Question

The area of an isosceles triangle having the base 24cm and length of one of the equal sides 20cm is using herons formula​

Answer 1

Given

• Base (BC) = 24cm
• Equal Side's (AB) = (AC) = 20cm

To Find

• Area of Triangle

We know,

$\bf \footnotesize{Semi- Perimeter = \dfrac{ Sum \: of \: all \: Sides}{2}}$

${ \implies\footnotesize \sf{Semi- Perimeter = \dfrac{ 24 + 20 + 20}{2}}}$

${ \implies\footnotesize \sf{Semi- Perimeter = \cancel\dfrac{64}{2}}}$

${ \implies\footnotesize \sf{Semi- Perimeter = 32}}$

$\sf {Now \: by \: \bf \pink{Herons \: Formula}}$

${ \sf\: Area \: of \: ∆ = \sqrt{s(s-a)(s-b)(s-c)}}$

${ \sf\: Area \: of \: ∆ = \sqrt{32(32-24)(32-20)(32-20)}}$

${ \sf\: Area \: of \: ∆ = \sqrt{32 \times 8 \times 12 \times 12}}$

${ \sf\: Area \: of \: ∆ = \sqrt{256 \times {12}^{2} }}$

${ \sf\: Area \: of \: ∆ = \sqrt{ {16}^{2} \times {12}^{2} }}$

${ \sf\: Area \: of \: ∆ = 16 \times 12 }$

${ \sf\: Area \: of \: ∆ = 192 }$

• Hance the area of Triangle be 192 cm².

Answer 2

and the

32 - 2× 2 ×2 × 2 ×2

8 = 2 × 2 × 2

12 = 2 ×3 × 2

12 = same as Frist