Std. 7 _Perimeter & Area

Question 1:

If the length and breadth of a rectangle are doubled, how many times the perimeter of the old rectangle will that of the new rectangle be?

ANSWER:

Let the length of old rectangle = l
Let the breadth of old rectangle = b
Perimeter of the old rectangle = 2(length + breadth) = 2(l + b)
When the length and the breadth of the rectangle are doubled, then
length of the new rectangle = 2l
breadth of the new rectangle = 2b
∴ Perimeter of new rectangle = 2(length + breadth)
= 2(2l + 2b)
= 2 × 2(l + b)
= 2 × perimeter of the old rectangle    [∵ perimeter of old rectangle = 2(l + b)]
Hence, the perimeter of the new rectangle will become two times of the perimeter of the old rectangle.

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Question 2:

If the side of a square is tripled, how many times the perimeter of the first square will that of the new square be?

ANSWER:

Let the length of each side of the old square = s
Then, perimeter of the old square = 4 × side = 4 × s = 4s
When the length of each side of the square is tripled, then
length of each side of the new square = 3s
∴ perimeter of the new square = 4 × side 
= 4 × 3s
= 3 × 4s
= 3 × perimeter of the old square    [∵ perimeter of the old square = 4s]
Hence, the perimeter of the new square will become three times of the perimeter of the old square. 

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Question 3:

Given alongside is the diagram of a playground. It shows the length of its sides. Find the perimeter of the playground.

ANSWER:

It is known that, perimeter of a polygon is equal to sum of the lengths of all the sides of the polygon.
Let us mark the vertices of the given polygon as A, B, C, D, E and F.
The given figure can be broken into rectangle and square by drawing a line CG || BA.

Now, ABCG is a rectangle and GDEF is a square.
Since, opposite sides of rectangle are equal, then
GC = AB = 10 m and AG = BC = 15 m
Since length of each sides of square is equal, then
GD = DE = EF = FG = 15 m
Perimeter of ABCDEFG = AB + BC + CD + DE + EF + FG + GA
= 10 + 15 + 5 + 15 + 15 + 15 + 15
= 90 m

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Question 4:

As shown in the figure, four napkins all of the same size were made from a square piece of cloth of length 1 m. What length of lace will be required to trim all four sides of all the napkins ?

ANSWER:

Length of each side of a square piece of cloth = 1 m
Now, four napkins all of same size were made from this square piece of cloth.
∴ Length of each side of a napkin, s = Length of each side of a square piece of cloth ÷ 2
= 1 ÷ 2
= 0.5 m
Length of lace needed to trim all the four sides of a napkin = perimeter of a napkin
= 4 × s
= 4 × 0.5
= 2 m
∴ Length of lace needed to trim all the four sides of four napkins = 4 × length of lace needed for a napkin
= 4 × 2
= 8 m

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Question 1:

If the side of a square is 12 cm, find its area.

ANSWER:

Length of each side of square, s = 12 cm
∴ Area of square = s²
= (12)²
= 12 × 12
= 144 cm²
Thus, the area of the square is 144 cm².

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Question 2:

If the length of a rectangle is 15 cm and breadth is 5 cm, find its area.

ANSWER:

Length of the rectangle, l = 15 cm
Breadth of the rectangle, b = 5 cm
∴ Area of rectangle = l × b
= 15 × 5
= 75 cm²
Hence, the area of rectangle is 75 cm².

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Question 3:

The area of a rectangle is 102 sqcm. If its length is 17 cm, what is its perimeter ?

ANSWER:

Area of rectangle = 102 cm²
Length of the rectangle, l = 17 cm
∴ Breadth of the rectangle, b = area of rectangle÷ length of rectangle  (∵ area of rectangle = length × breadth)

⇒b = 102÷17 = 6 cm
Perimeter of rectangle = 2(length + breadth)
= 2(17 + 6)
= 2 × 23
= 46 cm

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Question 4:

If the side of a square is tripled, how many times will its area be as compared to the area of the original square ?

ANSWER:

Let the length of each side of original square = s
Area of original square = side × side = s²
When the side of the square is tripled, then
length of each side of new square = 3s
Area of new square = side × side
= 3s × 3s
= 9s²
= 9 × s²
= 9 × area of original square    [∵ area of original square = s²]
Hence, the area of the new square will become nine times the area of the original square.

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Question 1:

A page of a calendar is 45 cm long and 26 cm wide. What is its area ?

ANSWER:

Length of the calendar, l = 45 cm
Breadth of the calendar, b = 26 cm
∴ Area of calendar = l × b
= 45 × 26
= 1170 cm²

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Question 2:

What is the area of a triangle with base 4.8 cm and height 3.6 cm ?

ANSWER:

We have, base of a triangle = 4.8 cm
Height of a triangle = 3.6 cm
∴ Area of a triangle = 12$\frac{1}{2}$ × base × height
= 12$\frac{1}{2}$ × 4.8 × 3.6
= 8.64 cm²

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Question 3:

What is the value of a rectangular plot of land 75.5 m long and 30.5 m broad at the rate of 1000 rupees per square metre ?

ANSWER:

Length of the rectangular plot of land, l = 75.5 m
Breadth of the rectangular plot of land, b = 30.5 m
∴ Area of rectangular plot of land = l × b
= 75.5 × 30.5
= 2302.75 m²
Cost of 1 m² of rectangular plot of land = 1000 rupees
∴ Cost of 2302.75 m² of rectangular plot of land = 1000 × 2302.75
= 2302750 rupees
Hence, the value of the rectangular plot of land is Rs 23,02,750.

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Question 4:

A rectangular hall is 12 m long and 6 m broad. Its flooring is to be made of square tiles of side 30 cm. How many tiles will fit in the entire hall ? How many would be required if tiles of side 15 cm were used?

ANSWER:

Length of the rectangular hall, l = 12 m
Breadth of the rectangular hall, b = 6 m
Area of the floor of the rectangular hall = l × b
= 12 × 6
= 72 m²
Length of each side of square tile used in the first case = 30 cm = 30÷ 100= 0.3 m    (∵ 100 cm = 1 m)
Area of each tile = side × side
= 0.3 × 0.3
= 0.09 m²
∴ Required number of square tiles each of side 30 cm = area of floor ÷ area of each tile 
= 72÷ 0.09
= 800
Length of each side of square tile used in the second case = 15 cm = 15÷ 100= 0.15 m   
 Area of each tile = side × side
= 0.15 × 0.15
= 0.0225 m²
∴ Required number of square tiles each of side 15 cm = area of floor÷ area of each tile
= 72÷ 0.0225
= 3200

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Question 5:

Find the perimeter and area of a garden with measures as shown in the figure alongside.

ANSWER:

Let us mark the vertices of the given polygon as A, B, C, D, E, F, G, H, I, J, K and L.
Perimeter of garden ABCDEFGHIJKL = sum of all the sides of polygon = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL
= 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13 + 13
= 143 m

The given figure of the garden can be broken into five squares as follows.



Now, length of each side of each of the five squares is equal to 13 m.
Area of each square = side × side
= 13 × 13
= 169 m²
∴ Area of the five squares = 5 × area of each square
= 5 × 169
= 845 m²
Hence, the area of the given garden is 845 m².

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Question 1:

Find the total surface area of cubes having the following sides.

(i) 3 cm (ii) 5 cm  (iii) 7.2 m (iv) 6.8 m (v) 5.5 m

ANSWER:

(i) Length of each side of cube, l = 3 cm
∴ Total surface area of the cube = 6 × l²
= 6 × (3)²
= 6 × 9
= 54 cm²

(ii) Length of each side of cube, l = 5 cm
∴ Total surface area of the cube = 6 × l²
= 6 × (5)²
= 6 × 25
= 150 cm²

(iii) Length of each side of cube, l = 7.2 m
∴ Total surface area of the cube = 6 × l²
= 6 × (7.2)²
= 6 × 51.84
= 311.04 m²

(iv) Length of each side of cube, l = 6.8 m
∴ Total surface area of the cube = 6 × l²
= 6 × (6.8)²
= 6 × 46.24
= 277.44 m²

(v) Length of each side of cube, l = 5.5 m
∴ Total surface area of the cube = 6 × l²
= 6 × (5.5)²
= 6 × 30.25
= 181.50 m²

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Question 2:

Find the total surface area of the cuboids of length, breadth and height as given below:

(i) 12 cm, 10 cm, 5 cm  (ii) 5 cm, 3.5 cm, 1.4 cm

(iii) 2.5 cm, 2 m, 2.4 m  (iv) 8 m, 5 m, 3.5 m

ANSWER:

(i) Length of the cuboid, l = 12 cm
Breadth of the cuboid, b = 10 cm
Height of the cuboid, h = 5 cm
∴ Total surface area of the cuboid = 2(l × b + b × h + h × l)
= 2(12 × 10 + 10 × 5 + 5 × 12)
= 2(120 + 50 + 60)
= 2 × 230
= 460 cm²

(ii) Length of the cuboid, l = 5 cm
Breadth of the cuboid, b = 3.5 cm
Height of the cuboid, h = 1.4 cm
∴ Total surface area of the cuboid =  2(l × b + b × h + h × l)
= 2(5 × 3.5 + 3.5 × 1.4 + 1.4 × 5)
= 2(17.5 + 4.9 + 7.0)
= 2 × 29.4
= 58.8 cm²

(iii) Length of the cuboid, l = 2.5 cm = 2.5100$\frac{2.5}{100}$ m = 0.025 m    [∵ 100 cm = 1 m]
Breadth of the cuboid, b = 2 m
Height of the cuboid, h = 2.4 m
∴ Total surface area of the cuboid =  2(l × b + b × h + h × l)
= 2(0.025 × 2 + 2 × 2.4 + 2.4 × 0.025)
= 2(0.05 + 4.8 + 0.06)
= 2 × 4.91
= 9.82 m²

(iv)  Length of the cuboid, l = 8 m
Breadth of the cuboid, b = 5 m
Height of the cuboid, h = 3.5 m
∴ Total surface area of the cuboid =  2(l × b + b × h + h × l)
= 2(8 × 5 + 5 × 3.5 + 3.5 × 8)
= 2(40 + 17.5 + 28)
= 2 × 85.5
= 171 m²

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Question 3:

A matchbox is 4 cm long, 2.5 cm broad and 1.5 cm in height. Its outer sides are to be covered exactly with craft paper. How much paper will be required to do so ?

ANSWER:

Length of the matchbox, l = 4 cm
Breadth of the matchbox, b = 2.5 cm
Height of the matchbox, h = 1.5 cm
∴ Surface area of the box = 2(l × b + b × h + h × l)
= 2(4 × 2.5 + 2.5 × 1.5 + 1.5 × 4)
= 2(10 + 3.75 + 6)
= 2 × 19.75
= 39.5 cm² 
Hence, 39.5 cm² of the craft paper will be needed to cover the matchbox.

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Question 4:

An open box of length 1.5 m, breadth 1 m, and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box ? The inside and outside surface of the box is to be painted with rust proof paint. At a rate of 150 rupees per sqm, how much will it cost to paint the box?

ANSWER:

Length of the open box, l = 1.5 m
Breadth of the open box, b = 1 m
Height of the open box, h = 1 m
Surface area of open box = Total surface area of the box − Area of the top
= 2(l × b + b × h + h × l) − l × b
= 2(1.5 × 1 + 1 × 1 + 1 × 1.5) − 1.5 × 1
= 2(1.5 + 1 + 1.5) − 1.5
= 2 × 4 − 1.5
= 8 − 1.5
= 6.5 m²
Hence, 6.5 m² of the sheet will be needed to make the open box.
Now, it is given that the inside and the outside surface of the open box is to be painted with rust proof paint.
∴ Total area of the box to be painted = 2 × Surface area of open box
= 2 × 6.5
= 13 m²
Now, cost of painting of 1 m² of area = 150 rupees
∴ Cost of painting of 13 m² of area = 13 × 150 = 1950 rupees
Hence, it will cost 1950 rupees to paint the open box from inside and outside with rust proof paint.

LIGHT (Std.6)

Question 1:

Choose an appropriate word and fill in the blanks.
(a) A ........... is a natural source of light.
(b) A ......... is an artificial source of light.
(c) When light passes through a prism, it gets separated into ....... colours.
(d) The image obtained in the pinhole camera is ......... .
(e) A shadow is formed when an ........... object comes in the way of light.
(f) When a ........... object comes in the way of light, light passes ......... it.
(options : seven, star, through, transparent, opaque, colors, shape, erect, inverted, luminous, candle)

ANSWER:

(a) A star is a natural source of light.
(b) A candle is an artificial source of light.
(c) When light passes through a prism, it gets separated into seven colours.
(d) The image obtained in the pinhole camera is inverted.
(e) A shadow is formed when an opaque object comes in the way of light.
(f) When a transparent object comes in the way of light, light passes through it.

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Question 2:

Write whether of the following objects are luminous or non-luminous.

Object	Luminous/Non-luminous
A book
A burning candle
A wax cloth
A pencil
A pen
A light bulb
A tyre
A torch

ANSWER:

Object	Luminous/Non-luminous
A book	Non-luminous
A burning candle	Luminous
A wax cloth	Non-luminous
A pencil	Non-luminous
A pen	Non-luminous
A light bulb	Luminous
A tyre	Non-luminous
A torch	Luminous

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Question 3:

Match the following.

Group A	Group B
(a) Mirror	(1) Non-luminous
(b) Firefly	(2) Inverted image
(c) Pinhole camera	(3) Reflection
(d) Moon	(4) Luminous

ANSWER:

Group A	Group B
(a) Mirror	(3) Reflection
(b) Firefly	(4) Luminous
(c) Pinhole camera	(2) Inverted image
(d) Moon	(1) Non-luminous

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Question 4:

Write the answers to the following.
(a) What things are necessary for the formation of a shadow ?
(b) When can an object be seen?
(c) What is a shadow?

ANSWER:

(a) The things necessary for the formation of a shadow are:

• A source of light
• An opaque object whose shadow has to be formed
• A screen on which shadow is to be formed

(b) An object is seen when the light rays reflected from its surface reaches our eyes.

(c) Shadow is a dark patch obtained on a screen where the light from a light source is blocked by an opaque object.

SOUND

Question 1: Fill in the blanks with the proper words. 

(a) The propagation of sound does not occur through a ........... . 

(b) Noise pollution is a .......... issue. 

(c) The sound which is disagreeable to the ears is called..... .

 (d) Noise has adverse effects on our .......... .

ANSWER: (a) The propagation of sound does not occur through a vacuum.

(b) Noise pollution is a social issue.

(c) The sound which is disagreeable to the ears is called noise.

 (d) Noise has adverse effects on our health.

 Page No 96: 

Question 2: 

What should we do ?

 (a) The silencer of a motorcycle is broken. (b) A factory in the surroundings is producing continuous loud noise.

 ANSWER: 

a) If the silencer of a motorcycle is broken, then take the motorcycle to a mechanic and get it reinstalled or repaired.

 (b) If a factory in the surroundings is producing continuous loud noise, then we can lodge a complaint against the factory to a local government body citing the troubles faced by the residents around and the danger it is causing to the environment.

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 Question 3:

 Write the answers in your own words.

 (a) What is meant by vibration ?

 (b) Explain with the help of practical examples how sound is propagated through solids.

 (c) What is meant by noise pollution? 

(d) What measures will you take to control noise pollution ? 

 ANSWER: (a) The to and fro motion of a body about its mean position is known as vibration.

 (b) Sound is propagated in the form of waves in solids. It can be explained clearly from the below examples: Place your ear on one end of a long table. Ask your friend to tap the table from the other end. You can hear the produced sound. Sound reaches your ear after travelling through the table. This indicates that sound can travel through solids and in the form of waves. ​Stethoscope is another example of sound travelling through solids. Doctors use stethoscopes to listen to our heartbeat.

 (c) Noise pollution is the excess of unwanted sound in the environment. It can adversely affect human and other living beings health.

 (d) Following measures can be taken to control noise pollution: To control noise pollution, we must control its source. Hence, silencers must be installed in vehicles such as motorcycles, cars, trucks, buses, and other noise producing machines. We should watch television and listen to music at a low volume. Also, use of loudspeakers as well as horns of buses and trucks should be minimized. Regular maintenance of automobiles should be done so that noise produced by them can be kept under check. All industrial work should be done away from residential areas. More trees should be planted in residential areas as they help in reducing noise.