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Test: Geometry- 2 - CAT MCQ


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15 Questions MCQ Test Quantitative Aptitude (Quant) - Test: Geometry- 2

Test: Geometry- 2 for CAT 2024 is part of Quantitative Aptitude (Quant) preparation. The Test: Geometry- 2 questions and answers have been prepared according to the CAT exam syllabus.The Test: Geometry- 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Geometry- 2 below.
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Test: Geometry- 2 - Question 1

Detailed Solution for Test: Geometry- 2 - Question 1



 

Test: Geometry- 2 - Question 2

A square is inscribed in a semi circle of radius 10 cm. What is the area of the inscribed square? (Given that the side of the square is along the diameter of the semicircle.)

Detailed Solution for Test: Geometry- 2 - Question 2

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Test: Geometry- 2 - Question 3

Two circles of an equal radii are drawn, without any overlap, in a semicircle of radius 2 cm. If these are the largest possible circles that the semicircle can accommodate, what is the radius (in cm) of each of the circles?

Detailed Solution for Test: Geometry- 2 - Question 3

Test: Geometry- 2 - Question 4

PQRS is a Trapezzium, in which PQ is Parralel to RS, and PQ = 3 (RS) . The diagnol of the Trapezzium intersect each other at X, then the ratio of,  ar ( ∆ PXQ)  : ar ( ∆ RXS)  is?

Detailed Solution for Test: Geometry- 2 - Question 4

In ∆ PXQ and ∆ RXS

=> angle P = angle R

    angle Q  =  angle S 

:- ∆ PXQ  ~ ∆ RXS  ( AA similarity rule)  

ar ( ∆ PXQ) /  ar ( ∆ RXS) = ( PQ / RS) ^ 2 

 =  ( 3 / 1 ) ^ 2

=     9 / 1 

Therefore,  ar ( ∆ PXQ) :  ar          ( ∆ RXS) 

=   9:1

Test: Geometry- 2 - Question 5

Let ABCDEF be a regular hexagon. What is the ratio of the area of the triangle ACE to that of the hexagon ABCDEF?

Detailed Solution for Test: Geometry- 2 - Question 5

Test: Geometry- 2 - Question 6

A pond 100 m in diameter is surrounded by a circular grass walk-way 2 m wide. How many square metres of grass is the on the walk-way?

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Test: Geometry- 2 - Question 7

The dimensions of a rectangular box are in the ratio of 1:2:4 and the difference between the costs of covering it with the cloth and a sheet at the rate of Rs 20 and Rs 20.5 per sq m respectively is Rs 126. Find the dimensions of the box.

Detailed Solution for Test: Geometry- 2 - Question 7

Test: Geometry- 2 - Question 8

The ratio of the area of a square inscribed in a semicircle to that of the area of a square inscribed in the circle of the same radius is

Detailed Solution for Test: Geometry- 2 - Question 8

Test: Geometry- 2 - Question 9

The ratio of the area of a square to that of the square drawn on the its the diagonal is

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Test: Geometry- 2 - Question 10

What is the area of the triangle in which two of its medians 9 cm and 12 cm long intersect at the right angles?

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Test: Geometry- 2 - Question 11

Four horses are tethered at four comers of a square plot of side 14 m so that the adjacent horses can just reach one another. There is a small circular pond of area 20 m2 at the centre. Find the ungrazed area.

Detailed Solution for Test: Geometry- 2 - Question 11

Total area of plot = 14 * 14 = 196m2
Horses can graze in quarter circle of radius = 7m
Grazed area = 4 * (pie r2)/4 = 154 m2
Area of plot when horses cannot reach = (196 - 154) = 42m2
Ungrazed area = 42 - 20 = 22m2

Test: Geometry- 2 - Question 12

Two sides of a triangle are 4 and 5. Then, for the area of the triangle, which one of the following bounds is the sharpest?

Detailed Solution for Test: Geometry- 2 - Question 12

Let AB = 4 and BC = 5 and AB is perpendicular to BC

then Area = 1/2 AB . AC = 1/2 . 4.5 = 10 

Test: Geometry- 2 - Question 13

A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the lengths of the largest to the smallest side of this rectangle is?

Detailed Solution for Test: Geometry- 2 - Question 13


Let the longer side of the rectangle is 2a and the shorter side is b as shown in the figure.

Using pythagoras theorem, we get
a2 + b2 = 22 = 4   ...(1)

Also, AM ≥ GM

⇒ a2 + b2 ≥ 2ab

⇒ 4 ≥ 2ab

⇒ 2 ≥ ab

∴ Highest value of ab = 2, when a = b.

⇒ Ratio of longer side to shorter side = 2a : b = 2 : 1.

Hence, option (a).

Test: Geometry- 2 - Question 14

The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number of possible lengths of the fourth side is

Detailed Solution for Test: Geometry- 2 - Question 14

In a quadrilateral, the longest side is less than the sum of other three sides and greater than the least of the difference of any 2 of the other three sides.

Let the fourthe sides of the quadrilateral be x.

⇒ 2 - 1 < x < 1 + 2 + 4
⇒ 1 < x < 7
∴ x can be 2, 3, 4, 5 or 6

Hence, x can take 5 integral values.

Hence, option (a).

Test: Geometry- 2 - Question 15

Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the mid point of CD, then the length of AT, in cm, is

Detailed Solution for Test: Geometry- 2 - Question 15

Side of the regular hexagon = 2 cm.
Consider the figure below.

Consider the isosceles ∆ATF

TU is the altitude from T to AF. 
We know, in a regular hexagon the distance between any two parallel sides = √3 × side.
∴ TU = √3 × 2 = 2√3 cm.

Since ATF is an isosceles triangle, U will be the mid-point of AF
∴ AU = 1 cm.

In ∆ATU
AT2 = TU2 + AU2
⇒ AT2 = (2√3)2 + 12
⇒ AT2 = 13
⇒ AT = √13

Hence, option (c).

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