Question

find the area of an esosceles triangle each of whose equal sides measures 13cm and whose base measures 20cm.

Answer 1

$let \: \\ a = 13 \\ b = 13\\c = 20 \\ \\ \\ let \: \\ s = semiperimeter = \frac{a + b + c}{2 } \\ = \frac{13 + 13 + 20}{2} \\ = \frac{46}{2} = 23 \\ \\ \\ area \: = \sqrt{s(s - a)(s - b)(s - c)} \\ here \\ \: s - a = 23 - 13 = 10 \\ s - b = 23 - 13 = 10 \\ s - c = 23 - 20 = 3 \\ \\ \\ \\ = \sqrt{23 \times 10 \times 10 \times 3 } \\ = \sqrt{23 \times 3 \times 5\times 2 \times 5 \times 2} \\ = \sqrt{(23 \times 3) \times {5}^{2} \times {2}^{2} } \\ = 10\sqrt{69}$
hey friend

i hope it is helpful.....


n mark as BRAINLIEST if i deserve.....


;-)

Answer 2

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.)