Question

The quadrilateral whose virtices are A(1,0),B(7,0),C(6,3),D(2,3).find it's area

Answer 1

Step-by-step explanation:

A = ( 1 , 0 )

B = ( 7 , 0 )

C = ( 6 , 3 )

D = ( 2 , 3 )

By Pythagoras theorem :-

Hyp² = opp² + adj²

Hyp² = ( 2 - 1 )² + 3²

Hyp² = 1 + 9

Hyp² = 10

Hyp = √10 units

[ If opposite sides are equal then it is a parallelogram ]

We know that

" Diagonals of parallelogram divides it into two congruent triangles "

Area of ∆ ABC = 1/2 l 1 ( 0 - 3 ) + 7 ( 3 - 0 ) + 6 ( 0 - 0 ) l

=> 1/2 l -3 + 21 + 0 l

=> 1/2 l 81 l

=> 9 units square

Area of parallelogram = Area of ∆ABC = Area of ∆ADC

Area of parallelogram = 2 × Area of ∆ABC

=> 2 × 9

=> 18 units square

:. Area of parallelogram is 18 square units.