NCERT Solutions for Class 7 Maths - Perimeter and Area- 1

Exercise 9.1

Q.1. Find the area of each of the following parallelograms: 

We know that the area of parallelogram = base x height
(a) Here base = 7 cm and height = 4 cm
∴ Area of parallelogram = 7x4 = 28 cm²
(b) Here base = 5 cm and height = 3 cm
∴ Area of parallelogram = 5x3 = 15 cm²
(c) Here base = 2.5 cm and height = 3.5 cm
∴ Area of parallelogram = 2.5 x 3.5 = 8.75 cm²
(d) Here base = 5 cm and height = 4.8 cm
∴ Area of parallelogram = 5 x 4.8 = 24 cm²
(e) Here base = 2 cm and height = 4.4 cm
∴ Area of parallelogram = 2 x 4.4 = 8.8 cm²

Q.2. Find the area of each of the following triangles: 

We know that the area of triangle

(a) Here, base = 4 cm and height = 3 cm
∴ Area of triangle

(b) Here, base = 5 cm and height = 3.2 cm
∴ Area of triangle

(c) Here, base = 3 cm and height = 4 cm
∴ Area of triangle

(d) Here, base = 3 cm and height = 2 cm
∴ Area of triangle

Q.3. Find the missing values:  

We know that the area of parallelogram = base x height
(a) Here, base = 20 cm and area = 246 cm²
∴ Area of parallelogram = base x height
⇒ 246 = 20 x height

(b) Here, height = 15 cm and area = 154.5 cm²
∴ Area of parallelogram = base x height
⇒ 154.5 = base x 15

(c) Here, height = 8.4 cm and area = 48.72 cm²
∴ Area of parallelogram = base x height
⇒ 48.72 = base x 8.4

(d) Here, base = 15.6 cm and area = 16.38 cm²
∴ Area of parallelogram = base x height
⇒ 16.38 = 15.6 x height

Thus, the missing values are:

 

Q.4. Find the missing values:  

We know that the area of triangle

In first row, base = 15 cm and area = 87 cm²

In second row, height = 31.4 mm and area = 1256 mm²

In third row, base = 22 cm and area = 170.5 cm²

Thus, the missing values are:

Q.5. PQRS is a parallelogram (Fig 11.23), QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm

Given:  
SR = 12 cm, QM = 7.6 cm, PS = 8 cm,
(a) Area of parallelogram = base x height
= 12 x 7.6 = 91.2 cm²
(b) Area of parallelogram = base x height

Q.6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, Find the length of BM and DL.

Given:  
Area of parallelogram = 1470 cm²
Base (AB) = 35 cm and base (AD) = 49 cm
Since Area of parallelogram = base x height
⇒ 1470 = 35 x DL
⇒ DL = 1470/35
⇒ DL = 42 cm
Again, Area of parallelogram = base x height
⇒ 1470 = 49 x BM
⇒ BM  = 1470/49
⇒ BM = 30 cm
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.

Q.7. ΔABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of Δ ABC. Also, find the length of AD. 

Q.8. ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of ΔABC. What will be the height from C to AB i.e., CE?

In ΔABC, AD = 6 cm and BC = 9 cm
Area of triangle

Again, Area of triangle

Thus, height from C to AB i.e., CE is 7.2 cm

Exercise 9.2

(b) 28 mm 

(c) 21 cm 

Sol:

(a) A circumference of the circle

(b) A circumference of the circle (c) A circumference of the circle 

Q2: Find the area of the following circles, given that: 
(a) radius = 14 mm 
(b) diameter = 49 m 
(c) radius = 5 cm 

Sol:

Q3: If the circumference of a circular sheet is 154 m, find its radius. Also, find the area of the sheet. 

Sol: 
Circumference of the circular sheet = 154 m 
⇒ 2πr = 154 m
⇒ r = 154/2π
⇒
Now, Area of circular sheet =πr²
= 22 x 3.5 x 24.5 = 1886.5 m²
Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m² respectively.

Q4: A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase if he makes 2 rounds of a fence. Also, find the costs of the rope, if it cost Rs 4 per meter. 

Sol:
Diameter of the circular garden = 21 m
∴ Radius of the circular garden = 21/2 m
Now, Circumference of a circular garden
= 22 x 3 = 66 m
The gardener makes 2 rounds of a fence so the total length of the rope of fencing
= 2 x 2πr
= 2 x 66 = 132 m
Since, the cost of 1 meter rope = Rs 4
Therefore, Cost of 132 meter rope = 4 x 132 = Rs 528

Q5: From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. 

Sol:
Radius of circular sheet (R) = 4 cm and radius of removed circle (r) = 3 cm
 Area of the remaining sheet = Area of a circular sheet - Area of removed circle

Thus, the area of the remaining sheet is 21.98 cm².

Q6: Saima wants to put lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs Rs 15. 

Sol:
Diameter of the circular table cover = 1.5 m
∴ Radius of the circular table cover = 1.5/2 m
Circumference of circular table cover = 2πr= 4.71 m
Therefore, the length of required lace is 4.71 m.
Now the cost of 1 m lace = Rs 15
Then the cost of 4.71 m lace = 15 x 4.71 = Rs 70.65
Hence, the cost of 4.71 m lace is Rs 70.65.

Q7: Find the perimeter of the adjoining figure, which is a semicircle including its diameter. 

Sol:
Diameter = 10 cm
Radius = 10/2 = 5 cm
According to question,
Perimeter of figure = Circumference of semicircle + diameter

Thus, the perimeter of the given figure is 25.71 cm.

Q8: Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs 15/m². (Take π = 3.14)

Sol:
Diameter of the circular table top = 1.6 m
Radius of the circular table top = 1.6/2 = 0.8 m
Area of circular table top = πr² = 3.14 x 0.8 x 0.8 = 2.0096 m²
Now, the cost of polishing 1 m² = Rs 15
Then cost of polishing 2.0096 m² = 15 x 2.0096 = Rs 30.14 (approx.)
Thus, The cost of polishing a circular table top is Rs 30.14 (approx.)

Q9: Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also, find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? 

Sol:
Total length of the wire = 44 cm
 ∴ The circumference of the circle = 2πr = 44 cm

Now Area of the circle = πr²

Now the wire is converted into the square.
The perimeter of square = 44 cm

Now, area of square = side x side = 11 x 11 = 121 cm²

Therefore, on comparing, the area of the circle is greater than that of a square, so the circle encloses more area.

Q10: From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the adjoining figure). Find the area of the remaining sheet. 

Sol:
Radius of circular sheet (R) = 14 cm and Radius of smaller circle (r) = 3.5 cm
Length of rectangle (l) = 3 cm and breadth of rectangle (b) = 1 cm
According to question,
Area of remaining sheet = Area of circular sheet- (Area of two smaller circle + Area of rectangle)

Therefore, the area of the remaining sheet is 536 cm².

Q11: A circle of radius 2 cm is cut out from a square piece of an aluminum sheet of side 6 cm. What is the area of the leftover aluminum sheet? 

Sol:
Radius of circle = 2 cm and side of aluminium square sheet = 6 cm
According to question,
Area of aluminum sheet left = Total area of aluminum sheet - Area of circle
= side x side - πr²

Therefore, the area of the aluminum sheet left is 23.44 cm².

Q12: The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. 

Sol:
The circumference of the circle = 31.4 cm

Then area of the circle = πr² = 3.14 x 5 x 5
= 78.5 cm²
Therefore, the radius and the area of the circle are 5 cm and 78.5 cm² respectively.

Q13: A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? 

Sol:
Diameter of the circular flower bed = 66 m
∴ Radius of circular flower bed (r) = 66/2 = 33 m
∴ Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m

According to question,
Area of path = Area of bigger circle - Area of smaller circle

= 3.14 x 70 x 4= 879.20 m²
Therefore, the area of the path is 879.20 m².

Q14: A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? 

Sol:
Circular area covered by the sprinkler = πr²
= 3.14 x 12 x 12
= 3.14 x 144
= 452.16 m²
Area of the circular flower garden = 314 m²
As Area of the circular flower garden is smaller than area with a sprinkler. Therefore, the sprinkler will water the entire garden.

Q15: Find the circumference of the inner and the outer circles, shown in the adjoining figure. 

Sol:
Radius of outer circle (r) = 19 m
Circumference of outer circle = 2πr = 2 x 3.14 x 19 = 119.32 m
Now radius of inner circle (r')  = 19 - 10 = 9 m
∴ Circumference of inner circle = 2πr' = 2 x 3.14 x 9 = 56.52 m
Therefore, the circumferences of inner and outer circles are 56.52 m and 119.32 m respectively.

Q16: How many times a wheel of radius 28 cm must rotate to go 352 m? 

Sol:
Let wheel must be rotated n times of its circumference.
Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm
∴ Distance covered by wheel = n x circumference of wheel

Thus, the wheel must rotate 200 times to go 352 m.

Q17: The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? 

Sol:
In 1 hour, minute hand completes one round means making a circle.
Radius of the circle (r) = 15 cm
A circumference of circular clock = 2πr
= 2 x 3.14 x 15
= 94.2 cm
Therefore, the tip of the minute hand moves 94.2 cm in 1 hour.

Q.1. A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also, find the area of the garden in hectares. 

Length of rectangular garden = 90 m and breadth of rectangular garden = 75 m
Outer length of rectangular garden with path = 90 + 5 + 5 = 100 m
Outer breadth of rectangular garden with path = 75 + 5 + 5 = 85 m

Outer area of rectangular garden with path = length x breadth = 100 x 85 = 8,500 m²
Inner area of garden without path = length x breadth = 90 x 75 = 6,750 m²
Now, Area of path = Area of garden with path – Area of garden without path
= 8,500 – 6,750
= 1,750 m²

Q.2.  A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path. 

Length of rectangular park = 125 m
Breadth of rectangular park = 65 m
Width of the path = 3 m
Length of rectangular park with path = 125 + 3 + 3 = 131 m
Breadth of rectangular park with path = 65 + 3 + 3 = 71 m

∴ Area of path = Area of park with path - Area of park without path= (AB x AD) - (EF x EH)
= (131 x 71) - (125 x 65)
= 9301 - 8125 = 1,176 m²
Thus, the area of a path around the park is 1,176 m².

Q.3.  A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin. 

Length of painted cardboard = 8 cm ,  breadth of painted card = 5 cm
Also,  there is a margin of 1.5 cm long from each of its side.
Therefore, reduced length = 8 – (1.5 + 1.5) = 8 – 3 = 5 cm
and reduced breadth = 5 - (1.5 + 1.5] = 5 - 3 = 2 cm

Area of margin = Area of cardboard (ABCD) - Area of cardboard (EFGH)= (AB x AD) - (EF x EH)
= (8 x 5) - (5 x 2)
= 40 - 10
= 30 cm²
Thus, total area of margin is 30 cm².

Q.4. A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find: 
(i) the area of the verandah. 
(ii) the cost of cementing the floor of the verandah at the rate of Rs 200 per m². 

(i) The length of room = 5.5 m
Width of the room = 4 m
The length of room with verandah = 5.5 + 2.25 + 2.25 = 10 m
The width of room with verandah = 4 + 2.25 + 2.25 = 8.5 m
Area of verandah = Area of room with verandah – Area of room without verandah
= Area of ABCD – Area of EFGH
= (AB x AD) – (EF x EH)
= (10 x 8.5) – (5.5 x 4)
= 85 – 22
= 63 m² 

(ii) The cost of cementing 1 m² the floor of verandah = Rs 200
The cost of cementing 63 m² the floor of verandah = 200 x 63 = Rs 12,600

Q.5. A path 1 m wide is built along the border and inside a square garden of side 30 m. Find: 
(i) the area of the path. 
(ii) the cost of planting grass in the remaining portion of the garden at the rate of Rs 40 per m². 

(i) Side of the square garden = 30 m and
Width of the path along the border = 1 m
Side of square garden without path = 30 – (1 + 1) = 30 – 2 = 28 m
Now Area of path = Area of ABCD - Area of EFGH
= (AB x AD) - (EF x EH)
= (30 x 30) - (28 x 28)
= 900 - 784
= 116 m²

(ii) Area of remaining portion = 28 x 28 = 784 m² 
The cost of planting grass in 1 m² of the garden = Rs 40
The cost of planting grass in 784 m² of the garden = Rs 40 x 784 = Rs 31,360

Q.6. Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. 
(i) Find the area of the roads. 
(ii) Also find the area of the park excluding cross roads. Give the answer in hectares. 

Here, PQ = 10 m and PS = 300 m, EH = 10 m and EF = 700 m And KL = 10 m and KN = 10 m

Area of roads = Area of PQRS + Area of EFGH - Area of KLMN
[∵ KLMN is taken twice, which is to be subtracted] = (PS x PQ) + (EF x EH)  - (KL x KN)
= (300 x 10) + (700 x 10) - (10 x 10)
= 3000 + 7000 - 100
= 9,900 m²
Area of road in hectares,

∴
Now, Area of park excluding cross roads
= Area of park - Area of road
= (AB x AD) - 9,900 = (700 x 300) - 9,900
= 2,10,000 - 9,900
= 2,00,100 m²

Q.7. Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find: 
(i) the area covered by the roads. 
(ii) the cost of constructing the roads at the rate of Rs 110 per m². 

(i) Here, PQ = 3 m and PS = 60 m, EH = 3 m and
EF = 90 m and KL = 3 m and KN = 3 m

Area of roads = Area of PQRS + Area of EFGH - Area of KLMN
[∵  KLMN is taken twice, which is to be subtracted]
= (PS x PQ) + (EF x EH) - (KL x KN)
= (60 x 3)+ (90 x 3) - (3 x 3)
= 180 + 270 - 9
= 441 m²

(ii) The cost of 1 m² constructing the roads = Rs 110
The cost of 441 m² constructing the roads = Rs 110 x 441 = Rs 48,510
Therefore, the cost of constructing the roads = Rs 48,510

Q.8. Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (Take π = 3.14)

Radius of pipe = 4 cm
Wrapping cord around circular pipe = 2πr
= 2 x 3.14 x 4 = 25.12 cm
Again, wrapping cord around a square = 4 x side
= 4 x 4 = 16 cm
Remaining cord = Cord wrapped on pipe – Cord wrapped on square
= 25.12 – 16
= 9.12 cm
Thus, she is left with 9.12 cm cord.

Q.9. The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: 

(i) the area of the whole land. 
(ii) the area of the flower bed. 
(iii) the area of the lawn excluding the area of the flower bed. 
(iv) the circumference of the flower bed. 

Length of rectangular lawn = 10 m, breadth of the rectangular lawn = 5 m
And radius of the circular flower bed = 2 m
(i) Area of the whole land = length x breadth = 10 x 5 = 50 m²
(ii) Area of flower bed = πr²
= 3.14 x 2 x 2 = 12.56 m²
(iii) Area of lawn excluding the area of the flower bed = area of lawn - area of flower bed
= 50 - 12.56
= 37.44 m²
(iv) The circumference of the flower bed = 2πr = 2 x 3.1 4 x 2 = 12.56 m

Q.10. In the following figures, find the area of the shaded portions: 

(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm
Area of shaded portion = Area of rectangle ABCD - (Area of Δ FAE + area of Δ EBC} 

(ii) Here, SR = SU + UR = 10 + 10 = 20 cm, QR = 20 cm
PQ = SR = 20 cm
PT = PS - TS =(20 - 10)cm
TS = 10 cm, SU = 10 cm, QR = 20 cm and UR = 10 cm
Area of shaded region
= Area of square PQRS - Area of ΔQPT - Area of ΔTSU - Area of ΔUQR

Q.11. Find die area of the equilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm and BM ⊥ AC, DN ⊥ AC.

Here, AC = 22 cm, BM = 3 cm, DN = 3 cm
Area of quadrilateral ABCDF = Area of Δ ABC + Area of Δ ADC

Thus, the area of quadrilateral ABCD is 66 cm².