Math, asked by molly2001, 6 months ago

Find the area of an isosceles triangle whose one side is 10 cm greater than its equal side and its perimeter

is 100 cm.​

Answers

Answered by Anonymous
1

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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Step-by-step explanation:

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