Question

fint the area of an isosceles triangle having base 8cm and length of one of the equal sides is 6cm

Answer 1

Step-by-step explanation:

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Answer 2

Concept:

The region that an object's form defines as its area. The area of a figure or any other two-dimensional roundness in a plane is how much interior it occupies. A polygon having three edges and three vertices is called a triangle. It is one of the fundamental geometric forms. Triangle $ABC$ is the designation for a triangle with vertices $A, B,$ and $C$. In Euclidean geometry, any three points that are not collinear produce a distinct triangle and a distinct plane.

Given:

Here it is given that the triangle having base $8cm$ and length of one of the equal sides is $6cm$.

Find:

We have to find the area of an isosceles triangle having base $8cm$ and length of one of the equal sides is $6cm$.

Solution:

According to the question,

$6^2+8^2=(AD)^2$

$= > (AD)=2\sqrt{5}$

Now,

$1/2*b*h=1/2*2\sqrt{5} *6\\=6\sqrt{5}$

Now, $s=(a+b+c)/2$

         $S=(6+6+8)/2\\=10$

So, the area of the triangle

$=\sqrt{10*(10-6)(10-6)(10-8)} \\=8\sqrt{5} cm^2$

Hence, the area of the triangle $=8\sqrt{5} cm^2$.

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