Question

find the area of an isosceles triangle whose equal sides are of length 15cm each and third side is 12cm.​

Answer 1

hii mate

using herons formula,

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

s = 21

s-a = 6 = s-b

s-c = 9

area=

$\sqrt{21 \times 6 \times 6 \times 9}$

$\sqrt{3 \times 7 \times 3 \times 3 \times 2 \times 6 \times 6}$

$6 \times 3 \sqrt{21}$

$18 \sqrt{21}$

$18 \times 4.55$

=82.48

Answer 2

Answer:

$\large{\underline{\boxed{\sf 18 \sqrt{21}}}}$

Step-by-step explanation:

Given that, the sides of isosceles triangle are 15 cm, 15 cm, and 12 cm.

• a = 15
• b = 15
• c = 12

Using heron's formula -

S = $\rm \dfrac{a + b + c}{2}$

S = $\rm \dfrac{15 + 15 + 12}{2}$

S = $\rm \dfrac{42}{2}$

S = 21

Now,

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

$\rm Area = \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \\ \rm Area = \sqrt{21 \times 6 \times 6 \times 9} \\\\ \rm Area = \sqrt{7 \times 3 \times 2 \times 3 \times 2 \times 3 \times 3 \times 3}\\ \\ \rm Area = 18 \sqrt{21} \: cm^{2}$

Hence, the area of isosceles triangle is 18√21 cm².