Find the area of an isosceles triangle whose one side is 10 cm greater than its equal side and its perimeter
is 100 cm.
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Answer:
Let both the equal sides be = x .
One side is greater by 10 cm .. so = x+10
Now the perimeter = 100 cm
So
The sum of all sides = 100
x + x + (x+10) = 100
x + x + x + 10 = 100
3x+10 = 100
3x = 100-10=90
x=90/3
x=30
So we found outh that the equal sides are 30 and 30 .
While the the third side is (30+10)= 40 .
To find the height divide the triangle into half by joining the top corner o the triangle to the mid point of the base , that is the side of 40 cm length .
Now two right triangles are created .By pythagoras theorem
height ^ 2 = 30^2-20^2 = 900-400=500
height = root of 500 = 10root5
So area = 1/2 base * height = 1/2 * 40 * 10root5 = 20 * 10root5
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