Question

if the sides of a triangle are 45cm,39cm and 42 cm find its are​

Answer 1

$\Huge {\mathbf {\color{purple}{Q}{\mathfrak {\color{plum}{uestion}}}}}$

If the sides of a triangle are 45cm, 39cm and 42cm find its area.

$\Huge {\mathbf {\color{Purple}{A}{\mathfrak {\color{Plum}{nswer}}}}}$

Sides of a triangle are 45cm, 39cm and 42cm.

Here, a = 45cm, b = 39cm, c = 42cm

Semi perimeter of triangle = s

$\small \bf \color{royalblue}{ = (\frac{1}{2})(a + b + c)}$

$\small \bf \color{royalblue}{=(\frac{1}{2})(45 + 39 + 42)}$

$\large \bf \color{royalblue}{ = \frac{126}{2} }$

$\small \bf \color{royalblue}{= 63}$

$\: \: \:$

Area of a triangle

$\: \:$

$\small \bf \blue{ = \sqrt{s(s -a)(s - b)(s - c)} }$

$\small \bf \blue{ = \sqrt{63(63 - 45)(63 - 39)(63 - 42)}}$

$\small \bf \blue{ = \sqrt{63 \times 18 \times 24 \times 21} }$

$\small\sf \blue{ = \sqrt{7 \times 9 \times 2 \times 9 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7}}$

$\small \bf \blue{ = \sqrt{ {7}^{2} \times {9}^{2} \times {2}^{2} \times {2}^{2} \times {3}^{2} } }$

$\small \bf \blue{ = 7 \times9 \times 2 \times 2 \times 3 }$

$\small {\bold {\boxed {\underline {\underline {\mathbf {\blue {\red{ 756sq.cm}}}}}}}}$

$\: \: \: \:$

$\small \sf{∴ The \: area \: of \: the \: triangle \: is \: 756 sq.cm}$

Answer 2

Answer:

$\huge\boxed{A_\triangle=756cm^2}$

Step-by-step explanation:

We have the lengths of the sides of the triangle: 45cm, 39cm, 42cm.

Use the Heron's formula:

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

where

$a,\ b,\ c\ -\ \text{sides of a triangle}\\\\s=\dfrac{a+b+c}{2}\ -\text{half of a primeter}$

Let $a=45cm,\ b=39cm,\ c=42cm$

Calculate $s$

$s=\dfrac{45+39+42}{2}=\dfrac{126}{2}=63\ (cm)$

Calculate the area:

$A_\triangle=\sqrt{63(63-45)(63-39)(63-42)}\\\\A_\triangle=\sqrt{(63)(18)(24)(21)\\}\\\\A_\triangle=\sqrt{571,536}\\\\A_\triangle=756\ (cm^2)$