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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.
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 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}UnattemptedLength 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
=15cmUnattemptedBy 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 = 45.25cmUnattemptedLet ‘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(sa)(sb)(sc))
Where, S = (a + b + D)/2; i.e.; (5 + 4 + 3)/2 = 6
⟹2 √(6(65)(64)(63))
⟹2√(6×1×2×3)
⟹2√36
⟹2 × 6
⟹ 12 cm^{2}IncorrectRequired area = 2√(s(sa)(sb)(sc))
Where, S = (a + b + D)/2; i.e.; (5 + 4 + 3)/2 = 6
⟹2 √(6(65)(64)(63))
⟹2√(6×1×2×3)
⟹2√36
⟹2 × 6
⟹ 12 cm^{2}UnattemptedRequired area = 2√(s(sa)(sb)(sc))
Where, S = (a + b + D)/2; i.e.; (5 + 4 + 3)/2 = 6
⟹2 √(6(65)(64)(63))
⟹2√(6×1×2×3)
⟹2√36
⟹2 × 6
⟹ 12 cm^{2}  Question 5 of 12
5. Question
1 pointsA hallroom 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
= 483UnattemptedSide 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 = 26.41cmUnattemptedIn 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 nonparallel 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(sk)(sc) (sd))
= (55 + 95 )/40 √(60(6040)(6046) (6034))
= 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(sk)(sc) (sd))
= (55 + 95 )/40 √(60(6040)(6046) (6034))
= 3.75 ×√(60 × 20 ×14 ×26)
= 3.75 × 660.90
= 2478.41 cm^{2}UnattemptedLet 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(sk)(sc) (sd))
= (55 + 95 )/40 √(60(6040)(6046) (6034))
= 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 mUnattemptedA = 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 = 320meterUnattemptedRequired 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 = 24.33cmUnattemptedIn 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
=9.22cmUnattemptedBy 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 = 33.94cmUnattemptedLet ‘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