Answer

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Hint: Determine the sides of the triangle first. And then use Heron’s formula for the area of a triangle.

According to the question, the perimeter of the triangle is $30cm$. Then the semi-perimeter will be:

$ \Rightarrow s = \dfrac{{30}}{2} = 15cm$.

Length of two equal sides of the triangle is $12cm$ (given in the question).

Let the length of the third side is $x$. Then we can use perimeter to find the value of $x$. We’ll get:

$

\Rightarrow 12 + 12 + x = 30, \\

\Rightarrow x = 30 - 24, \\

\Rightarrow x = 6 \\

$

So, we have $12cm, 12cm$ and $6cm$ as the length of three sides of a triangle whose semi-perimeter is $s = 15cm$.

Now, we can use Heron’s Formula to determine its area. We have:

$ \Rightarrow A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $, where $a,b,c$ are length of sides of triangle in any order.

So, putting values from above, we’ll get:

$

\Rightarrow A = \sqrt {15\left( {15 - 12} \right)\left( {15 - 12} \right)\left( {15 - 6} \right)} , \\

\Rightarrow A = \sqrt {15 \times 3 \times 3 \times 9} , \\

\Rightarrow A = 3 \times 3 \times \sqrt {15} , \\

\Rightarrow A = 9\sqrt {15} \\

$

Thus the area of the triangle is $9\sqrt {15} c{m^2}$.

Note:

We can also use $Area = \dfrac{1}{2} \times b \times h$ to find out its area.

For an isosceles triangle, base is always the unequal side. So, in this case $b = 6cm$.

And we can easily find out the height of an isosceles using Pythagoras Theorem.

$

\Rightarrow h = \sqrt {{{12}^2} - {{\left( {\dfrac{b}{2}} \right)}^2}} , \\

\Rightarrow h = \sqrt {144 - 9} , \\

\Rightarrow h = \sqrt {135} , \\

\Rightarrow h = 3\sqrt {15} \\

$

Now, we can use $\dfrac{1}{2} \times b \times h$. We’ll get the same result.

According to the question, the perimeter of the triangle is $30cm$. Then the semi-perimeter will be:

$ \Rightarrow s = \dfrac{{30}}{2} = 15cm$.

Length of two equal sides of the triangle is $12cm$ (given in the question).

Let the length of the third side is $x$. Then we can use perimeter to find the value of $x$. We’ll get:

$

\Rightarrow 12 + 12 + x = 30, \\

\Rightarrow x = 30 - 24, \\

\Rightarrow x = 6 \\

$

So, we have $12cm, 12cm$ and $6cm$ as the length of three sides of a triangle whose semi-perimeter is $s = 15cm$.

Now, we can use Heron’s Formula to determine its area. We have:

$ \Rightarrow A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $, where $a,b,c$ are length of sides of triangle in any order.

So, putting values from above, we’ll get:

$

\Rightarrow A = \sqrt {15\left( {15 - 12} \right)\left( {15 - 12} \right)\left( {15 - 6} \right)} , \\

\Rightarrow A = \sqrt {15 \times 3 \times 3 \times 9} , \\

\Rightarrow A = 3 \times 3 \times \sqrt {15} , \\

\Rightarrow A = 9\sqrt {15} \\

$

Thus the area of the triangle is $9\sqrt {15} c{m^2}$.

Note:

We can also use $Area = \dfrac{1}{2} \times b \times h$ to find out its area.

For an isosceles triangle, base is always the unequal side. So, in this case $b = 6cm$.

And we can easily find out the height of an isosceles using Pythagoras Theorem.

$

\Rightarrow h = \sqrt {{{12}^2} - {{\left( {\dfrac{b}{2}} \right)}^2}} , \\

\Rightarrow h = \sqrt {144 - 9} , \\

\Rightarrow h = \sqrt {135} , \\

\Rightarrow h = 3\sqrt {15} \\

$

Now, we can use $\dfrac{1}{2} \times b \times h$. We’ll get the same result.

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