FOUNDATION
Class: VIII, MATHEMATICS
5. MENSURATION
TEACHING TASK (JEE mains)
01. let 'a' be the dimensions of new cube
Dimensions of new cuboid
l = 3a, b = a, h = a
T.S.A. of new cuboid
= 2 (lb + bh + hl)
= 2 (3a² + a² + 3a²) = 14a²
Sum of surface areas of 3 cubes = 3 x 6a²
:. Ratio = 14a² / 3 x 6a² = 7/9
Ans. B
02 l = 3cm, b = 4cm, h = 5cm
length of the diagonal = √l² + b² + h²
= √9 + 16 + 25
= √50 = 5√2 Ans. B
03 a = 5
T.S.A. = 6a² = 6 x 5² = 150 cm² Ans. D
04 Area of walls to be painted
= Area of walls - (Area of 1 door + Area of 3 windows)
= 2 x 4 x (7+6.5) - (3 x 1.4 + 3 x 2 x 1)
= 97.8 mt²
Total cost of 15 rooms = 15 x 5.25 x 97.8
= 7702
Ans. B
05
Surface area of Tin = 2(lb + bh + hl)
= 2 ( 30 x 40 + 30 x 50 + 40 x 50 )
= 9400 cm²
Total Area (in m²) = 20 x 9400 cm²
= 188000 cm²
Total Area (in mt²) = 188000/10000 mt²
= 18.8 mt²
Total cost = 18.8 x 20
= Rs 376
Ans. B
06
Cylinder
r = 1.4/2 = 0.7 mt
h = 2 m
Surface area = 2πrh
= 2 x 22/7 x 7/10 x 2 = 88/10
Area if 5 resolution = 5 x 88/10 = 44 cmmt²
Ans: D
07 Cylinder
r = 7cm
h = 50 x 0.5 = 25 cm
T.S.A = 2πr(r+h)
= 2 x 22/7 x 7(7+25)
= 44 x 32
= 1408 cm²
Ans: A
08 Difference in C.S.A =
2πRh - 2πrh = 44 -> (1)
Volume of metal used
πR²h - πr²h = 99 -> (2)
Given h = 14
(1) => 2πh(R-r) = 44
=> 2 x 22/7 x 14 x (R-r) = 44
=> R-r = 0.5 -> (3)
πh(R²-r²) = 99
=> 22/7 x 14 (R²-r²) = 99
=> R²-r² = 2.25
=> (R+r)(R-r) = 2.25
=> (R+r)(0.5) = 2.25
=> R+r = 4.5 -> (4)
solving (3) & (4)
R = 2.5cm
r = 2cm
Ans: A
09. L.S.A of right Pyramid = 1/2 x p x l
=> 260 = 1/2 x 40 x l
=> l = 13cm
Ans: C
10 sphere
radius = 21/2 Cm.
Surface area of 5 ball = 5 x 4πr²
= 5 x 4 x 22/7 x 21/2 x 21/2
= 6930 cm²
Ans: D
11. let a be the edge of the cube.
Given T.S.A = 1/2 x Volume
=> 6a^2 = 1/2 x a^3
=> 12 = a or a = 6√2
Ans: C,D
12. Base perimeter = 4a
= 4 x 4
= 16 or 32/2
Ans: B, C, D
13. Statement I: 2 x π x r = 44
=> 2 x 22/7 x r = 44
=> r = 7 cm
Surface area = 4πr^2
= 4 x 22/7 x 7 x 7 = 616 cm^2 (true)
Statement II: Conceptual (true)
Ans: A
14. Statement I: Conceptual (true)
Statement II: Conceptual (true)
Ans: A
15. Assertion: Conceptual (True)
Reason: Conceptual (False)
l > h
Ans: C
16. Assertion: L.S.A of cube = 4a^2 (True)
Reason: Conceptual (true)
Ans: A
17. cone. d = 8 mt
r = 4 mt
h = 3 mt
∴ slant height = 5 mt
Ans: B
18. C.S.A = πrl
= 22/7 x 4 x 5 = 440/7 = 62 6/7
Ans: A
19. Conceptual
Ans: A
20. Conceptual
Ans: A
21. cylinder. r = 42cm, h = 120cm
Area of play ground = 500 x 2πrh
= 500 r 2 x 22/7 x 42 x 120
= 15,840,000 cm^2
= 1584 mt^2
Ans: 1584
22. TSA = 2(lb + bh + hl)
= 2(15 x 10 + 10 x 5 + 5 x 15)
= 2(275)
= 550 cm^2
Ans: 550
23 a) 100 cm^2 (r)
b) 1000000 mt^2 (p)
c) 100 mm^2 (s)
d) 100 cm^2 (q)
Ans: r, p, s, q
24. a) 6 (r)
b) 12 (t)
c) 8 (s)
d) rectangle (q)
Ans: r, t, s, q
LEARNERS MAKI (CUCs) 6
01. Cmceptral Ans: C
02. Cmcoptnral Ans: B
03 Cmcoptral Ans: C
04. Cmcoptral Ans: D
05 Cmcoptral Ans: B
06 Cmcoptral Ans: B
07 Cmcoptral Ans: C
08 Cmcoptral Ans: C
09 Cmcoptral Ans: D
10 Cmcoptral Ans: A
JEE MAINS
01. Cmcoptral Ans: C
02 8 Ans: B
03 L.S.A = 2h (l + b)
= 2x5 (10 +8)
= 180 cm² Ans: A
04 2(lb+bh+hl) = 100
2(7x6 + bx2 + 2x7) =100
=> b = 4cm Ans: B
05 2πr = 154 => r = 49/2 , h=15
T.S.A = 2πr (r+h)
= 2x22/7 x 49/2 (49/2 +15) = 6083cm² Ans: C
06 Conceptual
07 Hemisphere. r = 10 cm
T.S.A = 3πr²
= 3 × 22/7 × (10)²
= 942.85
Ans. B
08 l = 60cm, b = 40cm, h = 30cm
T.S.A = 2(lb + bh + hl)
= 2(60 × 40 + 40 × 30 + 30 × 60)
= 10800 cm²
Cost: 20 cm² -> 50 paise
1 cm² -> 50/20 paise
10800 cm² -> 5/2 × 10800 paise
= 27000 paise
= Rs 270.
Ans. C
09 Cone * Please discard this problem
T.S.A = 113 = 792/7 cm²
=> πr(r+l) = 792/7
=> 22/7 r(r+l) = 792/7
=> r² + rl = 36
Q => r² + r√(r² + 6²)
10. let a = 15cm, b = 20cm
h = √(a² + b²)
= √(15² + 20²) = 25
Now, 1/2 × a × b = 1/2 × h × r
=> 15 × 20 = 25 × r
=> r = 12cm
S.A = πr(a+b)
= 22/7 × 12 × (15+20) = 1320cm²
Ans: C
11.
4πr² = 154
=> 4 × 22/7 × r² = 154
=> r = 7/2 or 3.5
Ans: C, D
12. Cubi L.S.A = 4a², T.S.A = 6a²
L.S.A/T.S.A = 4a²/6a² = 2/3 or 4/6
T.S.A = 3/2 . L.S.A = 1.5 L.S.A
Ans: A, B, C
13. Statement I: Conceptual (Toul)
Statement II: Conceptual (Toul)
Ans: A
14. Statement I: Conceptual (False)
Statement II: Conceptual (Toul)
Ans: D
15. Assertion: Conceptual (Toul)
Reason: Conceptual (Toul)
Ans: A
16. Assertion: Conceptual (True)
Reason: Conceptual (True)
Ans. A
17. r = 7cm, R = 10.5cm
External Surface area = 2πR²
= 2 x 22/7 x (10.5)²
= 693 cm²
Ans. B
18 Area of Ring = π(R² - r²)
= 22/7 ((10.5)² - 7²)
= 22/7 x 17.5 x 3.5
= 192.5 cm²
Ans. A
19. Conceptual
Ans. C
20 Conceptual
Ans : A, C
21. Let the no. of bangles be n.
Height of the cylinder = n x 2mm = n x 0.2 cm
radius '' '' = 1.4cm.
C.S.A = 2πrh = 572
=> 2 x 22/7 x 1.4 x n x 0.2 = 572
=> n = 325
Ans 325
22. T.S.A = 4πr²
= 4 x 22/7 x 7 x 7 =
= 616
Ans: 616
23 A) cube(r)
B) cylinder(s)
C) pyramid(q)
D) Cone(p)
24 A) L.S.A prism = perimeter of the base x height
= 25 x 10
= 250 cm² (p)
B) Area = 2 x 30 = 60 cm² (q)
C) T.S.A = [unreadable]
Base perimeter = 25
5 x a = 25
=> a = 5 cm
= side of pentagonal Prism = a = 5cm
height = h = 10cm
Area of each rectangle = l x b = 5 x 10 = 50 cm²
2
Q L.S.A = 5 x 50 = 250 cm²
.. T.S.A = L.S.A + 2 x Base are
= 250 + 2 x 30 = 250+60 = 310cm²
(r)
d) No. of faces = 5+2 = 7(s)
Am. P,Q,r,s
=> THE END ⇐