Study MaterialsImportant QuestionsPerimeter and Area Class 7 Extra Questions Maths Chapter 11

Perimeter and Area Class 7 Extra Questions Maths Chapter 11

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Extra Questions Very Short Answer Type

Question 1.
The side of a square is 2.5 cm. Find its perimeter and area.
Solution:
Side of the square = 2.5 cm
Perimeter = 4 × Side = 4 × 2.5 = 10 cm
Area = (side)2 = (4)2 = 16 cm2

Question 2.
If the perimeter of a square is 24 cm. Find its area.
Solution:
Perimeter of the square = 24 cm
Side of the square = \(\frac { 24 }{ 4 }\) cm = 6 cm
Area of the square = (Side)2 = (6)2 cm2 = 36 cm2

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    Question 3.
    If the length and breadth of a rectangle are 36 cm and 24 cm respectively. Find
    (i) Perimeter
    (ii) Area of the rectangle.
    Solution:
    Length = 36 cm, Breadth = 24 cm
    (i) Perimeter = 2(l + b) = 2(36 + 24) = 2 × 60 = 120 cm
    (ii) Area of the rectangle = l × b = 36 cm × 24 cm = 864 cm2

    Question 4.
    The perimeter of a rectangular field is 240 m. If its length is 90 m, find:
    (i) it’s breadth
    (ii) it’s area.
    Solution:
    (i) Perimeter of the rectangular field = 240 m
    2(l + b) = 240 m
    l + b = 120 m
    90 m + b = 120 m
    b = 120 m – 90 m = 30 m
    So, the breadth = 30 m.
    (ii) Area of the rectangular field = l × b = 90 m × 30 m = 2700 m2
    So, the required area = 2700 m2

    Question 5.
    The length and breadth of a rectangular field are equal to 600 m and 400 m respectively. Find the cost of the grass to be planted in it at the rate of ₹ 2.50 per m2.
    Solution:
    Length = 600 m, Breadth = 400 m
    Area of the field = l × b = 600 m × 400 m = 240000 m2
    Cost of planting the grass = ₹ 2.50 × 240000 = ₹ 6,00,000
    Hence, the required cost = ₹ 6,00,000.

    Question 6.
    The perimeter of a circle is 176 cm, find its radius.
    Solution:
    The perimeter of the circle = 176 cm
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q6

    Question 7.
    The radius of a circle is 3.5 cm, find its circumference and area.
    Solution:
    Radius = 3.5 cm
    Circumference = 2πr
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q7

    Question 8.
    Area of a circle is 154 cm2, find its circumference.
    Solution:
    Area of the circle = 154 cm2
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q8

    Question 9.
    Find the perimeter of the figure given below.
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q9
    Solution:
    Perimeter of the given figure = Circumference of the semicircle + diameter
    = πr + 2r
    = \(\frac { 22 }{ 7 }\) × 7 + 2 × 7
    = 22 + 14
    = 36 cm
    Hence, the required perimeter = 36 cm.

    Question 10.
    The length of the diagonal of a square is 50 cm, find the perimeter of the square.
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q10
    Solution:
    Let each side of the square be x cm.
    x2 + x2 = (50)2 [Using Pythagoras Theorem] 2x2 = 2500
    x2 = 1250
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q10.1
    x = √1250 = \(\sqrt { 2\times 5\times 5\times 5\times 5 }\)
    x = 5 × 5 × √2 = 25√2
    The side of the square = 25√2 cm
    Perimeter of the square = 4 × side = 4 × 25√2 = 100√2 cm

    Perimeter and Area Class 7 Extra Questions Short Answer Type

    Question 11.
    A wire of length 176 cm is first bent into a square and then into a circle. Which one will have more area?
    Solution:
    Length of the wire = 176 cm
    Side of the square = 176 ÷ 4 cm = 44 cm
    Area of the square = (Side)2 = (44)2 cm2 = 1936 cm2
    Circumference of the circle = 176 cm
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q11
    Since 2464 cm2 > 1936 cm2
    Hence, the circle will have more area.

    Question 12.
    In the given figure, find the area of the shaded portion.
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q12
    Solution:
    Area of the square = (Side)2 = 10 cm × 10 cm = 100 cm2
    Area of the circle = πr2
    = \(\frac { 22 }{ 7 }\) × 3.5 × 3.5
    = \(\frac { 77 }{ 2 }\) cm2
    = 38.5 cm2
    Area of the shaded portion = 100 cm2 – 38.5 cm2 = 61.5 cm2

    Question 13.
    Find the area of the shaded portion in the figure given below.
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q13
    Solution:
    Area of the rectangle = l × b = 14 cm × 14 cm = 196 cm2
    Radius of the semicircle = \(\frac { 14 }{ 2 }\) = 7 cm
    Area of two equal semicircle = 2 × \(\frac { 1 }{ 2 }\) πr2
    = πr2
    = \(\frac { 22 }{ 7 }\) × 7 × 7
    = 154 cm2
    Area of the shaded portion = 196 cm2 – 154 cm2 = 42 cm2

    Question 14.
    A rectangle park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q14
    Solution:
    Length of the rectangular park = 45 m
    Breadth of the park = 30 m
    Area of the park = l × 6 = 45m × 30m = 1350 m2
    Length of the park including the path = 45 m + 2 × 2.5 m = 50 m
    Breadth of the park including the path = 30 m + 2 × 2.5 m = 30m + 5m = 35m
    Area of the park including the path = 50 m × 35 m = 1750 m2
    Area of the path = 1750 m2 – 1350 m2 = 400 m2
    Hence, the required area = 400 m2.

    Question 15.
    In the given figure, calculate:
    (а) the area of the whole figure.
    (b) the total length of the boundary of the field.
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q15
    Solution:
    Area of the rectangular portions = l × b = 80 cm × 42 cm = 3360 cm2
    Area of two semicircles = 2 × \(\frac { 1 }{ 2 }\) πr2 = πr2
    = \(\frac { 22 }{ 7 }\) × 21 × 21
    = 22 × 3 × 21
    = 1386 cm2
    Total area = 3360 cm2 + 1386 cm2 = 4746 cm2
    Total length of the boundary of field = (2 × 80 + πr + πr) cm
    = (160 + \(\frac { 22 }{ 7 }\) × 21 + \(\frac { 22 }{ 7 }\) × 21)
    = (160 + 132) cm
    = 292 cm
    Hence, the required (i) area = 4746 cm2 and (ii) length of boundary = 292 cm.

    Question 16.
    How many times a wheel of radius 28 cm must rotate to cover a distance of 352 m?
    (Take π = \(\frac { 22 }{ 7 }\))
    Solution:
    Radius of the wheel = 28 cm
    Circumference = 2πr = 2 × \(\frac { 22 }{ 7 }\) × 28 = 176 cm
    Distance to be covered = 352 m or 352 × 100 = 35200 m
    Number of rotation made by the wheel to cover the given distance = \(\frac { 35200 }{ 176 }\) = 200
    Hence, the required number of rotations = 200.

    Perimeter and Area Class 7 Extra Questions Long Answer Type

    Question 17.
    A nursery school playground is 160 m long and 80 m wide. In it 80 m × 80 m is kept for swings and in the remaining portion, there are 1.5 m wide path parallel to its width and parallel to its remaining length as shown in Figure. The remaining area is covered by grass. Find the area covered by grass. (NCERT Exemplar)
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q17
    Solution:
    Area of the playground = l × b = 160 m × 80 m = 12800 m2
    Area left for swings = l × b = 80m × 80m = 6400 m2
    Area of the remaining portion = 12800 m2 – 6400 m2 = 6400 m2
    Area of the vertical road = 80 m × 1.5 m = 120 m2
    Area of the horizontal road = 80 m × 1.5 m = 120 m2
    Area of the common portion = 1.5 × 1.5 = 2.25 m2
    Area of the two roads = 120 m2 + 120 m2 – 2.25 m2 = (240 – 2.25) m2 = 237.75 m2
    Area of the portion to be planted by grass = 6400 m2 – 237.75 m2 = 6162.25 m2
    Hence, the required area = 6162.25 m2.

    Question 18.
    Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle. (NCERT Exemplar)
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q18
    Solution:
    DE (Radius) = AE + AD = 8 cm + 5 cm = 13 cm
    DB = AC = 13 cm (Diagonal of a rectangle are equal)
    In right ∆ADC,
    AD2 + DC2 = AC2 (By Pythagoras Theorem)
    ⇒ (5)2 + DC2 = (13)2
    ⇒ 25 + DC2 = 169
    ⇒ DC2 = 169 – 25 = 144
    ⇒ DC = √144 = 12 cm
    Perimeter of rectangle ABCD = 2(AD + DC)
    = 2(5 cm + 12 cm)
    = 2 × 17 cm
    = 34 cm

    Question 19.
    Find the area of a parallelogram-shaped shaded region. Also, find the area of each triangle. What is the ratio of the area of shaded portion to the remaining area of the rectangle?
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q19
    Solution:
    Here, AB = 10 cm
    AF = 4 cm
    FB = 10 cm – 4 cm = 6 cm
    Area of the parallelogram = Base × Height = FB × AD = 6 cm × 6 cm = 36 cm2
    Hence, the required area of shaded region = 36 cm2.
    Area ∆DEF = \(\frac { 1 }{ 2 }\) × b × h
    = \(\frac { 1 }{ 2 }\) × AF × AD
    = \(\frac { 1 }{ 2 }\) × 4 × 6
    = 12 cm2
    Area ∆BEC = \(\frac { 1 }{ 2 }\) × b × h
    = \(\frac { 1 }{ 2 }\) × GC × BC
    = \(\frac { 1 }{ 2 }\) × 4 × 6
    = 12 cm2
    Area of Rectangle ABCD = l × b = 10 cm × 6 cm = 60 cm2
    Remaining area of Rectangle = 60 cm2 – 36 cm2 = 24 cm2
    Required Ratio = 36 : 24 = 3 : 2

    Question 20.
    A rectangular piece of dimension 3 cm × 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm × 5 cm. Find the ratio of the areas of the two rectangles. (NCERT Exemplar)
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q20
    Solution:
    Length of the rectangular piece = 6 cm
    Breadth = 5 cm
    Area of the sheet = l × b = 6 cm × 5 cm = 30 cm2
    Area of the smaller rectangular piece = 3 cm × 2 cm = 6 cm2
    Ratio of areas of two rectangles = 30 cm2 : 6 cm2 = 5 : 1

    Perimeter and Area Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

    Question 21.
    In the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region.
    (Take π = \(\frac { 22 }{ 7 }\))
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q21
    Solution:
    PQ = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) × 14 = 7 cm
    PQRS is a square with each side 7 cm
    Radius of each circle = \(\frac { 7 }{ 2 }\) cm
    Area of the quadrants of each circle = \(\frac { 1 }{ 4 }\) × πr2
    Area of the four quadrants of all circles
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q21.1
    Area of the square PQRS = Side × Side = 7 cm × 7 cm = 49 cm2
    Area of the shaded portion = 49 cm2 – 38.5 cm2 = 10.5 cm2
    Hence, the required area = 10.5 cm2.

    Question 22.
    Find the area of the following polygon if AB = 12 cm, AC = 2.4 cm, CE = 6 cm, AD = 4.8 cm, CF = GE = 3.6 cm, DH = 2.4 cm.
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q22
    Solution:
    BE = AB – AE
    = 12 cm – (AC + CE)
    = 12 cm – (2.4 cm + 6 cm)
    = 12 cm – 8.4 cm
    = 3.6 cm
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q22.1
    Perimeter and Area Class 7 Extra Questions Maths Chapter 11 Q22.2
    Area of the polygon AFGH = Area of ∆ACF + Area of rectangle FCEG + Area of ∆GEB + Area of ∆ABH
    = 3.6 cm2 + 4.32 cm2 + 21.6 cm 2 + 6.48 cm2 + 14.4 cm2
    = 50.40 cm2
    Hence, the required area = 50.40 cm2.

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