Question

An isosceles triangle has perimeter 20 cm and each of equal sides is 8 cm. Find area of

the triangle.​

Answer 1

Step-by-step explanation:

Given,

Perimeter of triangle = 20 cm

and, length of equal sides = 8+8 = 16 cm

Now,

third length = perimeter of triangle - sum of two sides

= (20 - 16) cm

= 4 cm

Let a=8, b=8, c=4

Semiperimeter = a+b+c/2

= 8+8+4/2

= 20/2

= 10

Using Heron's Formula:

Area of triangle = √s(s-a)(s-b)(s-c)

= √10(10-8)(10-8)(10-4)

= √10×2×2×6

= √2×5×2×2×2×3

= 2×2√5×3

= 4√15 cm^2

Hence, the area of triangle = 4√15 cm^2

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Answer 2

Given

Perimeter of isosceles ∆ = 20 cm.

Length of each equal side = 8 cm.

To find

Area of triangle

Solution

Here, equal sides = a = 8 cm.

We know that,

➥ Perimeter of ∆ = Sum of 3 sides.

Putting values :

⟼ 20 = 8 + 8 + Inequal side

⟼ 20 - 16 = Inequal side

⟼ Inequal side = b = 4 cm.

We also know that,

➥ Area of isosceles ∆ = (b/4)√[4a² - b²]

Putting values we get :

➻ Area = (4/4)√[4(8)² - 4²]

➻ Area = (1)√[4 × 64 - 16]

➻ Area = (1)√[256 - 16]

➻ Area = 1 √240

➻ Area = √(16 × 15)

➻ Area = √(4² × 15)

➻ Area = 4√15 cm² or,

➻ Area = 15.49 cm²

Therefore,

Area of isosceles ∆ = 15.49 cm²