Geometry

M. Velrajan

1. From the fig, find the value of ‘x

(a)30°      (b)60°                (c)90°                    (d)45°

6x = 180o

x = 300

2. In the given figure,   |B: |C  = 2:3. Find   |B

(a)1200              (b)520                       (c)780                       (d)1300

130 =  B + C = 2k + 3k

k = 26, B = 52

 

3. If the length of a rectangle  is decreased by 50%  and the breadth is increased by 80%, 

then the % change in the area of the rectangle is.   

(a)decreased by 10%   (b)increased by 10%   (c)decreased by 20%   (d)increased by 20%

If the length and breadth of a rectangle are changed by x% and y% respectively then the change

in the area is (x + y + (xy/100))%.

( - 50 + 80 + (-50) × 80/100)% = 30 - 40 = -10%.  Ans. (a)

Aliter:  Area of the rectangle = lb

After decrease by 50%, length = l - (50/100)l = 50l/100 = l/2 and 

increase by 80%, breadth = b + (80/100) b = b + ⅘b = 9b/5

New area = (l/2)(9b/5) = (9/10)lb

Area decreased and Decrease in area =lb - (9/10)lb = lb/10.

Percentage of decrease = ((lb/10) / lb)100 = 10%     Ans. (a)

4. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is

(a)40%                   (b)42%                     (c)44%                           (d)46%

If the length and breadth of a rectangle are changed by x% and y% respectively then the change

in the area is (x + y + (xy/100))%.

20 + 20 + 20 × 20/100 = 44%

Aliter:  Area of the rectangle = lb

After increase by 20%, length = l + (20/100)l = 6l/5 and 

increase by 20%, breadth = b + (20/100) b = 6b/5

New area = (6l/5)(6b/5) = (36/25)lb

Area increased and increase in area = (36/25)lb - lb = (11/25)lb.

Percentage of increase = ((11lb/25) / lb)100 = (11/25)100 = 44%     Ans. (c)

5. The length of a rectangle is increased by 60%. By what percent would the width have to be decreased

so as to maintain the same area.(a)37 ½%           (b)60%           (c)75%               (d)120%

If the length and breadth of a rectangle are changed by x% and y% respectively then the change

in the area is (x + y + (xy/100))%.

Since the same area to be maintained, 60 + y + 60y/100 = 0.

160y = - 6000

y = - 600/16 = - 37 ½ %

Width(breadth) have to be decreased by 37 ½ %

Aliter:  Area of the rectangle = lb

After increase by 60%, length = l + (60/100)l = 8l/5 and 

decrease by y% breadth = b - (y/100) b = (100 - y)b/100

New area = (8l/5)((100 - y)b/100) = (8(100 - y)/500)lb

Since the same area to be maintained, new area = area

(8(100 - y)/500)lb = lb

800 - 8y = 500

8y = 800 - 500 = 300

y = 300/8 = 37 ½

6. The perimeter of a circle is equal to the perimeter of a square. Then, their areas are in the ratioad

(a)4 : 1                     (b) 11 : 7                         (c)14 : 11                (d)22 : 7

Perimeter of a circle of radius r = 2πr

Perimeter of a square of side a = 4a

Given that 2πr = 4a.   

Area of the circle = πr2 

Area of the square = a2 

Ratio of their areas = πr2 / a2 = πr2 / (π2r2 / 4) (since a = πr/2)

                                                  = 4 / π = 4 (7/22) = 14 / 11     Ans.  (c)

7. If the perimeter of an isosceles right triangle is (6+3√2 ) m. then the area of the triangle is

    (a)4.5 m2                         (b)5.4m2                                      (c)9 m2                                          (d)81 m2

If a is the equal sides of the isosceles right triangle then the other side of the triangle√2 a and the

perimeter of the triangle = a + a + √2 a = (2 + √2 )a = 6+3√2 

Hence a = 3 and Area = ½ a2 = ½ 9 = 4.5       Ans. (a)

8. If the area of the ∆ABC is 68 sq. units and the vertices are A (6,7), B (-4,1) and C (a, -9) taken in

order. Find the value of a.     (a)-2       (b)3             (c)5           (d)2

Area of the triangle with vertices (x1, y1), (x2, y2), (x3, y3) is ½  [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)].

Hence ½ [ 6(1 - (-9)) - 4(-9 - 7) + a(7 - 1)] = 68

60 + 64 + 6a = 136

6a = 12

a = 2         Ans. (d)

9. A hollow cylindrical iron pipe is of length 35 cm. Its outer and inner diameters are 10 cm and 8 cm

respectively. Find the weight of the pipe if 1 cu.cm of iron weighs 7 gm.         

(a)6.93 kg      (b)9.90 kg     (c)7.53 kg      (d)7.9

Outer and inner radius are 5cm, 4cm

The volume of cylinder = (22/7)lr2

Volume of the hollow cylinder = volume of the outer cylinder - volume of the inner cylinder

                                                    = (22/7) × 35 × 52 - (22/7) × 35 × 42

                                                    = (22/7) × 35 × (52 - 42) = (22/7) × 35 × 9 cu.cm 

Since 1 cu.cm of iron weighs 7 gm, weight of the pipe =  (22/7) × 35 × 9 × 7 

                                                                                           = 22 × 35 × 9 = 6930 gm = 6.93 kg

10. A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Find the ratio

of their volumes.         (a)3 : 2: 1                (b)1: 2: 3            (c)3: 1: 2             (d)1: 3: 2

Since the cone, hemisphere and cylinder stand on equal bases and have the same height,

Radius of the cone = radius of the hemisphere = radius of the cylinder = r

Height of the cone = height of the cylinder = h = height of the hemisphere

But height of the hemisphere = radius of the hemisphere = r. Hence h = r 

Volume of a cone = ⅓ πr2h = ⅓ πr3

Volume of the hemisphere = ⅔ πr3

Volume of the cylinder = πr2h = πr3

Ratio of their volumes is ⅓ πr3 :  ⅔ πr3 : πr3 : : ⅓ : ⅔ : 1 

                                                                           : : 1 : 2 : 3

11. The radius of a wheel is 1.75m. How many revolutions will it make in travelling 11 km?

(a)10                   (b)100               (c)1000                           (d)10000

In one revolution, the distance travelling by the wheel = perimeter of the wheel  = 2πr

                                                                                            = 2 × (22/7) × (7/4) = 11m

For 11m of travelling, number of revolutions = 1

For 11km = 11000m of travelling, number of revolutions = 11000/11 = 1000

12. Find the height of a parallelogram whose area is 300 cm2 and base is 15 cm.    

(a)10 cm              (b)15 cm                       (c)20 cm                       (d)30 cm

Area of parallelogram = base × height

Hence height = area/base = 300/15 = 20

13. A circular ground of radius 7m has a path of width  7m around it on its outside. The area of the path

is approximately equal to __ sq.m.(π = 22/7)  (a)154          (b)308          (c)462              (d)616

Since the path is outside the ground, radius of the inner circle = 7m and 

radius of the outer circle = inner radius + path width = 7 + 7 = 14m

Area of outer circle = (22/7) × 142 

Area of inner circle = (22/7) × 72 

Area of the path = area of outer circle - area of inner circle 

                             = (22/7) × 142  - (22/7) × 72 = (22/7) × (142  - 72) = (22/7) × 21 × 7 = 22 × 21

Clearly it is more than 400 but less than 600, hence without multiplying, from the answer choices

the Ans. (c) 

(or) Area of outer circle = (22/7) × 142 = 22 × 14 × 2 = 616

 area of inner circle = (22/7) × 72 = 22 × 7 = 154

Area of the path = area of outer circle - area of inner circle = 616 - 154 = 462 sqm

14. If the height and the base area of a right circular cone are 5 cm and 48 sq.cm respectively, then the

volume of the cone is equal to    (a)240 cm3         (b)120 cm3           (c)80 cm3         (d)480 cm3

Volume = ⅓ × base area × height = ⅓ × 48 × 5 = 80

15. The length, breadth and height of a room are respectively 12 meters, 9 meters and 6 meters.

How many cubic boxes are needed to fill the room if the side of each box is 1.5 meters?

(a)1072    (b)648   (c)324    (d)192

Room volume = 12 × 9 × 6

Box volume = (3/2) × (3/2) × (3/2)

Number of boxes needed = (12 × 9 × 6) / ((3/2) × (3/2) × (3/2)) = 12 × 2 × 2 × 2 × 2 = 192 

16. If the radius of a circle is increased by 25% then its area is increased by

(a)50%                (b)25%                        (c)56.25%                              (d)46.25%

Suppose the radius of the circle is increased by a% then the increase in the radius = (a/100)r.

The increase in the area of the circle = π(r + (a/100)r)2 - πr2 = π(2r(a/100)r + ((a/100)r)2)

                                                                                                     = πr2(a/100)(2 + (a/100))

Percentage of increase of area = [πr2(a/100)(2 + (a/100)) / πr2] × 100 

                                                    = a(2 + (a/100)).

So, if the radius of the circle is increased by a% then 

the percentage of increase of area = a(2 + (a/100)).

Here a = 25

Hence percentage of increase in area = 25(2 + (25/100)) =  25(2 + ¼) = 25 × 2 + 25 × ¼ = 56.25 

17. In the adjoining figure the arcs BD, have  A and C as centres. Find the area of shaded portions.

                          (a)82                  (b)72                 (c)42                            (d)84

The arc DB(blue) together with the lines BA and AD is the quarter of the circle ADB with centre

A and radius AD = 14. 

And the area of the quarter circle ADB = ¼ of the area of the circle with radius 14

= ¼ × (22/7) × 14 × 14 = 22 × 7 = 154

It is clear that the area of the unshaded portion DCB

= area of  square ABCD - area of quarter circle ADB

= 14 × 14 - 154 = 196 - 154 = 42.

By symmetry, area of the unshaded portion ADB = area of the unshaded portion DCB = 42

Area of the shaded portion = area of the quarter circle ADB -  the unshaded portion ADB

= 154 - 42 = 112

Oh this answer is not in the answer choices!

What is wrong?

Yes, 

Total area of the unshaded portion= 42 + 42 = 84

This is in the answer choices.

So that the question had to be “Find the area of the unshaded portion”

18. Find the length of the altitude of an equilateral  triangle of side 3√3 cm

(a)27 cm               (b)93cm           (c)9 cm                      (d)4.5 cm

Altitude of the equilateral triangle of side a = (√3/2)a.

Here a =3√3, hence the altitude = (√3/2)3√3 = 9/2 = 4.5

19. The total surface area of a cube is 384 cm2, then its volume is

(a)521 cm3   (b)512 cm3   (c)412 cm3   (d)421 cm3

Total surface area = 6a2 = 384

Hence a2 = 384 / 6 = 64 and a = 8

Volume = a3 = 64 × 8 = 512

20.  The upper part of a tree broken by wind makes an angel of 300 with the ground and the distance from the root of the point where the top of the tree touches the ground is 10√3 m. What is the height of the tree?   (1) 30 m  (2) 40 m   (3) 50 m      (4) 60 m

Suppose A is the root of the tree and BC broke part of the tree.

Given AC = 10√3

AB/AC = tan 30 = 1/√3 and AB/BC = sin 30 = ½

AB = AC/√3 = 10 and BC = 2AB = 20

Tree height = AB + BC = 10 + 20 = 30 m

 

21. A cube painted blue on all faces is cut into 125 cubes of equal size. How many cubes are there which are painted on one face only?    (a) 8 (b) 16 (c) 36 (d) 54

To get 125 = 53 cubes of equal size each side of the cube has to be divided into 5 equal parts.

Among the 125 equal small cubes the small cubes with one of the 8 vertices of the original cube as

a vertex are painted on 3 sides, the other small cubes along with the edges of the original cube are painted on 2 sides (12 × 3 = 36 small cubes).

The rest of the small cubes on each side of the original cube are painted on only one side 

(6 × 9 = 54 small cubes). And 125 - (8 + 36 + 54) = 27 small cubes are the inner small cubes and no

sides of them are painted).

Aliter : Consider the cube formula N3 = (N - 2)3 + 6(N -2)2 + 12(N - 2) + 8. Note that 8 is the

number of vertices, 12 is the number of edges and 6 are the number of sides of a cube.

In general, if the original cube is divided in to N3 small cubes then since 

N3 = (N - 2)3 + 6(N -2)2 + 12(N - 2) + 8,

8 small cubes are painted on three sides, 12(N - 2) small cubes are painted on two sides, 6(N - 2)2

small cubes are painted on only one side and (N - 2) small cubes are painted on no side.

Taking N = 5 in the cube formula,

53 = 33 + 6 × 32 + 12 × 3 + 8

8 painted on 3 sides, 12 × 3 on 2 sides, 6 × 32 on one side and 33 on no side.