Std 8 maths practice work ex 11.1 q13
Answers
Answer:
1. A square and a rectangular filled with measurements as given in the figure have the same perimeter. Which field has a larger area?
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Ans: Given that,
Side of the square =60 m
Length of the rectangle =80 m
Both the square and rectangle have same perimeter
To find,
Which field has larger area
Firstly to find the breadth of the rectangle
We know that,
Perimeter of the square = Perimeter of the rectangle
4(side of the square)=2(length×breadth)
4(60)=2l+2b
240=2(80)+2(b)
2b=240−160
2b=80
b=40 m
Now area of the square=(side)2
=(60)2
=3600 m2
Area of the rectangle =length×breadth
=80×40
=3200 m2
Area of the square > Area of the rectangle
∴ The area of the square has a comparatively larger area than the area of the rectangle.
2. Mrs. Kaushik has a square plot with the measurement as shown in the following figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs.55 per m2.
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Ans: Given that,
Side of the square plot =25 m
Length of the rectangular house =20 m
Breadth of the rectangular house =15 m
Cost of developing garden per m2=Rs.55
To find,
The total cost of developing a garden around the house
Area of the square =(side)2
=(25)2
=625 m2
Area of the rectangular house =length×breadth
=20×15
=300 m2
From the diagram, it can be noted that,
Area of the garden = Area of the square plot – Area of the rectangular house
=625−300
=325 m2
Cost of developing garden per m2=Rs.55
Cost of developing 325 m2 garden =325×55
=Rs.17,875
∴ The total cost for developing garden around the house will be Rs.17,875
3. The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of the garden Length of rectangle is 20−(3.5+3.5) meters
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Ans: Given that,Total length of the garden =20 m
Length of the rectangular part =20−(3.5+3.5)
=20−7
=13 m
Breadth of the rectangular part =7 m
Diameter of semi-circular part=7 m
Radius of semi-circular part =72
=3.5 m
To find,
The perimeter of the garden and the area of the garden
Finding the perimeter of the garden
Perimeter of the garden = Perimeter of rectangular part + Perimeter of two semi-circular part
Perimeter of rectangular part =2(l+b)
=2(13+7)
=2(20)
=40 m
Perimeter of two semi-circular part =2(2πr2)
=2πr
=2×227×3.5
=22 m
Perimeter of the garden =40+22
=62 m
Area of the garden = Area of the rectangular part + Area of the two semi-circular part
Area of rectangular part =l×b
=13×7
=91 m2
Area of two semi-circular part =2(πr22)
=πr2
=227×3.5×3.5
=38.5 m2
Area of the garden =91+38.5
=129.5 m2
∴ The perimeter of the garden is 62 m and the area of the garden is 129.5 m2.
4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2?(If required you can split the tiles in whatever way you want to fill up the corners).
Ans: Given that,
Length of the parallelogram tile=24 cm
Breadth of the parallelogram tile=10 cm
Area of the floor with tiles =1080 m2
Area of the a single parallelogram tile=length×breadth
=24×10
=240 cm2
Number of tiles required =Total area of the floorArea of each tile
We know that,
1 m=100 cm
1 m2=10000 cm2
1080 m2=1080×10000 cm2
n=1080×10000240
=45,000 tiles
∴45,000 tiles are required to make a floor.
5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food piece would the ant have to take a longer round? Remember, the circumference of a circle can be obtained by using the expression c=2πr, where r is the radius of the circle.
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Ans:
(a) Given that,
Diameter of the semicircle =2.8 cm
Radius of the semi-circle=2.82
=1.4 cm
Circumference of the semi-circle =2πr2
=227×1.4
=4.4 cm
Total distance around the food piece (a) = Diameter+Perimeter of the semi-circle=2.8+4.4 =7.2 cm
Total distance covered by ant in the food piece (a) is 7.2 cm
(b) Given that,
Length of the given figure =2.8 cm
Breadth of the given figure =1.5 cm
Radius of semi-circular part =2.82
=1.4 cmPerimeter of the semicircular part =2πr2
=227×1.4
=4.4 cm
=10.2 cm
Total distance covered by ant in food piece (b) is 10.2 cm
(c) Given that,
Two sides of the triangle are 2 cm, 2 cm(c) Given that,
Two sides of the triangle are 2 cm, 2 cm
Diameter of the semi-circular part =2.8 cm
Radius of the semi-circular part =2.82
=1.4 cm
Circumference of semi-circular part =2πr2
=227×1.4
=4.4 cm
Total distance covered by ant in food piece (c)=2+2+4.4
=8.4 cm
∴ For the food piece (b), the ant have to take a longer time since the perimeter is comparatively larger for food piece (b)
- hope you understand this answer........