Question

Q1. A triangle and a parallelogram have the same base and same area. If the

sides of the triangle are 9 cm, 12 cm and 15 cm and the parallelogram

stands on the base 9 cm, find the height of the parallelogram.

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Answer 1

Step-by-step explanation:

ATQ,

The given statements are as under :-

The triangle and the parallelogram have same base, i.e., 9 cm.

Three sides of the Δ are given as under:-

Base = 9 cm

Others sides = 12 cm and 15 cm respectively.

Also,

Area of Δ = Area of llgm

which means,

Area of Δ = Base x  Height

Area of Δ = 9 x h

∴ h = 1/9 x ( Area of Δ ) ................ (eq.1)

Now,

Area of Δ = √s(s-a)(s-b)(s-c)

where a,b,c are the sides of the triangle and s is the semi-perimeter of the triangle.

s = (a+b+c)/2

  = (9+12+15)/2

  = 36/2

  = 18 cm

Substitute the value of s,a,b,c in the formula above.

Area of Δ = √18(18-9)(18-12)(18-15)

                = √18(9)(6)(3)

                = √18 x 18 x 9

                = √18² x 3²

                = 18 x 3

                = 54 cm²

Now substitute the value of Area of Δ in eq.1

h = 1/9 x ( Area of Δ )

h = 1/9 x 54

h = 6 cm

∴ Height of the parallelogram is 6 cm.

Answer 2

Answer:

Step-by-step explanation: 6 cm

Sides of the triangle = 9 cm , 12 cm , 15 cm .

Semiperimeter(Half-Perimeter) = $\frac{(9 + 12 + 15)}{2}$ = 18 cm

By Heron's Formula , we have area of a triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

Here s = 18 , a = 9 , b = 12 , c = 15

∴ $\sqrt{18(18 - 9)(18 - 12)(18 - 15)}$

=> $\sqrt{18*9*6*3}$

=> $\sqrt{2916}$

=> 54 cm²

Since the area of the parallelogram is same with the triangle , the area of the parallelogram is also 54 cm² .

Also area of the parallelogram = base x height .

Base = 9cm .

Let the height be x cm . Then :-

∴ 54 = 9x

=> x = 6

So the height of the parallelogram is 6 cm .

Hope this helps you .

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