Question

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


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Answer 1

Given: An isosceles triangle has Perimeter 30 cm. & each of the equal sides of triangle is 12 cm.

Need to find: The area of the triangle?

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So, we will use Heron's Formula to find the Area of Triangle (let's say ∆ABC).

✇ If the perimeter of given isosceles triangle is 30 cm then the semi perimeter of the isosceles triangle would be 15 cm. i.e ( s ) = 15 cm.

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( I ) Finding third Side :

» As We know that, Perimeter of triangle is equal to sum of all sides of triangle. & The perimeter is Given that is 30 cm. Therefore:

$:\implies\bf Perimeter_{\:(triangle)} = \Big\{a + b + c\Big\}\\\\\\:\implies\sf 30 = 12 + 12 + c\\\\\\:\implies\sf 30 = 24 + c\\\\\\:\implies\sf 30 - 24 = c\\\\\\:\implies\sf 6 = c$

∴ Hence, third side of the triangle is 6 cm.

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( II ) Area of Triangle :

$\star\:\underline{\boxed{\pmb{\sf{Area_{\;(triangle)} = \sqrt{s\Big(s - a\Big)\Big(s - b\Big)\Big(s - c\Big)}}}}}$

$\frak{Sides}\begin{cases}\sf{\quad a =\bf{12\;cm}}\\\sf{\quad b =\bf{12\;cm}}\\\sf{\quad c=\bf{6\;cm}}\\\sf{\quad s = \bf{15\:cm}}\end{cases}$

$:\implies\sf Area_{\;(triangle)} = \sqrt{15\Big(15 - 12\Big)\Big(15 - 12\Big)\Big(15 - 6\Big)}\\\\\\:\implies\sf Area_{\;(triangle)} = \sqrt{15 \times 3 \times 3 \times 9} \\\\\\:\implies\sf Area_{\:(triangle)} = \sqrt{15 \times 9 \times 9}\\\\\\:\implies{\pmb{\boxed{\frak{Area_{\;(triangle)} = 9\sqrt{15}\;cm^2}}}}\;\bigstar$

$\therefore{\underline{\sf{Hence,\;the\;area\;of\; isosceles\; triangle\;is\;{\pmb{\sf{9\sqrt{15}\;cm^2}}}.}}}$

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Answer 2

Question:

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

Answer:

• Area of the triangle is 9√15 cm².

Explanation:

Given that:

• Perimeter of isosceles ∆ = 30 cm
• Equal sides of ∆ = 12 cm

To Find:

• Area of ∆?

Solution:

• Firstly finding unknown side i.e, third side of ∆ ::

We know that,

✪ Perimeter of ∆ = Sum of all sides ✪

Putting values in formula we get,

➻ 12 + 12 + Third side = 30

➻ Third side + 24 = 30

➻ Third side = 30 - 24

➻ Third side = 6 cm

• Now finding semi perimeter (s) of ∆ ::

We know that,

✪ $\bf s = \dfrac{Perimeter}{2}$ ✪

Putting values in formula we get,

➻ $\sf s = {\cancel{\dfrac{30}{2}}}$

➻ s = 15 cm

• Now, we have all required values. So, let's find area of ∆ ::

According to heron's formula we know that,

✪ Area of ∆ = √[s(s - a)(s - b)(s - c)] ✪

• Where, s is semi perimeter and a, b, c are sides of ∆. We have, a = 12 cm, b = 12 cm and c = 6 cm.

Putting all values in formula we get,

➻ Area of ∆ = √[15(15 - 12)(15 - 12)(15 - 6)]

➻ Area of ∆ = √(15 × 3 × 3 × 9)

➻ Area of ∆ = √(15 × 9 × 9)

➻ Area of ∆ = 9√15 cm²

∴ Area of ∆ is 9√15 cm².

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Know more :-

• Area of rectangle = 2(L + B)
• Area of equilateral ∆ = √3/4 × (side)²
• Area of square = (side)²
• Area of circle = πr²
• Area of trapezium = ½ × (a + b) × h

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