NCERT Exemplar Problems Class 7 Maths – Perimeter and Area

ncert textbook

Created with Sketch.

NCERT Exemplar Problems Class 7 Maths – Perimeter and Area

Question 1
Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters
(b) Shapes 1 and 4 have the same areas as well as the same perimeters
(c) Shapes 1, 2 and 4 have the same areas
(d) Shapes 1, 3 and 4 have the same perimeters

Question 2:
A rectangular piece of dimensions 3 cm x 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm x 5 cm (see the figure). Area of remaining sheet of paper is

Question 3:
36 unit squares are joined to form a rectangle with the least perimeter. Perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units

Question 4:
A wire is bent to form a square of side 22 cm. If the wire is rebent to form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
(b) Given, side of a square = 22 cm
Perimeter of square and circumference of circle are equal, because the wire has same length.
According to the question,

Question 5:
Area of the circle obtained in Q.4 is
(a) 196 cm2 (b) 212 cm2 (c) 616 cm2 (d) 644 cm2

Question 6:
Area of a rectangle and the area of a circle are equal. If the dimensions of the rectangle are 14 cm x 11 cm, then radius of the circle is
(a) 21 cm (b) 10.5 cm (c) 14 cm (d) 17 cm

Question 7:
Area of shaded portion in the figure given below is

Question 8:
Area of parallelogram ABCD (see the figure) is not equal to
(a) We know that,
Area of parallelogram = Base x Corresponding Height
So, area of parallelogram ABCD = AD x BE = BCx BE [∵ AD = BC ]
or area of parallelogram ABCD = DC xBF

Question 9:
Area of Δ MNO in the figure is

Question 10:
Ratio of area of ΔMNO to the area of parallelogram MNOP in the same figure of Q.9 is
(a) 2 : 3 (b) 1 : 1
(c) 1:2 (d) 2 : 1

Question 11:
Ratio of areas of ΔMNO, ΔMOP and ΔMPQ in the given figure is

Question 12:
In the given figure, EFGH is a parallelogram, altitudes FK and FI are 8 cm and 4 cm, respectively. If EF = 10 cm, then the area of EFGH is
(c) In parallelogram EFGH, EF = HG = 10 cm [Given]
Area of parallelogram EFGH = Base x Corresponding height =10 x 4 = 40 cm2

Question 13:
In reference to a circle the value of π is equal to

Question 14:
Circumference of a circle is always
(a) more than three times of its diameter
(b) three times of its diameter
(c) less than three times of its diameter
(d) three times of its radius
(a) We know that,
Circumference of a circle = 2πr
∴ Circumference = 2 x 3.14 x r [∵ π = 3.14]
⇒ Circumference = 3.14 x d [∵ d = 2r]
So, circumference of circle is always more than three times of its diameter.

Question 15:
Area of Δ PQR is 100 cm2 as shown in the below figure. If altitude QT is 10 cm, then its base PR is

Question 16:
In the given figure, if PR = 12 cm, QR = 6 cm and PL = 8 cm, then QM is
(c) Given that, PR = 12 cm, QR = 6 cm and PL = 8 cm
Now, in right angled APLR, using Pythagoras theorem,
(Hypotenuse)2 = (Perpendicular)2 + (Base)2

Question 17:
In the given figure, ΔMNO is a right angled triangle. Its legs are 6 cm and 8 cm long. Length of perpendicular NP on the side MO is

Question 18:
Area of a right angled triangle is 30 cm2. If its smallest side is 5 cm, then its hypotenuse is
(a) 14 cm (b) 13 cm (c) 12 cm (d) 11 cm

Question 19:
Circumference of a circle of diameter 5 cm is
(a) 14 cm (b) 31.4 cm (c) 15.7 cm (d) 1.57 cm

Question 20:
Circumference of a circular disc is 88 cm. Its radius is
(a) 8 cm (b) 11 cm (c) 14 cm (d) 44 cm

Question 21:
Length of tape required to cover the edges of a semi-circular disc of radius 10 cm is
(a) 62.8 cm (b) 51.4 cm (c) 31.4 cm (d) 15.7 cm
(b) In order to find the length of tape required to cover the edges of a semi-circular disc, we have to find the perimeter of semi-circle

Question 22:
Area of a circular garden with diameter 8 m is
(a) 12.56m2 (b) 25.12m2 (c) 50.24m2 (d) 2000.96m2

Question 23:
Area of a circle with diameter (m), radius (n) and circumference (p) is
(a) 2πn (b) πm2 (c) πp2 (d) πn2
(d) Given, diameter = m, radius = n and circumference = p
∴ Area of circle = πr2 = πn2

Question 24:
A table top is semi-circular in shape with diameter 2.8 m. Area of this table top is
(a) 3.08 m2 (b) 6.16 m2 (c) 12.32 m2 (d) 24.64 m2

Question 25:
If 1 m2 = x mm2, then the value of x is
(a) 1000 (b) 10000 (c) 100000 (d) 1000000

Question 26:
If p squares of each side 1mm makes a square of side 1cm, then p is equal to
(a) 10 (b) 100
(c)1000 (d) 10000

Question 27:
12 m2 is the area of
(a) a square with side 12m (b) 12 squares with side 1 m each
(c) 3 squares with side 4 m each (d) 4 squares with side 3 m each

Question 28:
If each side of a rhombus is doubled, how much will its area increase?
(a) 1.5 times (b) 2 times
(c) 3 times (d) 4 times
(b) Let b be the side and h be the height of a rhombus.
∴ Area of rhombus = b x h [∵ area of rhombus = base x corresponding height]
If each side of rhombus is doubled, then side of rhombus = 2b
Now, area of rhombus = 2b x h = 2(b x h) = 2 times of original
Hence, the area of rhombus will be increased by 2 times.

Question 29:
If the sides of a parallelogram are increased to twice its original lengths, how much will the perimeter of the new parallelogram increase?
(a) 1.5 times (b) 2 times
(c) 3 times (d) 4 times
(b) Let the length and breadth of the parallelogram be l and b, respectively.
Then, perimeter = 2(1 + b) [∵ perimeter of parallelogram = 2 x (length + breadth)]
If both sides are increased twice, then new length and breadth will be 21 and 2b, respectively.
Now, new perimeter = 2(21 + 2b) = 2 x 2(l + b) = 2 times of original perimeter.
Hence, the perimeter of parallelogram will be increased 2 times.

Question 30:
If radius of a circle is increased to twice its original length, how much will the area of the circle increase?
(a) 1.4 times (b) 2 times (c) 3 times (d) 4 times
(d) Let r be the radius of the circle.
∴ Area of circle = π2
If radius is increased to twice its original length, then radius will be 2r.
Now, area of new circle = π(2r)2 = 4π2 = 4 times of original area
Hence, the area of circle will be increased by 4 times.

Question 31:
What will be the area of the largest square that can be cut-out of a circle of radius 10 cm?
(a) 100 cm2 (b) 200 cm2 (c) 300 cm2 (d) 400 cm2
(b) Given, radius of circle = 10 cm
The largest square that can be cut-out of a circle of radius 10 cm will have its diagonal equal to the diameter of the circle.
Let the side of a square be x

Question 32:
What is the radius of the largest circle that can be cut-out of the rectangle measuring 10 cm in length and 8 cm in breadth?
(a) 4 cm (b) 5 cm (c) 8 cm (d) 10 cm

Question 33:
The perimeter of the figure ABCDEFGHIJ is

Question 34:
The circumference of a circle whose area is 81πr2, is
(a) 9πr (b)18πr (c)3πr (d)81πr

Question 35:
The area of a square is 100 cm2. The circumference (in cm) of the largest circle cut of it is
(a) 5π (b) 10π (c) 15π (d) 20π

Question 36:
If the radius of a circle is tripled, the area becomes
(a) 9 times (b) 3 times (c) 6 times (d) 30 times
Solution :
(a) Let r be the radius of a circle.
∴ Area of circle = πr2
If radius is tripled, then new radius will be 3r.
∴ Area of new circle = π(3r)2 = 9π2 = 9 times of original
Hence, the area of a cirlce becomes 9 times to the original area.

Question 37:
The area of a semi-circle of radius 4r is
(a) 8πr2 (b) 4πr22 (c) 12πr2 (d) 2πr2
Solution :

Fill in the Blank

In questions 38 to 56, fill in the blanks to make the statements true.

Question 38:
Perimeter of a regular polygon = Length of one side x_________
Solution :
Perimeter of regular polygon = Length of one side x Number of sides.

Question 39:
If a wire in the shape of a square is rebent into a rectangle, then the________of both shapes remain same, but______may vary.
Solution :
When we change the shape, then the perimeter remains same as the length of wire is fixed, but area changes as shape changes.

Question 40:
Area of the square MNOP of the given figure is 144 cm2. Area of each triangle is______ .
Solution :
Given, area of square MNOP = 144 cm2
Since, there are 8 identical triangles in the given square MNOP.
Hence, area of each triangle =1/8 x Area of square MNOP= 1/8 x 144= 18 cm2

Question 41:
In the given figure, area of parallelogram BCEF is________ cm2, where ACDF is rectangle.
Solution :

Question 42:
To find area, any side of a parallelogram can be chosen as__________ of the parallelogram.
Solution :
While calculating the area of the parallelogram, we can choose any side as base.

Question 43:
Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding______of the base.
Solution :
Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding height/altitude of the base.

Question 44:
The distance around a circle is its______
Solution :
The distance around a circle is its circumference.
In case of circle, perimeter is known as circumference.

Question 45:
Ratio of the circumference of a circle to its diameter is denoted by symbol_______.
Solution :

Question 46:
If area of a triangular piece of cardboard is 90 cm2, then the length of altitude corresponding to 20 cm long base is_____cm.
Solution :

Question 47:
Value of n is____________ approximately.
Solution :
We know that, π = 22/7 = 3.14

Question 48:
Circumference C of a circle can be found by multiplying diameter d with_____.
Solution :
∵ Circumference = 2πr
Since, diameter (d) = 2r
So, C = π x d
Hence, π is the answer.

Question 49:
Circumference ‘C’ of a circle is equal to 2π x______.
Solution :
Circumference = 2π x r
Hence, r is the answer.

Question 50:
1 m2 =_____ cm2
Solution :
We know that, 1 m = 100 cm
∴ 1 m2 =(100)2cm2
⇒ 1m2 =10000 cm2

Question 51:
11 cm2 =_______mm2
Solution :
We know that, 1 cm = 10 mm
∴ 1 cm2 = (10)2 mm2 = 100 mm2

Question 52:
1 hectare =________ m2
Solution :
1 hectare = 10000 m2

Question 53:
Area of triangle = 1/2 x base x______.
Solution :
Area of triangle = 1/2 x Base x Height.

Question 54:
1 km2 =_______ m2
Solution :
We know that, 1 km = 1000 m
∴ 1 km2 = (1000)2 m2 = 1000000 m2

Question 55:
Area of a square of side 6 m is equal to the area of________________ squares of each side 1 cm.
Solution :
Let number of squares having side 1 cm = x
According to the question,
Area of side 6 m square = Area of side 1 cm square [∵ area of square = (side)2]
∴ (6 m)2 = x x (1 cm)2 [∵ 1 m = 100 cm]
⇒ (600 cm)2 = x x (1 cm)2
⇒ 360000 cm2 = x cm2
⇒ x = 360000

Question 56:
10 cm2 =______m2.
Solution :

Question 57:
In the given figure, perimeter of (ii) is greater than that of (i), but its area is smaller than that of (i).
Solution :
Perimeter is the sum of sides of any polygon and area is space that the polygon required. So, by observing the figures we can say that, perimeter of (ii) is greater than (i) and area is less than that of (i).

Question 58:
In the given figure,
(a) Area of (i) is the same as the area of (ii)
(b) Perimeter of (ii) is the same as (i).
(c) If (ii) is divided into squares of unit length, then its area is 13 unit squares.
(d) Perimeter of (ii) is 18 units.
Solution :
(a) True
Area of both figures is same, because in both number of blocks are same.
(b) False
Because 2 new sides are added in (ii). So, the perimeter of (ii) is greater than (i).
(c) False
∴ Area of 1 square = 1 x 1 = 1 unit squares
∵ Number of squares = 12 So, total area = 12 x 1 = 12 unit squares
(d) True
∵ Perimeter is the sum of all sides. So, it is 18 units.

Question 59:
If perimeter of two parallelograms are equal, then their areas are also equal.
Solution :
Their corresponding sides and height may be different. So, area cannot be equal.

Question 60:
All congruent triangles are equal in area.
Solution :
Congruent triangles have equal shape and size. Hence, their areas are also equal.

Question 61:
All parallelograms having equal areas have same perimeters.
Solution :
It is not necessary that all parallelograms having equal areas have same perimeters as their base and height may be different.

In questions 62 to 65, observe all the four triangles FAB, EAB, DAB and CAB as shown in the given figure.

Question 62:
All triangles have the same base and the same altitude.
Solution :
It is clear from the figure that all triangles have same base AB and all the vertices lie on the same line, so the distance between vertex and base of triangle (i.e. length of altitude) are equal.

Question 63:
All triangles are congruent.
Solution :
It is clear from the figure that all triangles have only base line is equal and no such other lines are equal to each other.

Question 64:
All triangles are equal in area.
Solution :
Because the triangles on same base and between same parallel lines have equal in area.

Question 65:
All triangles may not have the same perimeter.
Solution :
It is clear from the figure that all triangles may not have the same perimeter.

Question 66:
In the given figure, ratio of the area of ΔABC to the area of ΔACD is the same as the ratio of base BC of ΔABC to the base CD of ΔACD.
Solution :

Question 67:
Triangles having the same base have equal area.
Solution :
∵ Area of triangle = 1/2 x Base x Height
So, area of triangle does not only depend on base, it also depends on height. Hence, if triangles have equal base and equal height, then only their areas are equal.

Question 68:
Ratio of circumference of a circle to its radius is always Zπ : 1.
Solution :
∵ Circumference : Radius = 2πr: r = 2π : 1

Question 69:
5 hectare = 500 m2
Solution :
As we know that, 1 hectare = 10000 m2
So, 5 hectare = 5 x 10000 m2 = 50000 m2

Question 70:
An increase in perimeter of a figure always increases the area of the figure.
Solution :
This is not necessary. See the Q. 57.

Question 71:
Two figures can have the same area but different perimeters.
Solution :
See the Q. 58.

Question 72:
Out of two figures, if one has larger area, then its perimeter need not to be larger than the other figure.
Solution :

Question 73:
A hedge boundary needs to be planted around a rectangular lawn of size 72 m x 18 m. If 3 shrubs can be planted in a metre of hedge, how many shrubs will be planted in all?
Solution :
Here, length of rectangular lawn = 72 m and breadth of rectangular lawn = 18 m
∵ Perimeter of rectangle =2 x (Length + Breadth)
∴ Perimeter of rectangular lawn = 2 (72 + 18) = 2 (90) = 180 m
If 3 shrubs can be planted in a metre of hedge.
Then, number of shrubs = 3 x Perimeter of rectangular lawn = 3 x 180 = 540

Question 74:
People of Khejadli village take good care of plants, trees and animals. They say that plants and animals can survive without us, but we cannot survive without them. Inspired by her elders Amrita marked some land for her pets (camel and ox) and plants. Find the ratio of the areas kept for animals and plants to the living area. What value depicted here?

Solution :
We know that,
Area of rectangle =l x b and area of circle = πr2
From the given figure,
Area of total rectangular land = 15 m x 10 m=150 m2
Area of land covered by plants = 9 m x 1 m = 9 m2
Area of land covered by camel = 5 m x 3 m = 15 m2
∴ Region of land covered by ox is circular area.
So, diameter, d = 2.8 m

Question 75:
The perimeter of a rectangle is 40 m. Its length is four metres less than five times its breadth. Find the area of the rectangle.
Solution :

Question 76:
A wall of a room is of dimensions 5 m x 4 m. It has a window of dimensions 1.5 m x 1 m and a door of dimensions 2.25 m x 1 m. Find the area of the wall, which is to be painted.
Solution :
Given,a wall of a room is of dimensions 5 m x m.
∴ Length of the room = 5 m and breadth of the room = 4 m
∴ Area of the room = l x b=5 x 4 = 20 m2
Also, length of the window = 1.5 m and breadth of the window = 1 m [given]
∴ Area of the window = l x b = 1.5 x 1 = 1.5 m2
Now, length of the door = 2.25 m and breadth of the door = 1 m
∴ Area of the door = l x b = 2.25 x 1 = 2.25 m2
Now, area of the wall to be painted = Area of the room – (Area of the window + Area of the door)
= 20 – (1.5 + 2.25) = 20 – 3.75 = 16.25 m2

Question 77:
Rectangle MNOP is made up of four congruent rectangles. If the area of one of the rectangles is 8m2 and breadth is 2 m, then find the perimeter of MNOP.
Solution :

Question 78:
In the given figure, area of ΔAFB is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the altitude of the parallelogram to the base AB of length 10 cm. What is the area of ΔDAO, where 0 is the mid-point of DC
Solution :

Question 79:
Rat’o of the area of ΔWXY to the area of ΔWZY is 3 :4 in the given figure. If the area of ΔWXZ is 56 cm2 and WY = 8 cm, find the lengths of XY and YZ.
Solution :

Question 80:
Rani bought a new field that is next to one she already owns in the given figure. This field is in the shape of a square of side 70 m. She makes a semi-circular lawn of maximum area in this field.
(i) Find the perimeter of the lawn.
(ii) Find the area of the square field excluding the lawn.
Solution :

Question 81:
In the given figure, find the area of parallelogram ABCD, if the area of shaded triangle is 9 cm2.
Solution :

Question 82:
Pizza factory has comeout with two kinds of pizzas. A square pizza of side 45 cm costs Rs 150 and a circular pizza of diameter 50 cm cost Rs 160. Which pizza is a better deal?
Solution :

Question 83:
Three squares are attached to each other as shown in the figure given below. Each square is attached at the mid-point of the side of the square to its right.
Solution :

Question 84:
In the following figure, ABCD is a square with AB = 15 cm. Find the area of the square BDFE.
Solution :

Question 85:
In the given figures, perimeter of ΔABC = perimeter of ΔPQR. Find the area of ΔABC.
Solution :

Question 86:
Altitudes MN and MO of parallelogram MGHK are 8 cm and 4 cm long respectively in the below figure. One side GH is 6 cm long. Find the perimeter of MGHK.
Solution :

Question 87:
In the given figure, area of ΔPQR is 20 cm2 and area of ΔPQS is 44 cm2. Find the length RS, if PQ is perpendicular to QS and QR is 5 cm.
Solution :

Question 88:
Area of an isosceles triangle is 48 cm2. If the altitudes corresponding to the base of the triangle is 8 cm, find the perimeter of the triangle.
Solution :

Question 89:
Perimeter of a parallelogram shaped land is 96 m and its area is 270 m2. If one of the sides of this parallelogram is 18 m, find the length of the other side. Also, find the lengths of altitudes l and m in the given figure.
Solution :

Question 90:
Area of a ΔPQR right angled at Q is 60 cm2 in the figure. If the smallest side is 8 cm long, find the length of the other two sides.
Solution :

Question 91:
In the given figure, a rectangle with perimeter 264 cm is divided into five congruent rectangles. Find the perimeter of one of the rectangles.
Solution :

Question 92:
Find the area of a square inscribed in a circle whose radius is 7 cm in the below figure.
[Hint Four right angled triangles joined at right angles to form a square]
Solution :

Question 93:
Find the area of the shaded portion in question 92.
Solution :

In questions 94 to 97, find the area enclosed by each of the following figures.

Question 94:
Solution :

Question 95:
Solution :

Question 96:
Solution :

Question 97:
Solution :

In questions 98 and 99, find the areas of the shaded region.

Question 98:
Solution :

Question 99:
Solution :

Question 100:
A circle with radius 16 cm is cut into four equal parts and rearranged to form another shaped as shown in the below figure.
Does the perimeter change? If it does change, by how much does it increase or decrease?
Solution :
Yes, the perimeter changes. The perimeter is increased by 2r = 2 x 16= 32 cm.

Question 101:
A large square is made by arranging a small square surrounded by four congruent rectangles as shown in the given figure. If the perimeter of each of the rectangle is 16 cm, find the area of the large square.
Solution :

Question 102:
ABCD is a parallelogram in which AE is perpendicular to CD as shown in the given figure. Also, AC = 5 cm, DE = 4 cm and area of ΔAED = 6 cm2. Find the perimeter and area of parallelogram ABCD.
Solution :

Question 103:
Ishika has designed a small oval race track for her remote control car. Her design is shown in the given figure. What is the total distance around the track? Round your answer to the nearest whole centimetre.
Solution :

Question 104:
A table cover of dimensions 3 m 25 cm x 2 m 30 cm is spread on a table. If 30 cm of the table cover, is hanging all around the table, find the area of the table cover, which is hanging outside the top of the table. Also, find the cost of polishing the table top at Rs 16 per square metre.
Solution :

Question 105:
The dimensions of a plot are 200 m x 150 m. A builder builds 3 roads which are 3 m wide along the length on either side and one in the middle. On either side of the middle road he builds houses to sell. How much area did he get for building the houses?
Solution :

Question 106:
A room is 4.5 m long and 4 m wide. The floor of the room is to be covered with tiles of size 15 cm by 10 cm. Find the cost of covering the floor with tiles at the rate of Rs 4.50 per tile.
Solution :

Question 107:
Find the total cost of wooden fencing around a circular garden of diameter 28 m, if 1 m of fencing costs Rs 300.
Solution :

Question 108:
Priyanka took a wire and bent it to form a circle of radius 14 cm. She bent it into a rectangle with one side 24 cm long. What is the length of the wire? Which figure encloses more area, the .circle or the rectangle?
Solution :

Question 109:
How much distance, in metres, a wheel of 25 cm radius will cover, if it rotates 350 times?
Solution :

Question 110:
A circular pond is surrounded by a 2 m wide circular path. If outer circumference of circular path is 44 m, find the inner circumference of the circular path. Also, find area of the path.
Solution :

Question 111:
A carpet of size 5 m x 2 m has 25 cm wide red border. The inner part of the carpet is blue in colour (see the figure). Find the area of blue portion. What is the ratio of areas of red portion to blue portion?
Solution :

Question 112:
Use the following figure, showing the layout of a farm house.
(a) What is the area of land used to grow hay?
(b) It costs Rs 91 per m2 to fertilise vegetable garden. What is total cost?
(c) A fence is to be enclosed around the house. The dimensions of house are 18.7 m x 12.6 m. Atleast how many metres of fencing are needed?
(d) each banana tree required 1.25 m2 of ground space. How many banana trees can there be in the orchard?
Solution :

Question 113:
Study the layout given in the figure and answer the questions.
(a) Write an expression for w the total area covered by both the bedrooms and the kitchen.
(b) Write an expression to calculate the perimeter of the living room.
(c) If the cost of carpeting is Rs 50 per m2, write an expression for calculating the total cost of carpeting both the bedrooms and the living room.
(d) If the cost of tiling is Rs 30 per m2, write an expression for calculating the total cost of floor tiles used for the bathroom and kitchen floors.
(e) If the floor area of each bedroom is 35 m2, then find x.
Solution :

Question 114:
10 m long and 4 m wide rectangular lawn is in front of a house. Along its three sides, a 50 cm wide flower bed is there is shown in the given figure. Find the area of the remaining portion.
Solution :

Question 115:
school playground is divided by a 2 m wide path, which is parallel to the width of the playground and a 3 m wide path which is parallel to the length of the ground in the given figure. If the length and width of the playground are 120 m and 80 m respectively, find the area of the remaining playground.
Solution :

Question 116:
In a park of dimensions 20 m x 15 m, there is a L shaped 1 m wide flower bed as shown in the figure. Find the total cost of manuring for the flower bed at the rate of Rs 45 per m2.
Solution :

Question 117:
Dimensions of a painting are 60 cm x 38 cm. Find the area of the wooden frame of width 6 cm around painting as shown in given figure.
Solution :

Question 118:
A design is made up of four congruent right triangles as shown in the given figure. Find the area of the shaded portion.
Solution :

Question 119:
A square tile of length 20 cm has four quarter circles at each corner as shown in the figure (i). Find the area of shaded portion. Another tile with same dimensions has a circle in the centre of the tile in the figure (ii). If the circle touches all the four sides of the square tile, find the area of the shaded portion. In which tile, area of shaded portion will be more? (Take, π= 3.14)
Solution :

Question 120:
A rectangular field is 48 m long and 12 m wide. How many right triangular flower beds can be laid in this field, if sides including the right angle measure 2 m and 4 m, respectively?
Solution :
Given, dimenstions of a rectangular field = 48 m x 12 m
and dimensions of a right angled triangle = 2 m x 4 m

Question 121:
Ramesh grew wheat in a rectangular field that measured 32 metres long and 26 metres wide. This year he increased the area for wheat by increasing the length but not the width. He increased the area of the wheat field by 650 square metres. What is the length of the expanded wheat field?
Solution :

Question 122:
In the given figure, Δ AEC is right angled at E, B is a point on EC, BD is the altitude of Δ ABC, AC = 25 cm, BC = 7 cm and AE = 15 cm. Find the area of Δ ABC and the length of DB.
Solution :

Question 123:
Solution :

Question 124:
Calculate the area of shaded region in the given figure, where all of the short line segments are at right angles to each other and 1 cm long.
Solution :

Question 125:
The plan and measurement for a house are given in the figure. The house is surrounded by a path 1 m wide.
Find the following
(i) Cost of paving the path with bricks at rate of Rs 120 per m .
(ii) Cost of wooden flooring inside the house except the bathroom at the cost of Rs 1200 per m2.
(iii) Area of living room.
Solution :

Question 126:
Architects design many types of buildings. They draw plans for houses, such as the plant is shown in the following figure.
An architect wants to install a decorative moulding around the ceilings in the rooms. The decorative moulding costs Rs 500 per m.
(a) Find how much moulding will be needed for each room.
(i) family room (ii) living room
(iii) dining room (iv) bedroom 1
(v) bedroom 2
(b) The carpet costs Rs 200 per m2. Find the cost of carpeting each room.
(c) What is the total cost of moulding for all the five rooms?
Solution :

Question 127:
ABCD is a given rectangle with length as 80 cm and breadth as 60 cm. P, Q, R, S are the mid-points of sides AB, BC, CD, DA, respectively.
A circular rangoli of radius 10 cm is drawn at the centre as shown in the given figure. Find the area of shaded portion.
Solution :

Question 128:
4 squares each of the side 10 cm have been cut from each corner of a rectangular sheet of paper of size 100 cm x 80 cm. From the remaining piece of paper, an isosceles right triangle is removed whose equal sides are each of 10 cm length. Find the area of the remaining part of the paper.
Solution :
Area of each square =(10)2 cm2 = 100 cm2 [∵ area of square = (side)2]
Area of rectangular sheet = 100 x 80 cm2 [∵ area of rectangle = length x breadth]
= 8000 cm2

Question 129:
A dinner plate is in the form of circle. A circular region encloses a beautiful design as shown in the given figure. The inner circumference is 352 mm and outer is 396 mm. Find the width of circular design.
Solution :

Question 130:
The moon is about 384000 km from earth and its path around the earth is nearly circular. Find the length of path described by moon in one complete revolution. (Take,π = 3.14)
Solution :
Length of path described by moon in one complete revolution = 2πr
= 2 x 3.14 x 384000 [∵ radius = distance of moon from the earth]
= 2411520 km

Question 131:
A photograph of Billiard/Snooker table has dimensions as 1/10 th of its actual size as shown in the given figure.
The portion excluding six holes each of diameter 0.5 cm needs to be polished at rate of Rs 200 per m2. Find the cost of polishing.
Solution :


Leave a Reply

Your email address will not be published. Required fields are marked *