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Test: Geometry- 2

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

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This mock test of Test: Geometry- 2 for UPSC helps you for every UPSC entrance exam. This contains 15 Multiple Choice Questions for UPSC Test: Geometry- 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Geometry- 2 quiz give you a good mix of easy questions and tough questions. UPSC students definitely take this Test: Geometry- 2 exercise for a better result in the exam. You can find other Test: Geometry- 2 extra questions, long questions & short questions for UPSC on EduRev as well by searching above.
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:

Solution:

The two adjacent angles can be 30° and 60°, therefore others angle can be 150° and 120°

QUESTION: 2

Solution:

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.

Solution:

QUESTION: 4

Read the passage below and solve the questions based on it.
A 10 m long piece of wire is cut into two pieces, one of which is bent into a circle and the other into the square enclosing it.

Q.

The ratio of the radius of the circle to the perimeter of the square is

Solution:

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.)

Solution:

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?

Solution:

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?

Solution:

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

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?

Solution:

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?

Solution:

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.

Solution:

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

Solution:

QUESTION: 12

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

Solution:

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?

Solution:

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.

Solution:

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

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?

Solution:

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

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