Question

The lengths of the sides of a triangle are 52cm,56 cm and 60 cm. Then its semi perimeter 's' is

Answer 1

Answer:

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is

(a) 1322 cm2

(b) 1311 cm2

(c) 1344 cm2

(d) 1392 cm2

Thinking Process

(i) First, determine the semi-perimeter of a triangle by using the formula, s = (a + b + c)/2

(ii) Further, determine the area of triangle by using the formula, area of triangle (Fieron’s formula) = √s(s-a) (s-b) (s-c)

Answer 2

Step-by-step explanation:

Answer:

Sides are 52cm, 56cm, and 60 cm

Area of the Triangle = ?

By Using Heron's Formula,

The area of the given triangle is;

$$\begin{lgathered}\\ \bullet{\boxed{\sf{ Area= \sqrt{ s(s-a)(s-b)(s-c) } }}} \\\end{lgathered}$$

Where,

$$\begin{lgathered}\because {\sf{\bf{ s = \dfrac{a+b+c}{2} }}} \\\end{lgathered}$$

$$\begin{lgathered}\implies{\sf{ \dfrac{52+56+60}{2} }} \\ \\ \implies{\sf{ \dfrac{ \cancel{168}^{ \: \: 84}}{ \cancel{2}} }} \\ \\ \implies{\sf{ 84 \: cm}} \\\end{lgathered}$$

Solution:

$$\begin{lgathered}\begin{lgathered}\\ \implies{\sf{ A= \sqrt{ 84(84-52)(84-56)(84-60) } }} \\\end{lgathered}\end{lgathered}$$

$$\begin{lgathered}\\ \implies{\sf{ \sqrt{ 84 \times 32 \times 28 \times 24} }} \\\end{lgathered}$$

$$\begin{lgathered}\\ \implies{\sf{ \sqrt{1806336} }} \\\end{lgathered}$$

$$\begin{lgathered}\begin{lgathered}\\ \implies{\sf{ 1344 \: cm^2 }} \\\end{lgathered}\end{lgathered}$$

Hence,

$$\sf\pink{\bf{ The\:area\:of\: triangle\:is\:1344\:cm^2}} .$$

Hope it helps!!