Q. 145.0( 1 Vote )

# A field is in the

The figure is given below:

In the figure we have A = 90°

So we have in ΔDAB

By pythogares theorem

(BD)2 = (18)2 + (24)2

(BD)2 = 324 + 576 = 900

BD = = 30cm.

Area of Quad ABCD = Area ΔABD + Area ΔBCD

Area ΔABD =

Area ΔABD = 216cm2

Now for area of ΔBCD

We have BC = 40cm, CD = 50cm, BD = 30cm.

We can assume BC = b = 40cm and CD = c = 50cm and BD = d = 30cm.

By Heron’s formula we get area of triangle

=

(where s = s is called as the semi perimeter )…(1)

S =

Substituting the value of s, a, b and c in equation 1 we get

=

=

The area of ΔBCD is 600cm2

Now we have

Area of Quad ABCD = Area ΔABD + Area ΔBCD

Area of Quad ABCD = 216cm2 + 600cm2

Area of Quad ABCD = 816 cm2.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

A piece of land iRS Aggarwal & V Aggarwal - Mathematics

The question consRS Aggarwal & V Aggarwal - Mathematics

The dimensions ofNCERT Mathematics Exemplar

A hand fan is madRD Sharma - Mathematics

Find the area of NCERT Mathematics Exemplar