Q. 14.1( 468 Votes )

A park, in the sh

Answer :

Given: The figure is given below:

 

The dimensions of the park as follows:

Angle C = 90°,

AB = 9m,

BC = 12m,

CD = 5m

And, AD = 8m

Now, BD is joined.
Thus quadrilateral ABCD can now be divided into two triangles ABD and BCD
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

Pythagoras Theorem: It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In
ΔBCD by Pythagoras theorem,

BD2 = BC2 + CD2

BD2 = 122 + 52

BD2 = 169

BD = 13m


As triangle BCD is right angled triangle,

Now, Semi perimeter of triangle (ABD) = (8 + 9 + 13)/2

                                                         = 30/2  = 15m

Using Heron’s formula,
ar(ABD) =
where, s = semiperimeter of the triangle and a, b, and c are the sides of the triangle.


=

=

=

= 6√35 m2

= 6 × 5.91
= 35.46 m2

Area of quadrilateral ABCD = 35.46 + 30

Area of quadrilateral ABCD = 65.46 m2


Thus, the park acquires an area of  65.5 m2(approx).

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Heron's Formula- Full GyanHeron's Formula- Full GyanHeron's Formula- Full Gyan46 mins
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
caricature
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