find the area of an isosceles triangle whose perimeter is 32cm and non-equal side is 12 cm respectively.
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find the area of an isosceles triangle whose perimeter is 32cm and non-equal side is 12 cm respectively.
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Given:-
In an isosceles triangle,
The sum of the remaining sides is cm.
Each equal side is cm.
The isosceles triangle has three sides 12 cm, 10 cm, 10 cm.
If we draw a leg from a vertex angle, the triangle is divided into two congruent right triangles.
By Pythagorean theorem,
Hence, base and height are 12 cm and 8 cm.
The area of the triangle is,
½ × base × height
= ½ × 12 × 8
= 48 cm².
The area of the isosceles triangle is 48cm^2
It is given that,
Therefore,
The sum of the equal sides of the triangle
=(32-12)cm
= 20 Cm
Therefore,
Now, imagine a perpendicular through the mid point of the triangle such that it bisects the base.
Therefore,
we get 2 right angled triangles.
For each equal triangle:
therefore,
[Pythagoras theorem]
Therefore,
For the isosceles triangle we get,
Therefore,
Therefore,