detailed solution

Correct option is A

Given that, A triangle and a parallelogram have the same base and the same area

We know that, Heron's formula

Area of triangle = $\sqrt{s (s - a) (s - b) (s -c)}$

where, s is the semi-perimeter = half of the perimeter

and  a, b and c are the sides of the triangle 

Let ABCD is a parallelogram and ABE is a triangle having a common base with parallelogram ABCD.

For ∆ABE, a = 30 cm, b = 26 cm, c = 28 cm

Semi Perimeter = s  = Perimeter/2

s = (a $+$ b $+$ c)/2

= (30 $+$ 26 $+$ 28)/2

= 84/2

= 42 cm

Thus, we get

Area of a ΔABE = $\sqrt{s (s - a) (s - b) (s -c)}$

= $\sqrt{42(42 - 30)(42 - 28)(42 - 26)}$

= $\sqrt{42 \times 12 \times 14 \times 16}$

= 336 cm²

Hence we get,

Area of parallelogram ABCD = Area of ΔABE 

Base × Height = 336 cm²

28  × Height = 336 cm²

On rearranging, we get

Height = 336/28 cm = 12 cm

Thus, height of the parallelogram is 12 cm.

detailed solution

Correct answer is 1

Given that, A triangle and a parallelogram have the same base and the same area

We know that, Heron's formula

Area of triangle = $\sqrt{s (s - a) (s - b) (s -c)}$

where, s is the semi-perimeter = half of the perimeter

and  a, b and c are the sides of the triangle 

Let ABCD is a parallelogram and ABE is a triangle having a common base with parallelogram ABCD.

For ∆ABE, a = 30 cm, b = 26 cm, c = 28 cm

Semi Perimeter = s  = Perimeter/2

s = (a $+$ b $+$ c)/2

= (30 $+$ 26 $+$ 28)/2

= 84/2

= 42 cm

Thus, we get

Area of a ΔABE = $\sqrt{s (s - a) (s - b) (s -c)}$

= $\sqrt{42(42 - 30)(42 - 28)(42 - 26)}$

= $\sqrt{42 \times 12 \times 14 \times 16}$

= 336 cm²

Hence we get,

Area of parallelogram ABCD = Area of ΔABE 

Base × Height = 336 cm²

28  × Height = 336 cm²

On rearranging, we get

Height = 336/28 cm = 12 cm

Thus, height of the parallelogram is 12 cm.