Question

find the area of the given quadrilateral please it is so urgent​

Answer 1

AB=√(AD^2-BD^2)

=√(289-225)

=8

area of triangle ABD+area of triangle BCD

1/2*8*15+1/2*6*15

60+45

105 cm^2

Answer 2

Answer:

Area of the given figure is 105 cm^2.

Step-by-step explanation:

Pythagoras theorem : If a right-angled triangle has sides a , b and c, then a relation between the sides : c^2 = a^2 + b^2 , where c is the hypotenuse { side opposite to right angle }.

Here,

= > AB^2 + ( 15 cm )^2 = ( 17 cm )^2

= > AB^2 + 225 cm^2 = 289 cm^2

= > AB^2 = 289 cm^2 - 225 cm^2

= > AB^2 = 64 cm^2

= > AB = √64 cm^2 = 8 cm

= > Area of the given figure = Area of ∆ having AB + area of ∆ containing pt C.

• Area of ∆ = 1 / 2 x height x base

= > Area of the given figure = [ 1 / 2 x 8 cm x 15 cm ] + [ 1 / 2 x 15 cm x 6 cm ]

= > Area of the given figure = 60 cm^2 + 45 cm^2

= > Area of the given figure = 105 cm^2

Hence, area of the given figure is 105 cm^2.