Question

find the area of an isosceles triangle which has these 12 cm and equal sides are 10 cm each​

Answer 1

Step-by-step explanation:

Measure of equal sides = 10 cm

Measure of third side = 12 cm

so, we need to find the area of an isosceles triangle.

Semi perimeter s = \dfrac{a+b+c}{2}=\dfrac{10+10+12}{2}=\dfrac{32}{2}=16\ cm

2

a+b+c

=

2

10+10+12

=

2

32

=16 cm

So, Area of isosceles triangle would be

\begin{gathered}\sqrt{s(s-a)(s-b)(s-c)}\\\\=\sqrt{16(16-10)(16-10)(16-12)}\\\\=\sqrt{16\times 6\times 6\times 4}\\\\=\sqrt{2304}\\\\=48\ cm^2\end{gathered}

s(s−a)(s−b)(s−c)

= 16(16−10)(16−10)(16−12)

= 16×6×6×4

= 2304

=48 cm

Answer 2

Answer:

48 sq cm

Step-by-step explanation:

a = 12 cm

b = C = 10cm

semi perimeter= S= (a+b+C) /2

=(12+10+10) /2

= 32/2

= 16

S= 16

$area = \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{16 \times (16 - 12)(16 - 10(16 - 10)}$

$= \sqrt{16 \times 4 \times 6 \times 6}$

$= 4 \times 2 \times 6$

=48sq cm

area of the triangle= 48 sq cm