Question

Find the area of the triangle in which a = 13m b = 14m c = 15m

Answer 1

Answer :

• Area of triangle is 84m²

Given :

sides of triangle:

• a = 13cm
• b = 14cm
• c = 15cm

To find :

• Area of the triangle

Solution :

As we know that ,

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

where s is a + b + c/ 2

putting the value :

⇢ s = 13 + 14 + 15 / 2

⇢ s = 42/2

⇢ s = 21

Now we have to find the area of triangle :

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

where s is 21 , a is 13 , b is 14 and c is 15

⇢ √21(21 - 13) (21- 14) (21 - 15) m²

⇢ √21(8) (7) (6) m²

⇢ √7056

⇢ 84m²

or

⇢ √21(21 - 13) (21- 14) (21 - 15) m²

⇢ √21(8) (7) (6) m²

⇢ √7 × 3 × 2 × 4 × 7 × 2 × 3 m²

⇢ √7² × 3² × 2² × 2² m²

⇢ 7 × 3 × 2 × 2 m²

⇢ 84m²

Hence, Area of triangle is 84m²

Answer 2

$\dag\:\underline{\sf Given \::}$

• $\sf\:a=13\:m$

• $\sf\:b=14\:m$

• $\sf\:c=15\:m$

$\dag\:\underline{\sf To\:find\::}$

• $\sf\:Area\:of\:the\:triangle$

$\dag\:\underline{\sf Solution\::}$

We know,

$\large{\underline{\boxed{\mathrm\pink{Semi perimeter=(\frac{a+b+c}{2})}}}}$

‎

$\sf\:Semi-perimeter=\frac{a+b+c}{2}$

$\sf\:=\frac{13+14+15}{2}$

$\sf\:=\frac{42}{2}$

$\sf\:=21\:m$

‎

Now by Heron's formula :

‎ ‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎

$\sf\:Area\: of \:triangle :\sqrt{s(s - a)(s - b)(s - c)}m^{2}$

‎ ‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎

‎‎$\sf\: =\sqrt{21(21 - 13)(21 - 14)(21 - 15)}\:m^{2}$

‎ ‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎

‎$\sf\:=\sqrt{21(8)(7)(6)}\:m^{2}$

‎ ‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎

‎$\sf\: =\sqrt{21\times 8 \times 7 \times 6}\:m^{2}$

‎ ‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎ ‎ ‎ ‎‎‎‎‎‎ ‎‎‎

‎$\sf\: =\sqrt{7056}\:m^{2}$

‎‎‎‎$\sf\:=84\:m^{2}$

• Thnku :)