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 3 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) – [( 12 × 4 × 10) + ( 12 × 6 × 6)]
A = 16 + 36 – [20 + 18] = 14
! meters. !
The area of the region R is 14 square
3. 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 ft2
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:
L = 2 π 8/9 = 16π/9
The arc length is 16π/9 inches.
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:
42 + b2 = 62
16 + b2 = 36
b2 = 20
20 =
b=
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.
!
3m
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:
A = ( 12 × 4 × 6) + ( 12 π 3 2 ) = 12 + 92 π
The area is 12 + 92 π square meters.
!
!
!
!
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 = πr2. Therefore the area is 49π square centimeters.
"
Solution part c): The area of a sector of a circle is given by A = 360
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
360
π 72 =
1
20
!
π 49 = 49π/20
The area of the sector is 49π/20 cm2.
! each of the
! following if all arcs shown are semicircles. The
8. Find the perimeters of
measurements are in cm.
(a)
(b)
18
6
6
Solution part a): P = 6 + 6π + 3π + 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.
2
--------- 4 ---------
2
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) 3cm on a side.
3
2
3
4
Figures not drawn to scale
The 2 cm by 4 cm rectangle has perimeter 12 cm and area 8 cm2. The 3cm by 3cm
rectangle (or square) also has perimeter 12 cm, but its area is 9 cm2.
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 3 cm by 4 cm rectangle, but different perimeters.
2
3
6
4
Figures not drawn to scale
The 2 cm by 6 cm rectangle has area 12 cm2 and perimeter 16 cm. The 3 cm by 4
cm rectangle also has area 12 cm2, but perimeter 14 cm.