Question

11. A field in the form of a parallelogram has one
diagonal of 42 m long and the perpendicular
distance of this diagonal from either of the
outlying vertices is 10 m as shown in the figure
Find the area of the field.​

Answer 1

Refer to the attachment attached to this answer.

Given :-

• A field in the shape of a parallelogram ABCD, whose one diagonal (say AC) is 42 m long. Also, The perpendicular distance from either of the outlying vertices (Vertices that are not on the diagonal. Here, the outlying vertices are B and D) is 10 m.

To Find :-

• Area of the field

Solution :-

Diagonal AC divides the parallelogram into two triangles namely ∆ABC & ∆ACD. But, we know, The diagonal of a parallelogram divides the parallelogram into two equal triangles. Which means, Area(∆ABC) = Area(∆ACD)

Now,

⇒ Area of field = Area(∆ABC) + Area(∆ACD)

⇒ Area of field = 2 Area(∆ABC)

In triangle ABC, we have

• Base = AC = 42 m
• Height = BF = 10 m

We know, Area of triangle is given by,

• 1/2 × Base × Height

⇒ Area of field = 2 × 1/2 × AC × BF

⇒ Area of field = AC × BF

⇒ Area of field = 42 × 10

⇒ Area of field = 420 m²

Hence, The area of the field is 420 m².

Answer 2

Answer:

Given :-

A field in the form of a parallelogram has one

diagonal of 42 m long and the perpendicular

distance of this diagonal from either of the

outlying vertices is 10 m

To Find :-

Area

Solution :-

At first

Let the field be ABCD

And

Diagonal be DE and BF

Now,

Area of field = 2 × ∆ABC

$\bf \: Area = \dfrac{1}{2} \times b \times h$

$\sf \: Area = \dfrac{1}{2} \times 10 \times 42$

$\sf \: Area = 5 \times 42$

$\sf \: Area = 210 \: \: {m}^{2}$

For the field

Area = 2 × 210

Area = 420 m²