Rectangle Problems
Rectangle problems with detailed solutions.
Perimeter of a Rectangle
Perimeter = 2 W + 2 L , w is the width and L is the length of the rectangle.
Area of a Rectangle
Area = L * W , w is the width and L is the length of the rectangle.
We now present some problems with detailed solutions.
Problem 1: A rectangle has a perimeter of 320 meters and its length L is 3 times its width W. Find the dimensions W and L, and the area of the rectangle.
Solution to Problem 1:

Use the formula of the perimeter to write.
2 L + 2 W = 320

We now rewrite the statement "its length L is 3 times its width W" into a mathematical equation as follows:
L = 3 W

We substitute L in the equation 2 L + 2 W = 320 by 3 W.
2(3 W) + 2 W = 320

Expand and group like terms.
8 W = 320

Solve for W.
W = 40 meters

Use the equation L = 3 W to find L.
L = 3 W = 120 meters

Use the formula of the area.
Area = L W = 120 * 40 = 4800 meters^{ 2}
Problem 2: The perimeter of a rectangle is 50 feet and its area is 150 feet^{ 2}. Find the length L and the width W of the rectangle, such that L > W.
Solution to Problem 2:

Use the formula of the perimeter to write
2 L + 2 W = 50

and the formula of the area to write
L W = 150

Divide all terms in the equation 2 L + 2 W = 50 by 2 to obtain
L + W = 25

Solve the above for W
W = 25  L

Substitute W by 25  L in the equation L W = 150
L(25  L) = 150

Expand the above equation and rewrite with right term equal to zero.
L^{ 2} + 25 L  150 = 0

The above is a quadratic equations with two solutions.
L = 10 and L = 15

Use W = 25  L to find the corresponding values of W.
W = 15 and W = 10

Since L > W, the rectangle has the dimensions
L = 15 feet and W = 10 feet.

As an exercise, check that the area and perimeter of the rectangle are 150 and 50 respectively.
Problem 3: The diagonal d of a rectangle has a length of 100 feet and its length y is twice its width x (see figure below). Find its area.
Solution to Problem 3:

We first use Pythagora's theorem.
100^{ 2} = x^{ 2} + y^{ 2}

We rewrite the statement "its length y is twice its width x" as a mathematical equation.
y = 2 x

Substitute y by 2 x in the equation 100^{ 2} = x^{ 2} + y^{ 2}.
100^{ 2} = x^{ 2} + (2 x)^{ 2}

Expand and group like terms.
100^{ 2} = 5 x^{ 2}

Solve for x, with x positive.
x = 20 sqrt(5) feet
and
y = 40 sqrt(5) feet.

Area is given by
Area = y x = 40 sqrt(5) * 20 sqrt(5) = 4000 feet^{ 2}.
Problem 4: Are the points A(1 , 0), B(5 , 2), C(4 , 5) and D(2 , 3) the vertices of a rectangle?
Solution to Problem 4:

We first calculate the slopes and see if the opposite sides are parallel.
m_{AB} = (2  0) / (5  (1)) = 1 / 3
m_{BC} = (5  2) / (4  5) = 3
m_{CD} = (3  5) / (2  4) = 1 / 3
m_{DA} = (0  3) / (1  (2)) = 3

Note that the slope of sides AB and CD are equal to 1 / 3 and therefore these two sides are parallel. Also the slopes of BC and DA are equal and therefore these two sides are parallel. ABCD is a parallelogram. Also the product of the slopes of AB and BC is equal to 1 and therefore AB is perpendicular to BC. Therefore the quadrilateral ABCD is a rectangle.
More references on geometry.
Geometry Tutorials, Problems and Interactive Applets.

