Question

A (-5, 5), B (-5, -3), C (4, -3), D (0, 5).
Find the area of ABCD.

Answer 1

$\large\underline{\sf{Solution-}}$

Given coordinates are

Coordinates of A be (-5, 5)

Coordinates of B (- 5, - 3)

Coordinates of C (4, - 3)

Coordinates of D (0, 5).

[ Please see the attachment ].

From graph we concluded that

AB = 8 units

BC = 9 units

DA = 5 units

Also,

AB is Perpendicular to BC.

and

AD is parallel to BC.

As we know that,

In a quadrilateral, if one pair of opposite sides are parallel and other pair non - parallel, the quadrilateral is a trapezium.

So, this figure, ABCD is a trapezium.

Now,

Area of trapezium = 1/2 × ( sum of || sides ) × distance

= 1/2 × ( AD + BC ) × AB

= 1/2 × ( 5 + 9 ) × 8

= 14 × 4

= 56 square units.

Additional Information :-

Let's solve one more problem of same type!!!

Question :- Find the Perimeter of the figure having vertices (2, 3), (2, 1), (0 1) and (0, 3)

Answer :-

Given coordinates are

Coordinates of A be (2, 3)

Coordinates of B (2, 1)

Coordinates of C (0, 1)

Coordinates of D (0, 3).

[ Please see the attachment ].

From graph we concluded that

AB = 2 units

BC = 2 units

CD = 2 units

DA = 2 units

Also,

AB is Perpendicular to BC.

BC is perpendicular to CD

CD is perpendicular to DA

DA is perpendicular to AB.

It implies, AB = BC = CD = DA = 2 units.

As we know that,

In a quadrilateral, if all the 4 sides are equal and angle between the sides is 90°, then quadrilateral is a square.

So, this figure, ABCD is a square.

Now, Perimeter of a square = 4 × side = 4 × 2 = 8 units.