Question

An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm . Find

the area of triangle.

Or

A triangular park
​

Answer 1

• Area of triangle is 34.83 cm²(approx).

Step-by-step explanation:

Given:-

• Perimeter of Isosceles triangle is 30 cm.
• Measure of each of equal sides is 12 cm.

To find:-

• Area of triangle.

Solution:-

Let, third side of triangle be x.

Perimeter of triangle = Sum of all sides

So,

$\longrightarrow$ 30 = 12 + 12 + x

$\longrightarrow$ 30 = 24 + x

$\longrightarrow$ 30 - 24 = x

$\longrightarrow$ x = 6

Measure of third side of triangle is 6 cm.

Now,

Here, We will use Heron's formula for finding area of triangle because height of triangle is not given.

Heron's formula is :-

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

Where,

• s is semi-perimeter of triangle.
• a, b and c are sides of triangle.

So,

Semi- Perimeter = Perimeter of triangle/2

$\longrightarrow$ s = 30/2

$\longrightarrow$ s = 15

Semi-perimeter is 15 cm

Then,

$\longrightarrow$ Area of triangle = √15(15 - 12)(15 - 12)(15 - 6)

$\longrightarrow$ Area of triangle = √15 × 3 × 3 × 9

$\longrightarrow$ Area of triangle = √5 × 3 × 3 × 3 × 3×3

$\longrightarrow$ Area of triangle = 3 × 3 × √3 × 5

$\longrightarrow$ Area of triangle = 9 × √15

$\longrightarrow$ Area of triangle = 9 × 3.87

$\longrightarrow$ Area of triangle = 34.83

Therefore,

Area of triangle is 34.83 cm²(approx).

Answer 2

♣ɢɪᴠᴇɴ:–

• Perimeter of the isosceles triangle = 30 cm
• Length of the equal sides = 12 cm

♣ᴛᴏ ғɪɴᴅ:–

• The area of the triangle.

♣sᴏʟᴜᴛɪᴏɴ:–

Let:–

$\qquad \bull \: \sf{The \: base \: angle=\bf{x} ^{ \circ} }$

$\large\underbrace {\bf{We \: know \: that:-}}$

$\qquad \bull \: \sf{The \: sum \: of \: the\: sides \: of \: the \: \triangle= \bf{30°}}$

$\qquad \bull \: \sf{12 + 12 + x = 30}$

${:} \implies \sf{24 + x = 30}$

${:} \implies \sf{ x = 30 - 24}$

${:} \implies \boxed{\sf{ x = 6}}$

$\qquad \therefore\: \sf{The \: base \: angle \: of \: the \: isosceles \: triangle= \underline {\underline {\bf{6°}}} }$

~ Using Heron's formula:–

$\large\underbrace {\bf{First \: condition:-}} \:$

$\qquad \bull \: \sf{s = \dfrac{1}{2} \bigg \lgroup{a + b + c \bigg \rgroup} }$

Where:-

• a = 12 cm
• b = 12 cm
• c = 156 cm
• s = semicircle

$\qquad \bull \: \sf{s = \dfrac{1}{2} \bigg \lgroup{12+ 12 + 6 \bigg \rgroup} }$

${:} \implies \sf{s = \dfrac{1}{2} \times 30}$

${:} \implies \sf{s = \cancel \dfrac{30}{2} }$

${:} \implies \boxed{\sf{s = 15} }$

$\large\underbrace {\bf{Second \: condition:-}}\:$

$\qquad \bull \: \sf{Area = \sqrt{s\bigg(s - a \bigg) \bigg(s - b \bigg) \bigg(s - c \bigg)}}$

$\qquad \bull \: \sf{Area = \sqrt{15\bigg(15 - 12\bigg) \bigg(15 - 12 \bigg) \bigg(15- 6 \bigg)}}$

${:} \implies \sf{Area = \sqrt{15 \times 3 \times 3 \times 9} }$

${:} \implies \sf{Area = \sqrt{1215}}$

${:} \implies \boxed {\sf{Area = 34.8568501}}$

$\qquad \therefore\: \sf{The \: area \: of \: the \: isosceles \: triangle= \underline {\underline {\bf{34.8568501°(approx \: value)}}} }$