Question

Find the area of each parallelogram. What is the relationship between the areas?

Answer 1

Answer:

Area of parallelogram is given by:

$A = bh$

where b is the base and h is the height of parallelogram.

In parallelogram TQRS.

Coordinate of TQRS are;

T(8, 16), Q(4, 4), R(16, 4) and S(20, 16)

Coordinate of T'Q'R'S' are;

T'(2, 4), Q'(1, 1), R'(4, 1) and S'(5, 4)

Find the length of QR and PT:

Using distance(D) formula:

$D = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$QR = \sqrt{(4-16)^2+(4-4)^2} =\sqrt{(-12)^2+0}= \sqrt{144}= 12$ units

Similarly;

For PT:

From the graph:

P(8, 4) and T(8, 16), then

$PT = \sqrt{(8-8)^2+(4-16)^2} =\sqrt{(-0)^2+(-12)^2}= \sqrt{144}= 12$ units

In parallelogram TQRS

PT represents the height and QR represents the base of the parallelogram respectively.

then;

Area of parallelogram TQRS = $QR \cdot PT$

⇒Area of parallelogram TQRS = $12 \cdot 12 = 144$ unit square.

Now, in parallelogram T'Q'R'S'

Q'R' represents the base and P'T' represents the height of the parallelogram respectively.

here, P'(2, 1)

Find the length of Q'R' and P'T':

$Q'R' = \sqrt{(1-4)^2+(1-1)^2} =\sqrt{(-3)^2+0}= \sqrt{9}= 3$ units

$P'T' = \sqrt{(2-2)^2+(1-4)^2} =\sqrt{(-0)^2+(-3)^2}= \sqrt{9}= 3$ units

Then;

Area of parallelogram T'Q'R'S' = $Q'R' \cdot P'T'$

Area of parallelogram T'Q'R'S' = $9 \cdot 9= 81$ unit square.

Now, we have to find the relationship between the areas.

$\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}$

then;

the relationship between the areas of TQRS and T'Q'R'S' is:

$\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}$