Solutions Section J: Perimeter and Area 1. The 6 by 10 rectangle below has semi-circles attached on each end. 6 10 a) Find the perimeter of (the distance around) the figure above. b) Find the area enclosed by the figure above. Solution part a): Let P be the perimeter of the figure. Then, P = 10 + 10 + (1/2) 6 π + (1/2) 6 π = 20 + 6 π The perimeter is 20 + 6 π units of length Solution part b): The area A of the figure is the sum of the area of the rectangle and the areas of the two semi-circles. A = 6 10 + π 2 = 60 + 9π The area of the figure is 60 + 9π square units 2. Find the area of the shaded region R. There is more than one way to solve this problem. The strategy employed here is to calculate the sum of the areas of the two squares (4-by-4 and 6-by-6) and subtract from that area the sum of the areas of the two (white) right triangles lying within the entire figure, but outside of the shaded region R.
The result is the area A of R given in square meters: A = (4 4) + (6 6) [( 1 4 10) + ( 1 6 6)] 2 2 A = 16 + 6 [20 + 18] = 14 The area of the region R is 14 square meters.. What is the area of the parallelogram below? The area A of the prallelogram is the product of its base and height. A = 10 ft 4 ft = 40 ft 2 4. A circle is divided into 9 congruent sectors. If the radius of the circle is 8 inches, a) what is the area of one of the 9 sectors? b) what is the arc length of one of the 9 sectors? Solution part a): The area A in square inches of one of the sectors is one-ninth the area of the entire circle: A = π 8 2 /9 = 64π/9 The area is 64π/9 square inches. Solution part b): The arc length L in inches for one of the sectors is one-ninth the length of arc of the entire circle. In otherwords, L is one-ninth the circumference of the circle: The arc length is 16π/9 inches. L = 2 π 8/9 = 16π/9
5. Find the area of a right triangle whose hypotenuse is 6 inches long and with one of the other sides 4 inches long. The area of a right triangle is one-half the product of the lengths of its perpendicular sides. One of those sides has length 4 inches. To find the length b in inches of the other side, use the Pythagorean theorem: 4 2 + b 2 = 6 2 16 + b 2 = 6 b 2 = 20 b = 20 = 4 " 5 = 2 5 The lengths of the perpendicular legs of the right triangle are 4 inches and 2 5 inches. The area A in square inches of the triangle is therefore, A = 1 2 4 2 5 = 4 5 The area is 4 5 square inches. 6. What is the area of the figure below? The arc is a semicircle. m The area A of the figure is the sum of the areas of the triangle and the semi-circle. Expressed in square meters, A is given by: The area is 12 + 9 2 π square meters. A = ( 1 2 4 6) + ( 1 2 π 2 ) = 12 + 9 2 π
7. If the diameter of a circle is 14 cm, find each of the following. (a) The circumference of the circle (b) The area of the circle (c) The area of a sector of the circle that corresponds to a central angle of 18. Solution part a): Using the formula for circumference, C = π d, the answer is 14π cm. Solution part b): The radius r of the circle is 7 cm and the formula for the area of a circle is A = πr 2. Therefore the area is 49π square centimeters. Solution part c): The area of a sector of a circle is given by A = " 60 πr2, where θ is the measure in degrees of the central angle that determines the sector. With θ = 18 and r = 7 cm, in square centimeters, this gives, A = 18 60 π 72 = 1 20 The area of the sector is 49π/20 cm 2. π 49 = 49π/20 8. Find the perimeters of each of the following if all arcs shown are semicircles. The measurements are in cm. (a) (b) 18 6 6 ---------4--------- 2 2 Solution part a): P = 6 + 6π + π + 6π = 6 + 15π The perimeter for part a is 6 + 15π cm. Solution part b): P = 2 + 2 + 2π + 4π = 4 + 6π The perimeter for part b is 4 + 6π cm.
9. a) Use your ruler to draw two rectangles with the same perimeters, but different areas. Calculate the areas and perimeter. There are many possible answers. For example, a rectangle 2 cm by 4 cm has the same perimeter as a square (which is also a rectangle) cm on a side. 2 4 Figures not drawn to scale The 2 cm by 4 cm rectangle has perimeter 12 cm and area 8 cm 2. The cm by cm rectangle (or square) also has perimeter 12 cm, but its area is 9 cm 2. b) Use your ruler to draw two rectangles with the same areas, but different perimeters. Calculate the area and perimeters. Again, there are many possible answers. For example, a rectangle 2 cm by 6 cm has the same area as a cm by 4 cm rectangle, but different perimeters. 2 6 4 Figures not drawn to scale The 2 cm by 6 cm rectangle has area 12 cm 2 and perimeter 16 cm. The cm by 4 cm rectangle also has area 12 cm 2, but perimeter 14 cm.