Question

what is the area of the isoceles triangle whose base is 8 cm and the other 2 sides are 6 cm each? ​

Answer 1

Step-by-step explanation:

area of triangle =1/2×base×height

base=8cm

height=6cm

so, area will be = 1/2×8×6

= 1×8×3

=24cm.

Answer 2

Answer:

$8\sqrt{5}cm^{2}$

Step-by-step explanation:

In an isosceles triangle, the altitude (height) is the perpendicular bisector of the base, forming a right angle with the base and dividing it into two equal parts. Therefore, half the base, the altitude, and the leg form a right triangle. (This is depicted in the attachment below.)

Since a²+b²=c² (let a=height of the triangle):

a²+4²=6²

a²=20

a=$\sqrt{20}$

h=$\sqrt{20}$

h=2$\sqrt{5}$

The area of a triangle=$\frac{1}{2}$bh

Area=$\frac{1}{2}$(8cm)($2\sqrt{5}$cm)

Area=8$\sqrt{5}$cm²