irregular shapes ws.doc

8/17/2019 Irregular shapes WS.doc

1/108

1. Estimate area occupied by the figure, if the area of each square is 1 cm2.

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


2/108

3. Suzanne wants to put a fence around the garden using wooden planks. hefigure shown represents the shape of her garden. !ow many feet of woodenplanks will she need to fence it"

a. 25 ft

b. 20 ft

c. 14 ft

d. 18 ft

#. $ind the perimeter of the polygon shown.

a. 50 cm

b. 64 cm

c. 36 cm

d. 46 cm

%. &dentify the perimeter of the regular polygon shown.

a. 42 yd


3/108

b. 14 yd

c. 49 yd2

d. 49 yd

'. he perimeter of the rectangle is 22 cm. $ind the (alue of w .

a. 6 cm

b. 4 cm

c. 3 cm

d. 5 cm

). $ind the perimeter of the square.

a. 24 cm2

b. 24 cm

c. 36 cm

d. 12 cm


4/108

*. $ind the perimeter of the rectangle shown.

a. 50 cm2

b. 30 cm

c. 15 cm

d. 15 cm2

+. $ind the perimeter of the triangle.

a. 26 m

b. 34 m

c. 48 m

d. 52 m

1. Estimate the area of the figure, if each square is of 1 sq.cm.


5/108

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 cm2.

a. about 17 cm2

b. about 29 cm2

c. about 8 cm2


6/108

d. about 12 cm2

-nswer /a0

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

-nswer /c0

3. Suzanne wants to put a fence around the garden using wooden planks. hefigure shown represents the shape of her garden. !ow many feet of woodenplanks will she need to fence it"


7/108

a. 25 ft

b. 20 ft

c. 14 ft

d. 18 ft

-nswer /b0

#. $ind the perimeter of the polygon shown.

a. 50 cm

b. 64 cm

c. 36 cm

d. 46 cm

-nswer /a0

%. &dentify the perimeter of the regular polygon shown.

a. 42 yd

b. 14 yd

c. 49 yd2

d. 49 yd

-nswer /d0


8/108

'. he perimeter of the rectangle is 22 cm. $ind the (alue of w .

a. 6 cm

b. 4 cm

c. 3 cm

d. 5 cm

-nswer /c0

). $ind the perimeter of the square.

a. 24 cm2

b. 24 cm

c. 36 cm

d. 12 cm

-nswer /b0


9/108

*. $ind the perimeter of the rectangle shown.

a. 50 cm2

b. 30 cm

c. 15 cm

d. 15 cm2

-nswer /b0

+. $ind the perimeter of the triangle.

a. 26 m

b. 34 m

c. 48 m

d. 52 m

-nswer /c0


10/108

1. 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.

-nswer /b0

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

-nswer /a0


11/108

1. hat is the area of a parallelogram, if its base is + ft and its height is 1# ft"a. 46 ft2

b. 126 ft2

c. 46 ft

d. 63 ft2

2. $ind the area of the colored region in the figure.

a. 22 cm2

b. 20 cm2

c. 28 cm2

d. 26 cm2

3. hat is the area of -4 in the figure, if the height of -4 is 3 times theheight of 54 and area of 54 is # inch2"

a. 115 inch2

b. 120 inch2

c. 110 inch2


12/108

d. None of the above

#. he perimeter of an equilateral triangle is 3' yd. hat is the length of each

side of the triangle"a. 108 yd

b. 12 yd

c. 3 yd

d. 9 yd

%. hich of the figures ha(e equal areas"

a. i!ure 1 " 3

b. i!ure 1 " 2

c. i!ure 2 and 3

d. i!ure 1#2 " 3

'. 6ohn spent 2 3 of an hour watching a mo(ie and 1 6

of an hour cleaning his room. hat fraction of an hour did he spendin both watching the mo(ie and cleaning his room"

a.

3 9

b.

1 3

c.

2 3

d.

5 6


13/108

). hich two parallelograms ha(e same area but different perimeters"

a. i!ure 2 and i!ure 3

b. i!ure 1 and i!ure 3

c. i!ure 1 and i!ure 2

d. i!ure 2 and i!ure 4

*. hich two parallelograms ha(e same area but different perimeters"






14/108

+. hich two parallelograms ha(e same area but different perimeters"





1. hich statement about the figures is true"

a. $oth the fi!ures have the same area.

b. $oth the fi!ures have the same %en!th.

c. $oth the fi!ures have the same &erimeter.

d. $oth the fi!ures have the same 'idth.


15/108

11. hich two parallelograms ha(e same area but different perimeters"





12. hich of the following is true for a right triangle and a rectangle ha(ingequal bases and equal heights"

a. (he &erimeter of the trian!%e is equa% to the &erimeter of the rectan!%e

b. (he area of the trian!%e is equa% to the area of the rectan!%e

c. (he area of the trian!%e is ha%f the area of the rectan!%e

d. (he area of the rectan!%e is ha%f the area of the trian!%e

13. - parallelogram and a rectangle ha(e equal bases and equal heights.hat is the area of the rectangle, if the area of the parallelogram is ## cm2"

a. 44 cm2

b. 11cm2

c. 22 cm2

d. 88 cm2

1#. he height of a parallelogram is twice its base. hat are the measures ofthe base and the height, if the area of the parallelogram is 2** m2"

a. 6 m and 12 m


16/108

b. 24 m and 48 m

c. 12 m and 24 m

d. 12 m and 36 m

1%. hat is the height of the parallelogram, if the base is half of its height andits area is 12* cm2"

a. 13 cm

b. 14 cm

c. 26 cm

d. 16 cm

1'. he state of 7irginia is shaped like a triangle. &f it has a base length of #'miles and co(ers an area of about #,+* square miles, what will be its height"

a. about 202 mi%es

b. about 406 mi%es

c. about 812 mi%es

d. about 101 mi%es

1). $ind the area of the gi(en parallelogram -45, if the area of the triangle-E is )* cm2.

a. 156 cm2


17/108

b. 78 cm2

c. 150 cm2

d. 39 cm2

1*. $ind the area of the triangle -E, if the area of the parallelogram -45 is1 cm2.

a. 80 cm2

b. 50 cm2

c. 30 cm2

d. 90 cm2

1+. $ind the area of rectangle 89:S, if the area of the triangle 89 is * in.2.

a. 140 in.2

b. 120 in.2

c. 160 in.2

d. 80 in.2


18/108

2. $ind the area of the parallelogram -45.

a. 13.5 cm2

b. 15 cm2

c. 15.5 cm

d. 32 cm

21. hat is the area of the shaded region in the figure"

a. 180 cm2

b. 120 cm2

c. 130 cm2


22. hat is the area of the triangle -5 in the figure, if the area of aparallelogram is 3 ft2"

a. 15 ft2

b. 20 ft2


19/108

c. 17 ft2


23. hat is the area of the parallelogram, if the area of the triangle is 22m2".

a. 47 m2

b. 44 m2

c. 54 m2


2#. hat is the height of the triangle, if the base of the triangle is twice theheight, and its area is 22% ft2"

a. 16 ft

b. 19 ft

c. 15 ft


2%. hat is the height of the triangle, if the triangle and the parallelogramha(e the same base and the same area"


20/108

a.

h ) 2* b.

* ) h c.

2h ) *d. None of the above

2'. he area of a parallelogram is half of the area of the triangle. hat is theheight of the triangle, if they ha(e the same base length of ' cm and the area of the parallelogram is 3+ cm2"

a. 26 cm

b. 21 cm

c. 31 cm


2). hat is the base length of the parallelogram, if the height is half of thebase and its area is 12* in.2"

a. 21 in

b. 18 in

c. 16 in



21/108

2*. hat is the area of -5 in the figure"

a. 48 in.2

b. 12 in.2

c. 24 in.2


2+. Split the trapezoid into a triangle and a parallelogram and find the area ofthe trapezoid.

a. 20 m.2

b. 30 m.2

c. 25 m.2

d. 15 m.2

3. $ind the perimeter of the figure.


22/108

a. 24 in.

b. 48 in.

c. 12 in.

d. 10 in.

31. $ind the perimeter of -4 in the figure.

a. 12 in.

b. 18 in.

c. 21 in.

d. 3 in.

32. $ind the perimeter of the parallelogram shown.

a. 10 in.

b. 20 in.

c. 5 in.

d. 16 in.


23/108

33. $ind the length of ;< in the gi(en triangle, if the perimeter is 3% ft.

a. 7 ft

b. 11 ft

c. 6 ft

d. 8 ft

3#. $ind the perimeter of the shaded region in the parallelogram -45.

a. 12 cm

b. 18 cm

c. 11 cm

d. 15 cm

3%. $ind the perimeter of the figure.


24/108

a. 12.5 in.

b. 14.7 in.

c. 17.7 in.

d. 15.7 in.

3'. hat 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

3). hat is area of the shaded region"

a. 105 in.2

b. 120 in.2

c. 95 in.2


1. hat is the area of a parallelogram, if its base is + ft and its height is 1# ft"


25/108

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.]

= 12

[!ultipl".]

So# area of the parallelogram is 12 ft2.

2. $ind 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 $olored region is a parallelogram %&'F.

(he base length of the parallelogram %&'F = ) $m[From the figure.]

(he height of the parallelogram %&'F = 4 $m[From the figure.]


26/108

Area of the parallelogram %&'F = base * height[Formula.]

= ) * 4[Substitute the values.]

= 2+

So# area of the $olored region = area of the parallelogram %&'F = 2+ $m 2.

3. hat is the area of -4 in the figure, if the height of -4 is 3 times theheight of 54 and area of 54 is # inch2"

a. 115 inch2

b. 120 inch2

c. 110 inch2


Solution

Area of a triangle = 1 , 2 * base * height[Formula.]

-et h be the height of '%/

Area of '%/ = 1 , 2 * %/ * h.

0eight of A%/ = times the height of '%/ = h.

Area of A%/ = 1 , 2 * %/ * h

= * 12 * %/ * h3

= * Area of '%/3


27/108

Area of A%/ = * 4 = 12 in$h2.

#. he perimeter of an equilateral triangle is 3' yd. hat is the length of eachside of the triangle"

a. 108 yd

b. 12 yd

c. 3 yd

d. 9 yd

Solution

5n an e6uilateral triangle# all sides are e6ual.

So# the perimeter of an e6uilateral triangle = × measure of ea$h side.

= × measure of ea$h side[Substitute the values.]

, = × measure of ea$h side['ivide ea$h side b" .]

12 = measure of ea$h side[Simplif".]

So# the measure of ea$h side of the e6uilateral triangle is 12 "d.

%. hich of the figures ha(e equal areas"

a. i!ure 1 " 3

b. i!ure 1 " 2

c. i!ure 2 and 3

d. i!ure 1#2 " 3


28/108

Solution

Figure 1 represents a parallelogram 7ith its base length 4 units and height 2 units.

Area of the parallelogram = base × height = 4 × 2 = + s6uare units.

Figure 2 represents a re$tangle of length 4 units and 7idth 2 units.

Area of the re$tangle = length × 7idth = 4 × 2 = + s6uare units.

Figure represents a triangle 7ith a base length of units and a height of units.

Area of the triangle = 1 , 2 × base × height = 1 , 2 × × = 9 s6uare units.

So# the parallelogram in figure 1 and the re$tangle in figure 2 have e6ual areas.

'. 6ohn spent 2 3 of an hour watching a mo(ie and 1 6

of an hour cleaning his room. hat fraction of an hour did he spendin both watching the mo(ie and cleaning his room"

a.

3 9

b.

1 3

c.

2 3

d.

5 6

-nswer /d0


29/108

). hich two parallelograms ha(e same area but different perimeters"





Solution

Area of a parallelogram = %ase × 0eight 8erimeter of a parallelogram = 2 × %ase 0eight3

Area = : × 4 = 2 s6.units 8erimeter = 2 × : 43 = 1+ units


30/108

Area = 4 × 4 = 1 s6.units 8erimeter = 2 × 4 43 = 1 units

Area = + × 2 = 1 s6.units 8erimeter = 2 × + 23 = 2 units

Area = 4 × = 12 s6.units 8erimeter = 2 × 4 3 = 14 units

(herefore# parallelograms in Figure 2 and Figure have same area but differentperimeters.


31/108

*. hich two parallelograms ha(e same area but different perimeters"





Solution


Area = : × 4 = 2 s6.units 8erimeter = 2 × : 43 = 1+ units


32/108

Area = × = 1+ s6.units 8erimeter = 2 × 3 = 1+ units

Area = × 2 = 12 s6.units 8erimeter = 2 × 23 = 1 units


(herefore# parallelograms in Figure and Figure 4 have same area but differentperimeters.


33/108

+. hich two parallelograms ha(e same area but different perimeters"





Solution


Area = : × ) = : s6.units 8erimeter = 2 × : )3 = 24 units

Area = × = s6.units 8erimeter = 2 × 3 = 24 units


34/108

Area = 9 × 4 = s6.units 8erimeter = 2 × 9 43 = 2 units

Area = + × 4 = 2 s6.units 8erimeter = 2 × + 43 = 24 units

(herefore# parallelograms in Figure 2 and Figure have same area but differentperimeters.

1. hich statement about the figures is true"

a. $oth the fi!ures have the same area.

b. $oth the fi!ures have the same %en!th.

c. $oth the fi!ures have the same &erimeter.

d. $oth the fi!ures have the same 'idth.

Solution

8arallelogram in Figure 1 has a base of units and a height of 2 units.

8arallelogram in Figure 2 has a base of 4 units and a height of units.


35/108


Area = × 2 = 12 s6.units 8erimeter = 2 × 23 = 1 units


(herefore# the statement ;%oth the figures have the same area; is true.


36/108

11. hich two parallelograms ha(e same area but different perimeters"





Solution

Area of a parallelogram = %ase × 0eight# 8erimeter of a parallelogram = 2 × %ase 0eight3

Area = ) × 4 = 2+ s6.units# 8erimeter = 2 × ) 43 = 22 units


37/108

Area = 4 × : = 2 s6.units# 8erimeter = 2 × 4 :3 = 1+ units

Area = × + = 24 s6.units# 8erimeter = 2 × +3 = 22 units

Area = × 4 = 24 s6.units# 8erimeter = 2 × 43 = 2 units

(herefore# parallelograms in Figure and Figure 4 have same area but differentperimeters.

12. hich of the following is true for a right triangle and a rectangle ha(ingequal bases and equal heights"

a. (he &erimeter of the trian!%e is equa% to the &erimeter of the rectan!%e

b. (he area of the trian!%e is equa% to the area of the rectan!%e

c. (he area of the trian!%e is ha%f the area of the rectan!%e

d. (he area of the rectan!%e is ha%f the area of the trian!%e

Solution

Area of the triangle = 12 × b × h

= 12 × Area of the re$tangle


38/108

So# the area of the triangle is half the area of the re$tangle.

-et the length of the $ommon base be b

-et the length of the $ommon height be h

Area of the re$tangle = b × h

13. - parallelogram and a rectangle ha(e equal bases and equal heights.hat is the area of the rectangle, if the area of the parallelogram is ## cm2"

a. 44 cm2

b. 11cm2

c. 22 cm2

d. 88 cm2

Solution

5f the base and height of the parallelogram and the re$tangle are same# then thearea of the parallelogram is same as that of re$tangle.

So# the area of the re$tangle = the area of the parallelogram = 44 $m 2.

1#. he height of a parallelogram is twice its base. hat are the measures ofthe base and the height, if the area of the parallelogram is 2** m2"

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

-et b be the base of the parallelogram.

0eight of the parallelogram = (7i$e that of the base = 2b

Area of the parallelogram = base × height[Formula.]

2++ = b × 2b[Substitute the values.]


39/108

144 = b2

144 = b2# b = 12[(a


40/108

1'. he state of 7irginia is shaped like a triangle. &f it has a base length of #'miles and co(ers an area of about #,+* square miles, what will be its height"

a. about 202 mi%es

b. about 406 mi%es

c. about 812 mi%es

d. about 101 mi%es

Solution

Area of a triangle = 1 , 2 × base × height[Formula.]

Area $overed b" the state of irginia = 1 , 2 × 4 × h# 7here h is the height of thetriangle.

1 , 2 × 4 × h = 4#9+ h > 22⇒

So# the height of the triangle is about 22 miles.

1). $ind the area of the gi(en parallelogram -45, if the area of the triangle-E is )* cm2.

a. 156 cm2

b. 78 cm2

c. 150 cm2

d. 39 cm2


41/108

Solution

Area of a parallelogram = base × height Area of a triangle = 1 , 2 × base × height

5f the base and height of the parallelogram and the triangle are same then the areaof the parallelogram is t7i$e the area of the triangle.

(hus the area of the parallelogram A%/' = 2 × area of the triangle A%&

= 2 × )+ = 1:[Substitute the values.]

So# the area of the parallelogram A%/' is 1: $m2.

1*. $ind the area of the triangle -E, if the area of the parallelogram -45 is

1 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

5f the base and height of the parallelogram and the triangle are same then the areaof the triangle is half of the area of the parallelogram.

(hus the area of the triangle A%& = 1 , 2 × area of the parallelogram A%/'

= 12 × 1 = :[Substitute the values.]

So# the area of the triangle A%& is : $m2.


42/108

1+. $ind the area of rectangle 89:S, if the area of the triangle 89 is * in.2.

a. 140 in.2

b. 120 in.2

c. 160 in.2

d. 80 in.2

Solution

Area of a re$tangle = length × 7idth Area of a triangle = 1 , 2 × base × height

5f the base and height of the re$tangle and the triangle are same then the area of there$tangle is t7i$e the area of the triangle.

(hus the area of the re$tangle 8?@S = 2 × area of the triangle 8?(

= 2 × + = 1

[Substitute the values.]

So# the area of the re$tangle 8?@S is 1 in.2.

2. $ind the area of the parallelogram -45.

a. 13.5 cm2

b. 15 cm2

c. 15.5 cm

d. 32 cm


43/108

Solution

-ength of the base = $m.

[iven.] Area of the parallelogram = base length × height.[Formula]

= $m × 2.: $m[Substitute the respe$tive values in the formula.]

= 1: $m2.

So# the area of the parallelogram A%/' is 1: $m2.

21. hat is the area of the shaded region in the figure"

a. 180 cm2

b. 120 cm2

c. 130 cm2


Solution

From the figure# shaded region = A/'.

A/' is in triangular shape.

Area of the triangle A/' = 1 , 2 × base × height[Formula.]

From the figure base of the triangle A/' = 1: $m and height = 1 $m.


44/108

(he area of the triangle A/' = 1 , 2 × 1: × 1[Substitute the values.]

12 $m2 [Simplif".]

(he area of the triangle A/' = 12 $m2.

22. hat is the area of the triangle -5 in the figure, if the area of aparallelogram is 3 ft2"

a. 15 ft2

b. 20 ft2

c. 17 ft2


Solution

(he area of the triangle = 1 , 2 × area of the parallelogram

= 12× [Sin$e area of the parallelogram = .]

= 1:[Simplif".]

5f 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.(he area of the triangle = 1: ft2.


45/108

23. hat is the area of the parallelogram, if the area of the triangle is 22m2".

a. 47 m2

b. 44 m2

c. 54 m2


Solution

From the figure# the triangle and the parallelogram have same bases and the sameheights.

(he area of the parallelogram = 2 × area of the triangle.

= 2 × 22[Substitute the area of triangle = 22.]

= 44

(he area of the parallelogram = 44 m2.

2#. hat is the height of the triangle, if the base of the triangle is twice theheight, and its area is 22% ft2"

a. 16 ft

b. 19 ft

c. 15 ft


Solution

-et h is the height of the triangle.

iven# area of the triangle = 22: ft2


46/108

(he base = t7i$e the height = 2 × h.

Area of a triangle = 1 , 2 × base × height[Formula.]

22: = 1 , 2 × 2 × h × h[Substitute the values.]

22: = h2 [Simplif".]

22: = h2[(a


47/108

Area of the triangle = 1 , 2× base × height

Area of the parallelogram = base × height

iven# area of the triangle = area of the parallelogram

1 , 2 × base × height = base × height[Substitute the area formulae.]

1 , 2 × height = height[%ases are e6ual.]

1 , 2 × h = 0[Substitute the height values.]

h = 20

2'. he area of a parallelogram is half of the area of the triangle. hat is theheight of the triangle, if they ha(e the same base length of ' cm and the area of the parallelogram is 3+ cm2"

a. 26 cm

b. 21 cm

c. 31 cm


Solution

(he area of the parallelogram = 9 $m2

(he area of the parallelogram = 1 , 2 * area of the triangle[iven.]

9 = 1 , 2 * area of the triangle[Substitute the area of parallelogram.]

)+ = area of the triangle

[!ultipl" ea$h side b" 2.] Area of the triangle = )+ $m2

%ase length of the parallelogram = base length of the triangle = $m[iven.]

Area of the triangle = 1 , 2 * base * height[Formula.]


48/108

)+ = 1 , 2 * * height[Substitute the values.]

)+ = * height[!ultipl" 7ith 1 , 2.]

)+ , = *height , ['ivide ea$h side b" .]

0eight of the triangle = 2 $m[Simplif".]

2). hat is the base length of the parallelogram, if the height is half of thebase and its area is 12* in.2"

a. 21 in

b. 18 in

c. 16 in


Solution

-et b be the base length of the parallelogram.

0eight of the parallelogram = half of the base = b2

Area of the parallelogram = base × height[Formula.]

12+ = b × b2[Substitute the values.]

12+ × 2 = b22 × 2[!ultipl" ea$h side b" 2.]

2: = b2

[Simplif".]2: = b2[(a


49/108

2*. hat is the area of -5 in the figure"

a. 48 in.2

b. 12 in.2

c. 24 in.2


Solution

A'% is a triangle.

%ase of the triangle A'% = in. and the height of the triangle A'% = 4 in.

Area of the triangle A'% = 1 , 2 × base × height[Formula.]

1 , 2 × × 4[Substitute the values.]

= 12[Simplif".]

(he area of A'% = 12 in.2

2+. Split the trapezoid into a triangle and a parallelogram and find the area ofthe trapezoid.

a. 20 m.2

b. 30 m.2


50/108

c. 25 m.2

d. 15 m.2

Solution

'ra7 a line parallel to /% from the point '.

-et the line tou$h A% on the point &.

Bo7 the trapeCoid is split into a parallelogram '&%/ and a A'&.

%ase length of the parallelogram = %& = '/ = m.# height = 4 m.

Area of a parallelogram '&%/ = base × height = × 4 = 24 m.2

%ase length of the A'& = A& = A% D '/ = 9 D = m.

0eight of the A'& = 4 m.

Area of the A'& = 1 , 2 × base × length = 1 , 2 × × 4 = m.2

Area of the trapeCoid A%/' = Area of a parallelogram '&%/ Area of the A'&.

Area of the trapeCoid A%/' = 24 = m.2


51/108

3. $ind the perimeter of the figure.

a. 24 in.

b. 48 in.

c. 12 in.

d. 10 in.

Solution

8erimeter of a figure = sum of all the sides of the figure.

From the figure A% = + in.# %/ = 1 in.# A/ = in.

8erimeter of A%/ = A% %/ /A

= + 1 [Substitute the values.]

= 24

So# the perimeter of the figure is 24 in.

31. $ind the perimeter of -4 in the figure.

a. 12 in.

b. 18 in.


52/108

c. 21 in.

d. 3 in.

Solution

8erimeter of a triangle is the sum of all the sides of the triangle.

From the figure# A% = in.# %/ = 1 in. and A/ = + in.

8erimeter of A%/ = A% %/ /A

= 1 +[Substitute the values.]

= 21

[Add.]

(he perimeter of the A%/ is 21 in.

32. $ind the perimeter of the parallelogram shown.

a. 10 in.

b. 20 in.

c. 5 in.

d. 16 in.

Solution

8erimeter of a parallelogram = sum of all the sides of the parallelogram

From the figure# A% = : in. and A' = in.

As the lengths of the parallel sides are e6ual# A% = /' = : in.# A' = %/ = in.


53/108

8erimeter of the parallelogram = A% %/ /' 'A

= : : [Substitute the lengths.]

= 1[Add.]

So# the perimeter of the parallelogram given in the figure is 1 in.

33. $ind the length of ;< in the gi(en triangle, if the perimeter is 3% ft.

a. 7 ft

b. 11 ft

c. 6 ft

d. 8 ft

Solution

From the figure# !B = 1: ft# E! = 1 ft

8erimeter of a triangle = sum of all sides[Formula.]

8erimeter of the triangle !BE = !B BE E![Formula.]

: = 1: BE 1[Substitute the values.]

: = 2+ BE[Add 1: and 1.]

: D 2+ = 2+ D2+ BE[Subtra$t 2+ from ea$h side.]

) = BE[Simplif".]


54/108

(he length of BE of the triangle is ) ft.

3#. $ind the perimeter of the shaded region in the parallelogram -45.

a. 12 cm

b. 18 cm

c. 11 cm

d. 15 cm

Solution

From the figure# %&/ is the shaded area in the parallelogram A%/'.

From the figure# %/ = : $m# %& = $m# A% = ) $m# and '& = $m.

&/ = '/ D '& = A% D '& = ) D = 4 $m.[-engths of the parallel sides in a parallelogram are e6ual# A% = '/.]

8erimeter of the %&/ = %/ /& &%[Formula.]

= : 4 Substitute the values3

= 12[Add.]

So# the perimeter of the shaded region of the figure is 12 $m.


55/108

3%. $ind 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# %A = : in.# A/ = :.2 in.# /' = 2.: in.# %/ = in.# and '% = 2 in.

8erimeter of a figure = sum of all the sides of the figure[Formula.]

8erimeter of the figure = A/ /' '% %A

= :.2 2.: 2 :[Substitute the values.]

= 14 .)[Add.]

So# the perimeter of the figure is 14.) in.


56/108

3'. hat 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 = units.

0eight of the triangle = units.

Area of a triangle 7ith base b and height h = 1 , 2 × b × h

= 12 × × [Substitute the values.]

= 4.:[Simplif".]

So# the area of the triangle is 4.: s6uare units.

3). hat is area of the shaded region"

a. 105 in.2

b. 120 in.2

c. 95 in.2


57/108


Solution

Area of the shaded region is e6ual to the area of the triangle 7ith measures of 1: in.height and 14 in. as base.

Area of the triangle = 1 , 2 * base * height[Formula.]

= 12 * 14 * 1:[Substitute the values.]

= 1: in.2

[Simplif".]

Area of the shaded region in the figure = 1: in.2

1. $ind the perimeter of the figure shown below. =>i(en a ? #% and b

? 2).@

a. 144 cm

b. 90 cm

c. 99 cm

d. 63 cm


58/108

2. - steel wire in the form of a circular ring of radius # m is straightened and

cut into 2 equal pieces. $ind the length of each piece. =ake π ? 3.@a. 22 m

b. 12 m

c. 24 m

d. 4 m

3. $ind the area of the shaded portion of the figure if all the angles in thefigure are right angles.

a. 22 m2

b. 28 m

c. 20 m

d. 27 m2


59/108

#. $ind the area of the shaded portion of the figure if all the angles in thefigure are right angles.

a. 18 yd2

b. 24 yd2

c. 28 yd

d. 16 yd2

%. >raph the points and find the area and perimeter.-/1, 10, /1, %0, 4/', #0, 5/', 10

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

'. he perimeter of a rectangle is 2# cm and the base is # cm. hat is thearea"

a. 24 cm2

b. 32 cm2

c. 8 cm2

d. 16 cm2


60/108

). Estimate the perimeter of the co(er of the book.

a. 98 cm

b. 238 cm

c. 190 cm

d. 138 cm

*. Aary wants to buy a curtain for her door. $ind the area of the cloth requiredif the height and width of the door are ) ft and 3.% ft.

a. about 18.5 ft2

b. about 24.5 ft2

c. about 19 ft2

d. about 21 ft2

+. he length of a rectangle is increased by %B and the width is decreasedby 2%B. !ow is its area affected"

a. increases by 12.5+

b. increases by 50+

c. unchan!ed

d. decreases by 25+

1. $ind the area and the perimeter of the rectangle formed by the points

8/1,0, 9/C 2, 0, :/C 2, #0, S/x , y 0.a. 12 sq.units and 7 units

b. 12 sq.units and 14 units



61/108

d. data insufficient

11. hat is the perimeter of the figure shown"

a. 43 square units

b. 43 units

c. 96 square units

d. 48 units

12. $ind the length of the wire required to fence the park which is in theshape of a square. he wire is used for two rounds around the park.

a. 640 yd

b. 3200 yd2

c. 6400 yd2

d. 320 yd


62/108

13. $ind the perimeter of the regular polygon.

a. 63 yd

b. 72 yd

c. 64 yd

d. 54 yd

1#. $ind the perimeter of the regular polygon.

a. 121 ft

b. 110 ft

c. 22 ft

d. 55 ft

1%. $ind the perimeter of the regular polygon

a. 64 in.

b. 36 in.

c. 54 in.


63/108

d. 48 in.

1'. he perimeter of a regular pentagon is 1 cm. !ow long is each side"

a. 500 cm

b. 250 cm

c. 20 cm

d. 10 cm

1). he perimeter of a regular heDagon is 2# m. !ow long is each side"a. 144 m

b. 6 m

c. 4 m

d. 12 m

1*. hat is the perimeter of polygon -45E"=>i(en - ? # cm, 4 ? % cm, 45 ? % cm, E5 ? 11 cm and E- ? * cm@

a. 33 cm

b. 55 cm

c. 25 cm

d. 15 cm


64/108

1+. hat is the perimeter of rectangle 89:S"=>i(en a ? #.# cm and b

? ).# cm@

a. 11.8 cm

b. 32.5 cm

c. 23.6 cm

d. 33.6 cm

2. hich of the following is the perimeter of a sqaure"a.

2 √ dia!ona%

b.

22 √ dia!ona%

c.

42 √ dia!ona%

d. 2 dia!ona%

21. $ind the side of a square, if its perimeter is # cm.a. 10 cm

b. 14 cm

c. 5 cm

d. 20 cm

22. hat is the perimeter of a regular polygon ha(ing 1 sides each of length2 cm"

a. 20 cm

b. 25 cm

c. 30 cm

d. 4 cm


65/108

23. - steel wire in the form of a circular ring of radius 3 m is straightened andcut into # equal pieces. $ind the length of each piece.

a. 18.84 m

b. 4.71 m

c. 3 m

d. 14.71 m

2#. hat is the ratio of the circumference of a circle to its diameter"a.

e b.

22 7

c. 3.14

d.

π

2%. $ind the perimeter of the figure shown. ake π ? 3. =>i(en a

? 3, b ? 12, r ? '.@

a. 186 units

b. 48 units

c. 168 units

d. 120 units

2'. &f l and b are the length and the width of a rectangle, thenwhat is its perimeter"


66/108

a.

2,l - b b.

2l - b c.

2b - l d.

l - b

2). $ind the perimeter of the square A;F, if s ? 33 √.

a.

63 √ cm

b.

3 3 √ cm

c. 12 cm

d.

123 √ cm

2*. $ind the circumference of a circle of radius % cm.a. 10 cm

b.

5π cmc.

10π cmd.

20 π cm

2+. hat is the area of the rectangle -45" =>i(en a ? 17 8

and b ? 9 4 @


67/108

a.

289 64 sq. cm

b.

81 16 sq. cm

c.

153 32 sq. cm

d.

32 153 sq. cm

3. $ind 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

orksheets in this topic

• 8erimeter /ir$umferen$e and Area 8artD1

:elated orksheets

• 8erimeter /ir$umferen$e and Area Gor


68/108

a. 50 cm2

b. 25 cm2

c.

25π cm2

d.

50π cm2

32. he perimeter of a square is 13.2 cm. hat is its area"a. 13.20 cm2

b. 10.89 cm2

c. 14.89 cm2

d. 26.4 cm2

33. &f -1, -2, -3 represent the areas of respecti(e regions, then what is thearea of polygon 89:S"

a. /1 - /2 - /3

b. /1/2/3

c.

A 1 2 +A 2 2 +A 3 2 −

− − − − − − − − − − − −

√ d. None of the above

3#. - square and a rectangle ha(e equal perimeters. &f area of the square is #cm2, then find the perimeter of the rectangle.

a. 13 cm


69/108

b. 8 cm

c. 18 cm

d. 4 cm

3%. $or a gi(en perimeter, which of the following will ha(e maDimum area"a. square

b. rhombus

c. rectan!%e

d. circ%e

3'. hat is the length of the outer boundary of a semicircle of radius r "

a.

π r - 2r b.

2r c.

r - π d.

π r

3). $ind the area of a semiCcircle of radius #.' cm.a. 21.16 sq.cm

b.

4.6π sq.cmc. 10.58 sq.cm

d.

10.58π sq.cm

3*. here are % concentric circles. :adius of the smallest circle is % cm.&ncrease in radius of each consecuti(e circle is 1 cm. hat will be the area of acircle in sq. cm which has a circumference equal to the sum of thecircumferences of the % concentric circles"


70/108

a.

1225π sq.cm b. 1225 sq.cm

c.

70 π sq.cm

d.

35π sq.cm

1. $ind the perimeter of the figure shown below. =>i(en a ? #% and b

? 2).@

a. 144 cm

b. 90 cm

c. 99 cm


71/108

d. 63 cm

Solution

8erimeter of the figure = 4: :p 6[From the figure.]

:p = 4: $m p = 4:: = 9 $m⇒['ivide ea$h side b" :.]

6 = 2) $m 6 = 9 $m⇒['ivide ea$h side b" .]

8erimeter of the figure = 4: : × 9 × 9 = 144 $m[From steps and 4.]

(herefore the perimeter of the figure is 144 $m.

2. - steel wire in the form of a circular ring of radius # m is straightened and

cut into 2 equal pieces. $ind the length of each piece. =ake π ? 3.@a. 22 m

b. 12 m

c. 24 m

d. 4 m

Solution

-ength of the steel 7ire = $ir$umferen$e of the ring = 2 Ir [Formula.]

= 2I × 4


72/108

[Substitute.]

= 24 m[Simplif".]

(otal length of the rods = 24 m[Step 1.]

-ength of one rod = 242[(here are 2 rods of e6ual length.]

= 12 m.

3. $ind the area of the shaded portion of the figure if all the angles in thefigure are right angles.

a. 22 m2

b. 28 m

c. 20 m

d. 27 m2

Solution

Area of the re$tangle 1 = l × b = : m × 2 m = 1 m2


73/108

'ivide the shaded portion into parts.

Area of the shaded portion = area of the s6uare area of the re$tangle 1 area ofthe re$tangle 2

Area of the s6uare = s2

= 22

= 4 m2

Area of the re$tangle 2 = l × b = 4 m × 2 m = + m2

So# area of the shaded portion = 4 m2 1 m2 + m2 = 22 m2

#. $ind the area of the shaded portion of the figure if all the angles in thefigure are right angles.

a. 18 yd2

b. 24 yd2

c. 28 yd

d. 16 yd2

Solution


74/108

Area of the shaded portion = area of re$tangle D 4area of re$tangle 7ith sides 4 "dand "d3

Area of re$tangle = l × b = + × = 4+ "d2

[length = 2 and breadth = 2 2 2.]

Area of re$tangle 7ith sides "d and 2 "d = × 2 = "d2

Area of the shaded portion = 4+ D 43 = 24 "d2

[Substitute in step 1 and simplif".]

So# the area of the shaded portion is 24 "d2.

%. >raph the points and find the area and perimeter.-/1, 10, /1, %0, 4/', #0, 5/', 10

a. 20 sq.units and 9 units

b. 18 units and 20 units


d. 20 sq.units and 18 units

Solution

raph the points on the $oordinate plane.

(he figure obtained b" Joining the points is a re$tangle 7ith length : units andbreadth 4 units.

Area of the re$tangle = l × b = : × 4 = 2 s6.units

8erimeter of the re$tangle = 2l b3 = 2: 43 = 1+ units

So# the area and perimeter of the figure are 2 s6.units and 1+ units.

'. he perimeter of a rectangle is 2# cm and the base is # cm. hat is thearea"

a. 24 cm2

b. 32 cm2

c. 8 cm2

d. 16 cm2


75/108

Solution

(he perimeter of a re$tangle = 2l b3[Formula.]

2l b3 = 24 $m[iven# the perimeter of a re$tangle is 24 $m.]

2l 243 = 24 $m[iven# base is 1 $m.]

2l = 24 D + = 1 $m[Simplif".]

l = 1 , 2 = + $m[Simplif".]

Area of the re$tangle = l × b = + $m × 4 $m = 2 $m2

). Estimate the perimeter of the co(er of the book.

a. 98 cm

b. 238 cm

c. 190 cm

d. 138 cm

Solution

(he perimeter of the $over of the boo< = 22:3 423 243

= : + + = 1+

So# the perimeter of the $over of the boo< is 1+ $m.


76/108

*. Aary wants to buy a curtain for her door. $ind the area of the cloth requiredif the height and width of the door are ) ft and 3.% ft.

a. about 18.5 ft2

b. about 24.5 ft2

c. about 19 ft2

d. about 21 ft2

Solution

Area of a re$tangle = l × b

Area of the door = ) ft × .: ft = 24.: ft2

Area of the $urtain = Area of the door = 24.: ft2

So# !ar" needs about 24.: ft2 of $loth for her door.

+. he length of a rectangle is increased by %B and the width is decreasedby 2%B. !ow is its area affected"

a. increases by 12.5+

b. increases by 50+

c. unchan!ed

d. decreases by 25+

Solution

Area of a re$tangle = l × b[Formula.]

5f length is in$reased b" :K# then the ne7 length is l : , 1l

= l 12l = 2l

5f 7idth is de$reased b" 2:K# then the ne7 7idth is b D 2: , 1b

= b D 14b = 4b

Be7 area of a re$tangle = , 2l × , 4b

= 9+lb3 = 1.12: lb3[Simplif".]


77/108

5n$rease in area = 1.12: lb3 D lb3 = .12: lb3

8er$entage in$rease in area = .12: lb3lb × 1 = 12.:K

So# the area is in$reased b" 12.:K.

1. $ind the area and the perimeter of the rectangle formed by the points

8/1,0, 9/C 2, 0, :/C 2, #0, S/x , y 0.a. 12 sq.units and 7 units

b. 12 sq.units and 14 units


d. data insufficient

Solution

raph the points on the $oordinate plane.

As 8?@S is a re$tangle# the possible $oordinates for point @ are 1# 43.

Area of the re$tangle = l × b = 4 × = 12 s6.units

8erimeter of the re$tangle = 2l b3 = 24 3 = 14 units

So# the area and perimeter of the retangle are 12 s6.units and 14 units.


78/108

11. hat is the perimeter of the figure shown"

a. 43 square units

b. 43 units

c. 96 square units

d. 48 units

Solution

8erimeter of the figure = 4 4 + : 2 2

= 4+ units[Simplif".]

12. $ind the length of the wire required to fence the park which is in theshape of a square. he wire is used for two rounds around the park.

a. 640 yd

b. 3200 yd2

c. 6400 yd2

d. 320 yd


79/108

Solution

-ength of the 7ire re6uired to fen$e the par< = 8erimeter of the par<

8erimeter of the par< = 4 × side3 = 4+3 = 2 "d[8ar< is in s6uare shape.]

So# the length of the 7ire re6uired to fen$e the par< is 2 "d.

13. $ind the perimeter of the regular polygon.

a. 63 yd

b. 72 yd

c. 64 yd

d. 54 yd

Solution

(he regular pol"gon has sides# ea$h 7ith a length of 9 "d.

8erimeter of the pol"gon = number of sides × side length = × 9 "d = :4 "d.

1#. $ind the perimeter of the regular polygon.

a. 121 ft

b. 110 ft


80/108

c. 22 ft

d. 55 ft

Solution

(his regular pol"gon has : sides# ea$h 7ith a length of 11 ft.

8erimeter of the pol"gon = number of sides × side length = : × 11 ft = :: ft.

1%. $ind the perimeter of the regular polygon

a. 64 in.

b. 36 in.

c. 54 in.

d. 48 in.

Solution

(he regular pol"gon has + sides# ea$h 7ith a length of in.

8erimeter of the pol"gon = number of sides × side length = + × in. = 4+ in.

1'. he perimeter of a regular pentagon is 1 cm. !ow long is each side"a. 500 cm

b. 250 cm

c. 20 cm

d. 10 cm

Solution

8erimeter of the pol"gon = number of sides × side length = 1 $m


81/108

Bumber of sides in regular pol"gons = :

Side length of the regular pentagon = perimeternumber of sides = 1 $m: = 2 $m

1). he perimeter of a regular heDagon is 2# m. !ow long is each side"a. 144 m

b. 6 m

c. 4 m

d. 12 m

Solution

8erimeter of the pol"gon = number of sides × side length = 24 m

Bumber of sides in regular he*agon =

Side length of the regular he*agon = perimeternumber of sides = 24 m = 4 m

1*. hat is the perimeter of polygon -45E"=>i(en - ? # cm, 4 ? % cm, 45 ? % cm, E5 ? 11 cm and E- ? * cm@

a. 33 cm

b. 55 cm

c. 25 cm

d. 15 cm

Solution

8erimeter of A%/'& = A% %/ /' '& &A['efinition.]


82/108

8erimeter = 4 : : 11 +[Substitute.]

8erimeter of A%/'& = $m.[ Add ]

1+. hat is the perimeter of rectangle 89:S"=>i(en a ? #.# cm and b

? ).# cm@

a. 11.8 cm

b. 32.5 cm

c. 23.6 cm

d. 33.6 cm

Solution

8erimeter of re$tangle = 2-ength %readth3[Formula.]

8erimeter of 8?@S = 24.4 ).43[Substitute.]

8erimeter of 8?@S = 2. $m[Simplif".]

2. hich of the following is the perimeter of a sqaure"a.

2 √ dia!ona%

b.

22 √ dia!ona%c.

42 √ dia!ona%

d. 2 dia!ona%

Solution


83/108

8erimeter of a s6uare = 4 × side = 4× diagonal2['iagonal = 2side.]

= 22diagonal[Formula.]

31. $ind the area of the circle of radius %2 √ cm.a. 50 cm2

b. 25 cm2

c.

25π cm2

d.

50π cm2

Solution

Area of re6uired $ir$le = I × :232

[Area of $ir$le = I × radius2.]

Area of re6uired $ir$le = :I $m2

32. he perimeter of a square is 13.2 cm. hat is its area"a. 13.20 cm2

b. 10.89 cm2

c. 14.89 cm2

d. 26.4 cm2

Solution

Side of s6uare = perimeter4[Formula.]

Side = 1.2 , 4 $m[Substitute.]

Side of s6uare = . $m[Simplif".]

Area of s6uare = . $m × . $m


84/108

[Area of s6uare = side2.]

Area of s6uare = 1.+9 s6.$m.

33. &f -1, -2, -3 represent the areas of respecti(e regions, then what is thearea of polygon 89:S"

a. /1 - /2 - /3

b. /1/2/3

c.

A 1 2 +A 2 2 +A 3 2 −

− − − − − − − − − − − −

√ d. None of the above

Solution

(he area of a region is the sum of the areas of its nonDoverlapping parts.

Area of 8?@S( = A1 A2 A[Step 1.]

3#. - square and a rectangle ha(e equal perimeters. &f area of the square is #cm2, then find the perimeter of the rectangle.

a. 13 cm

b. 8 cm

c. 18 cm

d. 4 cm

Solution


85/108

Area of s6uare = side2 = 4 $m2

[Formula.]

Side = 4$m2 = 2 $m[Step 1 and simplif".]

8erimeter of s6uare = 4 ×2 $m = + $m[8erimeter of s6uare = 4 × side.]

8erimeter of re$tangle = + $m[8erimeter of re$tangle = perimeter of s6uare.]

3%. $or a gi(en perimeter, which of the following will ha(e maDimum area"a. square

b. rhombus

c. rectan!%e

d. circ%e

Solution

(he most s"mmetri$al figure 7ill have the ma*imum area for a given perimeter.

/ir$le 7ill have the ma*imum area.[/ir$le is the most s"mmetri$al.]

3'. hat is the length of the outer boundary of a semicircle of radius r "

a.

π r - 2r b.

2r c.

r - π d.

π r

Solution

@e6uired length = ar$ length A!% %E EA


86/108

@e6uired length = Ir r r = Ir 2r [Ar$ length A!% = $ir$umferen$e of the $ir$le2.]

3). $ind the area of a semiCcircle of radius #.' cm.a. 21.16 sq.cm

b.

4.6π sq.cmc. 10.58 sq.cm

d.

10.58π sq.cm

Solution

Area of re6uired semiD$ir$le = I4.322[Area of semiD$ir$le = Iradius322.]

Area of re6uired semiD$ir$le = 1.:+I s6.$m.[Simplif".]

3*. here are % concentric circles. :adius of the smallest circle is % cm.&ncrease in radius of each consecuti(e circle is 1 cm. hat will be the area of acircle in sq. cm which has a circumference equal to the sum of thecircumferences of the % concentric circles"

a.

1225π sq.cm b. 1225 sq.cm

c.

70 π sq.cmd.

35π sq.cm


87/108

Solution

@adii of the other 4 $ir$les are# $m# ) $m# + $m and 9 $m.[@adius of ea$h $onse$utive $ir$le is more b" 1 $m.]

Sum of the $ir$umferen$es of the $ir$les = 2I × : 2I × 2I × ) 2I × + 2I ×9[/ir$umferen$e = 2Ir.]

= 2I × : = ) I $m

(he radius of the $ir$le having same $ir$umferen$e of )I $m = : $m

Area of the $ir$le of : $m radius = I: × :3 = 122:I s6.$m

1. he product of π and the diameter d of a circle is the GGGGGGGGGG of the circle.

a. area

b. circumference

c. radius

d. none of these

Solution

/ir$umferen$e of $ir$le / = 2 × I × r = Id.[Substitute 2r = d.]

(he produ$t of I and the diameter d of a $ir$le is the $ir$umferen$e of the $ir$le.

2. hat is the area of the circle with a circumference of 13 ft" =Hse π ?3.1#.@

a. 28.45 sq. ft.

b. 13.45 sq. ft.

c. 21.45 sq. ft.

d. 18.45 sq. ft.

Solution


88/108

/ir$umferen$e of the $ir$le# $ = 1 ft.[iven.]

/ir$umferen$e of a $ir$le = 2 × I × r.[Formula.]

= 2 × .14 × r = 1[Substitute.]

r = 1 , 2×.143⇒['ivide ea$h side b" 2 × .143.]

= 2.) ft[Simplif".]

Area of the $ir$le = Ir 2 = .14 × 2.) × 2.)[Substitute r = 2.).]

= 1.4: s6.ft[Simplif".]

(he area of the $ir$le is 1.4: s6. ft.

3. 6ackson ran around a circular track. hich of the following is required tofind the distance co(ered by him"

a. radius

b. area

c. circumference

d. diameter

Solution

(he distan$e $overed b" La$


89/108

-nswer /d0

%. Estimate the length of a rope by which a cow must be tethered so that itcan graze an area of 1+2 yd.2.a. 16 yd

b. 8 yd

c. 9 yd

d. 17 yd

-nswer /b0

'. $ind 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

-nswer /b0

). - rectangular tin sheet, 1# in. long and % in. wide, is rolled along its lengthto form a cylinder by making the opposite edges meet. hat is the base radiusof the cylinder formed"

a. 14 in.

b. 2.23 in.

c. 11 in.


90/108

d. 10.2 in.

Solution

A re$tangular tin sheet is rolled along the length to form a $"linder.

/ir$umferen$e of the base of the $"linder = length of the re$tangular sheet = 14 in.

/ir$umferen$e of the base of the $"linder# 2Ir = 14 in.[Sin$e# the $ir$umferen$e of the $ir$le = 2Ir.]

14 = 2 × .14 × r[Substitute the values of I and the $ir$umferen$e.]

14 , .2+ = .2+r.2+['ivide ea$h side b" .2+.]

2.2 = r [Simplif".]

(he base radius of the $"linder = 2.2 in.

*. $ind the circumference of a circle whose radius is 33 cm. =Hse I ? 3.1#.@a. 103.62 cm

b. 414.48 cm

c. 217.24 cm

d. 207.24 cm

Solution

(he radius r3 of the $ir$le is $m.[iven.]

(he $ir$umferen$e of the $ir$le = 2 × I × r [Formula.]

= 2 × .14 × [Substitute I = .14 and r = $m.]

= 2).24 $m[Simplif".]

(he $ir$umferen$e of the $ir$le is 2).24 $m.


91/108

+. $ind the circumference of a basketball hoop if it has a diameter of 2' cm.=Hse I ? 3.1#.@

a. 81.64 cm

b. 29.14 cm

c. 163.28 cm

d. 8.28 cm

Solution

(he $ir$umferen$e of a $ir$le = I × diameter.[Formula.]

= .14 × 2

[Substitute I = .14 and diameter = 2 $m.]

= +1.4 $m.[Simplif".]

(he $ir$umferen$e of the bas


92/108

(he $ir$umferen$e of the $ir$le is 1:.+ in$hes.

11. he diameter of a small pizza is * inches. hat is its area" =Hse I ? 3.1#.@a. 52.2 in.2

b. 48.2 in.2

c. 54.2 in.2

d. 50.2 in.2

Solution

(he radius of the piCCa = 'iameter , 2 = + , 2

[Substitute d = +.]

= 4 in$hes

Area of a $ir$le = I r 2.[Formula.]

= I × 42


= .14 × 1[Substitute I = .14.]

= :.2 in.2

12. hat is the area of a circular field of radius 1.+ miles" =Hse I ? 3.1#.@a. 13.33 mi%es2

b. 9.33 mi%es2

c. 11.33 mi%es2

d. 14.73 mi%es2

Solution

(he area of a $ir$le = I r 2

I × 1.932


93/108

= .14 × .1[Substitute I = .14.]

= 11. miles2.

13. he circumference of the bottom of a circular swimming pool is %% ft.

hat is the area of the bottom of the pool" =Hse π ? 3.1#.@a. 240.40 ft2

b. 240.40 ft3

c. 2374.62 ft3

d. 9498.50 ft

Solution

/ir$umferen$e of a $ir$le = 2Ir

2 * .14 * r = ::[Substitute I = .14.]

r = ::2*.14['ivide ea$h side b" 2 * .143.]

= +.): ft.

Area of the bottom = Ir 2

= .14 * +.): * +.):[Substitute r = +.):.]

= 24.4 ft.2

(he area of the bottom is about 24.4 ft2.

1#. hat is the radius of the circle with circumference +2 cm." =Hse π ? 3.1#.@

a. 15.64 cm

b. 13.64 cm

c. 17.64 cm

d. 14.64 cm


94/108

Solution

/ir$umferen$e of the $ir$le = 2 I r

92 = 2 × .14 × r [Substitute the values.]

r = 922×.14['ivide ea$h side b" 2 × .143.]

= 14.4 $m.[Simplif".]

(he radius of the $ir$le is 14.4 $m.

1%. hat is the diameter of the circle whose circumference is1+' inches"

a. 65.62 inches

b. 62.42 inches

c. 64.42 inches

d. 60.42 inches

Solution

/ir$umferen$e of the $ir$le = I × d

= .14 × d[Substitute I = .14.]

/ir$umferen$e of the $ir$le = 19.[iven.]

.14 × d = 19

d = 19 , .14['ivide ea$h side b" .14.]

= 2.42[Simplif".]

(he diameter of the $ir$le is 2.42 in$hes.


95/108

1'. hat is the radius of the circle in the figure"

a.

25 √ ft.

b.

3 √ ft.

c.

32 √ ft.

d.

33 √ ft.

Solution

From the figure# A%/' is a s6uare of side ft. and E' is the radius of the $ir$le.

(he length of the diagonal %' = 2 × radius = 2r [Formula.]

From the figure# A%2 A'2 = %'2

[Sin$e A%' forms a right triangle.]

2 2 = 2r32


= 4r 2

)2 = 4r 2

1+ = r 2

['ivide ea$h side b" 4.]

2 = r [(a


96/108

1). hat is the area of the shaded region in the figure"

=Hse π ? 3.1#.@

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 $m. and 9 $m.respe$tivel".

5n a right triangle# 0"potenuse2 = %ase2 0eight2.[8"thagorean theorem.]

A/2 = A%2 %/2

A/2 = 122 92

A/2 = 144 +1 = 22:

A/ = 22:

A/ = 1:[22: = 1: ]

A/ is the diameter of the $ir$le.

So# radius of the $ir$le = A/ , 2 = 1: , 2 = ).:.

Area of the shaded region = Area of the $ir$le D Area of the triangle

= .14 × ).: × ).: = 1). $m.2

Area of the $ir$le = Ir 2

[Substitute r = ).: and I = .14 and simplif".]


97/108

Area of the triangle = 1 , 2 × base × height

= 12 × 12 × 9 = = :4 $m. 2

[Substitute base = 12 and height = 9 and simplif".]

(he area of the shaded region = 1). D :4 = 122. $m2.

1*. hat is the area of the shaded region in the figure"

=Hse π ? 3.1#.@

a. 100 cm2

b. 78.5 cm2

c. 21.5 cm2

d. 31.5 cm2

Solution

From the figure# the side of the s6uare is 1 $m and the radius of the $ir$le is : $m

respe$tivel".

Area of the shaded region = Area of the s6uare D Area of the $ir$le

Area of the s6uare = a2 = 1 × 1[Substitute a = 1.]

= 1 $m.2

= .14 × : × : Area of the $ir$le = Ir 2

[Substitute r = : and I = .14.]

= )+.: $m.2

(he area of the shaded region = 1 D )+.: = 21.: $m2.


98/108

1+. - goat is tied to a pole with a # m rope. he goat can graze to the fulllength of the rope and 3'o around the pole. !ow much area does the goat

ha(e to graze" =Hse π ? 3.1#.@a. 16 m2

b. 25.12 m2

c. 50.24 m2

d. 1440 m2

Solution

Area of the $ir$ular region 7here the goat $an graCe = Ir 2

= .14 × 4 × 4

[Substitute r = 4 and I = .14.]

= :.24 m.2

(he area the goat has to graCe is :.24 m2.

2. he diameter of a car wheel is 1* in. !ow far will the car go in *re(olutions of the wheel"

a. 144

b.

144 π c. 8

d. 18

Solution

(he diameter of the $ar 7heel = 1+ in.

/ir$umferen$e of the $ir$le $ar 7heel3 = I d

= 1+I in[Substitute d = 1+.]

For one revolution the $ar travels 1+I in.

'istan$e traveled b" the $ar in + revolutions = 1+I × +

= 144I in


99/108

After + revolutions# the $ar travels 144I in.

21. Each side of the square in the figure shown is 1' in. hat is the area ofthe shaded region if the four circles inscribed in the square are congruent"

=Hse π ? 3.1#.@

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 ea$h of the $ongruent $ir$les is 4 in.

Area of the shaded region = area of the s6uare D sum of the areas of the four $ir$les

Area of the s6uare = a2 = 1 × 1[Substitute a = 1.]

= 2: in.2

Area of one $ir$le = Ir 2

= .14 × 4 × 4

[Substitute r = 4 and I = .14.]

= :.24 in.2

Area of four $ir$les = 4 × :.24 = 2.9 in.2

(he area of the shaded region = 2: D 2.9 = ::.4 in.2


100/108

22. -n aluminum tube, % in. long, is bent to make a circular hoop. $ind thediameter of the hoop.

a. 50

b.

48 π c.

π 50

d.

50 π

Solution

/ir$umferen$e of the $ir$ular hoop = length of the aluminum tube = : in.

/ir$umferen$e of the $ir$ular hoop = Id

I × d = :

d = :I

23. he circumference of a circle is 2 r . $ind the circumference of the

circle, when the radius r is equal to 2.a. 4

b. 6

c. 3

d. 5

Solution

&valuate 2r for r = 2

2r = 223[Substitute r for 2.]

= 4[!ultipl".]

(he $ir$umferen$e of the $ir$le is 4.


101/108

2#. < is the center of the circle. he circle is di(ided into three equal partsand one of them is shaded. &f the radius of the circle is 12 ft, then what is thearea of the shaded sector"

a.

46π ft2

b.

50π ft2

c.

48π ft2

d.

49π ft2

Solution


= I × 122

[Substitute r = 12.]

= 144I[Simplif".]

Area of the shaded se$tor = @evolution of the se$tor × Area of the $ir$le

= 12 , × 144I[Substitute the values.]

= 4+I ft2.[Simplif".]

2%. he central angle of a sector is 2o. &f the area of the circle is 1* m.2, thenwhat is the area of the sector"

a. 10 m2

b. 20 m2

c. 12 m2

d. 15 m2


102/108

Solution

@evolution of the se$tor = 2MM

Area of the se$tor = @evolution of the se$tor × Area of the $ir$le

= 2MM × 1+[Substitute the values.]

= 1 m.2

[Simplif".]

2'. - square is inscribed in a circle whose diameter is 12 cm as shown in thefigure. $ind the area of the shaded portion in the figure.

a. 41.04 cm2

b. 40 cm2

c. 50 cm2

d. none of these

Solution

(he diameter of a $ir$le = 12 $m.

(he diameter of a $ir$le divides the ins$ribed s6uare into t7o $ongruent righttriangles.

-et S be the side of s6uare.

2S

2

= 144S2 = )2 Appl"ing 8"thagorean theorem for the triangle = S2 S2 = 122

['ivide ea$h side b" 2.]

Area of the s6uare# S2 = )2.



103/108

= I × 2 = 11.4 $m2

[Substitute r = .]

(he area of the shaded portion = 11.4 D )2 = 41.4 $m. 2.

2). - wire 1 cm long is bent to form a circle. $ind the area of the circleformed.

a.

2500 (πUnknown node type: sup) cm2

b.

π 2500 cm2

c. 2500 cm2

d.

2500 π cm2

Solution

/ir$umferen$e of the $ir$le = 2Ir = 1 $m.

2Ir = 1

Ir = 12['ivide ea$h side b" 2.]

r = : , I['ivide ea$h side b" I.]


= I × : , I32 [Substitute r = : , I.]

= 2: , I $m2.

2*. he diameter of the wheel of a truck is 1 cm. !ow many re(olutions willit take to tra(el ' cm."

a.

6π b.

6π c.

7π d.

5π


104/108

Solution

/ir$umferen$e of the 7heel = Id

I × 1[Substitute d = 1.]

1I $m.

(he distan$e $overed b" the 7heel in one revolution = /ir$umferen$e of the $ir$le

,1I = ,IBumber of revolutions of the 7heel 7hile travelling $m.

2+. - circular ground of radius # m has a path of width 2 m around it asshown in the figure. hat is the area of the path"

a.

10π m2

b.

20π m2

c.

30π m2

d. none of these

Solution

@adius of the inner $ir$le = 4 m.

@adius of the outer $ir$le = 4 2 = m.

Area of the inner $ir$le = Ir 2

I × 42

[Substitute r = 4.]

1I m2

Area of the outer $ir$le = Ir 2


105/108

I × 2

[Substitute r = .]

I m2

(he area of the path = I D 1I = 2 I m.2.

3. he circumference of two concentric circles is 13 m. and + m. hat isthe difference between their radii"

a.

20π m b.

20π mc.

π 20 m

d.45π m

Solution

-et r 1 and r 2 be the radii of outer and inner $ir$les.

/ir$umferen$e of the outer $ir$le# 2Ir 1 = 1 m.

r 1 = :,I['ivide ea$h side b" 2I.]

/ir$umferen$e of the inner $ir$le# 2Ir 2 = 9 m.

r 2 = 4:,I['ivide ea$h side b" 2I.]

= 2,I m.(he differen$e bet7een their radii = :,I D 4:,I

31. &f the diameter of a circle is increased by *B, by what percent will itscircumference increase"

a. 85+

b. 80+

c. 70+


106/108

d. 90+

Solution

-et d be the diameter of the $ir$le.

(he $ir$umferen$e of the $ir$le = Id.

(he diameter of a $ir$le is in$reased b" +K.

(he ne7 diameter of the $ir$le = 1 .+3 × d = 1.+d

(he ne7 $ir$umferen$e of the $ir$le = I × 1.+d = 1.+ I d

(he $ir$umferen$e of the $ir$le is in$reased b" +K.

32. &f the radius of the circle is decreased by 3B, by what percent will itscircumference decrease"

a. 30+

b. 65+

c. 55+

d. 60+

Solution

-et r be the radius of the $ir$le.

(he radius of a $ir$le is de$reased b" K.

(he ne7 radius of the $ir$le = 1 D .3 * r = .)r

/ir$umferen$e of the $ir$le = 2Ir

(he ne7 $ir$umferen$e of the $ir$le = 2 * I * .)r

= .) * 2Ir

= 1 D .3 * 2Ir

So# the $ir$umferen$e of the $ir$le is de$reased b" K.


107/108

33. &f the radius of a circle is increased by 3B, by what percent will its areaincrease"

a. 64+

b. 69+

c. 30+

d. 74+

Solution



@adius of the $ir$le is in$reased b" K.

(he radius of the ne7 $ir$le = 1 .3 r = 1.r

(he ne7 area of the $ir$le = Ir 2

= I * 1.r32

[Substitute r = 1.r.]

= 1.9Ir 23

(he area of the $ir$le is in$reased b" 9K.

3#. &f the area of a circle is decreased by )B, by about what percent will itsradius decrease"

a. 41+

b. 70+

c. 46+

d. 51+

Solution



(he ne7 area of the $ir$le = 1 D .)3Ir 2


108/108

= . * Ir 2

= I.:4r32

[Appro*imatel" N. = .:4.]

(he radius of the $ir$le is de$reased b" 4K.

3%. - circular cake is cut into equal pieces in such a way that each piecemakes an angle of #%o at the center. $ind the number of pieces.

a. 10

b. 8

c. 6

d. 12

Solution

(he $omplete angle at the $enter of a $ir$le is o.

Bumber of pie$es = $omplete angle , angle subtended b" ea$h pie$e at the $enter.

= o,4:o


(he total number of pie$es that are $ut from the $a

irregular shapes ws.doc

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