Mensuration - Perimeter, Area, Volume Grade 7 · 2016-03-01 · Mensuration - Perimeter, Area, Volume ... A wire frame is in rectangular shape which is 41 cm long and 20.6 cm broad

Grade 7Mensuration - Perimeter, Area, Volume

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Answer the questions

(1) A square park has an area of 4900 sq. m. It has to be f enced. The f encing will require use of awire that must be able to enclose the park 4 times, and each circuit of the wire would be 2 %greater than the perimeter of the park. What is the length of wire needed f or this?

(2) Find area of shaded region (All measurements are in cm)

(3) In the f igure below, ABCD is a square with sides 24. We draw the arc of a circle centered at Cf rom D to B, and another arc of a circle centered at A f rom B to D as seen in the picture below.What is the area of the space between those two arcs?

ID : au-7-Mensuration-Perimeter-Area-Volume [1]

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(4) The f ront wall of a f actory as shown below has to be painted. The area of the entrance shownin grey does not need to be painted. If the cost of painting is $10 per square m, then what willthe total cost f or painting be?

(5) If perimeter of f ollowing semi-circular shape is 108 cm, f ind its radius (assume π = 22/7).

(6) A wire f rame is in rectangular shape which is 41 cm long and 20.6 cm broad. If wire is reshaped inthe circular shape, f ind the radius of the circle. (assume π = 22/7)

Choose correct answer(s) from given choice

(7) Which if the f ollowing shapes represent the same area ?

a.

b.

c. d.



(8) Find the area of the f ollowing shape. (All measures are in meters).

a. 125m2 b. 169m2

c. 84m2 d. 441m2

(9) A wire of length 40 meter is to be f olded in the f orm of rectangle. If each side of rectangle hasto be an integer (measured in meters), what is the maximum number of rectangles that can bef ormed by f olding the wire.

a. 9 b. 20

c. 10 d. 5

(10) The length, width and height of a hall are 14, 9 and 4 meters respectively. Find the area of itsf our walls.

a. 194 m2 b. 164 m2

c. 204 m2 d. 184 m2

(11) The width of the circular track shown in the picture is 2 meters. If radius of the outer circle is 13meters, f ind area of the track. (assume π = 3 f or this question)

a. 172.8 m2 b. 158.4 m2

c. 144 m2 d. 144 m2

Fill in the blanks

(12) The perimeter of any rectilinear f igure is the of the sides.



(13) Perimeter of the triangle with sides 167 cm, 161 cm and 105 cm is cm .

(14) If diameter of a wheel of a truck is 68.6 cm. The truck will cover distance of km, if

wheel completes 700000 revolutions.

(15) One wall of the auditorium is 46 m long and 18 m high. If the cost of paint a sq. m. of the wall is

$ 61.70 , then the cost to paint the entire wall is $

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Answers

(1) 1142.4 m

Step 1

Let's assume that a be the length of the park.

Theref ore the area of a square park = a2 = 4900

⇒ a2 = 702

⇒ a = 70 m

Step 2

Now the perimeter of a square = 4a = 4 X 70 = 280 m

Step 3

Since each circuit of the wire be 2% greater than the perimeter of the park.Theref ore the length of the each circuit of the wire = perimeter of the square park + 2% ofthe perimeter of the square park

= 280 + 280 X 2

100

= 280 + 560

100

= 280 + 5.6= 285.6 m

Step 4

The f encing will require use of a wire that must be able to enclose the park 4 times.Theref ore the length of wire needed f or this = 4 X 285.6 = 1142.4 m.



(2) 350 cm2

Step 1

The area of the shaded region = The area of the rectangle - The area of the bothunshaded right angled triangles

Step 2

Area of the rectangle = 22 × 26 = 572 cm2

Step 3

Area of the right angled triangle with base 14 cm and height 16 cm = 1

2 (Base × Height)

= 1

2 (14 × 16)

= 112 cm2

Step 4

Area of the right angled triangle with base 22 cm and height 10 cm = 1

2 (Base × Height)

= 1

2 (22 × 10)

= 110 cm2

Step 5

Hence, the area of shaded region is: 572 - (112 + 110) = 350 cm2



(3) 288 π - 576

Step 1

Lets draw a line f rom point B to point D as shown in the f ollowing f igure.

According to question the side of the square is 24.In right angle triangle ΔABD

BD2 = AD2 + AB2

⇒ BD2 = 242 + 242

⇒ BD2 = 2 × 242

⇒ BD = √(2 × 242)⇒ BD = 24√2

Now the area of the right angle triangle ΔABD = AD × AB

2

= 24 × 24

2

= 288

Step 2

If you look at the f igure caref ully, you will notice that the radius of the circle is equal to theside of the square.

Now the area of the quarter(BCD) of the circle is = πr2

4

= π(24)2

4

= 576π

4

= 144π

Step 3

Now the space between two arc = 2(144π - 288) = 288 π - 576



(4) $ 1420

Step 1

If we look at the f ront wall of the f actory caref ully, we notice that,the length of the wall = 20 mThe width of the wall = 8 m

The area of the wall = 160 m2

Step 2

The height of the entrance = 3 mThe width of the entrance = 6 m

The area of the entrance = 18 m2

Step 3

The area of the entrance does not need to be painted,The area of the wall need to be painted = The area of the wall - The area of the entrance= 160 - 18

= 142 m2

Step 4

The cost of painting is $10 per square m,the total cost f or painting = 142 × 10 = $1420



(5) 21 cm

Step 1

Lets assume r and d are the radius and diameter of the semi-circular shape respectively.Diameter of the semi-circular shape = 2r,Perimeter of the semi-circular shape = πr + d= πr + 2r= r(π + 2)

= r( 22

7 + 2)

= r( 22

7 + 2)

= r( 22 + 14

7 )

= 36r

7

Step 2

According to question perimeter of the semi-circular shape is 108 cm.

Theref ore 36r

7 = 108

⇒ r = 108 × 7

36

⇒ r = 21

Step 3

Now radius of the semi-circular shape = 21 cm



(6) 19.6

Step 1

According to question the length and width of the rectangular wire f rame are 41 cm and20.6 cm respectively.The perimeter of the rectangular wire f rame = 2(Length + Width)= 2(41 + 20.6)= 123.2 cm

Step 2

Since the wire is reshaped in the circular shape, the perimeter of the rectangle is equal tothe perimeter of the circle.Theref ore 2πr = 123.2

or r = 123.2

2π

=

123.2

2 × 22

7

= 123.2 × 7

44

= 19.6 cm

Step 3

Now the radius of the circle is 19.6 cm.

(7)

a. d.



(8) b. 169m2

Step 1

We need to f ind the area of the shape below:

Step 2

Let us f irst divide this shape into two rectangles. Area of the give shape will be the sum ofareas of these two rectangles.

rectangle 1 rectangle 2Step 3

Let us now f ind the area of rectangles 1 and 2:

Area of rectangle 1 = Length × Width= 5 × 21

= 105 m2

Area of rectangle 2 = Length × Width= 16 × 4

= 64 m2

Step 4

Let us now add the areas of rectangle 1 and 2 to get the area of the original shape:Area of given shape = Area of rectangle 1 + Area of rectangle 2

= 105 m2 + 64 m2



= 169 m2

(9) c. 10

Step 1

We have to maximize the numbers of rectangles that can be created with a given length ofwire. Theref ore we need to keep size of rectangles as small as possible, so we can makemaximum number or rectangles

Step 2

It is given that each side has to be an integer (i.e. it cannot be 0.5m, 0.8m etc.), theref oresmallest possible length will be of 1 meter

Step 3

Perimeter of rectangle with each side of 1 meter = 4 × 1 = 4 meters

Step 4

Since 4 meters wire is needed f or 1 rectangle,Theref ore number of rectangles that can be created with 40 meter = 40 ÷ 4 = 10

(10) d. 184 m2

Step 1

Lets assume l, w and h are the length, width and height of the hall respectively, as shown inthe f ollowing f igure.

Step 2

According to question the length, width and height of the hall are 14, 9 and 4 metersrespectively.Theref ore l = 14 metersw = 9 metersh = 4 meters

Step 3

Now area of hall's f our walls = 2(l × h) + 2(w × h)= 2h(l + w)= 2 × 4(14 + 9)= 8(14 + 9)

= 184 m2

Step 4

Theref ore the area of the hall's f our walls is 184 m2.



(11) c. 144 m2

Step 1

Lets assume r1 and r2 be the inner and outer radius of the given circle respectively.

Step 2

According to question radius of the outer circle is 13 meters and width of the circular trackis 2 m.Theref ore r2 = 13 meters and r1 = 13 - 2 = 11 meters

Step 3

Now area of the track is = area of the outer circle - area of the inner circle

= π(r2)2 - π(r1)2

= π(r22 - r12)

= π(132 - 112)= π(169 - 121)= 3 × 48 [Since π = 3]

= 144 m2

Step 4

Area of the track is 144 m2.

(12) sum

(13) 433



(14) 1509.2

Step 1

If you look at the question caref ully, you will notice that diameter of a wheel of a truck = 68.6 cm Total revolutions of wheel= 700000

Step 2

The circumference of a circle (or wheel) is the distance around the outside.It is given by the f ormula:

C = πD (where, D is the diameter of wheel and π = 22

7 )

= 22

7 × 68.6

= 215.6So if you put a mark on the outer rim of the wheel, and you rotate the wheel once, thatspot will travel a distance of 215.6 cm.

Step 3

Total revolutions of wheel = 700000Distance covered by one revolution of wheel = 215.6 cmTotal distance covered by truck = Total revolutions of wheel × Distance covered by onerevolution of wheel = 700000 × 215.6 cm

= 700000 × 215.6

100000 km (since, 1cm = 1/100000km)

= 1509.2 km

Step 4

Theref ore the total distance covered by truck is 1509.2 km .

(15) 51087.6



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Mensuration - Perimeter, Area, Volume Grade 7 · 2016-03-01 · Mensuration - Perimeter, Area, Volume ... A wire frame is in rectangular shape which is 41 cm long and 20.6 cm broad