# Free Basic Geometry - 01 Practice Test - CAT

### Question 1

In the following figure, the radii of the largest and smallest circles are 20 and 5. Find the radius of the intermediate circle.

#### SOLUTION

Solution :B

Solution: The fig can be considered as:

Since the triangles are similar:

r1+r2r2+r3=r1−r2r2−r3Substituting r1= 20 and r3= 5 , r2 = 10. Option (b).

### Question 2

Find the height of the right cylinder whose volume is 511 cm3 and the area of the base is 36.5 cm2.

#### SOLUTION

Solution :C

Ans: c.

Soln: Volume of a cylinder = πr2h

Area of the base = π r2

Thus, height = (VolSurface Area)=51136.5= 14.

### Question 3

The ratio between the curved surface area and the total surface area of right cylinder is 2:3 and the total surface area is 924 cm2. What is the volume of the cylinder?

#### SOLUTION

Solution :A

Solution:- Total surface Area: 2πr2+2πrh2πrh=32⇒1+rh=32⇒rh=12⇒h=2r

TSA:(2πr2+2πrh) = 924 ⇒(2πr2+2πr×2r) = 924 ⇒ r = 7 and h = 14

Volume = πr2h=227×7×7×14 = 2156Option(a).

### Question 4

The radius of an iron rod is decreased to one fourth of its original radius. If its volume remains constant, then the length will become.

#### SOLUTION

Solution :D

Solution:- Volume of original rod =πr2h

Changed volume of changed rod = π(r4)2h1

But, Volume remains constant.

πr2h = π(r4)2h1⇒h1 = 16hOption(d).

### Question 5

If the diagonals of a rhombus are 18 cm and 24 cm respectively, then find its perimeter.

#### SOLUTION

Solution :C

Ans: c

Find length of sides by considering the half of diagonals

(pythagoras theorem):√92+122 = 15. Perimeter: 4 × 15. Option(c).

### Question 6

If the right circular cone is cut into three solids of volumes V1,V2 and V3 by two cuts which are parallel to the base and trisects the altitude, then V1:V2:V3 is:

#### SOLUTION

Solution :D

Solution: -

The resultant figure is made of three similar triangles. The height and radius will be in ratio 13:23:1 = 1:2:3.

r1 = 1, h1 = 1

r2 = 2, h2 = 2

r3 = 3, h3 = 3

Volume will be in the ratio of r\(^2\)h for the three circular cones.

Required Volumes are

V1 = r21 x h1 = 1

V2= r22 x h2 - V1 = 8 - 1 = 7

V3= r23 x h3 - (V1 + V2) = 27 - (7 + 1) = 19

Volumes will be in given ratio

= 1:7:19.Option (d).

### Question 7

Water flows at the rate of 10 m per mins from a cylindrical pipe of radius 2.5 mm. A conical vessel of diameter 40 cm and depth 24 cm is filled with water flowing from this pipe. The time taken to fill the conical vessel is:

#### SOLUTION

Solution :C

Area=πr2=6.25π

Rate of flow= (πr2x1000) mm3/min=6250π mm2/minVolume of cone=13 πr2h = 13 π (200)2x 240 mm3=80πx40000=3200000π mm3Thus, total time taken is:(Volume of cone in mm3)(Rate of water flow in mm3/min) = 51.2minThus answer option (c) is correct.

### Question 8

If the inner rectangle is 8cm x 6 cm. Find the area of the shaded region.

#### SOLUTION

Solution :D

Diagonal of rectangle: √(82+62), =10.

Half of diagonal = radius of circle = 5.

Area of shaded region: π52 - (8x6) = 30.57

### Question 9

In the figure find the area outside the trapezium, given square is of side 16 cm.

#### SOLUTION

Solution :A

Required Area = Area of triangle AFG + Area of triangle DEF + Area of triangle BCD =

(12×6×8)+(12×8×6)+(12×10×16)

= 128 cm2

### Question 10

The top of a conical container has a circumference of 308 m. Water flows in at a rate of 12 m3 every 2 secs. When will the container be half filled, if its depth is 12m.

#### SOLUTION

Solution :A

Ans:a 2πr=308 2 × (227)×r = 308 Find half volume of conical container.

Time reqd:vol of 12 containerflow rate=15078.286 = 2520; 252060 =42min

### Question 11

A well is dug 20 ft deep and the mud which came out is used to build a wall of width 1 ft around the well on the ground. If the height of the wall around the well is 5 ft, then what is the radius of the well?

#### SOLUTION

Solution :D

Solution: - Let radius = r feet.

Volume of mud from well: πr2(20) cubic feet.

Volume of wall around well: 5π((r+1)2-r2).

πr2(20) = 5π((r+1)2-r2) => 4r2 - 2r - 1=0 ⇒ r = (√5+1)4

### Question 12

A rectangular field is of dimension 15.4m x12.1 m. A circular well of 0.7 m radius and 3 m depth is dug in the field. The mud, dug out from the well, is spread in the field. By how much would the level of the field rise?

#### SOLUTION

Solution :B

Solution: - Volume of earth taken out: πr2h. = π x (0.7)2 x 3 = V.

Area of field without circle: (15.4x12.1)-(π(0.72)) = A (say).

Axh=V

h=2.5 cm

### Question 13

There is an equilateral triangle of side ‘a’ units. Now we join any two sides of the triangle to form a cone.What is the slant height of the cone formed?

#### SOLUTION

Solution :B

The cone will have slant height same as the triangle side.

### Question 14

There is an equilateral triangle of side ‘a’ units. Now we join any two sides of the triangle to form a cone.What is the radius of the cone formed?

#### SOLUTION

Solution :D

Circumfrence of the base of cone = 2πr = side of triangle = a

radius (r) = a2π

### Question 15

Sixteen cylindrical sprite cans are placed in a square carton such that each can either touches another can or a wall of the carton. Each can is of unit radius. What is the bottom area of the carton?

#### SOLUTION

Solution :B

Ans: b (Add the radii of cans in 1 row. We get 8 as length. Same for breadth. So, carton dimensions: 8 by 8. Bottom area: 8x8=64 sq units.)