 6^{2}}$
$= 8 c m$
Area $= \frac{1}{2} \times 12 \times 8 = 48 c m^{2}$