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

Answer:

$here \: s \: is \: half \: perimeter \\ and \: a \: b \: and \: c \: are \: sides\\ s = 23 \\ area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{23(23 - 13)(23 - 13)} (23 - 20) \\ = \sqrt{23 \times 10 \times 10 \times 3} \\ = \sqrt{6900} \\ = > 83.0662386$

Answer 2

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.