Rectangle problems with detailed solutions.

## Perimeter of a Rectangle
## Area of a Rectangle
## Problems with Solutions## Problem 1A 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 2The 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 3The 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 4Are 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.
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