Question

An isosceles triangle has a perimeter of 30 cm and its equal side is 12 cm in length. Find the area of this triangle.​

Answer 1

Answer:

$9 \sqrt{15} {cm}^{2}$

Step-by-step explanation:

let the unequal side be x

perimeter= sum of sides

30cm = 12+12+x

30cm = 24+x

30-24 = x

x= 6 cm

s.p.= 30/2 = 15cm

Now area of an isosceles triangle:

√s(s-a)(s-b)(s-c)

√15(15-12)(15-12)(15-6)

√15(3)(3)(9)

√5×3×3×3×3×3

3×3√5×3

9√15 cm^2

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Answer 2

First,let the third side be x.

It is given that the length of the equal sides us 12 cm and it's perimeter is 30 cm.

So,$30=12+12+x$

$⇒ 30 = 24 + x$

$⇒24 + x = 30$

$⇒ x= 30 - 24$

$⇒ x = 6$

So,the length of the third side is 6 cm.

Thus,the semi perimeter of the isosceles triangle (s) = 30/2 cm =15 cm

By using Heron's Formula,

Area of the triangle,

$= \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \: {cm}^{2}$

$= \sqrt{15 \times 3 \times 3 \times 9} \: {cm}^{2}$

$= 9 \sqrt{15} \: {cm}^{2}$