Question

find the area of an isosceles triangle whose base is 18 cm and one of its equal sides is 15 CM

Answer 1

$\large \tt AHOY!! \:$

let the sides of the isosceles triangle be a, b and c.

given the length of the sides of the isosceles triangle :-

a = 18cm

b = 15cm

c = 15cm

semi-perimeter of the isosceles triangle = (18 + 15 + 15)/2

= 48/2

= 24cm

area of the triangle = √s(s-a)(s-b)(s-c)

= √24(24-18)(24-15)(24-15)

= √(24 × 6 × 9 × 9)

= √(2 × 2 × 2 × 3 × 2 × 3 × 3 × 3 × 3 × 3)

= √(2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3)

= 2 × 2 × 3 × 3 × 3

= 108cm²

$\large \tt HOPE \: THIS \: HELPS!!$

Answer 2

$\huge \green{Answer}$
$= \red{ {108cm}^{2} }$

$\pink {solution}$

Let the sides of the isosceles triangle be

A , B .

In the questions given the length of the sides of the isosceles triangle are given bellow .

A = 15cm , B= 18cm

Semi-perimeter of the isosceles triangle

(18 + 15 + 15)/2

48/2

24cm

$Area \: of \: the \: isosceles \\ \: triangle = \sqrt{ {s(s - a)}^{2}(s - b) } \\ \\ \\ = \sqrt{ {24(24 - 15)}^{2}(24 - 18) } \\ \\ = \sqrt{24(81)(6)} \\ \\ = \sqrt{11664} \\ \\ = \sqrt{108 \times 108} \\ \\ = 108 {cm}^{2}$

Therefore the area of Isosceles triangle is
${108cm}^{2}$

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