Finish Test
0 of 25 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
Information
- Most Important Multiple Choice Questions
- Online Mensuration Exercise with Correct Answer Key and Solutions
- Useful for all Competitive Exams
You have already completed the quiz before. Hence you can not start it again.
Test is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 25 questions answered correctly
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
- Not categorized 0%
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- Answered
- Review
-
Question 1 of 25
1. Question
In a parallelogram, the length of one diagonal and the perpendicular dropped on that diagonal are 30 and 20 metres respectively. Find its area.
Hint
In a parallelogram,
Area = Diagonal × length of perpendicular on it
= 30 × 20 = 600 m²
-
Question 2 of 25
2. Question
ABCD is a parallelogram. P, Q, R and S are points on sides AB, BC, CD and DA, respectively such that AP = DR. If the area of the rectangle ABCD is 16 cm², then the area of the quadrilateral PQRS is:
Hint
Area of the quadrilateral PQRS
= Area of ∆SPR + Area of ∆PQR
(Since PR = AD and AP + PB = AB)
-
Question 3 of 25
3. Question
The cross section of a canal is a trapezium in shape. If the canal is 7 metres wide at the top and 9 metres at the bottom and the area of cross-section is 1280 square metres, find the length of the canal.
Hint
Let the length of canal = h m. Then,
area of canal
⇒
-
Question 4 of 25
4. Question
A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, then the altitude of the triangle is:
Hint
Let the common base be x m.
Now, area of the triangle
= area of the parallelogram
½ × x × altitude of the triangle = x × 100
Altitude of the triangle = 200 m
-
Question 5 of 25
5. Question
Two sides of a plot measure 32 metres and 24 metres and the angle between them is a perfect right angle. The other two sides measure 25 metres each and the other three angles
What is the area of the plot?Hint
(32 – y)² + (24 -x)² = 625 ….(1)
x² + y² = 625 …(2)
⇒ (24)² + (32)² – 64y – 48x = 0
From (1) & (2)
⇒ 64y – 48x = 576 + 1024
⇒ 4y + 3x = 36 + 64 = 100
⇒
∴
From (2)
⇒ -600x +16x² + 10000 + 9x² = 625 × 16
⇒ 25x² – 600x + 10000 – 625 × 16 = 0
⇒ x = 24 and y = 7
∴ Area = (24 × 25) + ½ 24 × 7 = 684
-
Question 6 of 25
6. Question
One diagonal of a rhombus is 24cm side is 13 cm Find the area of the rhombus.
Hint
(side)² = (½ × one diagonal)² + (½ × other diagonal)²
⇒ 13² = (½ × one diagonal)² + (½ × 24)²
⇒ 169 – 144 = (½ × diagonal)²
⇒ 25 = (½ × diagonal)²
⇒ 5 = ½ × diagonal
∴ diagonal = 10
∴ Area = ½ × 10 x 24
= 120 sq. cm.
-
Question 7 of 25
7. Question
When the circumference and area of a circle are numerically equal, then the diameter is numerically equal to
Hint
According to question, circumference of circle = Area of circle
⇒
[where d = diameter]∴ d = 4
-
Question 8 of 25
8. Question
The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it cover in 5 revolutions? [use π = ²²⁄₇]
Hint
Required area covered in 5 revolutions
= 5 × 2πrh
= 5 × 2 × ²²⁄₇ × 0.7 × 2 = 44 m²
-
Question 9 of 25
9. Question
A cow is tethered in the middle of a field with a 14 feet long rope. If the cow grazes 100 sq. ft. per day, then approximately what time will be taken by the cow to graze the whole field?
Hint
Area of the field grazed
=
sq. ft. = 616 sq. ft.
Number of days taken to graze the field
= 616/100 days
= 6 days (approx)
-
Question 10 of 25
10. Question
A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of a side of the square.
Hint
Length of the wire = Perimeter of the circle
= 2π × 28 = 176 cm²
Side of the square
= ¹⁷⁶⁄₄ = 4cm
-
Question 11 of 25
11. Question
Semi-circular lawns are attached to both the edges of a rectangular field measuring 42 m × 35 m. The area of the total field is:
Hint
Area of the field
= 1470 + 1386 + 962.5
= 3818.5 m²
-
Question 12 of 25
12. Question
A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. What is the area of the circle?
Hint
Perimeter of the circle
= 2πr = 2(18 + 26)
⇒
⇒ r = 14
∴ Area of the circle
=
. -
Question 13 of 25
13. Question
A horse is tethered to one corner of a rectangular grassy field 40 m by 24 m with a rope 14 m long. Over how much area of the field can it graze?
Hint
Area of the shaded portion
= 154 m²
-
Question 14 of 25
14. Question
How many plants will be there in a circular bed whose outer edge measure 30 cms, allowing 4 cm² for each plant?
Hint
Circumference of circular bed = 30 cm
Area of circular bed
Space for each plant = 4 cm²
∴ Required number of plants
-
Question 15 of 25
15. Question
From a square piece of a paper having each side equal to 10 cm, the largest possible circle is being cut out. The ratio of the area of the circle to the area of the original square is nearly :
Hint
Area of the square = (10)² = 100 cm²
The largest possible circle would be as shown in the figure below :
Area of the circle
Required ratio
= 0.785 ≈ 0.8 = ⅘
-
Question 16 of 25
16. Question
If the area of a circle decreases by 36%, then the radius of a circle decreases by
Hint
If area of a circle decreased by x % then the radius of a circle decreases by
=
-
Question 17 of 25
17. Question
A circular grass lawn of 35 metres in radius has a path 7 metres wide running around it on the outside. Find the area of path.
Hint
Radius of a circular grass lawn (without path) = 35 m
∴ Area = πr² = π (35)²
Radius of a circular grass lawn ( with path)
= 35 + 7 = 42 m
∴ Area = πr² = π(42)²
∴ Area of path = π(42)² – π(35)²
= π(42² – 35²)
= π( 42 + 35) (42 –35)
= π × 77 × 7
-
Question 18 of 25
18. Question
Four equal circles are described about the four corners of a square so that each touches two of the others. If a side of the square is 14 cm, then the area enclosed between the circumferences of the circles is:
Hint
The shaded area gives the required region.
Area of the shaded region
= Area of the square – area of four quadrants of the circles
= (14)² – 4 × ¼ π (7)²
-
Question 19 of 25
19. Question
In a special racing event, the person who enclosed the maximum area would be the winner and would get Rs 100 every square metre of area covered by him/her. Jonnson, who successfully completed the race and was the eventual winner, enclosed the area shown in the figure below. What is the prize money won? (Note : The arc from C to D makes a complete semi-circle).
AB = 3 m, BC = 10 m, CD = BE = 2 m
Hint
Area of the semicircle
= π/2 = 1.571m²
Area of ∆ ABE
= ½ AB × BE
= ½ 3 × 2 = 3m²
Area of rectangle BCDE
= 10 × 2 = 20 m²
Total covered area
= 1.571 + 3 + 20
= 24.571 m²
Prize money won
= 24.571 × 100 = Rs 2457
-
Question 20 of 25
20. Question
The figure shows a rectangle ABCD with a semi-circle and a circle inscribed inside it as shown. What is the ratio of the area of the circle to that of the semi-circle?
Hint
Let the radius of the semi- circle be R and that of the circle be r, then from the given data, it is not possible to express r in terms of R. Thus option (d) is the correct alternative.
-
Question 21 of 25
21. Question
In ∆ACD, AD = AC and ∠C = 2∠E,. The distance between parallel lines AB and CD is h.
Then
I. Area of parallelogram ABCD
II. Area of ∆ADEHint
∠A = ∠C 60° (alternative angles)
∠C = ∠D 60° ( since AC = AD and ∠A 60° )
∆ ACD is equilateral
so its area
(where x is side)Area of parallelogram ABCD
Area of ∆ ADE = ½ × AD × AE
= ½ × x × x tan 60°
Therefore we see,
area of parallelogram ABCD = Area of ∆ADE
-
Question 22 of 25
22. Question
Wheels of diameters 7 cm and 14 cm start rolling simultaneously from X and Y, which are 1980 cm apart, towards each other in opposite directions. Both of them make the same number of revolutions per second. If both of them meet after 10 seconds, the speed of the smaller wheel is:
Hint
Let each wheel make x revolutions per sec. Then,
⇒ 66x = 198
⇒ x = 3.
Distance moved by smaller wheel in 3 revolutions
cm = 66 cm.
∴ Speed of smaller wheel
= 66/3 cm/s = 22 cm/s.
-
Question 23 of 25
23. Question
In the adjoining the figure, points A, B, C and D lie on the circle. AD = 24 and BC = 12. What is the ratio of the area of the triangle CBE to that of the triangle ADE
Hint
AD = 24, BC = 12
In ∆BCE and ∆ADE
since ∠CBA = ∠CDA (Angles by same arc)
∠BCE = ∠DAE (Angles by same arc)
∠BEC = ∠DEA (Opposite angles)
BCE and DAE are similar ∆s
with sides in the ratio 1 : 2
Ratio of area = 1:4 ( i.e square of sides)
-
Question 24 of 25
24. Question
The figure shows a circle of diameter AB and radius 6.5 cm. If chord CA is 5 cm long, find the area of triangle ABC
Hint
In the figure ∠ACB is 90°
(angle subtended by diameter= 90°)
AC = 5, AB = 13
Using Pythagoras theorem,
AB² = AC² + CB²
Area of ∆ ABC = ½ × 5 × 12 = 30
-
Question 25 of 25
25. Question
There are two concentric circular tracks of radii 100 m and 102 m, respectively. A runs on the inner track and goes once round on the inner track in 1 min 30 sec, while B runs on the outer track in 1 min 32 sec. Who runs faster?
Hint
Radius of the inner track = 100 m
and time = 1 min 30 sec = 90 sec.
Also, Radius of the outer track = 102 m
and time = 1 min 32 sec = 92 sec.
Now, speed of A who runs on the inner track
=
And speed of B who runs on the outer track
=
Since, speed of A > speed of B
∴ A runs faster than B.
Share this Test