#### Question

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.

#### Solution

The sides of a triangle DCE are

DC = 15 cm, CE = 13 cm, ED = 14 cm

Let h be the height of parallelogram ABCD

Given,

Perimeter of ΔDCE

2s = DC + CE + ED

`⇒S=1/2(15+13+4)`

`⇒s=1/2(42)`

`⇒s=21cm`

π΄πππ ππ Δπ·πΆπΈ = `sqrt(s(s-dc)(s-ce)(s-ed))`

`=sqrt(21(21-15)(21-13)(21-14))`

`=sqrt(21xx7xx8xx6)`

`=sqrt(84xx84)`

`84 cm^2`

Given that

Area of Δππ π·πΆπΈ= ππππ ππ π΄π΅πΆπ·

= Area of parallelogram ABCD = =`84cm^2`

⇒ 24×β=84 [∴ Area of parallelogram = base × height]

⇒ h = 6 cm

Is there an error in this question or solution?

Solution A Triangle and a Parallelogram Have the Same Base and the Same Area. If the Sides of the Triangle Are 13 Cm, Concept: Area of a Triangle by HeronβS Formula.