Measuring Space: Perimeter and Area Class 9 Competency Questions

20 Questions | 78 Marks

A circular park has a radius of 21 meters. A path of width 3.5 meters runs around it on the outside. Calculate the area of the path. 5marks

An athletics track is in the form of a ring whose inner radius is 21 m and outer radius is 28 m. Find the area of the track. 3marks

Two circular pieces of equal radius are joined together such that each passes through the centre of the other. If the radius of each circle is 10 cm, find the perimeter of the shape formed by the two circles. 3marks

A rectangular field is 50 m long and 30 m wide. It has a gravel path 2.5 m wide all around it on the inside. Find the area of the path. 3marks

A garden is in the shape of a rectangle with a semi-circular area at one end. The rectangular part is 20 m long and 14 m wide. The semi-circular end has a diameter equal to the width of the rectangle. Calculate the total area of the garden. 5marks

A circular table cover has a diameter of 1.56 m. Find its radius. If it has to be stitched along the edge, find the length of the decorative lace required. 3marks

Assertion (A): The area of a parallelogram is equal to the product of its base and height.Reason (R): A parallelogram can be transformed into a rectangle of the same base and height by cutting a triangle from one side and placing it on the other. [2 marks](a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true.

A triangular park has sides 122 m, 155 m and 110 m. A sprinkler at the centroid of the park sprays water in a circle of radius 5 m. Find the area of the park that is watered by the sprinkler. 5marks

The sides of a triangle are in the ratio 3:5:7 and its perimeter is 300 m. Find the area of the triangle using Heron's formula. 5marks

The wheels of a car are of radius 42 cm. If the wheels are rotated 25000 times every minute, how fast is the car travelling in km per hour? 5marks

A rectangular sheet of paper has dimensions 44 cm × 18 cm. A circular hole of radius 7 cm is cut from the center. Find the area of the remaining paper. 3marks

Two concentric circles have radii 7 cm and 14 cm. Find the area of the ring (annulus) between them. 3marks

A farmer wants to fence a circular field of diameter 10.5 m. He needs to place a post every 3 m along the boundary. How many posts will he need? 3marks

Consider a square ABCD and a triangle ABE such that ABE is inscribed within the square.Assertion (A): If the side of the square is 'a' units, the area of the square is

a^{2}

sq. units.Reason (R): The area of triangle ABE, where E is the midpoint of CD, will be half the area of the square. [2 marks](a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true.

A plot of land is in the shape of a rhombus. The perimeter of the rhombus is 100 m and one of its diagonals is 14 m. Find the area of the rhombus. 5marks

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand between 9:00 AM and 9:30 AM. 3marks

A sector of a circle has an angle of 120° and radius 21 cm. Find the area of the sector and the length of the arc. 5marks

A park is in the form of a circle of radius 20 m. A path 2 m wide is constructed along the boundary inside the park. Find the area of the path. 3marks

Case Study: A school playground is rectangular in shape with a length of 100 m and a breadth of 50 m. A semi-circular grassy area is added to one of the breadth sides.(i) Calculate the area of the rectangular part of the playground. [2 marks](ii) Calculate the area of the semi-circular grassy area. 3marks

Assertion (A): For any triangle, a median divides it into two triangles of equal area.Reason (R): The area of a triangle is given by

\frac{1}{2} \times b a s e \times h e i g h t

. A median divides the base into two equal parts, and both triangles share the same height. [2 marks](a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true.

Answers

Radius of the circular park (

r_{1}

) = 21 mWidth of the path = 3.5 mOuter radius of the park including the path (

r_{2}

) =

r_{1}

+ width = 21 + 3.5 = 24.5 mArea of the path = Area of outer circle - Area of inner circleArea =

π r_{2}^{2} - π r_{1}^{2} = π (r_{2}^{2} - r_{1}^{2})

Area =

\frac{22}{7} ((24.5)^{2} - (21)^{2})

Area =

\frac{22}{7} (600.25 - 441)

Area =

\frac{22}{7} (159.25)

Area =

22 \times 22.75 = 500.5

^{2}

. 5marks

Inner radius of the ring (

r_{1}

) = 21 mOuter radius of the ring (

r_{2}

) = 28 mArea of the track = Area of outer circle - Area of inner circleArea =

π r_{2}^{2} - π r_{1}^{2} = π (r_{2}^{2} - r_{1}^{2})

Area =

\frac{22}{7} ((28)^{2} - (21)^{2})

Area =

\frac{22}{7} (784 - 441)

Area =

\frac{22}{7} (343)

Area =

22 \times 49 = 1078

^{2}

. 3marks

Let the radius of each circle be

r = 10

cm.The shape formed by two overlapping circles passing through each other's centres is composed of two circular segments.In triangle ABC (where A and B are centres and C is an intersection point), AB = BC = AC = r = 10 cm. Thus,

△ A B C

is equilateral.The angle

∠ C A B = 60^{\circ} = \frac{π}{3}

radians.The angle subtended by the arc at the centre is

2 \times 60^{\circ} = 120^{\circ} = \frac{2 π}{3}

radians.The length of the arc of one circle that forms part of the perimeter is the major arc.Length of major arc = Circumference - Length of minor arcLength of minor arc subtending

120^{\circ}

at the centre =

\frac{120}{360} \times 2 π r = \frac{1}{3} \times 2 π r = \frac{2 π r}{3}

.Length of major arc =

2 π r - \frac{2 π r}{3} = \frac{4 π r}{3}

.The perimeter of the combined shape is the sum of the lengths of the two major arcs.Perimeter =

2 \times \frac{4 π r}{3} = \frac{8 π r}{3}

.Perimeter =

\frac{8 π \times 10}{3} = \frac{80 π}{3}

cm.Using

π \approx \frac{22}{7}

: Perimeter =

\frac{80}{3} \times \frac{22}{7} = \frac{1760}{21} \approx 83.81

cm.(Alternatively, using the given text's example: Each dotted arc is 1/3 of the circumference of the circle. The total length of the two red arcs is

2 \times \frac{2}{3} \times 2 π r = \frac{8 π r}{3}

.The shape's perimeter is made of two such arcs, so perimeter =

2 \times \frac{2}{3} \times 2 π r = \frac{8 π r}{3}

.Perimeter =

2 \times \frac{2}{3} \times 2 π (10) = \frac{80 π}{3}

cm). 3marks

Length of the rectangular field = 50 mWidth of the rectangular field = 30 mWidth of the gravel path = 2.5 mThe path is inside the field.Inner length of the field (excluding path) =

50 - 2 \times 2.5 = 50 - 5 = 45

mInner width of the field (excluding path) =

30 - 2 \times 2.5 = 30 - 5 = 25

mArea of the path = Area of outer rectangle - Area of inner rectangleArea =

(50 \times 30) - (45 \times 25)

Area =

1500 - 1125 = 375

^{2}

. 3marks

Length of the rectangular part = 20 mWidth of the rectangular part = 14 mDiameter of the semi-circular area = width of the rectangle = 14 mRadius of the semi-circular area (

r

) =

\frac{14}{2} = 7

mArea of the rectangular part = Length

\times

Width =

20 \times 14 = 280

^{2}

.Area of the semi-circular area =

\frac{1}{2} π r^{2} = \frac{1}{2} \times \frac{22}{7} \times (7)^{2}

Area =

\frac{1}{2} \times \frac{22}{7} \times 49 = 11 \times 7 = 77

^{2}

.Total area of the garden = Area of rectangular part + Area of semi-circular areaTotal area =

280 + 77 = 357

^{2}

. 5marks

Diameter of the circular table cover = 1.56 mRadius of the table cover (

r

) =

\frac{1.56}{2} = 0.78

mThe length of the decorative lace required is the circumference of the table cover.Circumference =

2 π r = 2 \times \frac{22}{7} \times 0.78

Circumference =

\frac{44}{7} \times 0.78 = \frac{34.32}{7} \approx 4.903

m. 3marks

(a) Both A and R are true and R is the correct explanation of A. 2marks

Sides of the triangular park are

a = 122

b = 155

c = 110

m.Perimeter (

P

) =

a + b + c = 122 + 155 + 110 = 387

m.Semi-perimeter (

s

) =

\frac{P}{2} = \frac{387}{2} = 193.5

m.Area of the park (using Heron's formula) =

\sqrt{s (s - a) (s - b) (s - c)}

Area =

\sqrt{193.5 (193.5 - 122) (193.5 - 155) (193.5 - 110)}

Area =

\sqrt{193.5 \times 71.5 \times 38.5 \times 83.5}

Area

\approx \sqrt{45801315.1875} \approx 6767.66

^{2}

.The sprinkler sprays water in a circle of radius

r = 5

m.Area watered by the sprinkler =

π r^{2} = \frac{22}{7} \times 5^{2} = \frac{22}{7} \times 25 = \frac{550}{7} \approx 78.57

^{2}

.Since the radius of the spray (5 m) is much smaller than the dimensions of the park, the entire area sprayed by the sprinkler will be within the park.Thus, the area of the park watered by the sprinkler is the area of the circle it sprays. 5marks

Sides of the triangle are in the ratio 3:5:7.Perimeter = 300 m.Let the sides be

3 x, 5 x, 7 x

3 x + 5 x + 7 x = 300

15 x = 300

$x = 20$ m.
Sides are:
$a = 3 \times 20 = 60$ m
$b = 5 \times 20 = 100$ m
$c = 7 \times 20 = 140$ m
Semi-perimeter ( $s$ ) = $\frac{300}{2} = 150$ m.
Area of the triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Area = $\sqrt{150 (150 - 60) (150 - 100) (150 - 140)}$
Area = $\sqrt{150 \times 90 \times 50 \times 10}$
Area = $\sqrt{6750000}$
Area = $\sqrt{675 \times 10000} = 100 \sqrt{675} = 100 \sqrt{225 \times 3} = 100 \times 15 \sqrt{3}$
Area = $1500 \sqrt{3}$ m $^{2}$ . [5 marks]

Radius of the wheels (

r

) = 42 cm = 0.42 m.Distance covered in one revolution = Circumference of the wheel =

2 π r

.Distance =

2 \times \frac{22}{7} \times 42 = 2 \times 22 \times 6 = 264

m.Number of rotations per minute = 25000.Distance covered per minute =

25000 \times 264

m.Distance per minute =

6, 600, 000

m.Speed in meters per minute = 6,600,000 m/min.Speed in meters per hour =

6, 600, 000 \times 60

m/hour.Speed =

396, 000, 000

m/hour.Speed in km per hour =

\frac{396, 000, 000}{1000} = 396, 000

km/hour.

Re-check calculation based on standard speeds: The given numbers seem extremely high for a car. Let's assume the question meant 250 rotations per minute.
If rotations per minute = 250.
Distance covered per minute = $250 \times 264 = 66, 000$ m/min.
Distance covered per hour = $66, 000 \times 60 = 3, 960, 000$ m/hour.
Speed in km per hour = $\frac{3, 960, 000}{1000} = 3960$ km/hour. This is still very high.

Let's consider the context of the chapter and typical values:
If the question is taken literally:
Number of rotations per minute = 25000.
Distance per minute = $25000 \times 264 m = 6, 600, 000 m$ .
Distance per hour = $6, 600, 000 \times 60 m = 396, 000, 000 m$ .
Speed in km/h = $\frac{396, 000, 000}{1000} = 396, 000 km/h$ .

Let's assume there might be a typo and the radius is in meters or a smaller number of rotations. However, sticking to the provided numbers:
Distance per revolution = $2 π r = 2 \times \frac{22}{7} \times 42 cm = 264 cm = 2.64 m$ .
Distance travelled in 1 minute = $25000 \times 2.64 m = 66000 m$ .
Distance travelled in 1 hour = $66000 \times 60 m = 3, 960, 000 m$ .
Speed in km/h = $\frac{3, 960, 000}{1000} = 3960 km/h$ .
This is an extremely high speed. Assuming the question is as stated. [5 marks]

Length of the rectangular sheet = 44 cmWidth of the rectangular sheet = 18 cmRadius of the circular hole = 7 cmArea of the rectangular sheet = Length

\times

Width =

44 \times 18 = 792

^{2}

.Area of the circular hole =

π r^{2} = \frac{22}{7} \times 7^{2} = \frac{22}{7} \times 49 = 22 \times 7 = 154

^{2}

.Area of the remaining paper = Area of rectangular sheet - Area of circular holeArea =

792 - 154 = 638

^{2}

. 3marks

Inner radius (

r_{1}

) = 7 cmOuter radius (

r_{2}

) = 14 cmArea of the ring = Area of outer circle - Area of inner circleArea =

π r_{2}^{2} - π r_{1}^{2} = π (r_{2}^{2} - r_{1}^{2})

Area =

\frac{22}{7} (14^{2} - 7^{2})

Area =

\frac{22}{7} (196 - 49)

Area =

\frac{22}{7} (147)

Area =

22 \times 21 = 462

^{2}

. 3marks

Diameter of the circular field = 10.5 mRadius of the circular field (

r

) =

\frac{10.5}{2} = 5.25

mCircumference of the field =

2 π r = 2 \times \frac{22}{7} \times 5.25

Circumference =

2 \times 22 \times 0.75 = 44 \times 0.75 = 33

m.The farmer needs to place a post every 3 m along the boundary.Number of posts =

\frac{Circumference}{Distance between posts} = \frac{33}{3} = 11

. 3marks

(a) Both A and R are true and R is the correct explanation of A. 2marks

Perimeter of the rhombus = 100 m.Let the side of the rhombus be 'a'.

4 a = 100 \Rightarrow a = 25

m.One diagonal (

d_{1}

) = 14 m.In a rhombus, diagonals bisect each other at right angles. Let the diagonals be

d_{1}

and

d_{2}

(\frac{d_{1}}{2})^{2} + (\frac{d_{2}}{2})^{2} = a^{2}

(\frac{14}{2})^{2} + (\frac{d_{2}}{2})^{2} = 25^{2}

7^{2} + (\frac{d_{2}}{2})^{2} = 625

49 + \frac{d_{2}^{2}}{4} = 625

\frac{d_{2}^{2}}{4} = 625 - 49 = 576

d_{2}^{2} = 576 \times 4 = 2304

$d_{2} = \sqrt{2304} = 48$ m.
Area of the rhombus = $\frac{1}{2} \times d_{1} \times d_{2}$
Area = $\frac{1}{2} \times 14 \times 48 = 7 \times 48 = 336$ m $^{2}$ . [5 marks]

The minute hand completes a full circle (360°) in 60 minutes.Length of the minute hand (radius,

r

) = 14 cm.In 60 minutes, the angle swept is 360°.In 1 minute, the angle swept is

\frac{360}{60} = 6^{\circ}

.In 30 minutes (from 9:00 AM to 9:30 AM), the angle swept (

θ

) =

30 \times 6^{\circ} = 180^{\circ}

.Area swept by the minute hand = Area of the sector.Area =

\frac{θ}{360^{\circ}} \times π r^{2}

Area =

\frac{180^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (14)^{2}

Area =

\frac{1}{2} \times \frac{22}{7} \times 196

Area =

11 \times 28 = 308

^{2}

. 3marks

Angle of the sector (

θ

) = 120°Radius (

r

) = 21 cmArea of the sector =

\frac{θ}{360^{\circ}} \times π r^{2}

Area =

\frac{120^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (21)^{2}

Area =

\frac{1}{3} \times \frac{22}{7} \times 441

Area =

\frac{22}{3} \times 63 = 22 \times 21 = 462

^{2}

.Length of the arc =

\frac{θ}{360^{\circ}} \times 2 π r

Length =

\frac{120^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21

Length =

\frac{1}{3} \times 2 \times 22 \times 3

Length =

2 \times 22 = 44

cm. 5marks

Radius of the park (

r_{1}

) = 20 mWidth of the path = 2 mThe path is inside the park.Radius of the inner circle (area excluding path) (

r_{2}

) =

r_{1}

- width =

20 - 2 = 18

m.Area of the path = Area of outer circle - Area of inner circleArea =

π r_{1}^{2} - π r_{2}^{2} = π (r_{1}^{2} - r_{2}^{2})

Area =

\frac{22}{7} (20^{2} - 18^{2})

Area =

\frac{22}{7} (400 - 324)

Area =

\frac{22}{7} (76)

Area =

\frac{1672}{7} \approx 238.86

^{2}

. 3marks

Case Study:Length of the rectangular playground = 100 mBreadth of the rectangular playground = 50 m

(i) Area of the rectangular part = Length $\times$ Breadth
Area = $100 m \times 50 m = 5000 m^{2}$ . [2 marks]

(ii) The semi-circular grassy area is added to one of the breadth sides, so its diameter is equal to the breadth.
Diameter of the semi-circle = 50 m
Radius of the semi-circle ( $r$ ) = $\frac{50}{2} = 25$ m.
Area of the semi-circular grassy area = $\frac{1}{2} π r^{2}$
Area = $\frac{1}{2} \times \frac{22}{7} \times (25)^{2}$
Area = $\frac{1}{2} \times \frac{22}{7} \times 625$
Area = $\frac{11}{7} \times 625 = \frac{6875}{7} \approx 982.14$ m $^{2}$ . [3 marks]

(a) Both A and R are true and R is the correct explanation of A. 2marks

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Measuring Space: Perimeter and Area — Competency Based Questions

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