Question

an isosceles triangle has perimeter 30cm and each of the equal side is 12cm . find the area of the triangle​

Answer 1

Answer:

40.25cm2...........................

Answer 2

Given,

• Perimeter of isosceles triangle = 30 cm
• Measure of equal side of the ∆ = 12 cm

To find : Area of triangle

Here,

→ Sum of all side of the ∆ = Perimeter of ∆

→ Sum of equal sides & unequal side = Perimeter of isosceles ∆

So, Let the unequal side of ∆ be x

Then,

$\to \rm{}12 + 12 + x = 30 \\ \to \rm \: \: \: \: \: \: \: 24 + x = 30 \\ \to \rm \: \: \: \: \: \: \: x = 30 - 24 \\ \to \rm \: \: \: \: \: \: \: x = \green6 \: cm$

Now, Applying Heron's Formula, we get,

$\rm \sqrt{ \frac{perimeter}{2} \bigg\{ \frac{perimeter}{2} - a \bigg\} \bigg\{ \frac{perimeter}{2} - b \bigg\} \bigg\{ \frac{perimeter}{2} - c \bigg\} } \\ \\ \\ \\ \to \tt{ \sqrt{ \frac{30}{2} \bigg\{ \frac{30}{2} - 12 \bigg\} \bigg\{ \frac{30}{2} - 12 \bigg\}\bigg\{ \frac{30}{2} - 6 \bigg\}} } \\ \\ \\ \to \tt \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \\ \to \tt \sqrt{15 \times 3 \times 3 \times 9} \\ \to \tt \sqrt{3 \times 5 \times 3 \times 3 \times 3 \times 3 } \\ \to \tt9 \sqrt{3 \times 5} \\ \to \tt \red{9 \sqrt{15} } \: \: \rm {cm}^{2}$

Hence,

• Area of isosceles triangle = 9√15 cm²