Question

What is the area of parallelogram ABCD?

11 square units
13 square units
15 square units
16 square units

Answer 1

13 square units

Explanation:

First of all, you need to identify that ABCD is a rectangle (AB=CD and AD=BC).

The area of a rectangle is calculated by multiplying the length and the width.

Secondly, we use the Pythagoras’s theorem to calculate side CD and AD (the length and width). I’ve added some labels to your original diagram (see picture attached) so that it’s easier to understand.

The Pythagoras’s theorem is a^2 + b^2 = c^2 (c is the hypotenuse).

So, for side CD:

3^2 + 1^2 = (CD)^2

9 + 1 = (CD)^2

CD = √ 10

and for side AD:

4^2 + 1^2 = (AD)^2

16 + 1 = (AD)^2

AD = √17

Lastly, to calculate the area:

√10 x √17 = 13.04

Your answer is 13 square units.

Hope this helped :)

Answer 2

Option B. 13 square units

Explanation:

Area of a parallelogram is defined by the expression

A =
`(1)/(2)(\text{Sum of two parallel sides)}` × (Disatance between them)

Vertices of A, B, C and D are (3, 6), (6, 5), (5, 1) and (2, 2) respectively.

Length of AB =
`\sqrt{(x-x')^(2)+(y-y')^(2)}`

=
`\sqrt{(5-6)^(2)+(6-3)^(2)}`

=
`√(10)`

Since length of opposite sides of a parallelogram are equal therefore, length of CD will be same as
`√(10)`

Now we have to find the length of perpendicular drawn on side AB from point D or distance between parallel sides AB and CD.

Expression for the length of the perpendicular will be =
`\fracAx_(1)+By_(1)+C{\sqrt{A^(2)+B^(2)}}`

Slope of line AB (m) =
`(y-y')/(x-x')`

=
`(6-5)/(3-6)=-((1)/(3) )`

Now equation of AB will be,

y - y' = m(x - x')

y - 6 =
`-(1)/(3)(x-3)`

3y - 18 = -(x - 3)

3y + x - 18 - 3 = 0

x + 3y - 21 = 0

Length of a perpendicular from D to side AB will be

=
`\frac{\sqrt{1^(2)+3^(2)}}`

=
`(13)/(√(10))`

Area of parallelogram ABCD =
`(1)/(2)(AB+CD)* (\text{Distance between AB and CD})`

=
`(1)/(2)(√(10)+√(10))* ((13)/(√(10) ) )`

=
`√(10)* (13)/(√(10) )`

= 13 square units

Option B. 13 units will be the answer.