1. CAT 2022 QA Slot 1 | Geometry – Quadrilaterals & Polygons CAT Question
   A trapezium ABCD has side AD parallel to BC. ∠BAD = 90°, BC = 3 cm and AD = 8 cm. If the perimeter of this trapezium is 36 cm, then its area, in sq cm, is

EXPLANATION

 

2. CAT 2022 QA Slot 2 | Geometry – Quadrilaterals & Polygons CAT Question
   Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals

EXPLANATION

Correct Answer: 10

Let the number of sides of polygons A and B be n and 2n, respectively

The number of sides of polygon B is 2*5, i.e. 10

3. CAT 2022 QA Slot 3 | Geometry – Quadrilaterals & Polygons CAT Question
   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 o possible lengths of the fourth side is

(a) 5

(b) 4

(c) 3

(d) 6

EXPLANATION

The correct answer is A:5
To determine the possible lengths of the fourth side of the quadrilateral,we need to consider the triangle inequality theorem.According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.This principle also holds for quadrilaterals. 
Given the lengths of the three sides as 1cm, 2cm, and 4cm, we can consider all the possible combinations of these sides as two sides of the quadrilateral.Let’s check each case: 
1cm+2cm> 4cm (True) 
1cm+4cm> 2cm (True) 
2cm+4cm> 1 cm (True) 
Since all three combinations satisfy the triangle inequality theorem,any of them can form a valid triangle,and consequently,a quadrilateral with the given side lengths. 
Now,let’s consider the possible lengths of the fourth side for each combination: 
For 1cm+2cm> 4cm: 
The fourth side can have a length between |1cm-2cm|+1 and 1cm+2cm-1. 
This gives us a range of possible lengths: 1 to 2cm. 
For 1cm+4cm> 2cm: 
The fourth side can have a length between |1cm – 4cm|+1 and 1cm+ 4cm- 1. 
This gives us a range of possible lengths: 2 to 4cm. 
For 2cm+ 4cm> 1cm: 
The fourth side can have a length between |2cm- 4cm|+ 1 and 2cm+ 4cm- 1. 
This gives us a range of possible lengths: 3 to 5cm. 
Now,let’s combine the ranges from all three cases: 1-2cm,2-4cm, and 3-5cm. 
The possible integer lengths for the fourth side are: 1, 2, 3, 4, 5. 
Therefore, the total number of possible lengths for the fourth side is 5. 

4. CAT 2021 QA Slot 1 | Geometry – Quadrilaterals & Polygons CAT Question
   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

(a) √14

(b) √12

(c) √13

(d) √15

EXPLANATION

T is the midpoint of CD in the given regular hexagon ABCDEF

CT=DT=2/2=1

Draw AT, AD and AC

By the property of hexagon,

AD=2×BC=2×2=4=2×=2×2=4

∠ABC=∠BCD=120^∘∠=∠=120^∘

We have, AB=BC=CD=2

∠ACD=120^∘−30^∘=90^∘∠=120^∘−30^∘=90^∘

ACD is a right triangle and AT is its median

AC²=AD²−CD²= 4² – 2² = 12

Using Apollonius’s theorem,

AC²+AD²=2(AT²+DT²)

12 + 4² = 2(AT²+1²)

AT = √13

Ans: ≈3.605551 cm

5. CAT 2021 QA Slot 2 | Geometry – Quadrilaterals & Polygons CAT Question
   The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If th difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is

(a) 20

(b) 30

(c) 40

(d) 25

EXPLANATION

It is given that: 
DP =x 
∴ AB =x
Now DP=CP So CD = 2x

Now, if we denote the height of the trapezium as ‘h,’ 
we can express the area of parallelogram ABPD as ‘xh’

Now, based on the given condition

6. CAT 2021 QA Slot 2 | Geometry – Quadrilaterals & Polygons CAT Question
   If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is

(a) √13 + √12

(b) √37 + √13

(c) (√13 + √12)/2

(d) (√37 + √13)/2

EXPLANATION

Answer: A
All the sides of the rhombus are equal.
The area of a rhombus is 12 cm².
Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.

The area of a rhombus is

d1*d2 = 24.

7. CAT 2021 QA Slot 3 | Geometry – Quadrilaterals & Polygons CAT Question
   A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires alon its two diagonals, at the rate of ₹125 per m, is

EXPLANATION

8. CAT 2021 QA Slot 3 | Geometry – Quadrilaterals & Polygons CAT Question
   Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10 cm and 20 cm, respectively. If the angle ∠ADC is equal to 30° the the area of the parallelogram is sq. cm. is

(a) 25(√3 + √15)/2

(b) 25(√5 + √15)

(c) 25(√3 + √15)

(d) 25(√5 + √15)/2

EXPLANATION

9. CAT 2020 QA Slot 1 | Geometry – Quadrilaterals & Polygons CAT Question
   A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is

(a) 5π/18

(b) 2π/15

(c) 3π/25

(d) 6π/25

EXPLANATION

10. CAT 2020 QA Slot 2 | Geometry – Quadrilaterals & Polygons CAT Question
    The sum of the perimeters of an equilateral triangle and a rectangle is 90 cm. The area, T, of the triangle and the area, R, of the rectangle, both in s cm, satisfy the relationship R = T². If the sides of the rectangle are in the ratio 1 : 3, then the length, in cm, of the longer side of the rectangle, is

(a) 27

(b) 24

(c) 18

(d) 21

EXPLANATION

Let the sides of the rectangle be “a” and “3a” m.

Hence the perimeter of the rectangle is 8a. Let the side of the equilateral triangle be “m” cm.

Hence the perimeter of the equilateral triangle is “3m” cm. Now we know that 8a+3m=90……(1)

Hence a=9 and the longer side of the rectangle will be 3a=27cm

11. CAT 2020 QA Slot 3 | Geometry – Quadrilaterals & Polygons CAT Question
    In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC = 4 cm, the area of the trapezium in sq.cm is

EXPLANATION

Given, BC = DE = 4

CD = BE = 5

12.CAT 2019 QA Slot 2 | Geometry – Quadrilaterals & Polygons CAT Question
Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B is 3/2 times each interior angle of A, the each interior angle, in degrees, of a regular polygon with a + b sides is

EXPLANATION

13. CAT 2018 QA Slot 1 | Geometry – Quadrilaterals & Polygons CAT Question
    Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD an CG is longer than EB, then the ratio of length of EB to that of CG is

(a) 2 : 5

(b) 4 : 9

(c) 3 : 8

(d) 1 : 3

EXPLANATION

It is given that EFGH is also a square whose area is 62.5% of that of ABCD. Let us assume that E divides AB in x : 1. Because of symmetry we can se that points F, G and H divide BC, CD and DA in x : 1. Because of symmetry we can se that points F, G and H divide BC, CD and DA in x : 1.

Let us assume that ‘x+1’ is the length of side of square ABCD. 

14. CAT 2018 QA Slot 1 | Geometry – Quadrilaterals & Polygons CAT Question
    In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP i perpendicular to CD. Then the area, in sq cm, of triangle APD is

(a) 12√3

(b) 24√3

(c) 18√3

(d) 32√3

EXPLANATION

Area of the parallelogram ABCD = (base) (height)

= CD × AP = 72 sq.cm.

⇒ 72 = 9 × (AP) ⇒ AP = 8cm

15. CAT 2018 QA Slot 1 | Geometry – Quadrilaterals & Polygons CAT Question
    Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth o ABCD?

(a) 24, 12

(b) 24, 10

(c) 25, 10

(d) 25, 9

EXPLANATION

We know that AC is the diameter and ABC = 90°. AC = 2*13 = 26 cm.

In right angle triangle ABC,

Let us check with the options. Therefore, we can say that option A is the correct answer.

Therefore, we can say that option A is the correct answer.

16. CAT 2018 QA Slot 2 | Geometry – Quadrilaterals & Polygons CAT Question
    A parallelogram ABCD has area 48 sqcm. If the length of CD is 8 cm and that of AD is s cm, then which one of the following is necessarily true?

(a) s ≠ 6

(b) s ≥ 6

(c) 5 ≤ s ≤ 7

(d) s ≤ 6

EXPLANATION

Answer:

As,sin0 < 1

Step-by-step explanation:

Area of parellogram ABCD= 2*Area of triangle ACD

48= 2* area of triangle ACD

Area of triangle of ACD =24 because 48/2=24

(1/2) * CD * DA * SinADC =24

=(1/2) *8* DA *SinADC =24

Ad *SinADC =6

ANSWER

As,sin0 < 1

17. CAT 2018 QA Slot 2 | Geometry – Quadrilaterals & Polygons CAT Question
    The area of a rectangle and the square of its perimeter are in the ratio 1 ∶ 25. Then the lengths of the shorter and longer sides of the rectangle are i the ratio ?

(a) 1 : 3

(b) 2 : 9

(c) 1 : 4

(d) 3 : 8

EXPLANATION

Let ‘a’ and ‘b’ be the length of sides of the rectangle, (a > b)

Area of the rectangle = a*b

Perimeter of the rectangle = 2*(a+b)

=>a*b/(2*(a + b))² = 1/25

=> 25ab = 4(a + b)²

=> 4a² – 17ab + 4b² = 0

=> (4a – b)(a – 4b) = 0

=> a = 4b or b/4

We initially assumed that a > b, therefore a ≠ 4. Hence, a = 4b => b: a = 1: 4

18. CAT 2017 QA Slot 2 | Geometry – Quadrilaterals & Polygons CAT Question
    Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

(a) 3√2

(b) 3

(c) 4

(d) √3

EXPLANATION

AC is base of an isoceles triangle ABC where AB and BC are side lengths of a regular hexagon. Interior angles of hexagon are 120^∘ each, therefore base angles are 30^∘ each.

By cosine rule,

19. CAT 2017 QA Slot 2 | Geometry – Quadrilaterals & Polygons CAT Question
    ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 120 degrees and ∠BAC = 30 degrees, then the value of ∠BCD (in degrees) is

EXPLANATION

Following the conditions of question, we get the figure as:

∠COD = 120⁰ (Given)
Then, ∠DAC = 60⁰ (Half of ∠COD i.e. 120⁰)
Therefore, ∠BAD = 30⁰+60⁰=90⁰
Hence, ∠BCD = 180⁰ – ∠BAD (Opposite angles of a cyclic quadrilateral are supplementary angles).
∠BCD = 180⁰–90⁰=90⁰

20. CAT 2008 QA | Geometry – Quadrilaterals & Polygons CAT Question
    Consider a square ABCD with midpoints E, F, G, H of AB, BC, CD and DA respectively. Let L denote the line passing through F and H. Consider points and Q, on L and inside ABCD, such that the angles APD and BQC both equal 120°. What is the ratio of the area of ABQCDP to the remaining area insid ABCD?

(a)4√2 3

(b) 2 + √3

(c)10−3√3 9

(d) 1 + 1√3

(e) 2√3 − 1

EXPLANATION

21. CAT 2007 QA | Geometry – Quadrilaterals & Polygons CAT Question
    Each question is followed by two statements A and B. Answer each question using the following instructions.
    Mark (1) if the question can be answered by using statement A alone but not by using statement B alone. Mark (2) if the question can be answered by using statement B alone but not by using statement A alone.
    Mark (3) if the question can be answered by using both the statements together but not by using either of the statements alone. Mark (4) if the question cannot be answered on the basis of the two statements.
    Rahim plans to draw a square JKLM with a point O on the side JK but is not successful. Why is Rahim unable to draw the square?
    A. The length of OM is twice that of OL.
    B. The length of OM is 4 cm.

(a) 1

(b) 2

(c) 3

(d) 4

(e) 5

EXPLANATION

If the side of the square is x cm, then the maximum length of OM is √2​x

The minimum length of OL is x

So, OM can never be 2 times OL

So, using statement A alone, we can conclude that Rahim is unable to draw the square.

Using statement B alone, we cannot answer the question.

So, option a) is the correct answer.

22. CAT 2006 QA | Geometry – Quadrilaterals & Polygons CAT Question
    An equilateral triangle BPC is drawn inside a square ABCD. What is the value of the angle APD in degrees?

(a) 75

(b) 90

(c) 120

(d) 135

(e) 150

EXPLANATION

All Sides of Square are equal  and all angles are equal to 90°

All Sides of a equilateral triangle are equal  and all angles are equal to 60°

Angles opposites to equal sides of triangle  are equal

Sum of angles of  a triangle is 180°

ABCD is square  , APD is Equilateral triangle

AB = BC = CD = AD = BP = CP

∠ABC = 90°  ,  ∠CBP  =  60°

∠ABC = ∠ABP  + ∠CBP

Hence ∠ABP   = 30°

∠BAP = ∠BPA  as AB = BP

∠BAP + ∠BPA  + ∠ABP  = 180°

Hence ∠BAP = ∠BPA = 75°

Similarly ∠CPD = 75°

∠BPC = 60°

Sum of all angles at a point is 360°

Hence ∠APD = 150°

23. CAT 2003 QA – Leaked | Geometry – Quadrilaterals & Polygons CAT Question
    Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90° o concave if the internal angle is 270°. If the number of convex corners in such a polygon is 25, the number of concave corners must be

(a) 20

(b) 0

(c) 21

(d) 22

EXPLANATION

Let us assume that, the total number of 90° angle is x and the total number of 270° angle is y.

Than,

→ sum of all 90° angles = 90 * x = 90x

→ sum of all 270° angles = 270 * y = 270y .

and,

→ Total sides polygon = (x + y).

Now, we know that,

• sum of all interior angles of a polygon with n sides is equal to = (n – 2) * 180° .

So,

→ sum of all angles of polygon with total sides (x + y) = [x + y – 2] * 180° .

Therefore,

→ 90x + 270y = (x + y – 2) * 180°

→ 90x + 270y = 180x + 180y – 360°

→ 180x – 90x + 180y – 270y = 360°

→ 90x – 90y = 360°

→ 90(x – y) = 360°

dividing both sides by 90°

→ x – y = 4

Putting given value of x = 25 now,

→ 25 – y = 4

→ y = 25 – 4

→ y = 21. (Option C) (Ans.)

24. CAT 2003 QA – Leaked | Geometry – Quadrilaterals & Polygons CAT Question
    In the figure below, ABCDEF is a regular hexagon and ∠AOF = 90°. FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of th hexagon ABCDEF?

(a) 1/12

(b) 1/6

(c) 1/24

(d) 1/18

EXPLANATION

25. CAT 2003 QA – Retake | Geometry – Quadrilaterals & Polygons CAT Question
    A square tin sheet of side 12 inches is converted into a box with open top in the following steps – the sheet is placed horizontally. Then, equal size squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of box. If x is an integer, then what value of x maximizes the volume of the box?

(a) 3

(b) 4

(c) 1

(d) 2

EXPLANATION

Volume of box = (12 – 2x) * (12 -2x) *  x

= 144x – 48 + 4

differentiate => 144 – 96x + 12

solve for x

=> 144 -96x + 12 = 0

=>  – 8x + 12 = 0

=> x = 6, 2

put x in volume x = 6 , v = 0

x = 2, v = 128 ans

CAT 2003 QA – Retake | Geometry – Quadrilaterals & Polygons CAT Question

Question 26:
Let ABCDEF be a regular hexagon. What is the ratio of the area of the triangle ACE to that of the hexagon ABCDEF?

Options:
(a) 1/3
(b) 1/2
(c) 2/3
(d) 5/6

EXPLANATION

Question 27:
The length of the circumference of a circle equals the perimeter of a triangle of equal sides, and also the perimeter of a square. The areas covered by the circle, triangle, and square are c, t, and s, respectively. Then,

Options:
(a) s > t > c
(b) c > t > s
(c) c > s > t
(d) s > c > t

EXPLANATION

Circumference of circle = 2π (radius)

Perimeter of Equilateral Triangle = 3 ( side)

Perimeter of square = 4 ( side)

Area of circle =  π (radius)²

Area of Equilateral Triangle = √3 ( side)² / 4

Area of square =   ( side)²

Step 1:

Assume that Circumference of circle = Perimeter of Triangle = Perimeter of Square = P

Step 2

Find radius of circle from circumference and then find area of the circle

2π (radius) = P  

=> radius  = P/2 π

Area of circle =  π (radius)²

= π (P/2 π)²

= P²/4π

≈ 0.08P

Step 3:

Find side of Equilateral triangle from perimeter and then find area of the triangle

3 (side) = P  

=> side  = P/3

Area of Triangle = √3 ( side)² / 4

= √3 ( P/3)² / 4

= √3 P² / 36

≈ 0.05P

Step 4:

Find side of square from perimeter and then find area of the square

4 (side) = P  

=> side  = P/4

Area of Square =   ( side)²  

=  ( P/4)²  

=  P² / 16

= 0.0625P

Hence 0.05P < 0.0625P < 0.08P

Hence Triangle covered  the least area and Circle Covered maximum area

Question 28:
Let S1 be a square of side a. Another square S2 is formed by joining the mid-points of the sides of S1. The same process is applied to S2 to form yet another square S3, and so on. If A1, A2, A3… are the areas and P1, P2, P3… are the perimeters of S1, S2, S3… respectively, then the ratio P1+P2+P3+…..

Options:
(a) 2(1+√2)a
(b) 2(2−√2)a
(c) 2(2+√2)a
(d) 2(1+2√2)a

EXPLANATION

he correct option is C

Option (c)

From the given condition in question

Area and perimeter of S₁ = a², 4a

Area and perimeter of S₂ = 

Area and perimeter of S₃ =

Area and perimeter of S₄ 

These are 2 infinite GPs , which can be solved as follows

Required ratio

=

=

=  ^{→   →} 

Question 29:
In the following figure, the area of the isosceles right triangle ABE is 7 sq.cm. If EC = 3BE, then the area of rectangle ABCD (in sq. cm.) is

Options:
(a) 64
(b) 82
(c) 26
(d) 56

EXPLANATION

Question 30:
In the above diagram, ABCD is a rectangle with AE = EF = FB. What is the ratio of the area of the triangle CEF and that of the rectangle?

Options:
(a) 1/6
(b) 1/8
(c) 1/9
(d) None of these

EXPLANATION

The correct option is B 1:6
Option (B) is correct option.

Question 31:
Based on the figure below, what is the value of x, if y = 10?

Options:
(a) 0
(b) 11
(c) 12
(d) None of these

EXPLANATION

The correct option is B 11

AB2=(x+4)2+(x−3)2=2×2+25+2x
Since solving this equation is very difficult. So, it is a better approach (Time saving) to put the values given in the options and try to find out a solution.
Hence, trying out we get 11 as the value of x .

Question 32:
ABCD is a rhombus with the diagonals AC and BD intersecting at the origin on the x-y plane. The equation of the straight line AD is x + y = 1. What is the equation of BC?

Options:
(a) x + y = –1
(b) x – y = –1
(c) x + y = 1
(d) None of the above

EXPLANATION

The equation of line BC, x+y=-1

A is correct option.

Step-by-step explanation:

ABCD is a rhombus with diagonals AC and BD intersecting at the origin on x-y plane.

Please see the attachment for figure.

In the given, ABCD is rhombus. Side AD and BC are opposite sides of rhombus.

Opposite sides of rhombus are parallel. AD || BC

If the equation of line AD, x+y=1

As we know the slope of parallel line is equal.

Equation of line BC, x+y=k

Intersection of diagonals AC and BD at original. Coordinate A and C should be equi-distance from origin because diagonals bisect each other.

Therefore, Coordinate C: (0,-1)  and Coordinate B: (-1,0)

Hence, The equation of line BC, x+y=-1

Question 33:
The figure below shows two concentric circles with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle touching it at points B, C, D, and A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD?

Options:
(a) π/4
(b) 3π/2
(c) π/2
(d) π

EXPLANATION

The correct option is C π/2
Let the diagonal of PQRS be 2r.
Therefore, side =r√2.
Now, ABCD is a square. And side r√2/2×√2 = r.
Perimeter of ABCD = 4r.
Circumference of bigger circle = 2 πr.
Therefore, required ratio π/2

Question 34:
There is a circle of radius 1 cm. Each member of a sequence of regular polygons S1(n), n = 4, 5, 6, …, where n is the number of sides of the polygon, is circumscribing the circle, and each member of the sequence of regular polygons S2(n), n = 4, 5, 6, …, where n is the number of sides of the polygon, is inscribed in the circle. Let L1(n) and L2(n) denote the perimeters of the corresponding polygons of S1(n) and S2(n), then {L1(13)+2π} is

Options:
(a) Greater than π and less than 1
(b) Greater than 1 and less than 2
(c) Greater than 2
(d) Less than π

EXPLANATION

The perimeter of the circle is equal to 2π

The perimeter of the polygon inscribing the circle is always greater than the perimeter of the circle => L1(13) > 2π.

The perimeter of the polygon inscribed in the circle is always less than the perimeter of the circle => L2(13) < 2 π.

Question 35:
A rectangle PRSU is divided into two smaller rectangles PQTU and QRST by the line TQ. PQ = 10 cm, QR = 5 cm, and RS = 10 cm. Points A, B, F are within rectangle PQTU, and points C, D, E are within the rectangle QRST. The closest pair of points among the pairs (A, C), (A, D), (A, E), (F, C), (F, D), (F, E), (B, C), (B, D), (B, E) are 10√3 cm apart.

Options:
(a) The closest pair of points among the six given points cannot be (F, C)
(b) Distance between A and B is greater than that between F and C.
(c) The closest pair of points among the six given points is (C, D), (D, E), or (C, E).
(d) None of the above

EXPLANATION

The correct option is A The closest pair of points among the six given points cannot be (F, C).
The diagonal length of a rectancle PUSR = 5√13 = 18 (approx)

Among given eight pairs the shortest distance = 10√3

So, the six points A, B, F, C, D and E are near corner of rectangle PUSR.

So, (F,C) cannot be the shortest distance.

Question 36:
AB > AF > BF ; CD > DE > CE ; and BF = 6√5 cm. Which is the closest pair of points among all the six given points?

Options:
(a) B, F
(b) C, D
(c) A, B
(d) None of these

EXPLANATION

The correct option is D None of these

(C, E) is the closest pair.

Question 37:
Directions: Each question is followed by two statements I and II. Mark:

1. if the question can be answered by any one of the statements alone but cannot be answered by using the other statement alone.
2. if the question can be answered by using either statement alone.
3. if the question can be answered by using both the statements together but cannot be answered by using either statement alone.
4. if the question cannot be answered even by using both the statements together.

Mr. X starts walking northwards along the boundary of a field from point A on the boundary, and after walking for 150 m reaches B, and then walks westwards, again along the boundary, for another 100 m when he reaches C. What is the maximum distance between any pair of points on the boundary of the field?

I. The field is rectangular in shape.
II. The field is a polygon, with C as one of its vertices and A as the mid-point of a side.

Options:
(a) 1
(b) 2
(c) 3
(d) 4

EXPLANATION

Using only statement A, we know that the shape of th field is rectangle. We cannot infer anything else with only statement A.

Using only statement B, we know that C is one of the vertices, but we do not know how many vertices exist.

Using both A and B, we know that the field is rectangular in shape and the length and breadth are 300 metres and 100 metres.

Hence, we can find the value of the diagonal which is the maximum distance between any two points inside the field. Hence, option C is the answer.

Question 38:
Four identical coins are placed in a square. For each coin, the ratio of area to circumference is the same as the ratio of circumference to area. Then find the area of the square that is not covered by the coins.

Options:
(a) 16(π – 1)
(b) 16(8 – π)
(c) 16(4 – π)
(d) 16(4 − π)

EXPLANATION

The correct option is C 16(4−π)
Let r be the radius of each circle.
Then by given condition,
πR² / 2πR = 2πR / πR²

⇒R²=4⇒R=2
∴ The length of the side of the square = 8
Now the area covered by 4 coins =4×π(2)²=16π
and area of the square = 64
∴ The area which is not covered by the coins
= 64 – 16π = 16 (4 – π)

Question 39:
The adjoining figure shows a set of concentric squares. If the diagonal of the innermost square is 2 units, and if the distance between the corresponding corners of any two successive squares is 1 unit, find the difference between

the areas of the eighth and the seventh squares, counting from the innermost square.

Options:
(a) 10√2 sq. units
(b) 30 sq. units
(c) 35√2 sq. units
(d) None of these

EXPLANATION

Since the diagonal of innermost square =2 units,

∴ Diagonal of the 7th square =2×7=14 units

⇒ Diagonal of the 8th square =2×8=16 units

Area of 8th square − Area of 7th square=1 / 2× 16² − 1 / 2×14²

=128−98=30 sq. units.

Question 40:
In a rectangle, the difference between the sum of the adjacent sides and the diagonal is half the length of the longer side. What is the ratio of the shorter to the longer side?

EXPLANATION

Let the length of the longer side of the rectangle be ′b′
The length of the short side of the rectangle be ′a′
As the right angled triangle obeys the Pythagoras theorem, hence the length of the diagonal can be obtained.
The length of the diagonal of the rectangle is √a²+b²
The sum of the adjacent sides of the rectangle = a+b
Given that the difference between the sum of adjacent sides and the diagonal is half the length of longer side, is given below:

The ratio of the shorter to the longer side is 3:4.

Question 41:
In the given figure, EADF is a rectangle, and ABC is a triangle whose vertices lie on the sides of EADF, and AE = 22, BE = 6, CF = 16, and BF = 2. Find the length of the line joining the mid-points of the sides AB and BC.

Options:
(a) 4√2
(b) 5
(c) 3.5
(d) None of these

EXPLANATION

The correct option is B 5

Option (b)

EF = AD = 8

(EADF is a rectangle)

CD = (22 – 16) = 6

So in the right angled triangle ADC, AD = 8 and CD = 6.

Therefore AC = 10

Therefore length of the line joining the mid – points of

AB and BC = 12 (10) = 5

Question 42:
If ABCD is a square and BCE is an equilateral triangle, what is the measure of ∠DEC?

Options:
(a) 15°
(b) 30°
(c) 20°
(d) 45°

EXPLANATION

The correct option is A 15^∘
Given ABCD is a square and BCE is an equilateral triangle

∠BCD=90^∘ [Interior angle of a square]

∠BCE=60^∘ [Interior angle of an equilateral triangle]

∴∠DCE=90^∘+60^∘=150^∘

BC = CD (Sides of a square) —— (1)

BC = CE (Sides of an equilateral triangle) —— (2).

From (1) and (2),
DC = EC

In ΔDCE, DC = CE
⇒∠CDE=∠CED ——— (3) [Angles opposite equal sides]

Also, In ΔDCE
∠DCE+∠CDE+∠CED=180^∘ ——— (4) [Angle sum property]

From (3) and (4),
2∠DEC+∠CED=180^∘

⇒2∠DEC=180∘−150^∘=30^∘

∠DEC=1 / 2(30^∘) = 15^∘

Question 43:
The figure shows the rectangle ABCD with a semicircle and a circle inscribed inside it as shown. What is the ratio of the area of the circle to that of the semicircle?

Options:
(a) (√2 − 1)² : 1
(b) 2(√2 − 1)² : 1
(c) (√2 − 1)² : 2
(d) None of these

EXPLANATION

Question 44:
Which of the following values of x do not satisfy the inequality (x² – 3x + 2 > 0) at all?

Options:
(a) 1 ≤ x ≤ 2
(b) –1 ≥ x ≥ –2
(c) 0 ≤ x ≤ 2
(d) 0 ≥ x ≥ –2

EXPLANATION

The correct option is A 1 ≤ x ≤ 2
Given inequality is x² − 3x + 2 > 0
⇒x² − 2x − x + 2 > 0

⇒x (x−2) − 1(x−2) > 0

⇒(x−2)(x−1)>0⇒x<1,x>2

Question 45:
PQRS is a square. SR is a tangent (at point S) to the circle with center O, and TR = OS. Then the ratio of the area of the circle to the area of the square is

Options:
(a) π/3
(b) 11/7
(c) 3π
(d) 7/11

EXPLANATION

It is given that PQRS is a square and SR is tangent to circle with centre O.

Lets assume the radius of circle be r.

Now, we know that TR − OS = r

Therefore, OR = OT + TR = r + r = 2r

ΔOSR is right angled triangle, 

Hence, OR²−OS²=SR²

⇒SR²=4r²−r²

=3r²⇒SR=√3r

Now the side of square is √3r
Area of square, As=√3r²

⇒As=3s²

Area of circle: Ac=πr²
Area of circle / Area of square= πr² / 3r² = π / 3

Question 46:
In the adjoining figure, AC + AB = 5AD, and AC – AD = 8. Then the area of the rectangle ABCD is

Options:
(a) 36
(b) 50
(c) 60
(d) Cannot be answered

EXPLANATION

AB = x = CD,

AD = y = BC,

AC = z

By Pythagoras theorem

z² = x² + y² ——-(1)

Given

AC + AB = 5AD

2 + x = 5y

z = 5y−x ——-(2)

squaring on both sides

z² = 25y² + x² − 10xy  ——–(3)

Subtract equation (1) from (3)

we get z² = 25y² + x²−10xy − z²= −y² − x² − 0

24y² − 10xy = 0

x / y = 12 / 5

AC−AD = 8

z − y =8

z=8+5

z=13

Area = 12 × 5 = 60 sq.units

Question 47:
Four friends start from four towns, which are at the four corners of an imaginary rectangle. They meet at a point that falls inside the rectangle after traveling distances of 40, 50, and 60 meters. The maximum distance that the fourth could have traveled is (approximately)…

Options:
(a) 67 meters
(b) 52 meters
(c) 22.5 meters
(d) Cannot be determined

EXPLANATION

The maximum distance covered by D would be 70 meters. This is because the four friends are starting from the corners of a rectangle, and the distance between two opposite corners of a rectangle is the maximum distance that can be covered by one of the friends. Since the distances covered by A, B, and C are 40, 50, and 60 meters respectively, the distance between two opposite corners must be greater than those distances. The only remaining option is 70 meters.

Question 48:
Data is provided followed by two statements – I and II – both resulting in a value, say I and II.
As your answer, Type 1, if I > II. Type 2, if I < II. Type 3, if I = II. Type 4, if nothing can be said.
In ΔACD, AD = AC, and ∠C = 2∠E. The distance between parallel lines AB and CD is h. Then
I. Area of parallelogram ABCD
II. Area of ΔADE

EXPLANATION

To determine the relationship between the areas of parallelogram ABCD and triangle ΔADE, we need to consider the information provided in the question.

Given:

1. In ΔACD, AD = AC, and ∠C = 2∠E.
2. The distance between parallel lines AB and CD is h.

Now, let’s analyze the statements:

I. Area of parallelogram ABCD
To calculate the area of a parallelogram, we can use the formula: Area = base × height. In this case, the base is CD, and the height is h (the distance between AB and CD). So, the area of parallelogram ABCD is CD × h.

II. Area of ΔADE
To calculate the area of a triangle, we can use the formula: Area = (1/2) × base × height. In this case, the base is AD, and the height is h (the distance between AB and CD). So, the area of triangle ΔADE is (1/2) × AD × h.

Now, let’s compare I and II:

I. Area of parallelogram ABCD = CD × h
II. Area of ΔADE = (1/2) × AD × h

We are given that AD = AC, and in ΔACD, ∠C = 2∠E. This means that ΔACD is isosceles, and ∠C is twice the size of ∠E. In an isosceles triangle, the height bisects the base, so we can say that CD = 2h and AD = h.

Now, we can calculate the areas:

I. Area of parallelogram ABCD = CD × h = 2h × h = 2h^2
II. Area of ΔADE = (1/2) × AD × h = (1/2) × h × h = (1/2)h^2

Now, let’s compare I and II:

I. 2h^2
II. (1/2)h^2

2h^2 > (1/2)h^2

So, I is greater than II.

Therefore, the answer is Type 1: I > II.

Question 49:
Consider the five points comprising the vertices of a square and the intersection point of its diagonals. How many triangles can be formed using these points?

Options:
(a) 4
(b) 6
(c) 8
(d) 10

EXPLANATION

o form a triangle, 3 points out of 5 can be chosen in 5C3=10 ways.
But out of these, the three points lying on the 2 diagonals will be collinear.
∴ 10−2=8 triangles can be formed

Question 50:
Let the consecutive vertices of a square S be A, B, C & D. Let E, F & G be the mid-points of the sides AB, BC & AD, respectively, of the square. Then the ratio of the area of the quadrilateral EFDG to that of the square S is nearest to

Options:
(a) 1/2
(b) 1/3
(c) 1/4
(d) 1/8

EXPLANATION

(A)

Quadrilateral EFDG can be split into 2 triangles i.e. GEF and GDF Area of triangle GEF is half the area of AGFB [since all the 4 triangles in AGFB are similar] Area of triangle GDF is half the area of GDCF Hence, the area of quadrilateral EGDF is half the area of square ABCD.