Question

The area of a below quadrilateral is cm? ​

Answer 1

Question:-

The area of a below Quadrilateral is -

(a) 150 cm²

(b) 180 cm²

(c) 100 cm²

(d) none of these

Solution :-

In ΔCDE,    

CE = Hypothenuse =  17 cm

DE = Base = 8 cm

CD = Height = ?

By Pythagoras Theorem,

(Hypothenuse)² = (Base)² + (Height)²

(17 cm)² = (8 cm)² + (CD)²

289 cm² = 64 cm² + (CD)²

(CD)² = 289 cm² - 64 cm²

(CD)² = 225 cm²

CD = $\sqrt{225\:cm^{2} }$

CD = 15 cm                  ....i.)

$Area \:\:of\:\: triangle\:\:DCE = \frac{1}{2} \times Base \times Height\\\\= \frac{1}{2} \times 8 cm \times 15 cm\\\\= 4 cm \times 15 cm\\\\= 60 cm^{2}$

From the eq. i.), it can be concluded that ABCD is a rectangle.

In rectangle ABCD,

BC = AD = Breadth = 8 cm

CD = BA = Length = 15 cm

Area of rectangle ABCD = Length x Breadth

                                        = 15 cm x 8 cm

                                        = 120 cm²

Total Area of the Quadrilateral

= Area of rectangle ABCD + Area of triangle CDE

= 120 cm² + 60 cm²

= 180 cm²

Answer:-

(b) 180 cm² is the correct answer.

Answer 2

(b) 180 cm² is the correct answer.

Solution :-

In ΔCDE,    

CE = Hypothenuse =  17 cm

DE = Base = 8 cm

CD = Height = ?

By Pythagoras Theorem,

(Hypothenuse)² = (Base)² + (Height)²

(17 cm)² = (8 cm)² + (CD)²

289 cm² = 64 cm² + (CD)²

(CD)² = 289 cm² - 64 cm²

(CD)² = 225 cm²

CD = $\sqrt{225\:cm^{2} }$

CD = 15 cm                  ....i.)

$Area \:\:of\:\: triangle\:\:DCE = \frac{1}{2} \times Base \times Height\\\\= \frac{1}{2} \times 8 cm \times 15 cm\\\\= 4 cm \times 15 cm\\\\= 60 cm^{2}$

From the eq. i.), it can be concluded that ABCD is a rectangle.

In rectangle ABCD,

BC = AD = Breadth = 8 cm

CD = BA = Length = 15 cm

Area of rectangle ABCD = Length x Breadth

                                        = 15 cm x 8 cm

                                        = 120 cm²

Total Area of the Quadrilateral

= Area of rectangle ABCD + Area of triangle CDE

= 120 cm² + 60 cm²

= 180 cm²