Test: Geometry- 2 - UPSC MCQ

# Test: Geometry- 2 - UPSC MCQ

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## 15 Questions MCQ Test CSAT Preparation - Test: Geometry- 2

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Test: Geometry- 2 - Question 1

### A cyclic quadrilateral is such that two of its adjacent angles are divisible by 6 and 10 respectively. One of the remaining angles will necessarily be divisible by:

Detailed Solution for Test: Geometry- 2 - Question 1

We know that the sum of the opposite angles of a cyclic quadrilateral is 180 degrees. Let the four angles be A, B, C, and D, with A and B being the angles divisible by 6 and 10, respectively.

Since A is divisible by 6 and B is divisible by 10, we know that A = 6m and B = 10n for some integers m and n.

Now, consider the opposite angles. Since the sum of opposite angles is 180 degrees, we have:

C = 180 - B = 180 - 10n
D = 180 - A = 180 - 6m

We want to find which of the given options the angles C or D are necessarily divisible by. Let's examine each option:

1. 3: Since B is divisible by 10, it is possible that B is divisible by 5 but not 3 (e.g. B = 10). In this case, C = 180 - B would not be divisible by 3. Also, A is divisible by 6, so A is always divisible by 3, which means D = 180 - A would never be divisible by 3. So, this option is incorrect.

2. 4: Since A is divisible by 6, it is possible that A is divisible by 2 but not 4 (e.g. A = 6). In this case, D = 180 - A would not be divisible by 4. Also, B is divisible by 10, so B is always divisible by 2, which means C = 180 - B would never be divisible by 4. So, this option is also incorrect.

3. 8: If A is divisible by 6, then it can be even or odd multiples of 6 (e.g. A = 6, 12, 18, ...). D will be 180 - A, which means D can be both even and odd (e.g. D = 180 - 6 = 174, D = 180 - 12 = 168, D = 180 - 18 = 162, ...). Since D can be both even and odd, it is not necessarily divisible by 8. Similarly, C can also be both even and odd, so it is not necessarily divisible by 8. Thus, this option is also incorrect.

4. None of these: Since none of the previous options work, the correct answer is None of these.

So, the correct answer is option 4: None of these.

Test: Geometry- 2 - Question 2

### The volume of two spheres are in the ratio 27 : 125. The ratio of their surface area is?

Detailed Solution for Test: Geometry- 2 - Question 2

Option 4 : 9 : 25

Given

Ratio of volume of two spheres = 27 : 125

Formula used

surface area of sphere =4πr2

Volume of sphere = (4/3)πr3

Calculation

[(4/3)πR3/(4/3)πr3] = 27/125

Where R and r are radius of two sphere

⇒ R3/r3= 27/125

⇒ R = 3r/5

Surface area of 1st sphere/Surface area of 2nd sphere =[4π(3r/5)2]/4πr2

⇒ 9 : 25

∴ Ratio of surface area of two spheres is 9 : 25.

Test: Geometry- 2 - Question 3

### Read the passage below and solve the questions based on it fix) is the area of a square where, Jr is the side of a square. g(x) is the perimeter of square where, Jr is the side of a square. h(xt y) is the area of a rectangle where* is the length and y is the breadth. i(xt y) is the perimeter of a rectangle where Jt is the length and y is the breadth.

Detailed Solution for Test: Geometry- 2 - Question 3

Test: Geometry- 2 - Question 4

Detailed Solution for Test: Geometry- 2 - Question 4

Test: Geometry- 2 - Question 5

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 5

Test: Geometry- 2 - Question 6

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 6

Test: Geometry- 2 - Question 7

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 7

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 8

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 8

Test: Geometry- 2 - Question 9

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?

Detailed Solution for Test: Geometry- 2 - Question 9

Test: Geometry- 2 - Question 10

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 10

Test: Geometry- 2 - Question 11

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 11

Test: Geometry- 2 - Question 12

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

Detailed Solution for Test: Geometry- 2 - Question 12

Test: Geometry- 2 - Question 13

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

Detailed Solution for Test: Geometry- 2 - Question 13

Test: Geometry- 2 - Question 14

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 14

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 15

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 15

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

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

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