0 of 12 questions completed

Questions:

- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12

#### Information

**Quiz Description** :

**Name: **Surface and Area objective question answer quiz

**Subject: **Aptitude

**Topic: **Area

**Questions: **12 Objective type

**Time Allowed: **20 Minutes

**Important for: **Bank PO, SSC CGL Mains, PCS, Samiksha Adhikari, CTET, State TET, NDA, Railways, Engineering Entrance exam, Job Aptitude test and Class 8, 9 & 10th students.

You have already completed the Test before. Hence you can not start it again.

Test is loading...

You must sign in or sign up to start the Test.

You have to finish following quiz, to start this Test:

#### Congratulations!!!" Surface and Area objective question answer quiz "

0 of 12 questions answered correctly

Your time:

Time has elapsed

Your Final Score is : 0

You have attempted : 0

Number of Correct Questions : 0 and scored 0

Number of Incorrect Questions : 0 and Negative marks 0

Average score | |

Your score |

- Not categorized

You have attempted: 0

Number of Correct Questions: 0 and scored 0

Number of Incorrect Questions: 0 and Negative marks 0

It’s time to share this quiz with your friends on

**Facebook**,**Twitter**,**Google Plus**,**Whatsapp**or**LinkedIn**…

Pos. | Name | Entered on | Points | Result |
---|---|---|---|---|

Table is loading | ||||

No data available | ||||

- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12

- Answered
- Review

- Question 1 of 12
##### 1. Question

1 pointsWhat is the area of a rectangle if it is 35 meters 5 decimeters long and 19 meters and 1 decimeters 2 centimeters wide?

CorrectLength of the rectangle (l) = 35.5 meters

Breadth of the rectangle (b) = 19.12 meters

∴ Area of the rectangle (A) = (l × b) = (35.5 × 19.12) meter^{2}

=678.76 meter^{2}IncorrectLength of the rectangle (l) = 35.5 meters

Breadth of the rectangle (b) = 19.12 meters

∴ Area of the rectangle (A) = (l × b) = (35.5 × 19.12) meter^{2}

=678.76 meter^{2} - Question 2 of 12
##### 2. Question

1 pointsIn ΔABC,∠B = 90°, AB = 9cm and BC = 12cm. Calculate the length of AC.

CorrectBy Pythagoras Theorem, AC

^{2}= AB^{2}+ BC^{2}

= 9^{2}+ 12^{2}

= 81 +144

= 225

AC = √225

=15cmIncorrectBy Pythagoras Theorem, AC

^{2}= AB^{2}+ BC^{2}

= 9^{2}+ 12^{2}

= 81 +144

= 225

AC = √225

=15cm - Question 3 of 12
##### 3. Question

1 pointsA square has the perimeter 64cm. What is the sum of the diagonals?

CorrectLet ‘a’ be the length of the side of the square ∎ABCD and AC is a diagonal of the square.

Perimeter of square ∎ABCD = 4a units

4a = 64cm

a = 16

We know that in a square each angle is 90° and the diagonals are equal.

In ΔABC, AC^{2}= AB^{2}+ BC^{2}= 16^{2}+ 16^{2}= 256 + 256 = 512

∴AC = √512 = 22.62cm

Diagonal AC = Diagonal BD

Hence, Sum of the diagonals = 22.62cm + 22.62cm = 45.25cmIncorrectLet ‘a’ be the length of the side of the square ∎ABCD and AC is a diagonal of the square.

Perimeter of square ∎ABCD = 4a units

4a = 64cm

a = 16

We know that in a square each angle is 90° and the diagonals are equal.

In ΔABC, AC^{2}= AB^{2}+ BC^{2}= 16^{2}+ 16^{2}= 256 + 256 = 512

∴AC = √512 = 22.62cm

Diagonal AC = Diagonal BD

Hence, Sum of the diagonals = 22.62cm + 22.62cm = 45.25cm - Question 4 of 12
##### 4. Question

1 pointsThe two adjacent sides of a parallelogram are 5cm and 4 cm respectively, and if the respective diagonal is 3cm then find the area of the parallelogram?

CorrectRequired area = 2√(s(s-a)(s-b)(s-c))

Where, S = (a + b + D)/2; i.e.; (5 + 4 + 3)/2 = 6

⟹2 √(6(6-5)(6-4)(6-3))

⟹2√(6×1×2×3)

⟹2√36

⟹2 × 6

⟹ 12 cm^{2}IncorrectRequired area = 2√(s(s-a)(s-b)(s-c))

Where, S = (a + b + D)/2; i.e.; (5 + 4 + 3)/2 = 6

⟹2 √(6(6-5)(6-4)(6-3))

⟹2√(6×1×2×3)

⟹2√36

⟹2 × 6

⟹ 12 cm^{2} - Question 5 of 12
##### 5. Question

1 pointsA hall-room 39m 10cm long and 35m 70cm broad is to be paved with equal square tiles. Find the largest tile so that the tiles exactly fit and also find the number of tiles required.

CorrectSide of largest possible tile = H.C.F. of length and breadth of the room

= H.C.F. of 39.10 and 35.70 m

= 1.70mAlso, number of tiles required = (Lenght × breadth of the room)/〖(H.C.F.of length and breadth of the room)〗^2

= (39.10 × 35.70)/〖1.70〗^2

= 483IncorrectSide of largest possible tile = H.C.F. of length and breadth of the room

= H.C.F. of 39.10 and 35.70 m

= 1.70mAlso, number of tiles required = (Lenght × breadth of the room)/〖(H.C.F.of length and breadth of the room)〗^2

= (39.10 × 35.70)/〖1.70〗^2

= 483 - Question 6 of 12
##### 6. Question

1 pointsIn a parallelogram, the lengths of adjacent sides are 15cm and 18cm respectively. If the length of one diagonal is 20cm, find the length of the other diagonal.

CorrectIn a parallelogram,

The sum of the squares of the diagonals = 2 × (Sum of the squares of the two adjacent sides)

D_{1}^{2}+ D_{2}^{2}= 2 (a^{2}+ b^{2})

20^{2}+ D_{2}^{2}= 2(15^{2}+ 18^{2})

400 + D_{2}^{2}= 2(255 + 324)

D_{2}^{2}= 1098 – 400 = 698

∴ D_{2}= √698 = 26.41cmIncorrectIn a parallelogram,

The sum of the squares of the diagonals = 2 × (Sum of the squares of the two adjacent sides)

D_{1}^{2}+ D_{2}^{2}= 2 (a^{2}+ b^{2})

20^{2}+ D_{2}^{2}= 2(15^{2}+ 18^{2})

400 + D_{2}^{2}= 2(255 + 324)

D_{2}^{2}= 1098 – 400 = 698

∴ D_{2}= √698 = 26.41cm - Question 7 of 12
##### 7. Question

1 pointsIn a trapezium, parallel sides are 55 and 95 centimeters respectively and non-parallel sides are 46 and 34 centimeters respectively. Find its area.

CorrectLet k = difference between the parallel sides = 95 – 55 = 40 cm

Now, s= (k + c + d)/2 = (40 + 46 + 34)/2 = 120/2 = 60 cm

∴Area = (a + b )/k √(s(s-k)(s-c) (s-d))

= (55 + 95 )/40 √(60(60-40)(60-46) (60-34))

= 3.75 ×√(60 × 20 ×14 ×26)

= 3.75 × 660.90

= 2478.41 cm^{2}IncorrectLet k = difference between the parallel sides = 95 – 55 = 40 cm

Now, s= (k + c + d)/2 = (40 + 46 + 34)/2 = 120/2 = 60 cm

∴Area = (a + b )/k √(s(s-k)(s-c) (s-d))

= (55 + 95 )/40 √(60(60-40)(60-46) (60-34))

= 3.75 ×√(60 × 20 ×14 ×26)

= 3.75 × 660.90

= 2478.41 cm^{2} - Question 8 of 12
##### 8. Question

1 pointsA 5500 cm

^{2}trapezium has the perpendicular distance between the two parallel sides 50 m. If one of the parallel sides be 80 m then find the length of the other parallel side.CorrectA = 1/2 (a + b)h

5500 = 1/2 (80 + b)×50

∴ b= 140 mIncorrectA = 1/2 (a + b)h

5500 = 1/2 (80 + b)×50

∴ b= 140 m - Question 9 of 12
##### 9. Question

1 pointsHow many meters of 75 centimeter wide carpet will be required to cover the floor of a room which is 20 meters long and 12 meters broad?

CorrectRequired Length = (Lenght of room × Breadth of room)/(Widht of carpet)

∴ Required length = (20 × 12)/0.75 = 320meterIncorrectRequired Length = (Lenght of room × Breadth of room)/(Widht of carpet)

∴ Required length = (20 × 12)/0.75 = 320meter - Question 10 of 12
##### 10. Question

1 pointsIn a parallelogram, the lengths of adjacent sides are 13cm and 17cm respectively. If the length of one diagonal is 18cm, find the length of the other diagonal.

CorrectIn a parallelogram,

The sum of the squares of the diagonals = 2 × (Sum of the squares of the two adjacent sides)

D_{1}^{2}+ D_{2}^{2}= 2 (a^{2}+ b^{2})

18^{2}+ D_{2}^{2}= 2(13^{2}+ 17^{2})

324 + D_{2}^{2}= 2(169 + 289)

D_{2}^{2}= 916 – 324 = 592

∴ D_{2}= √592 = 24.33cmIncorrectIn a parallelogram,

The sum of the squares of the diagonals = 2 × (Sum of the squares of the two adjacent sides)

D_{1}^{2}+ D_{2}^{2}= 2 (a^{2}+ b^{2})

18^{2}+ D_{2}^{2}= 2(13^{2}+ 17^{2})

324 + D_{2}^{2}= 2(169 + 289)

D_{2}^{2}= 916 – 324 = 592

∴ D_{2}= √592 = 24.33cm - Question 11 of 12
##### 11. Question

1 pointsIn ΔABC,∠B = 90°, AB = 6cm and BC =7cm. Calculate the length of AC.

CorrectBy Pythagoras Theorem, AC

^{2}= AB^{2}+ BC^{2}

= 6^{2}+ 7^{2}

= 36 +49

= 85

AC = √85

=9.22cmIncorrectBy Pythagoras Theorem, AC

^{2}= AB^{2}+ BC^{2}

= 6^{2}+ 7^{2}

= 36 +49

= 85

AC = √85

=9.22cm - Question 12 of 12
##### 12. Question

1 pointsA square has the perimeter 48cm. What is the sum of the diagonals?

CorrectLet ‘a’ be the length of the side of the square ∎ABCD and AC is a diagonal of the square.

Perimeter of square ∎ABCD = 4a units

4a = 48cm

a = 12

We know that in a square each angle is 90° and the diagonals are equal.

In ΔABC, AC^{2}= AB^{2}+ BC^{2}= 12^{2}+ 12^{2}= 144 + 144 = 288

∴AC = √288 = 16.97cm

Diagonal AC = Diagonal BD

Hence, Sum of the diagonals = 16.97cm + 16.97cm = 33.94cmIncorrectLet ‘a’ be the length of the side of the square ∎ABCD and AC is a diagonal of the square.

Perimeter of square ∎ABCD = 4a units

4a = 48cm

a = 12

We know that in a square each angle is 90° and the diagonals are equal.

In ΔABC, AC^{2}= AB^{2}+ BC^{2}= 12^{2}+ 12^{2}= 144 + 144 = 288

∴AC = √288 = 16.97cm

Diagonal AC = Diagonal BD

Hence, Sum of the diagonals = 16.97cm + 16.97cm = 33.94cm