Question

The area of a parallelogram ABCD, with AB = 40 cm
and corresponding altitude DE = 20 cm, equals the
area of a triangle PQR whose base is 50 cm. Find the
altitude of the triangle.​

Answer 1

Answer:

altitude of ∆ le = 32 cm

Step-by-step explanation:

area of //gram = bh = 40×20 =800

area of ∆le = 800

1/2×bh =800

1/2×50×h= 800

25×h =800

h= 800 / 25

h = 32 cm

Answer 2

The altitude of the triangle is 32 cm

Step-by-step explanation:

Given Data

In a Parallelogram ABCD, AB = 40cm (Base)

Corresponding altitude DE = 20 cm

In a Triangle PQR , Base = 50 cm

FInd the altitude of the Triangle, if the area of the Parallelogram is equals to the area of the triangle.

Area of the parallelogram = Base × Height ( Square units)

Area of the parallelogram = 40 × 20 cm²

Area of the parallelogram = 800 cm²

$\text{Area of the triangle} = \frac{1}{2} \times Base \times Height$

$\text{Area of the triangle} = \frac{1}{2} \times 50 \times Height$

$\text{Area of the triangle} = \frac{1}{2} \times 50 \times H$

Area of the parallelogram ABCD = Area of the triangle PQR

$800 = \frac{1}{2} \times 50 cm \times H$

$H = \frac{800 \times 2}{50}$

$H = \frac{1600}{50}$

H = 32 cm

Therefore the altitude of the triangle is 32 cm when the area of the parallelogram (800 cm²) is equal to the area of the triangle with base 50 cm.

To Learn More ...

1)In parallelogram ABCD, the length of the altitude corresponding to AB is 8 cm. What is the length of the altitude corresponding to BC?

https://brainly.in/question/1256103

2) If area of parallelogram ABCD is 80 CM then what is an area of triangle ADP

https://brainly.in/question/2800137