Find the area of an isosceles triangle where perimeter is 11cm and base 5cm.
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Two sides of a triangle are same, so by simple linear equation, the other two sides of the triangle would be 3 cm each with 5 cm base.
Now by Heron's formula, we can easily take out the area of the triangle.
Semiperimeter= 11/2 cm
S-a = 11/2-3= 5/2cm= s-b
S-c = 11/2-5= 1/2 cm
Now Heron's formula is........
√S(S-a)(S-b)(S-c)= √276/16
276 is not a perfect square but we will take out the root up to two decimal places, i.e 16.61/4= 4.15 cm^2
Now by Heron's formula, we can easily take out the area of the triangle.
Semiperimeter= 11/2 cm
S-a = 11/2-3= 5/2cm= s-b
S-c = 11/2-5= 1/2 cm
Now Heron's formula is........
√S(S-a)(S-b)(S-c)= √276/16
276 is not a perfect square but we will take out the root up to two decimal places, i.e 16.61/4= 4.15 cm^2
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