Geometry Questions and Answers

Q1 :

The length of the side of a square whose area is exactly equal to the area of a rectangle whose side are 6.4 m and 2.5 m is

A

0.6 m

B

2.4 m

C

60 m

D

4 m

Check Answer

Right Answer: D

4 m

Description:

Side of the square

$=sqrt{ 6.4 times 2.5}$ $= 6.4 \times 2.5$
$=sqrt{frac { 64 times 2.5}{100}} = sqrt{16} = 4 text m$

Q2 :

The outer and inner radii of a metallic spherical shell are 2 cm and 1 cm respectively. If it is melted to make a solid sphere, then the radius of the sphere will be

A

$7frac{1}{2}$ cm

B

$9frac{1}{3} text{ cm }$

C

$7frac{1}{3} text{ cm}$

D

3 cm

Check Answer

Right Answer: C

$7frac{1}{3} text{ cm}$

Description:

The volume of the metallic shell

$= frac{4}{3} pi (2)^3 - frac{4}{3} pi(1)^3 = frac{28pi}{3}$

But by question,

Volume of sphere $= frac{28pi}{3}$

$Rightarrow frac{4}{3}pi r^3 = frac{28pi}{3} Rightarrow r^3 = 7 Rightarrow r = 7^frac{1}{3}$

Q3 :

A cone of base diameter 20 cm and height 5 cm is melted to form a sphere of radius 5 cm. The volume of the material that remains is

A

$2 text{ cm}^3$

B

$5 text{ cm}^3$

C

$pi text{ cm}^3$

D

zero

Check Answer

Right Answer: D

zero

Description:

$text{ Volume of the cone } = frac{1}{3} times frac{22}{7} times 10^2 times 5$

$= frac{11000}{21}$

$text{ Volume of sphere } = frac{4}{3}pi times 125$

$= frac{11000}{21}$

Q4 :

A rectangle of sides 10cm and 8 cm is folded to form a cylinder in two different ways. The volume of the bigger cylinder thus formed is

A

$frac{160}{pi} text{cu cm }$

B

$frac{175}{pi} text{cu cm }$

C

$frac{200}{pi} text{cu cm }$

D

$frac{150}{pi} text{cu cm }$

Check Answer

Right Answer: A

$frac{160}{pi} text{cu cm }$

Description:

For the bigger cylinder,
$text{ H } = 10 text{ cm }, 2 pi r = 8$

$therefore text{ V } = pi r^2 quadtext{ H } = pi (frac{16}{pi^2}) times 10 = frac{160}{pi} text{cu cm}$

Q5 :

If the length of a side of rhombus 13 cm and one of its diagonals is of length 24 cm, then the area of rhombus is

A

$240 text{cm}^2$

B

$156 text{cm}^2$

C

$130 text{cm}^2$

D

$120 text{cm}^2$

Check Answer

Right Answer: D

$120 text{cm}^2$

Description:

Area of rhombus

$= frac{1}{2} times text{ product of diagonal }$

$= frac{1}{2} times 24 times 10$

$Big[ because text{ small diagonal } = 2sqrt{13^2 - 12^2 } = 2 times 5 = 10 Big]$

$= 120 text{ sq. m }$

Q6 :

The volume of a right circular cylinder is $1100 text{ cm}^3$ and the radius of its base is 5cm. The area of its curved surface is

A

$440 text{ cm}^2$

B

$(440 + 25 pi) text{ cm}^2$

C

$(440 + 50pi) text{ cm}^2$

D

$450 text{ cm}^2$

Check Answer

Right Answer: A

$440 text{ cm}^2$

Description:

$text{Volume of cylinder } = pi r^2h$

$= 1100 = pi times 5 times 5 times h$

$Rightarrow h = frac{1100}{pi times 25} text{m} = 14text{m.}$

$therefore text {Are of curved surface} = 2pi rh$

$= 2pi times 5 times 14 = 440 text{cm}^2$

Q7 :

A square and an equilateral triangle have equal perimeters. If the area of the equilateral triangle is $16sqrt3 text{ cm}^2$ , then the side of the square is

A

4 cm

B

$4sqrt2 text{ cm}$

C

$6sqrt2 text{ cm}$

D

6 cm

Check Answer

Right Answer: D

6 cm

Description:

Let side of square be $x$ and that of triangle is $y$

$begin{aligned} therefore 4x &= 3y x &= frac{3}{4}y text{Area of triangle } &= frac{sqrt3}{4} times y^2 16sqrt3 &= frac{sqrt3}{4} times y^2 y &= 8 text{ and } x &= frac{3}{4} times 8 = 6 text{ cm} end{aligned}$

Q8 :

The height and base radius of a right circular cone are 4 and 3 cm respectively. The total surface area of the cone is

A

$15pi text{ cm}^2$

B

$18pi text{ cm}^2$

C

$21pi text{ cm}^2$

D

$24pi text{ cm}^2$

Check Answer

Right Answer: D

$24pi text{ cm}^2$

Description:

Total surface are of cone $= pi rl + pi r^2$
$= pi 3(3 + 5) = 24pi text{ cm}^2$

Q9 :

If both the radius and height of a cone are increased by 50%, then the volume of the cone will increase by

A

100%

B

200%

C

225.50%

D

237.50%

Check Answer

Right Answer: D

237.50%

Description:

$begin{aligned} text{Volume of cone } &= frac{1}{3}pi r^2h text{New volume of cone } &=frac{1}{3}pi Bigg(frac{3}{2}Bigg)^2 r^2 frac{3}{2}h &= frac{1}{3}pifrac{9}{8}r^2h = frac{27}{8}v text{ increase in volume } &= frac{27}{8} v - v frac{19}{8}v % text{ increase } &= frac{frac{19}{8}V times 100}{V} = 237.5 end{aligned}$

Q10 :

If a solid sphere of radius $frac{3}{4}$ m is melted and formed into a right circular cylinder of height 1 m, then the radius of the base of the cylinder will be

A

0.75 m

B

1.00 m

C

1.25 m

D

1.50 m

Check Answer

Right Answer: A

0.75 m

Description:

$begin{aligned} text{Volume of the sphere } &= frac{4}{3}pifrac{3}{4} times frac{3}{4} times frac{3}{4} text{ m}^3 text{By question } pi r^2h &= frac{4}{3}pifrac{3}{4} times frac{3}{4} times frac{3}{4} r^2 &= frac{3}{4} times frac{3}{4} (h = 1, text{ given}) therefore r &= frac{3}{4} = 0.75 text{ m } end{aligned}$

Q11 :

The radius of the base of a right circular cylinder is halved and the height is increase by 50%. The ratio of the volume of the original cylinder to that of the new cylinder will be

A

2 : 3

B

8 : 3

C

3 : 1

D

4 : 1

Check Answer

Right Answer: B

8 : 3

Description:

$dfrac{V_0}{V_n} = dfrac{pi r^3h}{pibiggl(dfrac{r}{2}biggr)^2 . dfrac{3}{2}h} = dfrac{8}{3}$

Q12 :

The orthocentre of the triangle whose vertices are (0, 0), (3, 0) and (0, 4) is

A

(0, 0)

B

(0, 2)

C

(2, 0)

D

(2, 2)

Check Answer

Right Answer: A

(0, 0)

Description:

$text{Here AB } = sqrt{(3 - 0)^2 + (0 - 0)^2 } = 3$

$text{Similary BC } = 5 text{ and CA } = 4$

$Rightarrow$ The given triangle is a right angle and hence all the three altitudes intersect at the point A (0, 0)

$therefore text{The orthocentre } = (0,0)$

Q13 :

If ABCD is a rhombus, then

A

$text{AC}^2 + text{BD}^2 = 6text{AB}^2$

B

$text{AC}^2 + text{BD}^2 = 5text{AB}^2$

C

$text{AC}^2 + text{BD}^2 = 4text{AB}^2$

D

$text{AC}^2 + text{BD}^2 = 3text{AB}^2$

Check Answer

Right Answer: C

$text{AC}^2 + text{BD}^2 = 4text{AB}^2$

Description:

ABCD is a rhombus, then
by pythagoras theorem

$text{AC}^2 + text{BD}^2 = (text{OA} + text{OC})^2 (text{OB} + text{OD})^2$

$= 4(text{OA}^2 + text{OB}^2) = 4 text{AB}^2$

Q14 :

The circle $x^2 + y^2 - 8x - 6y + 16 = 0$

A

Touches the x-axis

B

Touches the y-axis

C

Touches both the axes

D

Do not Touch any axis

Check Answer

Right Answer: A

Touches the x-axis

Description:

The centre of the circle (4, 3) and its radius is $r = sqrt{16 + 9 - 16} = 3$
If the circle

If the circle touches the $x$ -axis, then at that point $y$ must be zero.

$therefore quad$ Equation of the circle becomes $x^2 - 8x + 16 = 0$

$Rightarrow (x - 4)^2 = 0 Rightarrow x = 4$

If the circle touches the y axis, then at that point $x = 0$ .

$therefore quad$ Equation of the circle becomes

$y^2 - 6y + 16 = 0$

$Rightarrow y$ is imaginary

$therefore quad$ The given circle touches the $x$ -axis only at the point (4, 0)

Q15 :

The area of the circle drawn, with its diameter as the diagonal of the cube of side of length 1 cm each, is

A

$frac{4pi}{3} text{ sq cm}$

B

$frac{3pi}{2} text{ sq cm}$

C

$frac{3pi}{4} text{ sq cm}$

D

$frac{2pi}{3} text{ sq cm}$

Check Answer

Right Answer: C

$frac{3pi}{4} text{ sq cm}$

Description:

Diagonal $= sqrt{3} text{ cm}$

Area of the circle

$= pi r^2 = Big(frac{sqrt{3}}{2}Big)^2$

$=frac{3pi}{4}$

Q16 :

The perimeters of an equilateral triangle and a square are the same. Then

A

$frac{text{Area of triangle}}{text{Area of square}} = frac{4}{3}$

B

$frac{text{Area of triangle}}{text{Area of square}} = 1$

C

$frac{text{Area of triangle}}{text{Area of square}} = 1.5$

D

$frac{text{Area of triangle}}{text{Area of square}} lt 1$

Check Answer

Right Answer: D

$frac{text{Area of triangle}}{text{Area of square}} lt 1$

Description:

Let the side of an equilateral triangle be $x$ , then perimeter $=3x$ and area of triangle $=frac{sqrt{3}}{4}x^2$ and side of a square = y then perimeter $= 4y text{ and area} = y^2$

$= 4 y and area = y^{2}$

By question

$3x = 4y Rightarrow frac{x}{y} = frac{4}{3}$ $frac{text{Area of triangle}}{text{Area of square}$

$frac{text{Area of triangle}}{text{Area of square}} =dfrac{frac{sqrt{3}}{4}x^2}{y^2} = dfrac{sqrt{3}}{4}timesBig(frac{4}{3}Big)^2$

$= frac{9sqrt{3}}{64}lt 1$

Q17 :

The area of the largest triangle inscribed in a semi-circle of radius R is

A

$2 text{ R}^2$

B

$text{R}^2$

C

$frac{1}{2}text{R}^2$

D

$frac{3}{2}text{R}^2$

Check Answer

Right Answer: B

$text{R}^2$

Description:

Required area $= frac{1}{2} times 2r times r = r^2$

Q18 :

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

A

$text{A}_1 text{A}_3 lt 16 text{A}_2^2$

B

$text{A}_1 text{A}_3 gt 16 text{A}_2^2$

C

$text{A}_1 text{A}_3 = 16 text{A}_2^2$

D

$text{A}_1 text{A}_3 gt 2text{A}_2^2$

Check Answer

Right Answer: A

$text{A}_1 text{A}_3 lt 16 text{A}_2^2$

Description:

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

$text{A}_1 = pi text{R}^2, text{A}_2 = frac{pitext{R}^2}{4}$

$text{A}_3 = pitext{R}^2 - frac{pitext{R}^2}{4} = frac{3pitext{R}^2}{4}$

$therefore text{A}_1text{A}_3 = frac{3}{4}pi^2text{R}^4 = 3pitext{R}^2 times text{A}_2$

$= 12 text{A}_2^2 lt 16 text{A}_2^2$

Q19 :

If $h,c$ and $v$ are respectively the height, the curved surface area and volume of a cone, then $3pi quad vh^3 - c^2h^2 + 9v^2$ is equal to

A

0

B

1

C

2

D

3

Check Answer

Right Answer: A

0

Description:

$c = pi quad r sqrt{r^2 + h^2}, v = frac{1}{3}pi r^2 h$

Now $3pi vh^3 = pi^2 r^2 h^4$

$c^2h^2 = pi r^2 (r^2 + h^2) h^2 text{ and } 9v^2 = pi^2 r^4 h^2$

$Rightarrow 3pi vh^3 - c^2h^2 + 9v^2 = 0$

Q20 :

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

$2310 text{ cm}^3$

B

$3080 text{ cm}^3$

C

$39270 text{ cm}^3$

D

$48510 text{ cm}^3$

Check Answer

Right Answer: D

$48510 text{ cm}^3$

Description:

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

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

Now from figure

$frac{text{PD}}{text{PO}} = frac{text{QC}}{text{QO}} Rightarrow frac{28}{45 + h} = frac{7}{h} Rightarrow h = 15$

Now on putting the value of $h$ in equation (I)

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

Q21 :

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

A

$44sqrt3 text{ cm}$

B

$45sqrt3 text{ cm}$

C

$46sqrt3 text{ cm}$

D

$47sqrt3 text{ cm}$

Check Answer

Right Answer: D

$47sqrt3 text{ cm}$

Description:

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

= 47

$therefore quad d = sqrt{3} l$

$= 47sqrt3 text{ cm }$

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

Q22 :

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%

Check Answer

Right Answer: B

32.25%

Description:

Area of circle $= 2pi r$

$text{Given, } dfrac{2pi(r_1 - r)}{2pi r} times 100 = 15$

$Rightarrow quad r_1 = frac{23}{20}r$

$therefore$ % increase in area = $= frac{pi Big(frac{23}{20}rBig)^2 - pi r^2}{pi r^2} times 100 = 32.25 %$

Q23 :

If the side of a square is expanded by eight cm, its area expanded 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

Check Answer

Q24 :

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

A

$(p + n )r$

B

$(2p - n )r$

C

$pr$

D

$(p - n)$

Check Answer

Right Answer: C

$pr$

Description:

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

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

Area of polygon $= n times frac{1}{2} times frac{2text{P}}{n } times r$

= Pr

Q25 :

The number of spherical bullets, each bullet being 4 cm in diameter, that can be made out of a cube of lead whose edge is 44 cm, is:

A

2541

B

2551

C

2561

D

2571

Check Answer

Right Answer: A

2541

Description:

No. of Spherical bullets $= frac{text{volume of a cube}}{text{volume of each bullet}}$

$= dfrac{44 times 44 times 44}{frac{4}{3} times frac{22}{7} times 2 times 2 times 2}$

= 2541

Q26 :

If a square of area $frac{text{A}}{2}$ is cut off from a given square of area A, then the ratio of diagonal of the cut off square to that of the given square is

A

$1:sqrt{5}$

B

$1:5$

C

$1:sqrt{2}$

D

$1:2sqrt{5}$

Check Answer

Right Answer: C

$1:sqrt{2}$

Description:

Area of a bigger square $= a^2$

$therefore text{Diagonal } = sqrt{2}a = d_1$

Area of a smaller square $=frac{a^2}{2} = x^2$

$x$ = side of smaller square ... $x = frac{a}{sqrt2}$

Diagonal of a smaller square $= d_2 = sqrt{2}x$

$d_2 = sqrt{2} cdot frac{a}{sqrt2} = a$

$therefore$ Required Ratio $= d_2 : d_1 = a:sqrt{2}a = 1 :sqrt{2}$

Q27 :

If the diagonal AC, of a rectangle ABCD is of length 2d and it divides angle BAD in the ratio 1:2, then the area of the rectangle is equal to

A

$sqrt{2}cdot d^2$

B

$4d^2$

C

$2sqrt{2}cdot d^2$

D

$sqrt{3} cdot d^2$

Check Answer

Right Answer: D

$sqrt{3} cdot d^2$

Description:

$text{AC} = 2d, x + 2x = 90degree Rightarrow x = 30degree frac{text{AB}}{text{AC}} = text{cos }30degree$

$therefore text{AB } = frac{sqrt3}{2} times 2d = sqrt{3}d$

$frac{text{BC}}{text{AC}} = sin 30degree$

$therefore text{BC } = text{AC }times frac{1}{2} = 2d times frac{1}{2} = d$

Area of a rectangle $= text{AB } times text{BC} = sqrt{3}d cdot d = sqrt{3}d^2$

Q28 :

If the sides of a triangle are 15 cm, 16 cm and 17 cm, then the area of the triangle is equal to

A

$16 sqrt{21}$ sq cm

B

$18 sqrt{21}$ sq cm

C

$24 sqrt{21}$ sq cm

D

$30 sqrt{21}$ sq cm

Check Answer

Right Answer: C

$24 sqrt{21}$ sq cm

Description:

$text{S } = dfrac{15+16+17}{2} = 24$

Area of a $triangle = sqrt{s(s - a)(s-b)(s-c)}$

$= sqrt{24(24-15)(24-16)(24-17)} = 24sqrt{21} text{ sq. cm}$

Q29 :

The radius of a wheel is 84 cm. If the wheel makes five revolutions in 5 seconds, then the speed of the wheel, approximately, is

A

19 km/h

B

33 km/h

C

35 km/h

D

38 km/h

Check Answer

Right Answer: A

19 km/h

Description:

5 revolution in 5 sec
$therefore$ One revolution in 1 sec
Distance covered in 1 revolution

$=2pi r = 2 times frac{22}{7} times 84 =2 times 22 times 12 text{ cm}$

$text{Speed } = dfrac{text{D}}{text{T}} = dfrac{2 times 22 times 12}{1text{ sec}}text{ cm}$

$= frac{2 times 22 times 12}{100} times frac{18}{5} text{km/hr} =19 text{ km/hr app}$

Q30 :

If the measurement of each of the interior angles of polygon is $160degree$ , then the number of sides of the polygon would be equal to

A

9

B

12

C

18

D

27

Check Answer

Right Answer: C

18

Description:

$frac{(n - 2) 180 }{n} = 160degree Rightarrow n = 18$

No of sides = 18