Question

Find the area of triangle whose side is 42m,56m,and 70m​

Answer 1

Given :

• Sides of a triangle is 42 m,56 m, and 70 m.

To find :

• Area of triangle.

According to the question,

We will use Heron's formula,

$\bf : \implies{ \dfrac{a + b + c}{2} } \\ \\ : \implies \bf{ \dfrac{4 2 + 56 + 70}{2} } \\ \\ : \implies \bf{ \dfrac{168}{2} } = 84 \: m$

$\bf : \implies{ \sqrt{s(s - a)(s - b)(s - c)} }$

$\bf : \implies{ \sqrt{84(84 - 42)(84 - 56)(84 -70)} }$

$: \implies \bf{ \sqrt{84(42)(28)(14)} }$

$: \implies \bf{ \sqrt{14 \times 2 \times 3 \times 14 \times 3 \times 14 \times 2 \times 14} }$

$: \implies \bf{14 \times 14 \times 2 \times 3}$

$: \implies{ \underline{ \boxed{ \bf \green{1176 \: {m}^{2} }}}}$

Answer 2

Sides of the triangle = 42 m, 56 m, 70 m

∴ Semi-perimeter (s) = (a + b + c)/2

⇒s = [(42 + 56 + 70) m]/2

⇒ s = 168/2 m = 84 m

We know,

Area of triangle = √[s(s - a)(s - b)(s - c)]

∴ Area = √[84(84 - 42)(84 - 56)(84 - 70)] m²

⇒ Area = √(84 × 42 × 28 × 14)

= √(1176)² = 1176 cm²