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Geometry Questions and Answers

Q21 :

A circular disc of area A1text{A}_1 is given. With its radius as the diameter, a circular disc of area A2text{A}_2 is cut out of it. The area of the remaining disc is denoted by A3text{A}_3. Then

A
  

A1A3<16A22text{A}_1 text{A}_3 lt 16 text{A}_2^2

B
  

A1A3>16A22text{A}_1 text{A}_3 gt 16 text{A}_2^2

C
  

A1A3=16A22text{A}_1 text{A}_3 = 16 text{A}_2^2

D
  

A1A3>2A22text{A}_1 text{A}_3 gt 2text{A}_2^2

View Answer
Correct Answer: A

A1A3<16A22text{A}_1 text{A}_3 lt 16 text{A}_2^2

Description:

Let the radius of the disc of area A1text{A}_1 be R. Then

A1=πR2,A2=πR24text{A}_1 = pi text{R}^2, text{A}_2 = frac{pitext{R}^2}{4}

A3=πR2−πR24=3πR24text{A}_3 = pitext{R}^2 - frac{pitext{R}^2}{4} = frac{3pitext{R}^2}{4}

∴A1A3=34π2R4=3πR2×A2therefore text{A}_1text{A}_3 = frac{3}{4}pi^2text{R}^4 = 3pitext{R}^2 times text{A}_2

=12A22<16A22= 12 text{A}_2^2 lt 16 text{A}_2^2

Q22 :

If h,ch,c and vv are respectively the height, the curved surface area and volume of a cone, then 3πvh3−c2h2+9v23pi quad vh^3 - c^2h^2 + 9v^2 is equal to

A
  

0

B
  

1

C
  

2

D
  

3

View Answer
Correct Answer: A

0

Description:

c=πrr2+h2,v=13πr2hc = pi quad r sqrt{r^2 + h^2}, v = frac{1}{3}pi r^2 h

Now 3πvh3=π2r2h43pi vh^3 = pi^2 r^2 h^4

c2h2=πr2(r2+h2)h2 and 9v2=π2r4h2c^2h^2 = pi r^2 (r^2 + h^2) h^2 text{ and } 9v^2 = pi^2 r^4 h^2

⇒3πvh3−c2h2+9v2=0Rightarrow 3pi vh^3 - c^2h^2 + 9v^2 = 0

Q23 :

If the radii of the circular ends of a bucket of height 45 cm are 28 cm and 7 cm respectively, then the capacity of the bucket is

A
  

2310 cm32310 text{ cm}^3

B
  

3080 cm33080 text{ cm}^3

C
  

39270 cm339270 text{ cm}^3

D
  

48510 cm348510 text{ cm}^3

View Answer
Correct Answer: D

48510 cm348510 text{ cm}^3

Description:

Capacity of the bucket = Volume of cone ADO - Volume of cone BCO

=13×π×(28)2×(45+h)−13×π×(7)2×h...(I)= frac{1}{3} times pi times (28)^2 times (45 + h) - frac{1}{3} times pi times (7)^2 times h quadtext{...(I)}

Now from figure

PDPO=QCQO⇒2845+h=7h⇒h=15frac{text{PD}}{text{PO}} = frac{text{QC}}{text{QO}} Rightarrow frac{28}{45 + h} = frac{7}{h} Rightarrow h = 15

Now on putting the value of hh in equation (I)

We get required Volume =48510 cm3= 48510 text{ cm}^3

Q24 :

If the radius and height of a right circular cylinder are 21 cm and 35 cm, respectively, then the total surface area of the cylinder is:

A
  

7092 sq cm

B
  

7192 sq cm

C
  

7292 sq cm

D
  

7392 sq cm

View Answer
Correct Answer: D

7392 sq cm

Description:

Total surface are of a Cylinder

=2πr(r+h)= 2pi r(r + h)

=2×227×21(21+35)= 2 times frac{22}{7} times 21 (21 + 35)

= 7392 sq cm

Q25 :

If the surface area of a cube is 13254 cm2, then the length of its diagonal is:

A
  

443 cm44sqrt3 text{ cm}

B
  

453 cm45sqrt3 text{ cm}

C
  

463 cm46sqrt3 text{ cm}

D
  

473 cm47sqrt3 text{ cm}

View Answer
Correct Answer: D

473 cm47sqrt3 text{ cm}

Description:

Length of diagonal  (l)=132546(l) = sqrt{dfrac{13254}{6}}

= 47

∴d=3ltherefore quad d = sqrt{3} l

=473 cm = 47sqrt3 text{ cm }

[∵Surface are = 6l2 of a cube][because text{Surface are = } 6l^2 text{ of a cube}]

Q26 :

A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent in the form of a circle, then the area of the circle is:

A
  

121 cm2121 text{ cm}^2

B
  

242 cm2242 text{ cm}^2

C
  

154 cm2154 text{ cm}^2

D
  

60 cm260 text{ cm}^2

View Answer
Correct Answer: C

154 cm2154 text{ cm}^2

Description:

Area of Square (a2)=121(a^2) = 121

∴therefore a = 11 cm

∴ length wire =4atherefore text{ length wire } = 4a

= 44  cm

∴ perimeter of circle =44 cm therefore text{ perimeter of circle } = 44 text{ cm }

⇒2πr=44Rightarrow quad 2pi r = 44cm

∴r=7therefore quad r = 7 cm

∴r=7therefore quad r = 7

∴Area of circle=pr2=227×7×7therefore text{Area of circle} = text{p}r^2 = frac{22}{7} times 7 times 7

=154 cm2= 154 text{ cm}^2

Q27 :

A wire in the form of a circle of radius 98 cm is cut and bent in the form of a square. The side of the square, thus, formed is:

A
  

146 cm

B
  

152 cm

C
  

154 cm

D
  

156 cm

View Answer
Correct Answer: C

154 cm

Description:

Here

r=98r = 98 cm

Let side of square =a=a

Given, 4a=2πr4a = 2pi r

=2×227×98= 2 times frac{22}{7} times 98

∴a=154therefore quad a = 154 cm

Q28 :

If the radius of a circle is increased such that its circumference increases by 15%, then the area of the circle will increase by:

A
  

31.25%

B
  

32.25%

C
  

33.25%

D
  

34.25%

View Answer
Correct Answer: B

32.25%

Description:

Area of circle  =2πr= 2pi r

Given, 2π(r1−r)2πr×100=15text{Given, } dfrac{2pi(r_1 - r)}{2pi r} times 100 = 15

⇒r1=2320rRightarrow quad r_1 = frac{23}{20}r

∴therefore % increase in area = =π(2320r)2−πr2πr2×100=32.25%= frac{pi Big(frac{23}{20}rBig)^2 - pi r^2}{pi r^2} times 100 = 32.25 %

Q29 :

If the side of a square is increased by 8 cm, its area increases by 120 sq cm. The side of the square is:

A
  

2.5 cm

B
  

3.5 cm

C
  

4.5 cm

D
  

5.5 cm

View Answer
Correct Answer: B

3.5 cm

No Description Available
Q30 :

If a circle touching all n sides of a polygon of perimeter 2p has radius r, then the area of the polygon is:

A
  

(p+n)r(p + n )r

B
  

(2p−n)r(2p - n )r

C
  

prpr

D
  

(p−n)(p - n)

View Answer
Correct Answer: C

prpr

Description:

 Given the perimeter of polygon =2Ptext{ Given the perimeter of polygon } = 2text{P}

 then each side of a polygon =2Pntext{ then each side of a polygon } = frac{2text{P}}{n}

Area of polygon =n×12×2Pn×r= n times frac{1}{2} times frac{2text{P}}{n } times r

= Pr

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