Question

5) An isosceles triangle has perimeter 20 cm and each of the equal sides is 6 cm, then its area ​

Answer 1

$\huge\sf\pink{Answer}$

☞ Area of the triangle is 8√5 cm²

━━━━━━━━━━━━━

$\huge\sf\blue{Given}$

✭ ABC is an Isosceles triangle

✭ Perimeter of the triangle is 20 cm

✭ Equal sides are 6 cm each

━━━━━━━━━━━━━

$\huge\sf\gray{To \:Find}$

◈ Area of the triangle?

━━━━━━━━━━━━━

$\huge\sf\purple{Steps}$

We know that,

$\underline{\boxed{\textsf{Perimeter = Sum of all sides}}}$

Substituting the given values,

➝ $\sf Perimeter = 2(Equal \ Sides) + Base$

➝ $\sf 20 = 6+6+ Base$

➝ $\sf 20 = 12+Base$

➝ $\sf 20-12 = Base$

➝ $\sf \red{Base = 8 \ cm}$

Area of the triangle can be found with the help of Heron's Formula,i.e,

$\sf \underline{\boxed{\sf Area \ of \ Traingle = \sqrt{s(s-a)(s-b)(s-c)}}}$

◕ s = Half Perimeter = $\sf \dfrac{20}{2}$ = 10

◕ a = 6

◕ b = 6

◕ c = 8

Substituting the values,

➳ $\sf Area \ of \ Traingle = \sqrt{s(s-a)(s-b)(s-c)}$

➳ $\sf \sqrt{10(10-6)(10-6)(10-8)}$

➳ $\sf \sqrt{10(4)(4)(2)}$

➳ $\sf \sqrt{10(32)}$

➳ $\sf \sqrt{320}$

➳ $\sf \sqrt{2\times 2\times 2\times 2\times 2\times 2\times 5}$

➳ $\sf 2\times 2\times 2\sqrt{5}$

➳ $\sf \orange{Area = 8\sqrt{5} \ cm^2}$

━━━━━━━━━━━━━━━━━━