Question

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm​

Answer 1

Answer:

Area of triangle

$\sqrt{s(s - a)(9 - b)(s - 0)}$

Here, s is the semi- perimeter,

and a,b,c are the sides of the triangle

Given a=18 cm, b=10 cm

and perimeter =42 cm

semi- perimeters =

$\frac{perimeter}{2} = \frac{42}{2}$

s=21cm.

Area of triangle

∴c=42−(18+10)cm =14 cm

∴ Area of triangle =

$\sqrt{21.(21 - 18)(21 - 10)(21 - 41)}$

$= \sqrt{21.3.7.11 {cm}^{2} }$

$= \sqrt{3 \times 7 \times 3 \times 7 \times 11 {cm}^{2} }$

$\sqrt{(3 \times 7 {)}^{2} } \times 11 {cm}^{2}$

$21 \sqrt{ 11{cm}^{2} }$

∴Thus, the required area of the triangle is

$21 \sqrt{11 {cm}^{2} }$

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Answer 2

To Find

• The area of triangle

Given

• The perimeter is 42 cm
• One side of triangle is 18 cm
• Second side of triangle is 10 cm

Suppose the third side of triangle is 'x'

Now Three sides of triangles are

• 18 cm 10 cm and x cm.

$\implies\sf \: Now \: \: perimeter = 18+10+x \\ \implies \tt \: 42cm = 18 + 10 + x \\ \implies \tt \:42cm = 28 + x \\ \tt \:x = 42 - 28cm \\ \tt \:x = = 21cm$

Semi perimeter of triangle = 42/2 = 21cm

Using Heron's Formula..

$\boxed {\underline{\sqrt{s(s - a)(s - b)(s - c)} }}$

The area of triangle is :

$\sf {{\sqrt{s(s - a)(s - b)(s - c)} }} \\ \\ \sf \: \sqrt{21(21 - 18)(21 - 10)(21 - 14} \: {cm}^{2} \\ \\ \sf \: \sqrt{21 \times 3 \times 11 \times 7} \: {cm}^{2} \\ \\ 21\sqrt{11} \: \sf {cm}^{2}$

$\red{\sf \: Hence \: the \: area \: of \: triangle \: is \: 21 \sqrt{11 \: {cm}^{2} } }$