Question

25. Find the area of the quadrilateral ABCD in which AD = 24 cm, angle BAD = 90° and triangle BCD is an equilateral triangle having each side equal to 26 cm. Also, find the perimeter of the quadrilateral. [Given, v3 = 1.73.] ​

Answer 1

answer

In Δ ABD

Using the Pythagoras theorem

BD

2

=AB

2

+AD

2

By substituting the values

26

2

=AB

2

+24

2

On further calculation

AB

2

=676−576

By subtraction

AB

2

=100

By taking out the square root

AB=

100

So we get

Base = AB = 10cm

We know that area of ΔABD=

2

1

×b×h

By substituting the values

Area of ΔABD=

2

1

×10×24

On further calculation

Area of ΔABD=120cm

2

We know that the area of ΔBCD=

3

/4a

2

By substituting the values

Area of ΔBCD=(1.73/4)(26)

2

So we get Area of ΔBCD=292.37cm

2

So we get area of quadrilateral ABCD = Area of Δ ABD + Area of Δ BCD

By substituting the values

Area of quadrilateral ABCD = 120 + 29237

By addition

Area of quadrilateral ABCD = 412.37 cm

2

The perimeter of quadrilateral ABCD = AB + BC + CD + DA

By substituting the values

Perimeter = 10 + 26 + 26 + 24

So we get

Perimeter = 86cm

Therefore, the area is 412.37 cm

2

and perimeter is