Find the area of a quadrilateral ABCD, ∠ C = 90o, AB = 9 cm, BC = 12

1.	Find the area of a quadrilateral ABCD, ∠ C = 90o, AB = 9 cm, BC = 12 cm, CD = 5 cm, AD = 8 cm. No useless answers, please. Also, explain everything in detail if u want branliest! The best answer indeed will be marked Brainliest. It's very urgent plz!AND plz don't copy from google
Answer» *CONCEPTS* *used:- Pythagoras* *theorem* : P↑2 + B↑2 = H↑2 where P,B and H are perpendicular,Base and Hypotenuse of a triangle respectively. *Heron's Formula* : $\sqrt{s(s - a)(s - b)(s - c)}$ where s is half of perimeter,and a,b,c are sides of a triangle. *Solution*:- Quadrilateral ABCD is divided into two triangles** by a diagonal BD. Now,in ∆ *BCD* By *Pythagoras* *theorem* *BD↑2 = BC↑2 + CD↑2 or, BD = 13 cm Area* of ∆BCD = (1/2)×BC×CD =(1/2)×12×5 = 30 *cm↑2 Now,in ∆ BAD* s = (9+8+13)/2 = 15 cm *s-a* = 15-9=6 *s-b* =15-8 = 7 *s-c* = 15-13= 2 *Area* of ∆ *BAD* = $\sqrt{15 \times 7 \times 6 \times 2}$ =6√35 *cm↑2 therefore*, Area of QUADRILATERAL ABCD is equal to the sum of areas of two TRIANGLES Area of ABCD = (30+6√35) cm↑**2

Find the area of a quadrilateral ABCD, ∠ C = 90o, AB = 9 cm, BC = 12 cm, CD = 5 cm, AD = 8 cm. No useless answers, please. Also, explain everything in detail if u want branliest! The best answer indeed will be marked Brainliest. It's very urgent plz!AND plz don't copy from google

Answer»

CONCEPTS used:-

Pythagoras theorem : P↑2 + B↑2 = H↑2

where P,B and H are perpendicular,Base and Hypotenuse of a triangle respectively.

Heron's Formula :

$\sqrt{s(s - a)(s - b)(s - c)}$

where s is half of perimeter,and a,b,c are sides of a triangle.

Solution:-

Quadrilateral ABCD is divided into two triangles by a diagonal BD.

Now,in ∆ BCD

By Pythagoras theorem

BD↑2 = BC↑2 + CD↑2

or, BD = 13 cm

Area of ∆BCD = (1/2)×BC×CD

=(1/2)×12×5 = 30 cm↑2

Now,in ∆ BAD

s = (9+8+13)/2 = 15 cm

s-a = 15-9=6

s-b =15-8 = 7

s-c = 15-13= 2

Area of ∆ BAD =

$\sqrt{15 \times 7 \times 6 \times 2}$

=6√35 cm↑2

therefore, Area of QUADRILATERAL ABCD is equal to the sum of areas of two TRIANGLES

Area of ABCD = (30+6√35) cm↑2