Question

Find the area of a isosceles triangle whose the same side are 13 cm and base is of 20 CM can you do this question by Pythagoras theorem​

Answer 1

$\huge\star\underline\mathfrak\red{Answer:-}$

Let the vertices of the given isosceles triangle be ABC respectively.

As it is an isosceles triangle, so two sides AB = AC = 13 cm

Length of the base BC = 20 cm

Construction: Draw perpendicular AD .

In triangle ADB,

=>> BD = 10cm (As D is mid point of BC)

=>> AB = 13 cm

=>> AD² + BD² = AB² (Using Pythagoras theorem)

=>> AD² + 10² = 13²

=>> AD² = 13² - 10² = 8.3 cm

So,

area of triangle ABC = 1/2 × 20 × 8.3 cm²

area of triangle ABC = 83 cm².

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Answer 2

Step-by-step explanation:

the area of isosceles triangle is 83sq.cm