# 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.

## Problems with 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.
mAB = (2 - 0) / (5 - (-1)) = 1 / 3
mBC = (5 - 2) / (4 - 5) = -3
mCD = (3 - 5) / (-2 - 4) = 1 / 3
mDA = (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 and Links to Geometry

Geometry Tutorials, Problems and Interactive Applets.