1. Six unit squares are joined to create the following figures. Which of the figures in Fig. 6.4 has the smallest perimeter?
1. Six unit squares are joined to create the following figures. Which of the figures in Fig. 6.4 has the smallest perimeter?
Solution:
The perimeter is smallest in Figure I
A closed figure's perimeter is the distance traversed once along the figure's edge.
We are aware that a figure's perimeter equals its number of sides times its average side length.
Think about the numbers provided.
10 units make up figure I and each side's length is 1 cm.
hence, perimeter = 10 1
= 10 cm
Figure (ii) comprises 12 units and each side measures 1 cm in length.
hence, perimeter = 12 1
= 12 cm
Figure (iii) has 14 sides, each measuring 1 cm in length.
hence, perimeter = 14 1
= 14 cm
Figure (iv) has 14 sides, each measuring 1 cm in length.
hence, perimeter = 14 1
= 14 cm
Figure I has the shortest perimeter when the perimeters of all the other figures are compared.
2. Two identical rectangular flower beds, each measuring 10m 5m, are located in a square-shaped park ABCD with side 100m (Fig. 6.5). The remaining park's perimeter is one mile long.
(A)360m , (B)400m , (C)340m , (D) 460m
2. Two identical rectangular flower beds, each measuring 10m 5m, are located in a square-shaped park ABCD with side 100m (Fig. 6.5). The remaining park's perimeter is one mile long.
(A)360m , (B)400m , (C)340m , (D) 460m
Solution:
400m
We are aware that the boundary's perimeter equals its length.
Provided are two rectangular flowerbeds that are each 10 m 5 m in size.
Hence, the remaining park's perimeter is 90 + 5 + 10 + 95 + 90 + 5 + 10 + 95.
= 400 m
3. A square's side measures 10 cm. If one side of the square is doubled, how many times larger would the new perimeter be?
A) twice, B) four times, and C) six times (D) eight times
3. A square's side measures 10 cm. If one side of the square is doubled, how many times larger would the new perimeter be?
A) twice, B) four times, and C) six times (D) eight times
Solution:
(A) twice
We are aware that side 4 equals the square's circumference.
= 10 × 4
= 40 cm
Assuming that a square's side is 10 cm
The square's side is doubled, or 10 + 10.
= 20 cm
As a result, the square's new perimeter is double its side length.
4. A rectangular sheet of paper has dimensions of 20 cm in length and 10 cm in width. As seen in Fig. 6.6, a rectangular piece is cut from the sheet. Which of the following assertions applies to the final sheet correctly?
(A) The area varies, but the perimeter stays the same.
(B) The area doesn't change, but the perimeter does.
( C)Area and perimeter are both changing .
(D) The area and perimeter stay unchanged.
4. A rectangular sheet of paper has dimensions of 20 cm in length and 10 cm in width. As seen in Fig. 6.6, a rectangular piece is cut from the sheet. Which of the following assertions applies to the final sheet correctly?
(A) The area varies, but the perimeter stays the same.
(B) The area doesn't change, but the perimeter does.
( C)Area and perimeter are both changing .
(D) The area and perimeter stay unchanged.
Solution:
(A) The area varies, but the perimeter stays the same.
We are aware that a large rectangle's area equals its length times its breadth.
= 10 × 20
= 200 cm2
5 x 2 is the area of a tiny rectangle.
= 10 cm2
Rectangle perimeter equals 2 (length + width).
= 2(20 + 10)
= 2 × 30
= 60 cm
Hence, the new figure's perimeter is 20 + 8 + 5 + 2 + 15 + 10
= 60
Area of the new figure is equal to the sum of its area and that of the large rectangle.
= 200 – 10
= 190 cm2
Perimeter remains the same but the area changes when all the findings are compared.
5. As seen in Fig. 6.7, two regular hexagons with perimeters of 30 cm each are connected. The new figure's edge measures
A 65 cm, B 60 cm, C 55 cm, and D 50 cm
5. As seen in Fig. 6.7, two regular hexagons with perimeters of 30 cm each are connected. The new figure's edge measures
A 65 cm, B 60 cm, C 55 cm, and D 50 cm
Solution:
(D) 50cm
The answer to the question is that each hexagon's perimeter is 30 cm.
The hexagon has six sides, as we are well aware.
Hence, each hexagonal side's length is equal to 30/6.
= 5 cm
Think about two hexagons that are connected together.
The new figure's perimeter is then equal to the sum of its sides' lengths.
= 10 × 5
= 50 cm
6. Match the shapes (each side measures 2cm) in column I with the corresponding perimeters in column II
6. Match the shapes (each side measures 2cm) in column I with the corresponding perimeters in column II
Solution:
Assuming that each side is 2 cm long.
Fig. (A) has fourteen sides.
Figure perimeter equals the sum of the lengths of all its sides.
= 14 × 2
= 28 cm
Eight sides make up Figure (B).
Figure perimeter equals the sum of the lengths of all its sides.
= 8 × 2
= 16 cm
The shape (C) has ten sides.
Figure perimeter equals the sum of the lengths of all its sides.
= 10 × 2
= 20 cm
12 sides make up Figure (D).
Figure perimeter equals the sum of the lengths of all its sides.
= 12 × 2
= 24 cm
7. Match the following
7. Match the following
Solution:
(A) A rectangle's perimeter equals 2 (length + width).
= 2 × (6 + 4)
= 2 × 10
= 20
(B) A square's perimeter equals 4 times the length of its sides.
= 4 × 5
= 20
(C) Equilateral triangle perimeter equals three times the length of a side
= 3 × 6
= 18
(D) Isosceles triangle perimeter equals the total length of all sides
= 4 + 4 + 2
= 10
8. The area of a rectangle obtained when the length is cut in half and the width is doubled stays constant.
8. The area of a rectangle obtained when the length is cut in half and the width is doubled stays constant.
Solution:-
True.
We are aware that a rectangle's area equals its length times its width.
According to the condition stated in the question, a rectangle's length is cut in half, or l/2.
Double the breadth to get 2b.
Hence, new area = l/2 2b.
= l × b.
As a result, if a rectangle's length is cut in half and its width is doubled, the resulting area of the rectangle stays the same.
9. If one of the sides of a square is doubled, the area of the square also doubles.
9. If one of the sides of a square is doubled, the area of the square also doubles.
Solution:-
False
We are aware that a square's area is equal to its sides.
The square's side is doubled according to the question's condition, which is given as 2 side.
next, fresh region = 2 sides by 2 sides
= 4 side2
As a result, doubling a square's side results in a four-fold increase in the square's size.
10. A normal octagon with a 6 cm side has a 36 cm perimeter.
10. A normal octagon with a 6 cm side has a 36 cm perimeter.
Solution:-
False.
The perimeter of a regular octagon is equal to the sum of the lengths of its sides.
= 8 × 6
= 48 cm
11. An engineer needs to identify the compound's location if they intend to construct a wall around the entire residence.
11. An engineer needs to identify the compound's location if they intend to construct a wall around the entire residence.
Solution:-
False.
An engineer must determine the compound's perimeter if they intend to construct a compound wall around the entire structure.
12. We must determine the wall's perimeter in order to calculate the price of painting it.
12. We must determine the wall's perimeter in order to calculate the price of painting it.
False.
We must determine the wall's area in order to calculate the cost of painting it.
13. Drawing four regular hexagons results in the pattern seen in Fig. 6.11. Find the length of each side of the hexagon if the design's perimeter is 28 cm.
13. Drawing four regular hexagons results in the pattern seen in Fig. 6.11. Find the length of each side of the hexagon if the design's perimeter is 28 cm.
Solution:-
The answer to the question implies that the design's perimeter is 28 cm.
Assuming that the design has 14 sides, each side's length equals 28/14.
= 2 cm
Four regular hexagons are linked in the provided figure.
Hence, the hexagon's sides are each 2 cm long.
14. An isosceles triangle has a perimeter of 50 cm. Find the third side if one of the two equal sides is 18 cm.
14. An isosceles triangle has a perimeter of 50 cm. Find the third side if one of the two equal sides is 18 cm.
Solution:-
We are aware that an isosceles triangle has two sides of equal length.
One of the two equal sides is 18 cm, which may be inferred from the question.
Moreover, an isosceles triangle's perimeter is 50 cm.
We are aware that the isosceles triangle's perimeter equals the sum of its sides.
Assume that side three is x.
Then,
50 = 18 + 18 + x
50 = 36 + x
X = 50 – 36
X = 14 cm
Hence, the third side of an isosceles triangle is 14 cm long.
15. A rectangle's length is equal to its width times three. The rectangle has a 40 centimetre perimeter. Identify its width and length.
15. A rectangle's length is equal to its width times three. The rectangle has a 40 centimetre perimeter. Identify its width and length.
Solution:-
It is clear from the question that,
rectangle's perimeter is 40 cm.
A rectangle's length is equal to three times its width (3b).
The formula for a rectangle's perimeter is: 2 (length + breadth).
40 = 2 × (3b + b)
40 = 2 × 4b
40 = 8b
40/8 = b
b = 5 cm
Hence, the rectangle's width is 5 cm.
rectangle's length is 3b.
= 3 × 5
= 15 cm
16. 200 rods were used by Tahir to gauge the distance around a square field (lathi). Later, he discovered that this rod was 140 cm long. Determine the field's side in metres.
16. 200 rods were used by Tahir to gauge the distance around a square field (lathi). Later, he discovered that this rod was 140 cm long. Determine the field's side in metres.
Solution:-
It is clear from the question that,
200 rods were used by Tahir to gauge the distance around a square field.
The rod is 140 cm long.
Hence, the square field's overall length is equal to 200 x 140.
= 28000 cm
Thus, one side of the square field's length is equal to 28000/4.
= 7000 cm
Knowing that 1 m equals 100 cm
side of field therefore equals 7000/100
= 70 m
17. A rectangular field has a length that is equal to its width. Jamal accomplished a 6km distance by jogging around it four times. How long is the field exactly?
17. A rectangular field has a length that is equal to its width. Jamal accomplished a 6km distance by jogging around it four times. How long is the field exactly?
Solution:-
It is clear from the question that,
A rectangular field's length is equal to twice its width, or 2b.
Rectangular field perimeter: 6 km/4
= 1.5 km
The formula for a rectangle's perimeter is: 2 (length + breadth).
1.5 = 2 × (2b + b)
1.5 = 2 × 3b
1.5 = 6b
b = 1.5/6
b = 0.25 km
field's length is equal to 2 divided by 0.25, or 0.5km.
= 0.5 × 1000
= 500 m
18. A rectangular field has a width of 150 metres and a length of 250 metres. Anuradha completes three laps of this field. She ran for how far? How many times should she circle the field in order to cover a 4km distance?
18. A rectangular field has a width of 150 metres and a length of 250 metres. Anuradha completes three laps of this field. She ran for how far? How many times should she circle the field in order to cover a 4km distance?
Solution:
It is clear from the question that,
A rectangular field is 250 metres long.
Rectangular field width is 150 metres.
Rectangle perimeter equals 2 (length + width).
= 2 × (250 + 150)
= 2 × 400
= 800 m
Considering that, Anuradha circles this field three times, or three times 800.
= 2400 m
We are aware that one kilometre equals one thousand metres.
So, 2400/1000
= 2.4 km, or 2 km and 400 m.
To reach a distance of 4 kilometres, Anuradha must circle the field 4000/800 times.
= 5 times
As a result, Anuradha should run five times around the field to cover a distance of 4km.
19. In front of Meena's home is a rectangular lawn that is 10 metres long and 4 metres broad (Fig. 6.12). Two of the shorter sides and one of the longer sides are walled, leaving a gap of one metre for the entrance. Determine the fencing's length.
19. In front of Meena's home is a rectangular lawn that is 10 metres long and 4 metres broad (Fig. 6.12). Two of the shorter sides and one of the longer sides are walled, leaving a gap of one metre for the entrance. Determine the fencing's length.
Solution:-
It is clear from the question that,
The rectangular lawn is 10 metres long.
rectangular lawn width = 4 m
Then,
Fencing perimeter equals fencing length
Fencing perimeter equals three sides: two shorter sides, one longer side, and a gap in the longer side.
= 4 + 4 + 10 – 1
= 18 – 1
= 17 m
Thus, the fencing is 17 metres long.
20. What is the length of the park's exterior boundary, as depicted in the figure? What will it cost to fence it in overall, at a rate of Rs. 20 per metre? The park's centre is marked by a rectangular flowerbed. At a rate of Rs. 50 per square metre, calculate the cost of maintaining the flowerbed.
20. What is the length of the park's exterior boundary, as depicted in the figure? What will it cost to fence it in overall, at a rate of Rs. 20 per metre? The park's centre is marked by a rectangular flowerbed. At a rate of Rs. 50 per square metre, calculate the cost of maintaining the flowerbed.
Solution:
The park's perimeter measures (200 + 300 + 80 + 300 + 200 + 260) m, or 1340 m. At a cost of Rs. 20 per metre, the cost to fence the park would be Rs. (20 1340) = Rs. 26800.
The flowerbed's size is (100 80) m2, or 8000 m2.
At the rate of Rs. 50 per square metre, the price to maintain the flowerbed is
= Rs. (50 8000).
=Rs. 400000
