Question

find the area of an isosceles triangle each of whose equal sides is 13 cm and base 24 cm

Answer 1

We have three sides- 13cm,13cm and 20cm, you can easily find the perimeter by using heron's formula

 First of all- we need to get the semi-perimeter
s=a+b+c/2=13+13+20/2
=46/2=23
now, put the heron's formula i.e. √s(s-a)(s-b)(s-c)
=√23(23-13)(23-13)(23-20)
=√23*10*10*3
=√23*2*5*2*5*3
=10√23*3
=10√69
=83.07cm^2 (approx.)

Hope it helped you. Plz mark me as brainliest.

Answer 2

$\underline{\underline{\bold{Question:}}}$

Find the area of an isosceles triangle each of whose equal sides is 13 cm and base 24 cm.

$\tt{Solution:}$

$\boxed{\bold{Area\:of\:an\:isosceles\:triangle=\dfrac{b}{4}\sqrt{4a^2-b^2}}}$

Where b = base of the triangle = 24 cm.

and a = equal sides of the triangle = 13 cm.

So

$\tt{Area=\dfrac{b}{4}\sqrt{4a^2-b^2}}\\\\\\\bold{=\dfrac{24}{4}\sqrt{4*13^2-24^2}}\\\\\\\bold{=6*\sqrt{676-576}}\\\\\\\bold{=6*\sqrt{100}=6*10=60\:cm^2.}\\\\\\\boxed{\boxed{\bold{Area\:of\:triangle=60\:cm^2.}}}$

$\underline{\bold{Important\:formula:}}\\\\\\\bold{Height\:of\:an\:isosceles\:triangle=\dfrac{1}{2}\sqrt{4a^2-b^2}}\\\\\\\bold{Heron\:formula:}\\\\\\\bold{Area\:of\:triangle=\sqrt{S(S-a)(S-b)(S-c)}}\\\\\bold{Where\quad a\:,b\:,c=sides}\\\\\bold{S=\dfrac{a+b+c}{2}.}$