Free Basic Geometry - 02 Practice Test - CAT

Question 1

Areas of three adjacent sides of a cuboid are a,b,c. Volume of the cuboid is N. The value of N $^{2}$ equals to?

A. abc

B. (ab+ac+bc)

\sqrt{(a^{3} + b^{3} + c^{3})}

D. None of these

SOLUTION

Solution : A

Assume sides to be x,y,z. Then xy=a, yz=b, xz=c.

Multiplying: (xyz) $^{2}$ = abc; =N $^{2}$ . (Vol=xyz)

Hence Ans: (a)

Question 2

A string of certain length n makes one perfect turn around a cube of side A, starting from corner X and ending at Y (as in the fig). Find the length of the string.

\sqrt{2} A

\sqrt{17} A

C. A

\sqrt{17}

(A + 1)

SOLUTION

Solution : B

The string is wrapped equally around the cube 4 times.

So the part ab = A/4. The hypotenuse of the triangle $\sqrt{(A^{2}) + \frac{A^{2}}{16}}$ The length of the string is four times this hypotenuse.

$\frac{4 \times (\sqrt{17} A)}{4} = \sqrt{17} A$

Hence Ans: b

Question 3

Three coins are placed in such a way that one coin touches the other two (in the same plane). If the radius of each coin is r, what is the area of the triangle that circumscribes this arrangement of coins.

\frac{\sqrt{3} (2 r + 2 r \sqrt{3})^{2}}{16}

\frac{\sqrt{3} (r + 2 r \sqrt{3})^{2}}{4}

\frac{\sqrt{3} (2 r + 2 r \sqrt{3})^{2}}{4}

\frac{\sqrt{3} (2 r + r \sqrt{3})^{2}}{16}

SOLUTION

Solution : B

Soln:

The angle shown is 30 degrees. Thus AB=r $\sqrt{3}$ (30-60-90 triangle).

Thus side of a triangle: 2r + 2r $\sqrt{3}$ .

Area of the equilateral triangle: $\frac{\sqrt{3} (r + 2 r \sqrt{3})^{2}}{4}$ Hence Ans: b

Question 4

A sphere of radius 13 cm is cut by a plane whose distance from the centre of the sphere is 5cm. What is the circumference of the section so obtained?

A. 10

π

B. 12

π

C. 24

π

D. 26

π

SOLUTION

Solution : C

From the fig: X = $\sqrt{13^{2} - 5^{2}}$ = 12x Thus, circumference= 24 $π$

Hence Ans: c

Question 5

Two spheres of radii 6 cm and 1 cm are inscribed inside a right circular cone. The bigger sphere touches the smaller sphere and also the base of the cone. What is the height of the cone?

A. 14 cm

B. 12 cm

\frac{85}{6}

\frac{72}{5}

SOLUTION

Solution : D

Soln: the triangles shown are similar. Hence $\frac{x + (1 + 6)}{x} = \frac{1}{6}$ ;x= $\frac{7}{5}$ ; Height of cone:13 + $(\frac{7}{5}) = \frac{72}{5}$ .

Hence, Ans: d)

Question 6

A square hole of cross sectional area 4 cm $^{2}$ is drilled across a cube with its length parallel to a side of the cube. If an edge of the cube measures 5 units, what is the total surface area of the structure so formed?

A. 158 cm

^{2}

B. 190 cm

^{2}

C. 166 cm

^{2}

D. 182 cm

^{2}

SOLUTION

Solution : D

Ans: d) 182 cm $^{2}$

New SA=(original SA)+(SA of square channel formed through the cube)-(squares on 2 faces)

= (25x6)+(2x5x4)-(4+4) ;=182 sq units

Question 7

What’s the area of the shaded region? Given that the side of the square = 1 unit.

\frac{π}{2}

\frac{1}{2}

C. (

\frac{π}{4}

) - (

\frac{1}{2}

)

D. (

\frac{π}{2}

) - 1

SOLUTION

Solution : D

Ans: d)

Soln: Consider the double shaded area.

Area: (Area of quadrant) – (Area of triangle).

( $\frac{π}{4}$ )-( $\frac{1}{2}$ x 1 x 1). Total shaded area is twice of this

= 2 ( $\frac{π}{4}$ ) - ( $\frac{1}{2}$ x 1 x 1); = $\frac{π}{2}$ -1,

Question 8

A goat is tied to two poles P and Q with 15 m ropes. P and Q are 20 m apart. What is the area the goat is likely to graze.

SOLUTION

Solution : C

Ans: c.

Question 9

An equilateral triangle plate is to be cut into n number of identical small equilateral triangle plates. Which of the following can be a value of n.

A. 212

B. 216

C. 256

D. 312

SOLUTION

Solution : C

Ans: c…. 256

Solution: The total no of plates has to be a perfect square, you can try proving this by using the unitary approach (i.e. by using smaller numbers)

Question 10

A fan has 3 equally spaced blades. The central dial has an area of 3 $π$ cm $^{2}$ . Length of each blade is (20- $\sqrt{3}$ )cm. If a triangle is drawn from the tips of the blades, find its area.

A. 900 sq cm

B. 300

\sqrt{3}

sq cm

C. 900 sq cm

D. 400

\sqrt{3}

sq units

SOLUTION

Solution : B

Ans: b) 300 $\sqrt{2}$ sq cm

Soln: Radius of the dial at the centre: $\sqrt{\frac{3 π}{π}}$

total length of the blade from the centre: (20- $\sqrt{3}$ )+ $\sqrt{3}$ ; = 20.

Half of each side: 30 cos 30. = 10 $\sqrt{3}$ . Total length of each side: 20 $\sqrt{3}$ .

Thus , area: $(\frac{\sqrt{3}}{4}) (20 \sqrt{3})^{2}$ . =300 $\sqrt{2}$ sq cm

Question 11

If (9p $^{2}$ +4), (2p+7) and (8p $^{2}$ -1) are three sides of a triangle. Which of the following CANNOT be a value for p?

A. 2

\sqrt{2}

C. 2

\sqrt{2}

D. None of these

SOLUTION

Solution : C

Ans: c 2 $\sqrt{2}$ .

Soln: Reverse Gear Approach: Use answer options

The largest side should be smaller than sum of other 2 sides.
The sides are 76, 4 $\sqrt{2}$ +7, 63; Here the sum of last two sides is 75.6 which is lesser than 76

Question 12

AB is a line segment with P as midpoint. Three semicircles are drawn with AP,PB and AB as diameters. Another circle is drawn so as to touch all three semicircles. What’s the radius of this smaller circle?

\frac{1}{4}

\frac{1}{6}

\frac{1}{7}

\frac{1}{8}

SOLUTION

Solution : B

Ans: b) $\frac{1}{6}$ AB.

Soln:

Let PB=r=PQ. QR=x. In triangle PRB, RB=2r-x. PR= r+x.

Since PRB is right angled, $(r + x)^{2} = (2 r - x)^{2} + r^{2}$ .

Solving we get x= $\frac{r}{3}$ = $\frac{A B}{6}$

Question 13

If a, b, c are sides of a triangle such that $a^{2} + b^{2} + c^{2}$ =ab+bc+ac. Then the triangles are:

A. equilateral

B. isosceles

C. right angled

D. obtuse angled

SOLUTION

Solution : A

Ans: a) equilateral.

Substitute values and check. 2,2,2 for equilateral; 1,1,2 for isosceles, 3,4,5 for right angled. The equation holds only for equilateral triangles

Question 14

Find the radius of the smallest circle if the equilateral triangle shown is of side 3 units.

\sqrt{3}

\frac{\sqrt{3} + 1}{2}

\frac{\sqrt{3} - 1}{2}

D. None of these

SOLUTION

Solution : D

Answer- (d) None of these

Solution:- Calculate the altitude of the equilateral triangle. Then according to the conditions given,
Diameter of the circle circumscribing the equilateral triangle= Altitude of the equilateral triangle + Diameter of the smaller cirlce.
Solve the above equation for the required answer.

Alternate Soln: The triangle’s dimension is used to calculate the radii of the outermost and the second circle.

Radius of the bigger circle: R= $\frac{(1.5)}{c o s 30}$ ; = $\sqrt{3}$ units.

Radius of smaller circle:

r = 1.5 tan 30. = $\frac{\frac{3}{2}}{\sqrt{3}}$ r= $\frac{\sqrt{3}}{2}$ .

Radius of smallest circle:

$\frac{((r a d i u s o f b i g g e s t) - (r a d i u s o f 2^{n d}))}{2}$ ;

= $\frac{\sqrt{3}}{4}$

Hence ans: d (none of these)

Question 15

Which of the following is not true?
a) Given that the sides of a scalene triangle are integers, then one of the sides can =1
b) The 3 medians of a triangle divide it into 6 similar triangles
c) An isosceles triangle can never be obtuse

A. Only A

B. only A and C

C. Only C

D. A, B and C

SOLUTION

Solution : B

option (b)

a is false as if one side is 1, another side is a, then the third side < a+1. this is not possible as it needs to be an integer value and since it’s a scalene triangle, the third side cannot be 1+a

b is true

c is false as there is a possibility of 120 $^{\circ}$ , 30 $^{\circ}$ , 30 $^{\circ}$

Free Basic Geometry - 02 Practice Test - CAT

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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