Question

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

Answer 1

Answer :-

• Area of Triangle = 21√11 cm²

Given :-

• First Side (Side a) = 18 cm
• Second Side (Side b) = 10 cm
• Perimeter of triangle = 42 cm

To Find :-

• Area of Triangle = ?

Explanation :-

In order to find the area of the triangle first we need to find the Third Side (Side c) of the Triangle.

Let the Third Side (Side c) to be 'x' cm.

$\underline { \boxed { \bf \implies Perimeter \: of \: Triangle = Sum \: of \: all \: Sides }}$

$\rm \implies Side \: 1 + Side \: 2 + Side \: 3 = 42 \: cm$

$\rm \implies 18 + 10 + x = 42 \: cm$

$\rm \implies 28 + x = 42 \: cm$

$\rm \implies x = 42 - 28 \: cm$

$\underline { \boxed { \bf \implies x = 14 \: cm }}$

Now, Finding the Area of Triangle using Heron's Formulae.

$\red{ \underline { \boxed { \bf \implies Heron's \: Formulae = \sqrt{s (s -a) (s - b) (s-c) } }}}$

Where,

• s = Semi - Perimeter of the Triangle
• a = First side of the Triangle
• b = Second side of the Triangle
• a = Third side of the Triangle

$\boxed { \bf \implies Semi \: Perimeter \: of \: Triangle = \frac{a + b + c}{2} }$

$\rm \implies \dfrac{18 + 10 + 14}{2}$

$\rm \implies \dfrac{42}{2}$

$\bf \implies 21 \: cm$

Now Substituting the values,

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

$\rm \implies \sqrt{21 (21 - 18) (21 - 10) (21 - 14) }$

$\rm \implies \sqrt{ 21 \times 3 \times 11 \times 7}$

$\blue { \underline { \boxed { \bf \star \: \implies 21 \sqrt{11} \: cm^2 \: \star}}}$

Hence, the area of the triangle is 21√11 cm².

@Agamsain

Answer 2

Answer:

$perimeter = 42cm \\ let \: the \: unknown \: side \: be \: x \: \\ perimeter \: = sum \: of \: all \: sides \\ 42 = 18 + 10 + x \\ x = 42 - 28 \\ x = 14 cm$

Area of Triangle =

$\sqrt{s(s - a)(s - b)(s - c)} \\ where \: s \: is \: semi \: perimeter \\ s = \frac{42}{2} = 21cm \\ \\ \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\ \\ \sqrt{21 \times 3 \times 11 \times 7 } \\ \\ \sqrt{4851 } = 21 \sqrt{11} sq \: cm$

Area of triangle =

$21 \sqrt{11} sqcm$