Question

7. The area of an isosceles triangle having base 2 cm and length of one of equal sides is 4 cm is
15 cm?
15
cm2
V2.
02/15 cm?
0 4V15 cm​

Answer 1

Question :-

The area of an isosceles triangle having base 2 cm and length of equal side is 4 cm ?

To Find :-

The Area of the Isosceles Triangle.

Given :-

• Base of the Isosceles Triangle = 2 cm.

• Length of equal side = 4 cm.

We Know :-

Area of an Isosceles TrianGle :

$\implies \bf{\underline{A = \dfrac{1}{4}b\sqrt{4a^{2} - b^{2}}}}$

Where :-

• a = Equal side of the Isosceles Triangle.

• b = Base of the Isosceles Triangle .

• A = Area of the Isosceles Triangle

Solution :-

By using the formula for Area of an Isosceles Triangle and Substituting the values in it , we get :

$\implies \bf{A = \dfrac{1}{4}b\sqrt{4a^{2} - b^{2}}} \\ \\ \\ \implies \bf{A = \dfrac{1}{4} \times 2 \times \sqrt{4 \times 4^{2} - 2^{2}}} \\ \\ \\ \implies \bf{A = \dfrac{1}{\not{4}} \times \not{2} \times \sqrt{4 \times 4^{2} - 2^{2}}} \\ \\ \\ \implies \bf{A = \dfrac{1}{2} \times \sqrt{4 \times 16 - 4}} \\ \\ \\ \implies \bf{A = \dfrac{1}{2} \times \sqrt{64 - 4}} \\ \\ \\ \implies \bf{A = \dfrac{1}{2} \times \sqrt{60}} \\ \\ \\ \implies \bf{A = \dfrac{1}{2} \times 2\sqrt{15}} \\ \\ \\ \implies \bf{A = \dfrac{1}{\not{2}} \times \not{2}\sqrt{15}} \\ \\ \\ \implies \bf{A = \sqrt{15}} \\ \\ \\ \therefore \purple{\bf{A = \sqrt{15}}}$

Hence, the area of the Isosceles Triangle is √15 cm.

Answer 2

Answer:

Question :-

The area of an isosceles triangle having base 2 cm and length of equal side is 4 cm ?

To Find :-

The Area of the Isosceles Triangle.

Given :-

Base of the Isosceles Triangle = 2 cm.

Length of equal side = 4 cm.

We Know :-

Area of an Isosceles TrianGle :

$</p><p>\implies \bf \green{\underline{A = \dfrac{1}{4}b\sqrt{4a^{2} - b^{2}}}}</p><p>$

Where :-

A = Equal side of the Isosceles Triangle.

B = Base of the Isosceles Triangle .

A = Area of the Isosceles Triangle

Solution :-

By using the formula for Area of an Isosceles Triangle and Substituting the values in it , we get :

$\begin{gathered}\implies \bf \pink{A = \dfrac{1}{4}b\sqrt{4a^{2} - b^{2}}} \\ \\ \\ \implies \bf{A = \dfrac{1}{4} \times 2 \times \sqrt{4 \times 4^{2} - 2^{2}}} \\ \\ \\ \implies \bf{A = \dfrac{1}{\not{4}} \times \not{2} \times \sqrt{4 \times 4^{2} - 2^{2}}} \\ \\ \\ \implies \bf{A = \dfrac{1}{2} \times \sqrt{4 \times 16 - 4}} \\ \\ \\ \implies \bf{A = \dfrac{1}{2} \times \sqrt{64 - 4}} \\ \\ \\ \implies \bf{A = \dfrac{1}{2} \times \sqrt{60}} \\ \\ \\ \implies \bf{A = \dfrac{1}{2} \times 2\sqrt{15}} \\ \\ \\ \implies \bf{A = \dfrac{1}{\not{2}} \times \not{2}\sqrt{15}} \\ \\ \\ \implies \bf{A = \sqrt{15}} \\ \\ \\ \therefore \red{\bf{A = \sqrt{15}}}\end{gathered}$

Hence, the area of the Isosceles Triangle is √15 cm.