Question 1 1 / -0
The perimeter of a rectangular field is 512 m and the ratio between the length and breadth is 5 : 3. Find the area
Solution
Let length and breadth be 5x and 3x2
Question 2 1 / -0
In a map, scale is defined as further 35 km to 1 cm. If area of a region is 4410 km2 , how much area it will occupy
Solution
Area should be 44100 km
2 Side of area = square root of 44100 = 210 km
Side in cm =
= 6
Area on map = 6 × 6 = 36 cm
2
Question 3 1 / -0
It needs 50 ml paint for painting a picture with dimensions 50 cm × 25 cm. How much paint is needed to paint a picture with dimensions 100 cm × 50 cm?
Solution
50 ml paint is used for painting 50 cm × 25 cm.
Area of painting = 1250 cm
2 Quantity of paint required to paint 1250 cm
2 area = 50 ml
Quantity of paint required to paint 5000 cm
2 area =
× 5000 = 200 ml
Question 4 1 / -0
A marginal walk all around the inside of an oblong piece of ground 40 m × 35 m occupies 650 m2 . Find the width of the walk.
Solution
Let the width of the walk be y m.
Question 5 1 / -0
Four circles of radii 2 cm placed in such a way that each touches the other as shown in figure. Find the area of the shaded region.
Solution
Area of shaded region = 4 cm2 - (4 × 90 × π × 4)/(360)2
Question 6 1 / -0
Find the radius of a circular field whose area is equal to the sum of the area of the four circles having radii 4 m, 8 m, 7 m, 10 m respectively.
Solution
Area of 4 circles = π(4)
2 + π(8)
2 + π(7)
2 + π(10)
2 Let radius be = r m
Then πr
2 = π(4)
2 + π(8)
2 + π(7)
2 + π(10)
2 R =
Question 7 1 / -0
The wheel of a bicycle has radius of 21 cm. Find how far it travels in making 250 revolutions?
Solution
Wheel of a bicycle has radius = 21 cm
250 revolution
Distance = 250 × 2 ×
= 33,000 cm = 300 m
Question 8 1 / -0
Perimeter of a rhombus is 52 cm. If one of the diagonal is 24 cm, find the area of the rhombus.
Solution
Perimeter of a Rhombus = 52 cm
Side = 13 cm
In ΔAOB, AB = 13 cm
OB = 12 cm
AO = 5 cm by Pythagoras
AC = 10 cm
Area of rhombus =
× 24 × 10
= 120 cm
2
Question 9 1 / -0
Area of parallelogram is 1280 square metres. If the height of the parallelogram is 32 m, find the length of its base.
Solution
Total area = 1280 m
2 base =
m = 40 m
Hence, answer option 2 is correct.
Question 10 1 / -0
A regular hexagon is inscribed in a circle whose radius is 4 cm. What is the area of the hexagon?
Solution
Area of hexagon = 6 × area of an equilateral tringle
= 6 ×
× 4 × 4 = 24
(radius = side of equilateral triangle
Question 11 1 / -0
The height of a room is 1/4 of the sum of the length and breadth and the cost of painting its walls at a rate of 50 paise per m2 is Rs. 400. Find its height.
Solution
Let height be h m.
Let length be
m.
Let breadth be b m.
Then, h =
………………. (1)
Cost of painting walls at 50 paise per m
2 = Rs. 400
Now, wall area = 2h(
+ b) m
2 So, 2h(
+ b) =
…………….. (2)
From (1) and (2):
2h(4h) =
h = 10
Thus, height of the room is 10 m.
Hence, answer option 3 is correct.
Question 12 1 / -0
The ratio of the length and breadth of a rectangular field is 16 : 9 and its area is 3600 m2 . Find its perimeter.
Solution
Let length and breadth be 16x and 9x, respectively.
Area = 3600 m
2 16x × 9x = 3600
144x
2 = 3600
x
2 =
x = 5
Perimeter = 2(16 × 5 + 9 × 5) = 2(80 + 45) = 250 m
Question 13 1 / -0
The radii of two circular fields are in the ratio 3 : 5. The area of the first field is what percent less than the area of the second?
Solution
Radius of two circular fields are in the ratio 3 : 5
Let radius be 3x and 5x
Area of first field = π(3x)
2 Area of second field = π(5x)
2 Percentage less area of first than second =
= 64%
Question 14 1 / -0
How many metres of wallpaper 45 cm wide will be required to cover the area of a wall 15 m × 60 m?
Solution
Area of the wall = 15 m × 60 m = 900 m
2 Let the required length of the wallpaper be
m.
Width of the wallpaper = 0.45 m
Now, area of total wallpaper = Area of the wall
× 0.45 = 900
= 2000 m
Question 15 1 / -0
The area of the circle inscribed in an equilateral triangle is 48π cm2 . What is the perimeter of the triangle?
Solution
Area of circle
Inscribed in an equilateral Δ is 48π
cm
2 Let radius of circle be r cm
Than πr
2 = 48π
R =
cm =
cm
In ΔOBE, since OB is bisector of
ABE
from symmetry of figure.
OBE = 30°
Also OE ⊥ BC
OEB = 90°
In ΔOBE,
tan 30° =
BE = 4 × 3 = 12 cm
BC = 2 × BE = 24 cm
Perimeter = 24 × 3 = 72 cm
Question 16 1 / -0
What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high?
Solution
The length of the longest pole is equal to the longest diagonal of the room.
The dimensions are 12 m, 4 m, 3 m
Longest diagonal =
m
=
m
= 13 m
Question 17 1 / -0
A straight wire, of length 110 cm, is bent to form a circle. What will be the radius of the circle?
Solution
Perimeter of circle = length of wire
2πr = 110 cm (Where r is radius)
r =
= 17.5 cm
Question 18 1 / -0
The area of a circle is halved when its radius is decreased by x cm. What is the radius of the circle?
Solution
Let radius of the circle be r cm.
Then, radius is reduced by x cm.
π(r - x)
2 =
2(r - x)
2 = r
2 2r
2 + 2x
2 - 4rx = r
2 r
2 - 4rx + 2x
2 = 0
r = 2x
=
= x(2 +
) cm
Question 19 1 / -0
Find the cost of fencing a rectangular field 225 m × 310 m at 45 paise per meter.
Solution
Perimeter of field = 2[225 m + 310 m]
Question 20 1 / -0
The area of the circular fields is in the ratio 16 : 49. If the radius of the later is 21 m, what is the radius of the former one?
Solution
Area of to field in the ratio = 16 : 49
Radius of latter = 21, Let radius of former be x m
Then
x = 12 m
Question 21 1 / -0
What is the cost of fencing a semicircular field of radius 14 meters at the rate of Rs. 1.50 per meter?
Solution
Radius of semicircular field = 14 m
Question 22 1 / -0
The cost of cutting the crops of a circular field of radius 14 cm is Rs. 27,440. What shall be the cost of cutting crops, at the same rate, on another circular field whose radius is 1/7 of the first?
Solution
Radius of field = 14 cm
Area of field = π × (14)
2 = 616 cm
2 Cost of cutting crops = Rs. 27,440
Rate of cutting per cm
2 =
= Rs. 44.54/cm
2 Radius of another circular field =
× 14 = 2 cm
Area =
× 4 = 12.57 cm
2 Total cost of cutting = Rs. 44.54 cm × 12.57 cm
2 = Rs. 560 (approx.)
Question 23 1 / -0
The perimeter of the rhombus is 100 cm. If one of the diagonals of rhombus is 48 cm, find the area of the rhombus.
Solution
Perimeter of a rhombus = 100 cm
One diagonal is d, = 48 cm
Then side of rhombus =
= 25 cm
Let other diagonal = 2x cm
In
AED = AE ⊥ ED
By Pythagoras (25)
2 = (24)
2 + (x)
2 625 = 576 + (x)
2 x = 7 cm
Other diagonal = d
2 = 14 cm
Area of parallelogram =
Question 24 1 / -0
The area of a rhombus is 96 m2 and one of its diagonals is of length 24 m. What is the side of the rhombus?
Solution
Area of a rhombus = 96 m
2 Diagonal d
1 = 24 m
Other diagonal d
2 = ?
Then area =
96 =
d
2 = 8 m
Let side b x m
Then in ΔAED AE ⊥ ED
AD
2 = AE
2 + DE
2 x
2 = 9
2 + 12
2 = 16 + 144 = 160
x =
=
x =
Question 25 1 / -0
The perimeter of a square is the same as the circumference of a circle of radius 12.6 cm. Find the side of the square.
Solution
Perimeter of square = 4 x a where a is side
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