Question

each of the equal side of an isosceles triangle is 13 cm and its base is 24cm . The area of triangle is ​

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\sf\underline{\pink{\:\:\:\:\:\:\:\: Given:-\:\:\:\:\:\:}}$

$\sf\bullet sides\:of\:isosceles \triangle\\ \sf:\implies 13cm\:,13cm\:and\:24cm$

• We have to find the Area of ∆

• By heron's Formula we can find our answer

$\boxed{\sf{Area=\sqrt{s(s-a)(s-b)(s-c)}}}$

$\sf\bullet where\:\: s \:\:is\:semi\: perimeter\\ \\ \sf:\leadsto s=\dfrac{a+b+c}{2}$

$\sf\underline{\red{\:\:\:\:\:\:\:\: A.T.Q:-\:\:\:\:\:\:}}$

• a= 13 cm

• b=13cm

• c= 24cm

$\sf:\implies s=\dfrac{13+13+24}{2}\\ \\ \sf:\implies s=\cancel{\dfrac{50}{2}}=25\\ \\ \sf:\implies {\blue{semi\: perimeter=25cm}}$

$\sf:\implies Area=\sqrt{s(s-a)(s-b)(s-c)}\\ \\ \sf:\implies Area=\sqrt{25(25-13)(25-13)(25-24)}\\ \\ \sf:\implies Area=\sqrt{25\times 12\times 12\times 1}\\ \\ \sf:\implies Area=\sqrt{5\times 5\times 12\times 12}\\ \\ \sf:\implies Area=5\times 12\\ \\ \sf:\implies Area\:of\:triangle=60cm{}^{2}$

$\boxed{\sf{\blue{Area\:of\:triangle=60cm{}^{2}}}}$

Answer 2

Plz refers to the attachments

Given:

Draw AD perpendicular bisector of BC

Therefore, BD=DC =>12cm

AB=AC=>13Cm [isosceles triangle]

$\mathbf \red{by \: pythagoras \: therome }$

AD²=BD²-AB²

AD²=(13)²-(12)²

AD²=169-144

AD²=25

AD=5cm

Now ,

Area of triangle =1/2×base ×height

=1/2×24×5

=12×5

=60cm²