Answer :

**Given: AB = 3 cm, BC = 4 cm, CD = 4 cm, DA =5 cm and AC = 5 cm**

**To find: Area of quadrilateral ABCD.**

Now we can divide the quadrilateral into two separate triangles and then calculate their area.

Now, in Δ ABC,

Using Pythagoras theorem,

AC^{2} = BC^{2} + AB^{2}

5^{2} = 3^{2} + 4^{2}

25= 16 + 9

25 = 25

Thus,

Triangle is right angled triangle and since AB and BC are smaller sides so they work as legs.

And we know that,

^{}

Now,

Perimeter of Δ ACD = 5 + 5 + 4

Semi perimeter of Δ ACD =

=

= 7 cm

Using Heron’s formula,

Area of =

where, s = semi perimeter of the triangle and a, b, and c are the sides of the triangle=

=

= cm^{2}

**Take √21 = 4.58**

= 9.16 cm^{2} (approx.)

**Area of quadrilateral ABCD = Area of Δ ABC+ Area of Δ ACD**

= 6 + 9.16

**= 15.16 cm ^{2}By rounding off,= 15.2 cm^{2} ( approx)**

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