Question

find the area of an isosceles triangle whose base is 24 cm and one of its equal side is 13​

Answer 1

Answer:

you can use heron's theorum

√s(s-1st side)(s-2nd side)(s-3rd side)

s=perimeter/2

Answer 2

Given:

• Find the area of an isosceles triangle whose base is 24 cm and one of its equal side is 13

Diagram:

• Refer to the attachment please!

Solution:

Let ABC be the isosceles triangle where,

• AB = AC = 13 cm

We draw AD ⊥ BC.

∴⠀⠀⠀⠀⠀⠀⠀⠀BD = DC = 12 cm

In right angled △ABD, by Pythagoras theorem,

We have ⠀⠀⠀⠀⠀⠀AB² = AD² + BD²

$\implies \: \: \: \: \: \: \sf \: {13}^{2} = AD {}^{2} + 12 \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AD {}^{2} = 13 {}^{2} - 12 {}^{2} \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: AD {}^{2} = 169 {}^{} - 144 {}^{} \\ \sf \: \: \: \: \: AD {}^{2} = 25 {}^{2} \\ \sf \: \: \: AD² = 5 {}^{2} \\ \sf \: \: \: \: \: \: \: \:AD = 5 \: cm$

$\\ \sf \: \: \: \: \: \: \: Area \: of \: △ABC \: = \frac{1}{2} \: base \times altitude \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times BC \times AC \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times24 \times 5 \: sq.cm \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \boxed{ \underline{ \bigstar{ \pink{ \bf{60 \: sq.cm}}}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀Hence Solved!!

⠀⠀⠀⠀⠀⠀⠀Hope it helps you!!

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