5 Marks Questions

Question 15 Marks

A rectangular paper chit of dimension 6 cm × 4 cm is cut as shown into two equal pieces. These two pieces are joined in different ways.

For example, the arrangement a. has a perimeter of 28 cm.
Find out the length of the boundary (i.e., the perimeter) of each of the other arrangements below.

Answer

(b)

∴ Length of boundary = AB + BC + CD + DE + EF + FG + GA
= 6 + 2 + 6 + 2 + 4 + 6 + 2
= 28 cm
(c)

Total length of boundary = AB + BC + CD + DE +EF + FG + GH + HA
= 2 + 6 + 2 + 2 + 6 + 2 + 6 + 2
= 28 cm
(d)

Total length of boundary = AB + BC + CD + DE + EF + FG + GH + HA
= 6 + 2 + 3 + 2 + 6 + 2 + 3 + 2
= 26 cm

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Question 25 Marks

Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalize your understanding of the perimeter of other regular polygons.

Answer

Some common objects with regular shapes and calculating their perimeters:
Here are a few examples:
1. Square Table:

Shape: Square
Side Length = 1 meter
Perimeter = 4 × 1 = 4 meters
2. Equilateral Triangle Clock:

Shape: Equilateral Triangle
Side Length = 30 cm
Perimeter = 3 × 30 = 90 cm
3. Hexagonal Tile:

Shape: Regular Hexagon
Side Length = 10 cm
Perimeter = 6 × 10 = 60 cm
In general, the Perimeter of a Regular Polygon = (Number of sides) × (Side length of a polygon) units.

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Question 35 Marks

Use your understanding from previous grades to calculate the area of any closed figure using grid paper and-

Q.1. Find the area of the blue triangle BAD.
Q.2. Find the area of the red triangle ABE.

Answer

1. Area of blue triangle BAD

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	6	6 × 1 = 6
Half-filled squares	2	2 × $\frac{1}{2}$ = 1
More than half-filled squares	3	3 × 1 = 3
Less than half-filled squares	3	0

∴ Total area of BAD = 6 + 1 + 3 = 10 sq. units
2. Area of red triangle ABE

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	5	5 × 1 = 5
Half-filled squares	2	2 × $\frac{1}{2}$ = 1
More than half-filled squares	4	4 × 1 = 4
Less than half-filled squares	3	0

∴ Total area of ABE = 5 + 1 + 4 = 10 sq. units
Area of rectangle ABCD = Number of fully tilled squares
= 20 × 1
= 20 sq. units

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Question 45 Marks

whether the two triangles overlap each other exactly. Do they have the same area?

Answer

Two triangles overlap each other exactly, which means they are congruent. Congruent triangles have the same shape and size, which implies that they also have the same area.

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Question 55 Marks

Akshi and Toshi start running along the rectangular tracks as shown in the figure. Akshi runs along the outer track and completes 5 rounds. Toshi runs along the inner track and completes 7 rounds. Now, they are wondering who ran more. Find out who ran the longer distance.

Answer

Here, perimeter of rectangular track PQRS = 2 × (l + b)
= 2 × (70 + 40)
= 2 × 110
= 220 m
and perimeter of rectangular track ABCD = 2 × (60 + 30)
= 2 × 90
= 180 m

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Question 65 Marks

On a squared grid paper (1 square = 1 square unit), make as many rectangles as you can whose lengths and widths are a whole number of units such that the area of the rectangle is 12 square units.
(a) Which rectangle has the greatest perimeter?
(b) Which rectangle has the least perimeter?
(c) If you take a rectangle of area 32 sq cm, what will your answers be? Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter? Give examples and reasons for your answer.

Answer

Perimeter of (a) = 2(1 + 24) = 2 × 25 = 50 units
Perimeter of (b) = 2(2 + 12) = 2 × 14 = 28 units
Perimeter of (c) = 2(4 + 6) = 2 × 10 = 20 units
Perimeter of (d) = 2(3 + 8) = 2 × 11 = 22 units
(a) Clearly rectangle (a) has the greatest perimeter.
(b) Obviously rectangle (c) has the least perimeter.
(c) Yes, it is possible to predict the shape of a rectangle with the greatest and least perimeter for a given area. Here’s how:
Greatest Perimeter: For a given area, the rectangle with the greatest perimeter will have one side as small as possible. This essentially means that the rectangle becomes very elongated.
For example, if the area is 24 square units, a rectangle with dimensions 1 unit by 24 units will have the greatest perimeter.
Example: Area = 24 square units
Dimensions = 1 unit by 24 units
Perimeter = 2(1 + 24) = 50 units
Least Perimeter: The rectangle with the least perimeter for a given area will be as close to a square as possible. This is because a square has the smallest perimeter for a given area among all rectangles.
Example: Area = 24 square units
Dimensions = 4 units by 6 units (since 4 × 6 = 24)
Perimeter = 2(4 + 6) = 20 units
Reasoning
Greatest Perimeter: When one side is minimized, the other side must be maximized to maintain the same area. This increases the sum of the sides, thus increasing the perimeter.
Least Perimeter: A square or a shape close to a square minimizes the sum of the sides for a given area, thus minimizing the perimeter.

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Question 75 Marks

Find the area of the following figures.

Answer

(i)

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	3	$3 \times 1=3$
Half-filled squares	2	2 × $ \frac{1}{2} $ = 1

∴ Total area of the figure = 3 + 1 = 4 sq. units
(ii)

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	6	$6 \times 1=6$
Half-filled squares	6	6 × $ \frac{1}{2} $ = 3

∴ Total area of the figure = 6 + 3 = 9 sq. units
(iii)

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	7	$7 \times 1=7$
Half-filled squares	6	6 × $ \frac{1}{2} $ = 3

∴ Total area of the figure = 7 + 3 = 10 sq. units
(iv)

Covered Area	Number	Area Estimated (sq. units)
Fully-filled squares	8	$8 \times 1=8$
Half-filled squares	6	6 × $ \frac{1}{2} $ = 3
More than half of a square	0

∴ Total area of the figure = 8 + 3 = 11 sq. unit

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Question 85 Marks

A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here?
(a) The area of each rectangle is larger than the area of the square.
(b) The perimeter of the square is greater than the perimeters of both the rectangles added together.
(c) The perimeters of both the rectangles added together are always $1 \frac{1}{2}$ times the perimeter of the square.
(d) The area of the square is always three times as large as the areas of both rectangles added together.

Answer

Now in the above square piece
side of square = 1 unit
area of square = 1 × 1 = 1 sq. unit.
and perimeter of square = 1 + 1 + 1 + 1 = 4 units.
Now after folding the above square piece in half becomes 2 rectangles

Perimeter of rectangle $R _1=1+\frac{1}{2}+1+\frac{1}{2}=3$ units.
Area of rectangle $R _1=\frac{1}{2} \times 1=\frac{1}{2}$ sq. unit.
Perimeter of rectangle $R _2=1+\frac{1}{2}+1+\frac{1}{2}=3$ units.
Area of rectangle $R _2=\frac{1}{2} \times 1=\frac{1}{2}$ sq. unit.
(a) Now, area of rectangle $R_1=$ area of rectangle $R_2=\frac{1}{2}<1$.
Hence, option (a) is not true.
(b) Here perimeter of square = 4 units
and perimeters of both the rectangles = 3 + 3 = 6 units.
which is greater than 4 units.
Hence option (b) is not true.
(c) Here perimeters of both the rectangles = 6 units
and perimeter of square $=4$ units $\times 1 \frac{1}{2}=4 \times \frac{3}{2}=6$ units.
The perimeters of both the rectangles added together are $1 \frac{1}{2}$ times the perimeter of the square.
Hence, option (c) is true.
(d) Here, the area of the square = 4 units
and areas of both the rectangles $=\frac{1}{2}+\frac{1}{2}=1$ unit.
The area of the square is four times the area of both rectangles.
Hence, option (d) is not true.

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Question 95 Marks

Four flower beds having sides 2 m long and 1 m wide are dug at the four comers of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?

Answer

We have, length of the garden = 15 m
and width of the garden = 12 m
∴ The area of the garden = Length × Width
= 15 × 12
= 180 sq. m
Now, also given that length of a flower bed = 2 m
and width of a flower bed = 1 m
The area of a flower bed = Length × Width
= 2 × 1
= 2 sq. m
Total area of 4 flower beds = 4 × 2
= 8 sq. m
The area available for laying down a lawn
= Area of the garden – Area of 4 flower beds
= 180 – 8
= 172 sq. m

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Question 105 Marks

Find the areas of the figures below by dividing them into rectangles and triangles.

Answer

(a) Figure have 20 full rectangles + 4 more than half rectangles + 4 less than half rectangles
= 20 × 1 + 4 × 1 + 4 × 0
= 20 + 4
= 24 sq. units
(b) Figure have 24 full rectangles, 2 half rectangles, 3 more than half and 3 less than half
∴ Area of figure $=24+1+2 \times \frac{1}{2}+3 \times 1+3 \times 0$
= 24 + 1 + 3 + 0
= 28 sq. units
(c) Figure have 36 full rectangles, 2 half rectangles, 9 more than half, and 10 less than half rectangles
∴ Area of figure $=36 \times 1+2 \times \frac{1}{2}+9 \times 1+10 \times 0$
= 36 + 1 + 9 + 0
= 46 sq. units.
(d) Figure have 13 full rectangles, 1 half, 2 more than half and 2 less than half.
∴ Area of figure $=13 \times 1+1 \times \frac{1}{2}+2 \times 1+2 \times 0$
= 13 + 0.5 + 2
= 15.5 sq. units
(e) Figure have 5 full rectangles, 5 half, 3 more than half and 4 less than half.
∴ Area of figure $=5 \times 1+5 \times \frac{1}{2}+3 \times 1+4 \times 0$
= 5 + 2.5 + 3
= 10.5 sq. units

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Question 115 Marks

By splitting the following figures into rectangles, find their areas (all measures are given in meters):

Answer

(a) Splitting the given figure into I, II, III, and IV rectangles as shown in the figure below, we get

Here, the area of rectangle I = length × breadth
= 4 cm × 3 cm
= 12 sq. cm
Area of rectangle II = length × breadth
= 3 cm × 2 cm
= 6 sq. cm
Area of rectangle III = length × breadth
= 4 cm × 1 cm
= 4 sq. cm
Area of rectangle IV = length × breadth
= 3 cm × 2 cm
= 6 sq. cm
The total area of the whole figure = 12 sq. cm + 6 sq. cm + 4 sq. cm + 6 sq. cm = 28 sq. cm.
Therefore, the total area of Figure (a) is 28 sq. cm.
(b) Similarly, by splitting figure (b) into I, II, and III rectangles as shown in the figure below, we get

Area of the rectangle I = length × breadth
= 3 cm × 1 cm
= 3 sq. cm
Area of rectangle II = length × breadth
= 3 cm × 1 cm
= 3 sq. cm
Area of rectangle III = length × breadth
= 3 cm × 1 cm
= 3 sq. cm
The total area of the figure = 3 sq. cm + 3 sq. cm + 3 sq. cm = 9 sq. cm.
Therefore, the total area of Figure (b) is 9 sq. cm.

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Question 125 Marks

Think and mark the positions as directed:
(a) Mark ‘A’ at the point where Akshi will be after she runs 250 m.
(b) Mark ‘B’ at the point where Akshi will be after she runs 500 m.
(c) Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘C’.
(d) Mark ‘X’ at the point where Toshi will be after she runs 250 m.
(e) Mark ‘Y’ at the point where Toshi will be after she runs 500 m.

Answer

(a) Distance covered by Akshi in 1 complete round = 220 m.
so, to cover ‘250 m Akshi have to move 30 m more = 220 + 30 = 250 m.
(b) Since the perimeter of the outer rectangle is 220 m, it implies that the distance covered till the starting point in 2 rounds will be 2 × 220 = 440 m
Now, the point where Akshi will cover 500 m will be 60 m away from this point.
(c) Distance covered by Akshi $=1000 m$
No. of rounds $=\frac{ 1 0 0 0 }{ 2 2 0 }=4$ complete rounds and 120 m more.
Thus, she'll be 120 m away from the starting point.
(d) Distance covered by Toshi in 1 round $=180 m$
So, to cover 250 m , she'll be 70 m away from the starting point.
(e) Since, the perimeter of the inner rectangle is 180 m , so to cover 500 m , Toshi will be taking 2 complete rounds and then she'll have to move 140 m more i.e. 140 m away from the starting point.

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Question 135 Marks

Each track is a rectangle. Akshi’s track has length 70 m and breadth 40 m. Running one complete round on this track would cover 220 m, i.e., 2 × (70 + 40) m = 220 m. This is the distance covered by Akshi in one round.
Question :
Q.1. Find out the total distance Akshi has covered in 5 rounds.
Q.2. Find out the total distance Toshi has covered in 7 rounds.
Who ran a longer distance?

Answer

1. Akshi runs on a rectangular track with a length of 70 metres and a breadth of 40 metres.
∴ Perimeter of track = 2 × (length + breadth)
= 2 × (70 + 40) = 220 m
Since, the distance covered in one round = 220 m
∴ Total distance covered in 5 rounds = 5 × 220 m
= 1100 m
2. Toshi runs on a rectangular track with a length of 60 m and breadth of 30 m.
∴ Perimeter of track = 2 × (length + breadth)
= 2 × (60 + 30) = 180 m
Since, the distance covered in ope round = 180 m
∴ Total distance covered in 7 rounds = 7 × 180 m
= 1260 m
So, Toshi ran a longer distance.

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