Question

each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm find the area of the triangle​

Answer 1

Answer:

Base- 12cm

Sides-2 more

Answer- Don't know...

Answer 2

In this question it's given that,

• Each equal sides of isosceles triangle is 2 cm more than the height.
• Base of the isosceles triangle is 12 cm.

We need to calculate the area of triangle from given values.

Let us assume the height of triangle as h cm.

• Now, according to the question, each equal sides of the isosceles triangle is 2 cm more than it's height. So,

$\longrightarrow\sf{Equal\: sides=(h+2)cm}$

We know, that Area of an isoceles triangle is,

$\twoheadrightarrow\sf{{Area}_{(Isoceles\: triangle)}=\dfrac{1}{4}b\sqrt{4{a}^{2}-{b}^{2}}}$

Also, Area of triangle is half of the base times height. That is,

$\twoheadrightarrow\sf{{Area}_{(Triangle)}=\dfrac{1}{2}(b\times h)}$

Now,

$\longrightarrow\sf{{Area}_{(Triangle)}={Area}_{(Isosceles\: triangle)}}$

Using Formula,

$\longrightarrow\sf{\dfrac{1}{2}(b\times h)=\dfrac{1}{4}b\sqrt{4{a}^{2}-{b}^{2}}}$

Substituting all the values, we get:

$\longrightarrow\sf{\dfrac{1}{2}(12\times h)=\dfrac{1}{4}12\sqrt{4{(h+2)}^{2}-{12}^{2}}}$

Simplifying, we get:

$\longrightarrow\sf{6h=3\sqrt{4{(h+2)}^{2}-{12}^{2}}}$

Using Identity:

ㅤㅤㅤㅤㅤㅤ★ (a+b)² = a² + b² + 2ab

$\longrightarrow\sf{6h=3\sqrt{4({h}^{2}+{2}^{2}+4h)-144}}$

Transposing 3 from RHS to LHS,

$\longrightarrow\sf{\dfrac{6h}{3}=\sqrt{{4h}^{2}+16+16h-144}}$

Performing division on LHS side.

$\longrightarrow\sf{2h=\sqrt{{4h}^{2}+16+16h-144}}$

Subtracting, 144 by 16, we get:

$\longrightarrow\sf{2h=\sqrt{4h}^{2}+16h-128}$

Taking square on both the sides.

$\longrightarrow\sf{{(2h)}^{2}={\left(\sqrt{{4h}^{2}+16h-128}\right)}^{2}}$

Simplifying,

$\longrightarrow\sf{{4h}^{2}={4h}^{2}+16h-128}$

Transposing 4h² and 16h from RHS to LHS,

$\longrightarrow\sf{{4h}^{2}-{4h}^{2}-16h=-128}$

Cancelling +4h² and -4h² on LHS,

$\longrightarrow\sf{-16h=-128}$

Transposing, -16 from LHS to RHS and performing division.

$\longrightarrow\sf{h=\dfrac{-16}{-128}}$

$\longrightarrow\bold{h=8}$

We've calculated the height of the triangle that is 8 cm. Now, we are given that base of the triangle is 12 cm. Substituting this values in Area of triangle Formula.

$\longrightarrow\sf{{Area}_{(Triangle)}=\dfrac{1}{2}(b\times h)}$

$\longrightarrow\sf{{Area}_{(Triangle)}= \dfrac{1}{2}(12\times 8)}$

Multiplying, 12 and 8.

$\longrightarrow\sf{{Area}_{(Triangle)}=\dfrac{1}{2}(96)}$

Dividing, 96 by 2.

$\longrightarrow\bold{{Area}_ {(Triangle)}=48\: {cm}^{2}}$

❝ Therefore, Area of the triangle is 48 cm². ❞