Quantitative Aptitude – Geometry – Circles – A chord of length 5 cm subtends an

Quantitative Aptitude – Geometry - Circles – A chord of length 5 cm subtends an

Slot -2 – Quantitative Aptitude – Geometry – Circles – A chord of length 5 cm subtends an

A chord of length 5 cm subtends an angle of 60° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120° at the centre of the same circle is?

a) 6√2

b) 8

c) 4√2

d) 5√3

Answer: d) 5√3

Solution:

In triangle ODA, OA=AD cosec 30=5

So if the angle AOB is 120 degree, then angle AOD will be 120/2=60

In the case length of chord = 2×OA Sin 60=5√3

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Quantitative Aptitude – Geometry – Triangles – A triangle ABC has area 32 sq units

Quantitative Aptitude – Geometry - Triangles – A triangle ABC has area 32 sq units

Slot -2 – Quantitative Aptitude – Geometry – Triangles – A triangle ABC has area 32 sq units

A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0,0) is?

a) 4√2 units

b) 8 units

c) 4 units

d) 2√2 units

Answer: c) 4 units

Solution: The distance OA will be minimum when the perpendicular from A on BC will pass through O.

Area of triangle ABC = 1/2 BC×AD = 1/2×8×(AO+OD ) = 4×(AO+4)
32 = 4AO+16
AO = 4
So required minimum distance = 4

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Quantitative Aptitude – Geometry – Mensuration – The area of a rectangle and the square

Quantitative Aptitude – Geometry - Mensuration – The area of a rectangle and the square

Slot -2 – Quantitative Aptitude – Geometry – Mensuration – The area of a rectangle and the square

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 in the ratio?

a) 1:3

b) 3:8

c) 2:9

d) 1:4

Answer: d) 1 : 4

Solution:
Given ratio of areas of rectangle and square = 1:25 = 4∶ 100=(1×4):(10×10)
Thus possible ratio of perimeter =(1+4)/(10+10) = 5/20= 1∶4

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Quantitative Aptitude – Geometry – Mensuration – A parallelogram ABCD has area 48 sqcm

Quantitative Aptitude – Geometry - Polygons – A parallelogram ABCD has area 48 sqcm

Slot -2 – Quantitative Aptitude – Geometry – Mensuration – A parallelogram ABCD has area 48 sqcm

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) 5≤s≤7

b) s≤6

c) s≥6

d) s≠6

Answer: c) s≥6

Solution: 1

Solution: As the area of ABCD = 48=s×h ————1)
In right-angled triangle CKD, DK ≤ CD ( CD is hypotenuses )
So h ≤8
Thus s ≥6——-(from eq 1)

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Quantitative Aptitude – Geometry – Triangles – On a triangle ABC, a circle with diameter BC

Quantitative Aptitude – Geometry - Circles – On a triangle ABC, a circle with diameter BC

Slot -2 – Quantitative Aptitude – Geometry – Triangles – On a triangle ABC, a circle with diameter BC

On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is?

Answer: 24 cm

Solution:

As CP is perpendicular to AB and BQ is perpendicular to AC. So

AB×CP=AC×BQ

30×20=25×BQ

BQ = 24 cm

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Quantitative Aptitude – Geometry – Rectangle – From a rectangle ABCD of area 768 sq cm

Quantitative Aptitude – Geometry - Rectangle – From a rectangle ABCD of area 768 sq cm

Slot -2 – Quantitative Aptitude – Geometry – Rectangle – From a rectangle ABCD of area 768 sq cm

From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is?

a) 88 + 12π

b) 82 + 24π

c) 80 + 16π

d) 86 + 8π

Answer: a) 88 + 12π

Solution:

Given area of semicircle = 72π
Or (πr^2)/2 = 72π

r = 12
So AB = 24

Thus 24×AD = 768

AD = 32
So Required perimeter = AD+CD+CB+πr = 32+24+32+12π = 88+12π

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Quantitative Aptitude – Geometry – Polygons – In a parallelogram ABCD of area 72 sq cm

Quantitative Aptitude – Geometry - Polygons – In a parallelogram ABCD of area 72 sq cm

Slot -1 – Quantitative Aptitude – Geometry – Polygons – In a parallelogram ABCD of area 72 sq cm

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 is perpendicular to CD. Then the area, in sq cm, of triangle APD is?

a) 24√3

b) 12√3

c) 32√3

d) 18√3

Answer: c) 32√3

Solution:

As given, Area of parallelogram ABCD = 72
So AB×AP=72

9AP=72
AP = 8 cm

in right-angled triangle APD, AP^2 + PD^2 = AD^2
So PD^2 = AD^2-AP^2 = 16^2-8^2 = 64×3

PD=8√3
Thus Area of triangle APD = 1/2 AP×PD = 1/2×8×8√3 = 32√3

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Quantitative Aptitude – Geometry – Mensuration – Let ABCD be a rectangle inscribed in a circle

Quantitative Aptitude – Geometry - Circles – Let ABCD be a rectangle inscribed in a circle

Slot -1 – Quantitative Aptitude – Geometry – Mensuration – Let ABCD be a rectangle inscribed in a circle

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 of ABCD?

a) 25, 10

b) 24, 12

c) 25, 9

d) 24, 10

Answer: d) 24, 10

Solution:

As ABCD is a rectangle angles A,B,C and D will be 90°. Thus AC will be diameter of circle of length 13*2 = 26

So length , width and 26 will form a Pythagorean triplet . From given options ,

only option d) 24, 10 satisfy this condition as
10^2+24^2=26^2

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Quantitative Aptitude – Geometry – Circles – In a circle with center O and radius 1 cm

Quantitative Aptitude – Geometry - Circles – In a circle with center O and radius 1 cm

Slot -1 – Quantitative Aptitude – Geometry – Circles – In a circle with center O and radius 1 cm

In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is?

a) (π/6)^(1/2)

b) (π/(4√3))^(1/2)

c) (π/(3√3))^(1/2)

d) (π/4)^(1/2)

Answer: c)  (π/(3√3))^(1/2)

Solution:

As OC =OD so angle OCD = angle ODC =(180 -60)/2 = 60

So triangle OCD is an equilateral triangle,

Area of OCD = √3/4 OC^2 = 1/6×π×1^2
OC^2 = π/(3√3)

So OC= (π/(3√3))^(1/2)

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Quantitative Aptitude – Geometry – Mensuration – Given an equilateral triangle T1 with side 24 cm

Quantitative Aptitude – Geometry - Mensuration – Given an equilateral triangle T1 with side 24 cm

Slot -1 – Quantitative Aptitude – Geometry – Mensuration – Given an equilateral triangle T1 with side 24 cm

Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,… will be?

a) 164√3

b) 188√3

c) 248√3

d) 192√3

Answer: d) 192√3

Solution:

As P,Q and R are mid points of AC,AB and BC so both triangle PQR and CBA will be similar and PQ = BC/2
So if the area of triangle ABC = A then area of triangle PQR = A/4
Thus
Sum of Areas = A+A/4+A/16+A/64+ ——–=A/(1-1/4)=4/3 A=4/3×√3/4×24^2=192√3

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