Question

find the area of an isosceles triangle whose equal sides measure 13 cm and base measures 24 cm
*plz write step by step solution otherwise no need to write only answer*​

Answer 1

answer is 60^2 cm

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Answer 2

Attached figure

$\sf{\huge{\underline{\green{Given :-}}}}$

• An isosceles triangle whose equal sides measure 13 cm and base measures 24 cm .

$\sf{\huge{\underline{\green{To\:Find :-}}}}$

• The area of isosceles triangle .

$\sf{\huge{\underline{\green{Answer :-}}}}$

According to the question,

We have an isosceles triangle .

It's Equal sides = 13 cm.

Base measures = 24 cm .

According to the attached Fig.

AB = AC = 13 cm

BC = 24 cm

We know that, the area of an isosceles triangle is

Draw, perpendicular AD on BC,

Such that BD = CD

Let's take AD as the height,

➝ BC = BD + CD

➝ BC = BD + BD

➝ BC = 2 BD

➝ 24 = 2 BD

➝ BD = 24/2

➝ BD = 12 cm

In right ΔADC , using PGT

AC² = AD² + DC²

AD² = AC² - DC²

➝ AD² = (13)² - (12)²

➝ AD² = 169 - 144

➝ AD² = 25

➝ AD = √25

➝ AD = 5 cm

The area of isosceles triangle = ½ × base × height

➝ ½ × 12 × 5

➝ 6 × 5

➝ 30 cm²

The area of isosceles triangle is 30 cm² .

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