Question

The perimeter of an isosceles triangle is 32cm. The ratio of the equal sides to its base is 3 is to 2. find the area of the triangle

Answer 1

let equal sides be x and base be y.

perimeter of an isosceles triangle = 32cm

=) 2x+y = 32

given, x/y = 3/2

=) x = 3y/2

=) 2×3y/2 + y = 32cm

=) 3y + y = 32cm

=) 4y = 32cm

=) y = 32/4 = 8cm

hence x = 12cm.

area of an isosceles triangle = y/4(√4x² - y²)

= 8/4 (√4(12)² - 8²)

= 2(√4(144) - 64)

= 2(√576-64)

= 2(√512)

= 32√2 cm².

Answer 2

Answer:

64 cm²

Step-by-step explanation:

$\text{Ratio of equal sides to base = 3:2}\\\\\text{Let each of the equal sides be 3x cm and the unequal side be 2x cm}\\\\\text{Perimeter = Sum of sides = 3x + 3x + 2x = 8x cm}\\\\\text{But, given perimeter = 32 cm}\\\\\implies 8x = 32\\\\\implies x = \dfrac{32}{8}\\\\\\\implies x = \boxed{4}\\\\\therefore \text{Equal sides = 3x = 3(4) = 12 cm.}\\\\\text{And, Unequal side (Base) = 2(4) = 8 cm.}\\\\\text{Area of triangle by heron's formula}\\\\= \sqrt{s(s-a)(s-b)(s-c)},$

$\text{Where a, b, and c are the three sides of triangle and s is the semiperimeter}\\\\= \dfrac{a+b+c}{2}\\\\= \dfrac{12+12+8}{2}\\\\= \dfrac{32}{2}\\\\= \boxed{16 \: cm}\\\\\therefore \text{Area of triangle}\\\\= \sqrt{16(16-12)(16-12)(16-8)}\\\\= \sqrt{16 \times 4 \times 4 \times 8}\\\\= \sqrt{4 \times 4 \times 4 \times 4 \times 4 \times 2}\\\\= 32 \sqrt{2} = \\\\  \large \underline{\boxed{\boxed{\bold{32 \sqrt{2} \: cm^2}}}}$

Hope it helps.