1. Estimate area occupied by the figure, if the area of each square is 1 cm 2.
a.
about 17 cm2
b.
about 29 cm2
c.
about 8 cm2
d.
about 12 cm2
2. Estimate the area of the shaded region in the following figure.
a.
42 square units
b.
49 square units
c.
41 square units
d.
40 square units
3. Suzanne wants to put a fence around the garden using wooden planks. The figure shown represents the shape of her garden. How many feet of wooden planks will she need to fence it?
a.
25 ft
b.
20 ft
c.
14 ft
d.
18 ft
4. Find the perimeter of the polygon shown.
a.
50 cm
b.
64 cm
c.
36 cm
d.
46 cm
5. Identify the perimeter of the regular polygon shown.
a.
42 yd
b.
14 yd
c.
49 yd 2
d.
49 yd
6. The perimeter of the rectangle is 22 cm. Find the value of w
a.
6 cm
b.
4 cm
c.
3 cm
d.
5 cm
7. Find the perimeter of the square.
a.
24 cm2
b.
24 cm
c.
36 cm
d.
12 cm
.
8. Find the perimeter of the rectangle shown.
a.
50 cm2
b.
30 cm
c.
15 cm
d.
15 cm2
9. Find the perimeter of the triangle.
a.
26 m
b.
34 m
c.
48 m
d.
52 m
10. Estimate the area of the figure, if each square is of 1 sq.cm.
a.
28 sq.cm.
b.
24 sq.cm.
c.
20 sq.cm.
d.
30 sq.cm.
11. Estimate the area of the figure, if each square is 1 square inch.
a.
50 square inches
b.
70 square inches
c.
40 square inches
d.
60 square inches
1. Estimate area occupied by the figure, if the area of each square is 1 cm 2.
a.
about 17 cm2
b.
about 29 cm2
c.
about 8 cm2
d.
about 12 cm2
Answer: (a) 2. Estimate the area of the shaded region in the following figure.
a.
42 square units
b.
49 square units
c.
41 square units
d.
40 square units
Answer: (c) 3. Suzanne wants to put a fence around the garden using wooden planks. The figure shown represents the shape of her garden. How many feet of wooden planks will she need to fence it?
a.
25 ft
b.
20 ft
c.
14 ft
d.
18 ft
Answer: (b) 4. Find the perimeter of the polygon shown.
a.
50 cm
b.
64 cm
c.
36 cm
d.
46 cm
Answer: (a) 5. Identify the perimeter of the regular polygon shown.
a.
42 yd
b.
14 yd
c.
49 yd2
d.
49 yd
Answer: (d)
6. The perimeter of the rectangle is 22 cm. Find the value of w
a.
6 cm
b.
4 cm
c.
3 cm
d.
5 cm
Answer: (c) 7. Find the perimeter of the square.
a.
24 cm2
b.
24 cm
c.
36 cm
d.
12 cm
Answer: (b)
.
8. Find the perimeter of the rectangle shown.
a.
50 cm2
b.
30 cm
c.
15 cm
d.
15 cm2
Answer: (b) 9. Find the perimeter of the triangle.
a.
26 m
b.
34 m
c.
48 m
d.
52 m
Answer: (c)
10. Estimate the area of the figure, if each square is of 1 sq.cm.
a.
28 sq.cm.
b.
24 sq.cm.
c.
20 sq.cm.
d.
30 sq.cm.
Answer: (b) 11. Estimate the area of the figure, if each square is 1 square inch.
a.
50 square inches
b.
70 square inches
c.
40 square inches
d.
60 square inches
Answer: (a)
1. What is the area of a parallelogram, if its base is 9 ft and its height is 14 ft? a.
46 ft2
b.
126 ft2
c.
46 ft
d.
63 ft2
2. Find the area of the colored region in the figure.
a.
22 cm2
b.
20 cm2
c.
28 cm2
d.
26 cm2
3. What is the area of ΔABC in the figure, if the height of ΔABC is 3 times the height of ΔDBC and area of ΔDBC is 40 inch2?
a.
115 inch2
b.
120 inch2
c.
110 inch2
d.
None of the above
4. The perimeter of an equilateral triangle is 36 yd. What is the length of each side of the triangle? a.
108 yd
b.
12 yd
c.
3 yd
d.
9 yd
5. Which of the figures have equal areas?
a.
Figure 1 & 3
b.
Figure 1 & 2
c.
Figure 2 and 3
d.
Figure 1,2 & 3
6. John spent 2
3
of an hour watching a movie and 1
6
of an hour cleaning his room. What fraction of an hour did he spend in both watching the movie and cleaning his room? a. 3
9
1
3
2
3
5
6
b. c. d.
7. Which two parallelograms have same area but different perimeters?
a.
Figure 2 and Figure 3
b.
Figure 1 and Figure 3
c.
Figure 1 and Figure 2
d.
Figure 2 and Figure 4
8. Which two parallelograms have same area but different perimeters?
a.
Figure 1 and Figure 2
b.
Figure 2 and Figure 4
c.
Figure 1 and Figure 3
d.
Figure 3 and Figure 4
9. Which two parallelograms have same area but different perimeters?
a.
Figure 2 and Figure 4
b.
Figure 1 and Figure 2
c.
Figure 2 and Figure 3
d.
Figure 1 and Figure 3
10. Which statement about the figures is true?
a.
Both the figures have the same area.
b.
Both the figures have the same length.
c.
Both the figures have the same perimeter.
d.
Both the figures have the same width.
11. Which two parallelograms have same area but different perimeters?
a.
Figure 1 and Figure 3
b.
Figure 3 and Figure 4
c.
Figure 1 and Figure 4
d.
Figure 2 and Figure 4
12. Which of the following is true for a right triangle and a rectangle having equal bases and equal heights? a.
The perimeter of the triangle is equal to the perimeter of the rectangle
b.
The area of the triangle is equal to the area of the rectangle
c.
The area of the triangle is half the area of the rectangle
d.
The area of the rectangle is half the area of the triangle
13. A parallelogram and a rectangle have equal bases and equal heights. What is the area of the rectangle, if the area of the parallelogram is 44 cm 2? a.
44 cm2
b.
11cm2
c.
22 cm2
d.
88 cm2
14. The height of a parallelogram is twice its base. What are the measures of the base and the height, if the area of the parallelogram is 288 m 2? a.
6 m and 12 m
b.
24 m and 48 m
c.
12 m and 24 m
d.
12 m and 36 m
15. What is the height of the parallelogram, if the base is half of its height and its area is 128 cm2? a.
13 cm
b.
14 cm
c.
26 cm
d.
16 cm
16. The state of Virginia is shaped like a triangle. If it has a base length of 406 miles and covers an area of about 40,980 square miles, what will be its height?
a.
about 202 miles
b.
about 406 miles
c.
about 812 miles
d.
about 101 miles
17. Find the area of the given parallelogram ABCD, if the area of the triangle ABE is 78 cm2.
a.
156 cm2
b.
78 cm2
c.
150 cm2
d.
39 cm2
18. Find the area of the triangle ABE, if the area of the parallelogram ABCD is 100 cm2.
a.
80 cm2
b.
50 cm2
c.
30 cm2
d.
90 cm2
19. Find the area of rectangle PQRS, if the area of the triangle PQT is 80 in. 2.
a.
140 in.2
b.
120 in.2
c.
160 in.2
d.
80 in.2
20. Find the area of the parallelogram ABCD.
a.
13.5 cm2
b.
15 cm2
c.
15.5 cm
d.
32 cm
21. What is the area of the shaded region in the figure?
a.
180 cm2
b.
120 cm2
c.
130 cm2
d.
None of the above
22. What is the area of the triangle ABD in the figure, if the area of a parallelogram is 30 ft2?
a.
15 ft2
b.
20 ft2
c.
17 ft2
d.
None of the above
23. What is the area of the parallelogram, if the area of the triangle is 22m 2?.
a.
47 m2
b.
44 m2
c.
54 m2
d.
None of the above
24. What is the height of the triangle, if the base of the triangle is twice the height, and its area is 225 ft2? a.
16 ft
b.
19 ft
c.
15 ft
d.
None of the above
25. What is the height of the triangle, if the triangle and the parallelogram have the same base and the same area?
a.
h
= 2H
b. H=h c. d.
2h =H None of the above
26. The area of a parallelogram is half of the area of the triangle. What is the height of the triangle, if they have the same base length of 6 cm and the area of the parallelogram is 39 cm2? a.
26 cm
b.
21 cm
c.
31 cm
d.
None of the above
27. What is the base length of the parallelogram, if the height is half of the base and its area is 128 in.2? a.
21 in
b.
18 in
c.
16 in
d.
None of the above
28. What is the area of ADB in the figure?
a.
48 in.2
b.
12 in.2
c.
24 in.2
d.
None of the above
29. Split the trapezoid into a triangle and a parallelogram and find the area of the trapezoid.
a.
20 m.2
b.
30 m.2
c.
25 m.2
d.
15 m.2
30. Find the perimeter of the figure.
a.
24 in.
b.
48 in.
c.
12 in.
d.
10 in.
31. Find the perimeter of ΔABC in the figure.
a.
12 in.
b.
18 in.
c.
21 in.
d.
3 in.
32. Find the perimeter of the parallelogram shown.
a.
10 in.
b.
20 in.
c.
5 in.
d.
16 in.
33. Find the length of NO in the given triangle, if the perimeter is 35 ft.
a.
7 ft
b.
11 ft
c.
6 ft
d.
8 ft
34. Find the perimeter of the shaded region in the parallelogram ABCD.
a.
12 cm
b.
18 cm
c.
11 cm
d.
15 cm
35. Find the perimeter of the figure.
a.
12.5 in.
b.
14.7 in.
c.
17.7 in.
d.
15.7 in.
36. What is the area of the triangle?
a.
11.5 square units
b.
1.5 square units
c.
4.5 square units
d.
5 square units
37. What is area of the shaded region?
a.
105 in.2
b.
120 in.2
c.
95 in.2
d.
None of the above
1. What is the area of a parallelogram, if its base is 9 ft and its height is 14 ft?
a.
46 ft2
b.
126 ft2
c.
46 ft
d.
63 ft2
Solution: Area of a parallelogram = base × height [Formula.] = 9 × 14 [Substitute the values of base and height.] = 126 [Multiply.] So, area of the parallelogram is 126 ft2.
2. Find the area of the colored region in the figure.
a.
22 cm2
b.
20 cm2
c.
28 cm2
d.
26 cm2
Solution: From the figure, the colored region is a parallelogram BEDF. The base length of the parallelogram BEDF = 7 cm [From the figure.] The height of the parallelogram BEDF = 4 cm [From the figure.]
Area of the parallelogram BEDF = base x height [Formula.] =7x4 [Substitute the values.] = 28 So, area of the colored region = area of the parallelogram BEDF = 28 cm 2.
3. What is the area of ΔABC in the figure, if the height of ΔABC is 3 times the height of ΔDBC and area of ΔDBC is 40 inch2?
a.
115 inch2
b.
120 inch2
c.
110 inch2
d.
None of the above
Solution: Area of a triangle = 1 / 2 x base x height [Formula.] Let h be the height of ΔDBC Area of ΔDBC = 1 / 2 x BC x h. Height of ΔABC = 3 times the height of ΔDBC = 3h. Area of ΔABC = 1 / 2 x BC x 3h = 3 x (12 x BC x h) = 3 x (Area of ΔDBC)
Area of ΔABC = 3 x 40 = 120 inch2.
4. The perimeter of an equilateral triangle is 36 yd. What is the length of each side of the triangle? a.
108 yd
b.
12 yd
c.
3 yd
d.
9 yd
Solution: In an equilateral triangle, all sides are equal. So, the perimeter of an equilateral triangle = 3 × measure of each side. 36 = 3 × measure of each side [Substitute the values.] 36 / 3 = 3 × measure of each side3 [Divide each side by 3.] 12 = measure of each side [Simplify.] So, the measure of each side of the equilateral triangle is 12 yd.
5. Which of the figures have equal areas?
a.
Figure 1 & 3
b.
Figure 1 & 2
c.
Figure 2 and 3
d.
Figure 1,2 & 3
Solution: Figure 1 represents a parallelogram with its base length 4 units and height 2 units. Area of the parallelogram = base × height = 4 × 2 = 8 square units. Figure 2 represents a rectangle of length 4 units and width 2 units. Area of the rectangle = length × width = 4 × 2 = 8 square units. Figure 3 represents a triangle with a base length of 6 units and a height of 3 units. Area of the triangle = 1 / 2 × base × height = 1 / 2 × 6 × 3 = 9 square units. So, the parallelogram in figure 1 and the rectangle in figure 2 have equal areas.
6. John spent 2
3
of an hour watching a movie and 1
6
of an hour cleaning his room. What fraction of an hour did he spend in both watching the movie and cleaning his room? a. 3
9
1
3
2
3
5
6
b. c. d.
Answer: (d)
7. Which two parallelograms have same area but different perimeters?
a.
Figure 2 and Figure 3
b.
Figure 1 and Figure 3
c.
Figure 1 and Figure 2
d.
Figure 2 and Figure 4
Solution: Area of a parallelogram = Base × Height Perimeter of a parallelogram = 2 × (Base + Height)
Area = 5 × 4 = 20 sq.units Perimeter = 2 × (5 + 4) = 18 units
Area = 4 × 4 = 16 sq.units Perimeter = 2 × (4 + 4) = 16 units
Area = 8 × 2 = 16 sq.units Perimeter = 2 × (8 + 2) = 20 units
Area = 4 × 3 = 12 sq.units Perimeter = 2 × (4 + 3) = 14 units Therefore, parallelograms in Figure 2 and Figure 3 have same area but different perimeters.
8. Which two parallelograms have same area but different perimeters?
a.
Figure 1 and Figure 2
b.
Figure 2 and Figure 4
c.
Figure 1 and Figure 3
d.
Figure 3 and Figure 4
Solution: Area of a parallelogram = Base × Height Perimeter of a parallelogram = 2 × (Base + Height)
Area = 5 × 4 = 20 sq.units Perimeter = 2 × (5 + 4) = 18 units
Area = 6 × 3 = 18 sq.units Perimeter = 2 × (6 + 3) = 18 units
Area = 6 × 2 = 12 sq.units Perimeter = 2 × (6 + 2) = 16 units
Area = 4 × 3 = 12 sq.units Perimeter = 2 × (4 + 3) = 14 units Therefore, parallelograms in Figure 3 and Figure 4 have same area but different perimeters.
9. Which two parallelograms have same area but different perimeters?
a.
Figure 2 and Figure 4
b.
Figure 1 and Figure 2
c.
Figure 2 and Figure 3
d.
Figure 1 and Figure 3
Solution: Area of a parallelogram = Base × Height Perimeter of a parallelogram = 2 × (Base + Height)
Area = 5 × 7 = 35 sq.units Perimeter = 2 × (5 + 7) = 24 units
Area = 6 × 6 = 36 sq.units Perimeter = 2 × (6 + 6) = 24 units
Area = 9 × 4 = 36 sq.units Perimeter = 2 × (9 + 4) = 26 units
Area = 8 × 4 = 32 sq.units Perimeter = 2 × (8 + 4) = 24 units Therefore, parallelograms in Figure 2 and Figure 3 have same area but different perimeters.
10. Which statement about the figures is true?
a.
Both the figures have the same area.
b.
Both the figures have the same length.
c.
Both the figures have the same perimeter.
d.
Both the figures have the same width.
Solution: Parallelogram in Figure 1 has a base of 6 units and a height of 2 units. Parallelogram in Figure 2 has a base of 4 units and a height of 3 units.
Area of a parallelogram = Base × Height Perimeter of a parallelogram = 2 × (Base + Height)
Area = 6 × 2 = 12 sq.units Perimeter = 2 × (6 + 2) = 16 units
Area = 4 × 3 = 12 sq.units Perimeter = 2 × (4 + 3) = 14 units Therefore, the statement 'Both the figures have the same area' is true.
11. Which two parallelograms have same area but different perimeters?
a.
Figure 1 and Figure 3
b.
Figure 3 and Figure 4
c.
Figure 1 and Figure 4
d.
Figure 2 and Figure 4
Solution: Area of a parallelogram = Base × Height, Perimeter of a parallelogram = 2 × (Base + Height)
Area = 7 × 4 = 28 sq.units, Perimeter = 2 × (7 + 4) = 22 units
Area = 4 × 5 = 20 sq.units, Perimeter = 2 × (4 + 5) = 18 units
Area = 3 × 8 = 24 sq.units, Perimeter = 2 × (3 + 8) = 22 units
Area = 6 × 4 = 24 sq.units, Perimeter = 2 × (6 + 4) = 20 units Therefore, parallelograms in Figure 3 and Figure 4 have same area but different perimeters.
12. Which of the following is true for a right triangle and a rectangle having equal bases and equal heights? a.
The perimeter of the triangle is equal to the perimeter of the rectangle
b.
The area of the triangle is equal to the area of the rectangle
c.
The area of the triangle is half the area of the rectangle
d.
The area of the rectangle is half the area of the triangle
Solution: Area of the triangle = 12 × b × h = 12 × Area of the rectangle
So, the area of the triangle is half the area of the rectangle. Let the length of the common base be b Let the length of the common height be h Area of the rectangle = b × h
13. A parallelogram and a rectangle have equal bases and equal heights. What is the area of the rectangle, if the area of the parallelogram is 44 cm 2? a.
44 cm2
b.
11cm2
c.
22 cm2
d.
88 cm2
Solution: If the base and height of the parallelogram and the rectangle are same, then the area of the parallelogram is same as that of rectangle. So, the area of the rectangle = the area of the parallelogram = 44 cm 2.
14. The height of a parallelogram is twice its base. What are the measures of the base and the height, if the area of the parallelogram is 288 m 2? a.
6 m and 12 m
b.
24 m and 48 m
c.
12 m and 24 m
d.
12 m and 36 m
Solution: Let b be the base of the parallelogram. Height of the parallelogram = Twice that of the base = 2b Area of the parallelogram = base × height [Formula.] 288 = b × 2b [Substitute the values.]
144 = b2 144 = b2, b = 12 [Taking square root on both sides.] Base of the parallelogram = 12 m Height of the parallelogram = 2b = 2(12) = 24 m Therefore, the base and height of the parallelogram are 12 m and 24 m respectively.
15. What is the height of the parallelogram, if the base is half of its height and its area is 128 cm2? a.
13 cm
b.
14 cm
c.
26 cm
d.
16 cm
Solution: Let h be the height of the parallelogram. Base of the parallelogram = Half of the height = h2 The area of the parallelogram = base × height 128 = h2 × h [Substitute the values.] 128 × 2 = h2 [Multiply each side by 2.] 16 = h [Take square root on both the sides.] Therefore, the height of the parallelogram is 16 cm.
16. The state of Virginia is shaped like a triangle. If it has a base length of 406 miles and covers an area of about 40,980 square miles, what will be its height?
a.
about 202 miles
b.
about 406 miles
c.
about 812 miles
d.
about 101 miles
Solution: Area of a triangle = 1 / 2 × base × height [Formula.] Area covered by the state of Virginia = 1 / 2 × 406 × h, where h is the height of the triangle. 1 / 2 × 406 × h = 40,980 ⇒ h ≈ 202 So, the height of the triangle is about 202 miles.
17. Find the area of the given parallelogram ABCD, if the area of the triangle ABE is 78 cm2.
a.
156 cm2
b.
78 cm2
c.
150 cm2
d.
39 cm2
Solution: Area of a parallelogram = base × height Area of a triangle = 1 / 2 × base × height If the base and height of the parallelogram and the triangle are same then the area of the parallelogram is twice the area of the triangle. Thus the area of the parallelogram ABCD = 2 × area of the triangle ABE = 2 × 78 = 156 [Substitute the values.] So, the area of the parallelogram ABCD is 156 cm2.
18. Find the area of the triangle ABE, if the area of the parallelogram ABCD is 100 cm2.
a.
80 cm2
b.
50 cm2
c.
30 cm2
d.
90 cm2
Solution: Area of a parallelogram = base × height Area of a triangle = 1 / 2 × base × height If the base and height of the parallelogram and the triangle are same then the area of the triangle is half of the area of the parallelogram. Thus the area of the triangle ABE = 1 / 2 × area of the parallelogram ABCD = 12 × 100 = 50 [Substitute the values.] So, the area of the triangle ABE is 50 cm2.
19. Find the area of rectangle PQRS, if the area of the triangle PQT is 80 in. 2.
a.
140 in.2
b.
120 in.2
c.
160 in.2
d.
80 in.2
Solution: Area of a rectangle = length × width Area of a triangle = 1 / 2 × base × height If the base and height of the rectangle and the triangle are same then the area of the rectangle is twice the area of the triangle. Thus the area of the rectangle PQRS = 2 × area of the triangle PQT = 2 × 80 = 160 [Substitute the values.] So, the area of the rectangle PQRS is 160 in. 2.
20. Find the area of the parallelogram ABCD.
a.
13.5 cm2
b.
15 cm2
c.
15.5 cm
d.
32 cm
Solution: Length of the base = 6 cm. [Given.] Area of the parallelogram = base length × height. [Formula] = 6 cm × 2.5 cm [Substitute the respective values in the formula.] = 15 cm2. So, the area of the parallelogram ABCD is 15 cm 2.
21. What is the area of the shaded region in the figure?
a.
180 cm2
b.
120 cm2
c.
130 cm2
d.
None of the above
Solution: From the figure, shaded region = ACD. ACD is in triangular shape. Area of the triangle ACD = 1 / 2 × base × height [Formula.] From the figure base of the triangle ACD = 15 cm and height = 16 cm.
The area of the triangle ACD = 1 / 2 × 15 × 16 [Substitute the values.] 120 cm2 [Simplify.] The area of the triangle ACD = 120 cm2.
22. What is the area of the triangle ABD in the figure, if the area of a parallelogram is 30 ft2?
a.
15 ft2
b.
20 ft2
c.
17 ft2
d.
None of the above
Solution: The area of the triangle = 1 / 2 × area of the parallelogram = 12× 30 [Since area of the parallelogram = 30.] = 15 [Simplify.] If a parallelogram and the triangle have same base and same height then the area of the triangle is half of the area of the parallelogram. From the figure, the parallelogram and the triangle have same base and same height. The area of the triangle = 15 ft2.
23. What is the area of the parallelogram, if the area of the triangle is 22m 2?.
a.
47 m2
b.
44 m2
c.
54 m2
d.
None of the above
Solution: From the figure, the triangle and the parallelogram have same bases and the same heights. The area of the parallelogram = 2 × area of the triangle. = 2 × 22 [Substitute the area of triangle = 22.] = 44 The area of the parallelogram = 44 m2.
24. What is the height of the triangle, if the base of the triangle is twice the height, and its area is 225 ft2? a.
16 ft
b.
19 ft
c.
15 ft
d.
None of the above
Solution: Let h is the height of the triangle. Given, area of the triangle = 225 ft2
The base = twice the height = 2 × h. Area of a triangle = 1 / 2 × base × height [Formula.] 225 = 1 / 2 × 2 × h × h [Substitute the values.] 225 = h2 [Simplify.] 225 = h2 [Taking square root on both sides.] 15 = h The height of the triangle = 15 ft.
25. What is the height of the triangle, if the triangle and the parallelogram have the same base and the same area?
a.
h
= 2H
b. H=h c. d.
2h =H None of the above
Solution: From the figure, the triangle and the parallelogram have same base. Height of the parallelogram = H Height of the triangle = h.
Area of the triangle = 1 / 2× base × height Area of the parallelogram = base × height Given, area of the triangle = area of the parallelogram 1 / 2 × base × height = base × height [Substitute the area formulae.] 1 / 2 × height = height [Bases are equal.] 1/2×h=H [Substitute the height values.] h = 2H
26. The area of a parallelogram is half of the area of the triangle. What is the height of the triangle, if they have the same base length of 6 cm and the area of the parallelogram is 39 cm2? a.
26 cm
b.
21 cm
c.
31 cm
d.
None of the above
Solution: The area of the parallelogram = 39 cm2 The area of the parallelogram = 1 / 2 x area of the triangle [Given.] 39 = 1 / 2 x area of the triangle [Substitute the area of parallelogram.] 78 = area of the triangle [Multiply each side by 2.] Area of the triangle = 78 cm2 Base length of the parallelogram = base length of the triangle = 6cm [Given.] Area of the triangle = 1 / 2 x base x height [Formula.]
78 = 1 / 2 x 6 x height [Substitute the values.] 78 = 3 x height [Multiply 3 with 1 / 2.] 78 / 3 = 3xheight / 3 [Divide each side by 3.] Height of the triangle = 26 cm [Simplify.]
27. What is the base length of the parallelogram, if the height is half of the base and its area is 128 in.2? a.
21 in
b.
18 in
c.
16 in
d.
None of the above
Solution: Let b be the base length of the parallelogram. Height of the parallelogram = half of the base = b2 Area of the parallelogram = base × height [Formula.] 128 = b × b2 [Substitute the values.] 128 × 2 = b22 × 2 [Multiply each side by 2.] 256 = b2 [Simplify.] 256 = b2 [Taking square root on both sides.] 16 = b The base length of the parallelogram = 16 in.
28. What is the area of ADB in the figure?
a.
48 in.2
b.
12 in.2
c.
24 in.2
d.
None of the above
Solution: ADB is a triangle. Base of the triangle ADB = 6 in. and the height of the triangle ADB = 4 in. Area of the triangle ADB = 1 / 2 × base × height [Formula.] 1/2×6×4 [Substitute the values.] = 12 [Simplify.] The area of ADB = 12 in.2
29. Split the trapezoid into a triangle and a parallelogram and find the area of the trapezoid.
a.
20 m.2
b.
30 m.2
c.
25 m.2
d.
15 m.2
Solution: Draw a line parallel to CB from the point D. Let the line touch AB on the point E.
Now the trapezoid is split into a parallelogram DEBC and a Δ ADE. Base length of the parallelogram = BE = DC = 6 m., height = 4 m. Area of a parallelogram DEBC = base × height = 6 × 4 = 24 m. 2 Base length of the Δ ADE = AE = AB - DC = 9 - 6 = 3 m. Height of the Δ ADE = 4 m. Area of the Δ ADE = 1 / 2 × base × length = 1 / 2 × 3 × 4 = 6 m. 2 Area of the trapezoid ABCD = Area of a parallelogram DEBC + Area of the Δ ADE. Area of the trapezoid ABCD = 24 + 6 = 30 m.2
30. Find the perimeter of the figure.
a.
24 in.
b.
48 in.
c.
12 in.
d.
10 in.
Solution: Perimeter of a figure = sum of all the sides of the figure. From the figure AB = 8 in., BC = 10 in., AC = 6 in. Perimeter of ΔABC = AB + BC + CA = 8 + 10 + 6 [Substitute the values.] = 24 So, the perimeter of the figure is 24 in.
31. Find the perimeter of ΔABC in the figure.
a.
12 in.
b.
18 in.
c.
21 in.
d.
3 in.
Solution: Perimeter of a triangle is the sum of all the sides of the triangle. From the figure, AB = 3 in., BC = 10 in. and AC = 8 in. Perimeter of ΔABC = AB + BC + CA = 3 + 10 + 8 [Substitute the values.] = 21 [Add.] The perimeter of the ΔABC is 21 in.
32. Find the perimeter of the parallelogram shown.
a.
10 in.
b.
20 in.
c.
5 in.
d.
16 in.
Solution: Perimeter of a parallelogram = sum of all the sides of the parallelogram From the figure, AB = 5 in. and AD = 3 in. As the lengths of the parallel sides are equal, AB = CD = 5 in., AD = BC = 3 in.
Perimeter of the parallelogram = AB + BC + CD + DA =5+3+5+3 [Substitute the lengths.] = 16 [Add.] So, the perimeter of the parallelogram given in the figure is 16 in.
33. Find the length of NO in the given triangle, if the perimeter is 35 ft.
a.
7 ft
b.
11 ft
c.
6 ft
d.
8 ft
Solution: From the figure, MN = 15 ft, OM = 13 ft Perimeter of a triangle = sum of all sides [Formula.] Perimeter of the triangle MNO = MN + NO + OM [Formula.] 35 = 15 + NO + 13 [Substitute the values.] 35 = 28 + NO [Add 15 and 13.] 35 - 28 = 28 -28 + NO [Subtract 28 from each side.] 7 = NO [Simplify.]
The length of NO of the triangle is 7 ft.
34. Find the perimeter of the shaded region in the parallelogram ABCD.
a.
12 cm
b.
18 cm
c.
11 cm
d.
15 cm
Solution: From the figure, ΔBEC is the shaded area in the parallelogram ABCD. From the figure, BC = 5 cm, BE = 3 cm, AB = 7 cm, and DE = 3 cm. EC = DC - DE = AB - DE = 7 - 3 = 4 cm. [Lengths of the parallel sides in a parallelogram are equal, AB = DC.] Perimeter of the ΔBEC = BC + CE + EB [Formula.] =5+4+3 (Substitute the values) = 12 [Add.] So, the perimeter of the shaded region of the figure is 12 cm.
35. Find the perimeter of the figure.
a.
12.5 in.
b.
14.7 in.
c.
17.7 in.
d.
15.7 in.
Solution: From the figure, BA = 5 in., AC = 5.2 in., CD = 2.5 in., BC = 3 in., and DB = 2 in. Perimeter of a figure = sum of all the sides of the figure [Formula.] Perimeter of the figure = AC + CD + DB + BA = 5.2 + 2.5 + 2 + 5 [Substitute the values.] = 14 .7 [Add.] So, the perimeter of the figure is 14.7 in.
36. What is the area of the triangle?
a.
11.5 square units
b.
1.5 square units
c.
4.5 square units
d.
5 square units
Solution: From the figure, the base length of the triangle = 3 units. Height of the triangle = 3 units. Area of a triangle with base b and height h = 1 / 2 × b × h = 12 × 3 × 3 [Substitute the values.] = 4.5 [Simplify.] So, the area of the triangle is 4.5 square units.
37. What is area of the shaded region?
a.
105 in.2
b.
120 in.2
c.
95 in.2
d.
None of the above
Solution: Area of the shaded region is equal to the area of the triangle with measures of 15 in. height and 14 in. as base. Area of the triangle = 1 / 2 x base x height [Formula.] = 12 x 14 x 15 [Substitute the values.] = 105 in.2 [Simplify.] Area of the shaded region in the figure = 105 in. 2
1. Find the perimeter of the figure shown below. [Given = 27.]
a.
144 cm
b.
90 cm
c.
99 cm
d.
63 cm
a
= 45 and b
2. A steel wire in the form of a circular ring of radius 4 m is straightened and cut into 2 equal pieces. Find the length of each piece. [Take π a.
22 m
b.
12 m
c.
24 m
d.
4m
= 3.]
3. Find the area of the shaded portion of the figure if all the angles in the figure are right angles.
a.
22 m2
b.
28 m
c.
20 m
d.
27 m2
4. Find the area of the shaded portion of the figure if all the angles in the figure are right angles.
a.
18 yd 2
b.
24 yd 2
c.
28 yd
d.
16 yd 2
5. Graph the points and find the area and perimeter. A(1, 1), B(1, 5), C(6, 4), D(6, 1) a.
20 sq.units and 9 units
b.
18 units and 20 units
c.
18 sq.units and 9 units
d.
20 sq.units and 18 units
6. The perimeter of a rectangle is 24 cm and the base is 4 cm. What is the area? a.
24 cm2
b.
32 cm2
c.
8 cm2
d.
16 cm2
7. Estimate the perimeter of the cover of the book.
a.
98 cm
b.
238 cm
c.
190 cm
d.
138 cm
8. Mary wants to buy a curtain for her door. Find the area of the cloth required if the height and width of the door are 7 ft and 3.5 ft. a.
about 18.5 ft2
b.
about 24.5 ft2
c.
about 19 ft2
d.
about 21 ft2
9. The length of a rectangle is increased by 50% and the width is decreased by 25%. How is its area affected? a.
increases by 12.5%
b.
increases by 50%
c.
unchanged
d.
decreases by 25%
10. Find the area and the perimeter of the rectangle formed by the points P(1,0), Q(- 2, 0), R(- 2, 4), S(x a.
12 sq.units and 7 units
b.
12 sq.units and 14 units
c.
24 sq.units and 14 units
,y
).
d.
data insufficient
11. What is the perimeter of the figure shown?
a.
43 square units
b.
43 units
c.
96 square units
d.
48 units
12. Find the length of the wire required to fence the park which is in the shape of a square. The wire is used for two rounds around the park.
a.
640 yd
b.
3200 yd 2
c.
6400 yd 2
d.
320 yd
13. Find the perimeter of the regular polygon.
a.
63 yd
b.
72 yd
c.
64 yd
d.
54 yd
14. Find the perimeter of the regular polygon.
a.
121 ft
b.
110 ft
c.
22 ft
d.
55 ft
15. Find the perimeter of the regular polygon
a.
64 in.
b.
36 in.
c.
54 in.
d.
48 in.
16. The perimeter of a regular pentagon is 100 cm. How long is each side? a.
500 cm
b.
250 cm
c.
20 cm
d.
10 cm
17. The perimeter of a regular hexagon is 24 m. How long is each side? a.
144 m
b.
6m
c.
4m
d.
12 m
18. What is the perimeter of polygon ABCDE? [Given AB = 4 cm, BC = 5 cm, CD = 5 cm, ED = 11 cm and EA = 8 cm]
a.
33 cm
b.
55 cm
c.
25 cm
d.
15 cm
19. What is the perimeter of rectangle PQRS?[Given
a
= 4.4 cm and b
= 7.4 cm]
a.
11.8 cm
b.
32.5 cm
c.
23.6 cm
d.
33.6 cm
20. Which of the following is the perimeter of a sqaure? a.
2
√
diagonal
b. 22
√
diagonal
42
√
diagonal
c. d.
2 diagonal
21. Find the side of a square, if its perimeter is 40 cm. a.
10 cm
b.
14 cm
c.
5 cm
d.
20 cm
22. What is the perimeter of a regular polygon having 10 sides each of length 2 cm? a.
20 cm
b.
25 cm
c.
30 cm
d.
4 cm
23. A steel wire in the form of a circular ring of radius 3 m is straightened and cut into 4 equal pieces. Find the length of each piece. a.
18.84 m
b.
4.71 m
c.
3m
d.
14.71 m
24. What is the ratio of the circumference of a circle to its diameter? a.
e b. 22
c.
7
3.14
d.
π
25. Find the perimeter of the figure shown. Take π = 30, b
= 12, r
a.
186 units
b.
48 units
c.
168 units
d.
120 units
26. If l and b what is its perimeter?
= 3. [Given a
= 6.]
are the length and the width of a rectangle, then
a. 2(l
+b
)
b. 2l
+b
2b
+l
c. d.
l
+b
27. Find the perimeter of the square MNLK, if s .
= 33
√
a. 63
√
cm
33
√
cm
√
cm
b. c.
12 cm
d. 123
28. Find the circumference of a circle of radius 5 cm. a.
10 cm
b. 5π
cm
c. 10π
cm
20 π
cm
d.
29. What is the area of the rectangle ABCD? [Given and b
=9
4
]
a
= 17
8
a. 289
64
sq. cm
b. 81
16
sq. cm
c. 153
32
sq. cm
32
153
sq. cm
d.
30. Find the area of the square of side 3.2 cm.
a.
20.24 cm2
b.
10.24 cm2
c.
12.8 cm2
d.
6.4 cm2
Worksheets in this topic » Perimeter Circumference and Area Part-1
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Related Questions 24 take test 1geometry volume area circumference perimeter what is the area and circumference of an ellipse? Is the circumference of a circle the same as the area of a circle?
31. Find the area of the circle of radius 52
√
cm.
a.
50 cm2
b.
25 cm2
c. 25π
cm2
50π
cm2
d.
32. The perimeter of a square is 13.2 cm. What is its area? a.
13.20 cm2
b.
10.89 cm2
c.
14.89 cm2
d.
26.4 cm2
33. If A1, A2, A3 represent the areas of respective regions, then what is the area of polygon PQRST?
a.
A1 + A2 + A3
b.
A1A2A3
c.
A
1
−
−
+A
2
−
−
−
2
2
−
−
+A −
3
−
−
2
−
−
−
√ d.
None of the above
34. A square and a rectangle have equal perimeters. If area of the square is 4 cm2, then find the perimeter of the rectangle. a.
13 cm
b.
8 cm
c.
18 cm
d.
4 cm
35. For a given perimeter, which of the following will have maximum area? a.
square
b.
rhombus
c.
rectangle
d.
circle
36. What is the length of the outer boundary of a semicircle of radius ?
r
a.
r
π
+ 2r
b. 2r c.
r
+π
π
r
d.
37. Find the area of a semi-circle of radius 4.6 cm. a.
21.16 sq.cm
b. c.
4.6π 10.58 sq.cm
sq.cm
d. 10.58π
sq.cm
38. There are 5 concentric circles. Radius of the smallest circle is 5 cm. Increase in radius of each consecutive circle is 1 cm. What will be the area of a circle in sq. cm which has a circumference equal to the sum of the circumferences of the 5 concentric circles?
a. 1225π 1225 sq.cm
b.
sq.cm
c. 70 π
sq.cm
35π
sq.cm
d.
1. Find the perimeter of the figure shown below. [Given = 27.]
a.
144 cm
b.
90 cm
c.
99 cm
a
= 45 and b
d.
63 cm
Solution:
Perimeter of the figure = 45 + 5p + 6q [From the figure.] 5p = 45 cm ⇒ p = 455 = 9 cm [Divide each side by 5.] 3q = 27 cm ⇒ q = 9 cm [Divide each side by 3.] Perimeter of the figure = 45 + 5 × 9 + 6 × 9 = 144 cm [From steps 3 and 4.] Therefore the perimeter of the figure is 144 cm.
2. A steel wire in the form of a circular ring of radius 4 m is straightened and cut into 2 equal pieces. Find the length of each piece. [Take π a.
22 m
b.
12 m
c.
24 m
d.
4m
Solution: Length of the steel wire = circumference of the ring = 2 πr [Formula.] = 2π × 4
= 3.]
[Substitute.] = 24 m [Simplify.] Total length of the rods = 24 m [Step 1.] Length of one rod = 242 [There are 2 rods of equal length.] = 12 m.
3. Find the area of the shaded portion of the figure if all the angles in the figure are right angles.
a.
22 m2
b.
28 m
c.
20 m
d.
27 m2
Solution: Area of the rectangle 1 = l × b = 5 m × 2 m = 10 m 2
Divide the shaded portion into parts. Area of the shaded portion = area of the square + area of the rectangle 1 + area of the rectangle 2 Area of the square = s2 = 22 = 4 m2 Area of the rectangle 2 = l × b = 4 m × 2 m = 8 m 2 So, area of the shaded portion = 4 m2 + 10 m2 + 8 m2 = 22 m2
4. Find the area of the shaded portion of the figure if all the angles in the figure are right angles.
a.
18 yd2
b.
24 yd2
c.
28 yd
d.
16 yd2
Solution:
Area of the shaded portion = area of rectangle - 4(area of rectangle with sides 4 yd and 3 yd) Area of rectangle = l × b = 8 × 6 = 48 yd 2 [length = 3 + 2 + 3 and breadth = 2 + 2 + 2.] Area of rectangle with sides 3 yd and 2 yd = 3 × 2 = 6 yd 2 Area of the shaded portion = 48 - 4(6) = 24 yd 2 [Substitute in step 1 and simplify.] So, the area of the shaded portion is 24 yd 2.
5. Graph the points and find the area and perimeter. A(1, 1), B(1, 5), C(6, 4), D(6, 1) a.
20 sq.units and 9 units
b.
18 units and 20 units
c.
18 sq.units and 9 units
d.
20 sq.units and 18 units
Solution:
Graph the points on the coordinate plane. The figure obtained by joining the points is a rectangle with length 5 units and breadth 4 units. Area of the rectangle = l × b = 5 × 4 = 20 sq.units Perimeter of the rectangle = 2(l + b) = 2(5 + 4) = 18 units So, the area and perimeter of the figure are 20 sq.units and 18 units.
6. The perimeter of a rectangle is 24 cm and the base is 4 cm. What is the area? a.
24 cm2
b.
32 cm2
c.
8 cm2
d.
16 cm2
Solution: The perimeter of a rectangle = 2(l + b) [Formula.] 2(l + b) = 24 cm [Given, the perimeter of a rectangle is 24 cm.] 2l + 2(4) = 24 cm [Given, base is 10 cm.] 2l = 24 - 8 = 16 cm [Simplify.] l = 16 / 2 = 8 cm [Simplify.] Area of the rectangle = l × b = 8 cm × 4 cm = 32 cm 2
7. Estimate the perimeter of the cover of the book.
a.
98 cm
b.
238 cm
c.
190 cm
d.
138 cm
Solution: The perimeter of the cover of the book = 2(25) + 4(20) + 2(4) = 50 + 80 + 8 = 138 So, the perimeter of the cover of the book is 138 cm.
8. Mary wants to buy a curtain for her door. Find the area of the cloth required if the height and width of the door are 7 ft and 3.5 ft. a.
about 18.5 ft2
b.
about 24.5 ft2
c.
about 19 ft2
d.
about 21 ft2
Solution: Area of a rectangle = l × b Area of the door = 7 ft × 3.5 ft = 24.5 ft 2 Area of the curtain = Area of the door = 24.5 ft 2 So, Mary needs about 24.5 ft2 of cloth for her door.
9. The length of a rectangle is increased by 50% and the width is decreased by 25%. How is its area affected? a.
increases by 12.5%
b.
increases by 50%
c.
unchanged
d.
decreases by 25%
Solution: Area of a rectangle = l × b [Formula.] If length is increased by 50%, then the new length is l + 50 / 100l = l + 12l = 32l If width is decreased by 25%, then the new width is b - 25 / 100b = b - 14b = 34b New area of a rectangle = 3 / 2l × 3 / 4b = 98(lb) = 1.125 (lb) [Simplify.]
Increase in area = 1.125 (lb) - (lb) = 0.125 (lb) Percentage increase in area = 0.125 (lb)lb × 100 = 12.5% So, the area is increased by 12.5%.
10. Find the area and the perimeter of the rectangle formed by the points P(1,0), Q(- 2, 0), R(- 2, 4), S(x a.
12 sq.units and 7 units
b.
12 sq.units and 14 units
c.
24 sq.units and 14 units
d.
data insufficient
,y
).
Solution:
Graph the points on the coordinate plane.
As PQRS is a rectangle, the possible coordinates for point R are (1, 4). Area of the rectangle = l × b = 4 × 3 = 12 sq.units Perimeter of the rectangle = 2(l + b) = 2(4 + 3) = 14 units So, the area and perimeter of the retangle are 12 sq.units and 14 units.
11. What is the perimeter of the figure shown?
a.
43 square units
b.
43 units
c.
96 square units
d.
48 units
Solution: Perimeter of the figure = 4 + 6 + 3 + 4 + 8 + 5 + 3 + 2 + 3 + 3 + 2 = 48 units [Simplify.]
12. Find the length of the wire required to fence the park which is in the shape of a square. The wire is used for two rounds around the park.
a.
640 yd
b.
3200 yd2
c.
6400 yd2
d.
320 yd
Solution: Length of the wire required to fence the park = Perimeter of the park Perimeter of the park = 4 × (side) = 4(80) = 320 yd [Park is in square shape.] So, the length of the wire required to fence the park is 320 yd.
13. Find the perimeter of the regular polygon.
a.
63 yd
b.
72 yd
c.
64 yd
d.
54 yd
Solution: The regular polygon has 6 sides, each with a length of 9 yd. Perimeter of the polygon = number of sides × side length = 6 × 9 yd = 54 yd.
14. Find the perimeter of the regular polygon.
a.
121 ft
b.
110 ft
c.
22 ft
d.
55 ft
Solution: This regular polygon has 5 sides, each with a length of 11 ft. Perimeter of the polygon = number of sides × side length = 5 × 11 ft = 55 ft.
15. Find the perimeter of the regular polygon
a.
64 in.
b.
36 in.
c.
54 in.
d.
48 in.
Solution: The regular polygon has 8 sides, each with a length of 6 in. Perimeter of the polygon = number of sides × side length = 8 × 6 in. = 48 in.
16. The perimeter of a regular pentagon is 100 cm. How long is each side? a.
500 cm
b.
250 cm
c.
20 cm
d.
10 cm
Solution: Perimeter of the polygon = number of sides × side length = 100 cm
Number of sides in regular polygons = 5 Side length of the regular pentagon = perimeternumber of sides = 100 cm5 = 20 cm
17. The perimeter of a regular hexagon is 24 m. How long is each side? a.
144 m
b.
6m
c.
4m
d.
12 m
Solution: Perimeter of the polygon = number of sides × side length = 24 m Number of sides in regular hexagon = 6 Side length of the regular hexagon = perimeternumber of sides = 24 m6 = 4 m
18. What is the perimeter of polygon ABCDE? [Given AB = 4 cm, BC = 5 cm, CD = 5 cm, ED = 11 cm and EA = 8 cm]
a.
33 cm
b.
55 cm
c.
25 cm
d.
15 cm
Solution: Perimeter of ABCDE = AB + BC + CD + DE + EA [Definition.]
Perimeter = 4 + 5 + 5 + 11 + 8 [Substitute.] Perimeter of ABCDE = 33 cm. [ Add ]
19. What is the perimeter of rectangle PQRS?[Given
a
= 7.4 cm]
a.
11.8 cm
b.
32.5 cm
c.
23.6 cm
d.
33.6 cm
Solution: Perimeter of rectangle = 2(Length + Breadth) [Formula.] Perimeter of PQRS = 2(4.4 + 7.4) [Substitute.] Perimeter of PQRS = 23.6 cm [Simplify.]
20. Which of the following is the perimeter of a sqaure? a.
2
√
diagonal
b. 22
√
diagonal
42
√
diagonal
c. d.
Solution:
2 diagonal
= 4.4 cm and b
Perimeter of a square = 4 × side = 4× diagonal2 [Diagonal = 2side.] = 22diagonal [Formula.]
31. Find the area of the circle of radius 52 a.
50 cm2
b.
25 cm2
√
c. 25π
cm2
50π
cm2
d.
Solution: Area of required circle = π × (52)2 [Area of circle = π × radius2.] Area of required circle = 50π cm2
32. The perimeter of a square is 13.2 cm. What is its area? a.
13.20 cm2
b.
10.89 cm2
c.
14.89 cm2
d.
26.4 cm2
Solution: Side of square = perimeter4 [Formula.] Side = 13.2 / 4 cm [Substitute.] Side of square = 3.30 cm [Simplify.] Area of square = 3.30 cm × 3.30 cm
cm.
[Area of square = side2.] Area of square = 10.89 sq.cm.
33. If A1, A2, A3 represent the areas of respective regions, then what is the area of polygon PQRST?
a.
A1 + A2 + A3
b.
A1A2A3
c.
A
1
−
2
−
−
+A
2
−
−
+A
2
−
−
−
3
−
−
2
−
−
−
√ d.
None of the above
Solution: The area of a region is the sum of the areas of its non-overlapping parts. Area of PQRST = A1 + A2 + A3 [Step 1.]
34. A square and a rectangle have equal perimeters. If area of the square is 4 cm2, then find the perimeter of the rectangle. a.
13 cm
b.
8 cm
c.
18 cm
d.
4 cm
Solution:
Area of square = side2 = 4 cm2 [Formula.] Side = 4cm2 = 2 cm [Step 1 and simplify.] Perimeter of square = 4 ×2 cm = 8 cm [Perimeter of square = 4 × side.] Perimeter of rectangle = 8 cm [Perimeter of rectangle = perimeter of square.]
35. For a given perimeter, which of the following will have maximum area? a.
square
b.
rhombus
c.
rectangle
d.
circle
Solution: The most symmetrical figure will have the maximum area for a given perimeter. Circle will have the maximum area. [Circle is the most symmetrical.]
36. What is the length of the outer boundary of a semicircle of radius ? a.
π
r
+ 2r
b. 2r c.
r
+π
π
r
d.
Solution:
Required length = arc length AMB + BO + OA
r
Required length = πr + r + r = πr + 2r [Arc length AMB = circumference of the circle2.]
37. Find the area of a semi-circle of radius 4.6 cm. a.
21.16 sq.cm
b. c.
4.6π 10.58 sq.cm
sq.cm
d. 10.58π
sq.cm
Solution: Area of required semi-circle = π(4.6)22 [Area of semi-circle = π(radius)22.] Area of required semi-circle = 10.58π sq.cm. [Simplify.]
38. There are 5 concentric circles. Radius of the smallest circle is 5 cm. Increase in radius of each consecutive circle is 1 cm. What will be the area of a circle in sq. cm which has a circumference equal to the sum of the circumferences of the 5 concentric circles?
a. b.
1225π 1225 sq.cm
sq.cm
c. 70 π
sq.cm
35π
sq.cm
d.
Solution: Radii of the other 4 circles are, 6 cm, 7 cm, 8 cm and 9 cm. [Radius of each consecutive circle is more by 1 cm.] Sum of the circumferences of the circles = 2π × 5 + 2π × 6+ 2π × 7 + 2π × 8 + 2π × 9 [Circumference = 2πr.] = 2π × 35 = 70 π cm The radius of the circle having same circumference of 70π cm = 35 cm Area of the circle of 35 cm radius = π(35 × 35) = 1225π sq.cm
1. The product of π __________ of the circle. a.
area
b.
circumference
c.
radius
d.
none of these
and the diameter d
of a circle is the
Solution: Circumference of circle C = 2 × π × r = πd. [Substitute 2r = d.] The product of π and the diameter d of a circle is the circumference of the circle.
2. What is the area of the circle with a circumference of 13 ft? [Use π 3.14.] a.
28.45 sq. ft.
b.
13.45 sq. ft.
c.
21.45 sq. ft.
d.
18.45 sq. ft.
Solution:
=
Circumference of the circle, c = 13 ft. [Given.] Circumference of a circle = 2 × π × r. [Formula.] = 2 × 3.14 × r = 13 [Substitute.] ⇒ r = 13 / (2×3.14) [Divide each side by (2 × 3.14).] = 2.07 ft [Simplify.] Area of the circle = πr2 = 3.14 × 2.07 × 2.07 [Substitute r = 2.07.] = 13.45 sq.ft [Simplify.] The area of the circle is 13.45 sq. ft.
3. Jackson ran around a circular track. Which of the following is required to find the distance covered by him? a.
radius
b.
area
c.
circumference
d.
diameter
Solution: The distance covered by Jackson = Circumference of the circular track So, the circumference is required to find the distance covered by Jackson.
4. Estimate the diameter of a circle, if the circumference of the circle is 2.4 cm. a.
0.7 cm
b.
1.0 cm
c.
0.9 cm
d.
0.8 cm
Answer: (d) 5. Estimate the length of a rope by which a cow must be tethered so that it can graze an area of 192 yd. 2. a.
16 yd
b.
8 yd
c.
9 yd
d.
17 yd
Answer: (b) 6. Find the area of the circle to the nearest whole number.
a.
9 sq.cm
b.
28 sq.cm
c.
18 sq.cm
d.
108 sq.cm
Answer: (b) 7. A rectangular tin sheet, 14 in. long and 5 in. wide, is rolled along its length to form a cylinder by making the opposite edges meet. What is the base radius of the cylinder formed?
a.
14 in.
b.
2.23 in.
c.
11 in.
d.
10.2 in.
Solution: A rectangular tin sheet is rolled along the length to form a cylinder. Circumference of the base of the cylinder = length of the rectangular sheet = 14 in. Circumference of the base of the cylinder, 2πr = 14 in. [Since, the circumference of the circle = 2πr.] 14 = 2 × 3.14 × r [Substitute the values of π and the circumference.] 14 / 6.28 = 6.28r6.28 [Divide each side by 6.28.] 2.23 = r [Simplify.] The base radius of the cylinder = 2.23 in.
8. Find the circumference of a circle whose radius is 33 cm. [Use π = 3.14.] a.
103.62 cm
b.
414.48 cm
c.
217.24 cm
d.
207.24 cm
Solution: The radius (r) of the circle is 33 cm. [Given.] The circumference of the circle = 2 × π × r [Formula.] = 2 × 3.14 × 33 [Substitute π = 3.14 and r = 33 cm.] = 207.24 cm [Simplify.] The circumference of the circle is 207.24 cm.
9. Find the circumference of a basketball hoop if it has a diameter of 26 cm. [Use π = 3.14.] a.
81.64 cm
b.
29.14 cm
c.
163.28 cm
d.
8.28 cm
Solution: The circumference of a circle = π × diameter. [Formula.] = 3.14 × 26 [Substitute π = 3.14 and diameter = 26 cm.] = 81.64 cm. [Simplify.] The circumference of the basketball hoop is 81.64 cm.
10. What is the circumference of a circle with diameter 49 inches? [Use π = 3.14.] a.
160.86 inches
b.
163.86 inches
c.
153.86 inches
d.
307.72 inches
Solution: The diameter of the circle is 49 inches. [Given.] The circumference of a circle = π × diameter. [Formula.] The circumference of the circle = 3.14 × 49. [Substitute π = 3.14 and diameter = 49 inches.] = 153.86 inches
The circumference of the circle is 153.86 inches.
11. The diameter of a small pizza is 8 inches. What is its area? [Use π = 3.14.] a.
52.2 in.2
b.
48.2 in.2
c.
54.2 in.2
d.
50.2 in.2
Solution: The radius of the pizza = Diameter / 2 = 8 / 2 [Substitute d = 8.] = 4 inches Area of a circle = π r2. [Formula.] = π × 42 [Substitute the values.] = 3.14 × 16 [Substitute π = 3.14.] = 50.2 in.2
12. What is the area of a circular field of radius 1.9 miles? [Use π = 3.14.] a.
13.33 miles2
b.
9.33 miles2
c.
11.33 miles2
d.
14.73 miles2
Solution: The area of a circle = π r2 π × (1.9)2
= 3.14 × 3.61 [Substitute π = 3.14.] = 11.33 miles2.
13. The circumference of the bottom of a circular swimming pool is 55 ft. What is the area of the bottom of the pool? [Use π a.
240.40 ft2
b.
240.40 ft3
c.
2374.62 ft3
d.
9498.50 ft
= 3.14.]
Solution: Circumference of a circle = 2πr 2 x 3.14 x r = 55 [Substitute π = 3.14.] r = 552x3.14 [Divide each side by (2 x 3.14).] = 8.75 ft. Area of the bottom = πr2 = 3.14 x 8.75 x 8.75 [Substitute r = 8.75.] = 240.40 ft.2 The area of the bottom is about 240.40 ft 2.
14. What is the radius of the circle with circumference 92 cm.? [Use π = 3.14.] a.
15.64 cm
b.
13.64 cm
c.
17.64 cm
d.
14.64 cm
Solution: Circumference of the circle = 2 π r 92 = 2 × 3.14 × r [Substitute the values.] r = 922×3.14 [Divide each side by (2 × 3.14).] = 14.64 cm. [Simplify.] The radius of the circle is 14.64 cm.
15. What is the diameter of the circle whose circumference is196 inches? a.
65.62 inches
b.
62.42 inches
c.
64.42 inches
d.
60.42 inches
Solution: Circumference of the circle = π × d = 3.14 × d [Substitute π = 3.14.] Circumference of the circle = 196. [Given.] 3.14 × d = 196 d = 196 / 3.14 [Divide each side by 3.14.] = 62.42 [Simplify.] The diameter of the circle is 62.42 inches.
16. What is the radius of the circle in the figure?
a. 25
√
ft.
b.
3
√
ft.
c. 32
√
ft.
33
√
ft.
d.
Solution: From the figure, ABCD is a square of side 6 ft. and OD is the radius of the circle. The length of the diagonal BD = 2 × radius = 2r [Formula.] From the figure, AB2 + AD2 = BD2 [Since ABD forms a right triangle.] 62 + 62 = (2r)2 [Substitute the values.] 36 + 36 = 4r2 72 = 4r2 18 = r2 [Divide each side by 4.] 32 = r [Take the square root on both sides.] So, the radius of the circle, r = 32 ft.
17. What is the area of the shaded region in the figure? [Use π
= 3.14.]
a.
122.63 cm2
b.
176.63 cm2
c.
54 cm2
d.
117.63 cm2
Solution: From the figure, the base and height of the right triangle are 12 cm. and 9 cm. respectively. In a right triangle, Hypotenuse 2 = Base2 + Height2. [Pythagorean theorem.] AC2 = AB2 + BC2 AC2 = 122 + 92 AC2 = 144 + 81 = 225 AC = 225 AC = 15 [225 = 15 ] AC is the diameter of the circle. So, radius of the circle = AC / 2 = 15 / 2 = 7.5. Area of the shaded region = Area of the circle - Area of the triangle = 3.14 × 7.5 × 7.5 = 176.63 cm.2 Area of the circle = πr2 [Substitute r = 7.5 and π = 3.14 and simplify.]
Area of the triangle = 1 / 2 × base × height = 12 × 12 × 9 = = 54 cm.2 [Substitute base = 12 and height = 9 and simplify.] The area of the shaded region = 176.63 - 54 = 122.63 cm 2.
18. What is the area of the shaded region in the figure? [Use π
= 3.14.]
a.
100 cm2
b.
78.5 cm2
c.
21.5 cm2
d.
31.5 cm2
Solution: From the figure, the side of the square is 10 cm and the radius of the circle is 5 cm respectively. Area of the shaded region = Area of the square - Area of the circle Area of the square = a2 = 10 × 10 [Substitute a = 10.] = 100 cm.2 = 3.14 × 5 × 5 Area of the circle = πr2 [Substitute r = 5 and π = 3.14.] = 78.5 cm.2 The area of the shaded region = 100 - 78.5 = 21.5 cm 2.
19. A goat is tied to a pole with a 4 m rope. The goat can graze to the full length of the rope and 360o around the pole. How much area does the goat have to graze? [Use π a.
16 m2
b.
25.12 m2
c.
50.24 m2
d.
1440 m2
= 3.14.]
Solution: Area of the circular region where the goat can graze = πr 2 = 3.14 × 4 × 4 [Substitute r = 4 and π = 3.14.] = 50.24 m.2 The area the goat has to graze is 50.24 m2.
20. The diameter of a car wheel is 18 in. How far will the car go in 8 revolutions of the wheel? a.
144π
b. 144
c.
8π
d.
18π
π
Solution: The diameter of the car wheel = 18 in. Circumference of the circle (car wheel) = π d = 18π in [Substitute d = 18.] For one revolution the car travels 18π in. Distance traveled by the car in 8 revolutions = 18π × 8 = 144π in
After 8 revolutions, the car travels 144π in.
21. Each side of the square in the figure shown is 16 in. What is the area of the shaded region if the four circles inscribed in the square are congruent? [Use π
= 3.14.]
a.
200.96 in.2
b.
55.04 in.2
c.
51.04 in.2
d.
59.04 in.2
Solution: From the figure, the radius of each of the congruent circles is 4 in. Area of the shaded region = area of the square - sum of the areas of the four circles Area of the square = a2 = 16 × 16 [Substitute a = 16.] = 256 in.2 Area of one circle = πr2 = 3.14 × 4 × 4 [Substitute r = 4 and π = 3.14.] = 50.24 in.2 Area of four circles = 4 × 50.24 = 200.96 in. 2 The area of the shaded region = 256 - 200.96 = 55.04 in. 2
22. An aluminum tube, 50 in. long, is bent to make a circular hoop. Find the diameter of the hoop. a.
50π
b. 48
π
c. π
50
d. 50
π
Solution: Circumference of the circular hoop = length of the aluminum tube = 50 in. Circumference of the circular hoop = πd π × d = 50 d = 50π
23. The circumference of a circle is 2r circle, when the radius r a.
4
b.
6
c.
3
d.
5
is equal to 2.
Solution: Evaluate 2r for r = 2 2r = 2(2) [Substitute r for 2.] = 4 [Multiply.] The circumference of the circle is 4.
. Find the circumference of the
24. O is the center of the circle. The circle is divided into three equal parts and one of them is shaded. If the radius of the circle is 12 ft, then what is the area of the shaded sector?
a. 46π
ft2
50π
ft2
48π
ft2
49π
ft2
b. c. d.
Solution: Area of the circle = πr2 = π × 122 [Substitute r = 12.] = 144π [Simplify.] Area of the shaded sector = Revolution of the sector × Area of the circle = 120 / 360 × 144π [Substitute the values.] = 48π ft2. [Simplify.]
25. The central angle of a sector is 20o. If the area of the circle is 180 m.2, then what is the area of the sector? a.
10 m2
b.
20 m2
c.
12 m2
d.
15 m2
Solution: Revolution of the sector = 20°360° Area of the sector = Revolution of the sector × Area of the circle = 20°360° × 180 [Substitute the values.] = 10 m.2 [Simplify.]
26. A square is inscribed in a circle whose diameter is 12 cm as shown in the figure. Find the area of the shaded portion in the figure.
a.
41.04 cm2
b.
40 cm2
c.
50 cm2
d.
none of these
Solution: The diameter of a circle = 12 cm. The diameter of a circle divides the inscribed square into two congruent right triangles. Let S be the side of square. 2S2 = 144 S2 = 72 Applying Pythagorean theorem for the triangle = S 2 + S2 = 122 [Divide each side by 2.] Area of the square, S2 = 72. Area of the circle = πr2
= π × 62 = 113.04 cm2 [Substitute r = 6.] The area of the shaded portion = 113.04 - 72 = 41.04 cm. 2.
27. A wire 100 cm long is bent to form a circle. Find the area of the circle formed. a. 2500
cm2
(πUnknown node type: sup)
b. π
c.
2500
2500π cm
2
2500
π
cm2
d. cm2
Solution: Circumference of the circle = 2πr = 100 cm. 2πr = 100 πr = 1002 [Divide each side by 2.] r = 50 / π [Divide each side by π.] Area of the circle = πr2 = π × (50 / π)2 [Substitute r = 50 / π.] = 2500 / π cm2.
28. The diameter of the wheel of a truck is 100 cm. How many revolutions will it take to travel 600 cm.? a. 6/π b. 6π c. 7/π d. 5/π
Solution: Circumference of the wheel = πd π × 100 [Substitute d = 100.] 100π cm. The distance covered by the wheel in one revolution = Circumference of the circle 600/100π = 6/π Number of revolutions of the wheel while travelling 600 cm.
29. A circular ground of radius 4 m has a path of width 2 m around it as shown in the figure. What is the area of the path?
a. 10π
m2
20π
m2
b. c. d.
30π m2 none of these
Solution: Radius of the inner circle = 4 m. Radius of the outer circle = 4 + 2 = 6 m. Area of the inner circle = πr2 π × 42 [Substitute r = 4.] 16π m2 Area of the outer circle = πr2
π × 62 [Substitute r = 6.] 36π m2 The area of the path = 36π - 16π = 20 π m.2.
30. The circumference of two concentric circles is 130 m. and 90 m. What is the difference between their radii? a. 20π
m
20/π
m
b. c.
π
/20 m
d. 45/π
m
Solution: Let r1 and r2 be the radii of outer and inner circles. Circumference of the outer circle, 2πr 1 = 130 m. r1 = 65/π [Divide each side by 2π.] Circumference of the inner circle, 2πr 2 = 90 m. r2 = 45/π [Divide each side by 2π.] = 20/π m. The difference between their radii = 65/π - 45/π
31. If the diameter of a circle is increased by 80%, by what percent will its circumference increase? a.
85%
b.
80%
c.
70%
d.
90%
Solution: Let d be the diameter of the circle. The circumference of the circle = πd. The diameter of a circle is increased by 80%. The new diameter of the circle = (1 + 0.8) × d = 1.8d The new circumference of the circle = π × 1.8d = 1.8 π d The circumference of the circle is increased by 80%.
32. If the radius of the circle is decreased by 30%, by what percent will its circumference decrease? a.
30%
b.
65%
c.
55%
d.
60%
Solution: Let r be the radius of the circle. The radius of a circle is decreased by 30%. The new radius of the circle = (1 - 0.3) x r = 0.7r Circumference of the circle = 2πr The new circumference of the circle = 2 x π x 0.7r = 0.7 x 2πr = (1 - 0.3) x 2πr So, the circumference of the circle is decreased by 30%.
33. If the radius of a circle is increased by 30%, by what percent will its area increase? a.
64%
b.
69%
c.
30%
d.
74%
Solution: Let r be the radius of the circle. Area of the circle = πr2 Radius of the circle is increased by 30%. The radius of the new circle = (1+ 0.30) r = 1.30r The new area of the circle = πr2 = π x (1.30r)2 [Substitute r = 1.30r.] = 1.69(πr2) The area of the circle is increased by 69%.
34. If the area of a circle is decreased by 70%, by about what percent will its radius decrease? a.
41%
b.
70%
c.
46%
d.
51%
Solution: Let r be the radius of the circle. Area of the circle = πr2 The new area of the circle = (1 - 0.70)πr 2
= 0.30 x πr2 = π(0.54r)2 [Approximately √0.30 = 0.54.] The radius of the circle is decreased by 46%.
35. A circular cake is cut into equal pieces in such a way that each piece makes an angle of 45o at the center. Find the number of pieces. a.
10
b.
8
c.
6
d.
12
Solution: The complete angle at the center of a circle is 360 o. Number of pieces = complete angle / angle subtended by each piece at the center. = 360o/45o [Substitute the values.] The total number of pieces that are cut from the cake is 8.
