Question

find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4cm​

Answer 1

Answer:

answer is $\sqrt{15\\}$cm^2

Step-by-step explanation:

base=b=2cm

equal sides=a=4cm

area=b/4*$\sqrt{4a^2-b^2\\}$=2/4*64-4=1/2*2$\sqrt{15}$=$\sqrt{15}$

Answer 2

Hi there!

Your Answer :-

Using Herons Formula.

Semi Perimeter = 4+4+2/2 = 5.

Heron's Formula :-

$\sf\sqrt{s(s - a)(s - b)(s - c)}$

Put the values.

>$\sf \sqrt{5(5 - 4)(5 - 4)(5 - 2)}$

>$\sf\sqrt{5 \times 1 \times 1 \times 3}$

>$\sf\sqrt{15} c {m}^{2}$

So, your answer is root(15)cm².