# 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. 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 on geometry. Geometry Tutorials, Problems and Interactive Applets.