Test: Geometry - CAT MCQ

The below figure shows the graph of both the given expressions

The required area = 4 × the area of the shaded triangle.
Area of the shaded triangle =
Therefore, the required area =

Since AB  is a diameter, therefore ∠AQB and ∠APB are angles in the semi–circle and hence are right–angles.
i.e. ∠A PB = ∠AQB = 90°
Let AQ  = x, then AP = 2x
In right ΔAPB,
AP² = AB² – BP² ⇒ AP² = 10² – 6² = 8²
⇒ AP = 8
⇒ 2x = 8
⇒ x = 4
In right ΔAQB
BQ² = AB² – AQ²
⇒ BQ² = 10² – 4² = 84
⇒ BQ = √84 ≈ 9.1

Let A, B  and C represent the coordinates (1, 1), (8, 10) and (4, 6) respectively.

The number of ways to go from A to B via C
= (The number of ways to go from A to C) × (The number of ways to go from C to B)

56 × 70 = 3920

Clearly, the triangle will be right angled triangle and its hypotenuse would be the diameter of its circumcircle.
Given equation, 3x + 5y – 45 = 0
When y = 0, Then x = 15
When x = 0, Then y = 9
Hence, A(15, 0) and B(0, 9) and O(0, 0) are the vertices of the right triangle formed whose hypotonuse is the diameter of the ΔOAB.


∴ Now, Radius(L) =
Hence, the integer closest to L = 9.

In the figure, AB and CD are two chords of a circle, intersects each other at point E.
∴ AE × BE = CE × DE
⇒ AE × BE = 7 × 15 = 105
(AE – BE)² = (AE + BE)² – 4 × AE × BE = (AB)² – 4 
× 105
⇒ (AE – BE)² = (20.5)² – 420
⇒ (AE – BE)² = 420.25 – 420 = 0.25
⇒ AE – BE = 0.5

In ΔABC, let AC = a and AB = b
Then by pythagoras theorem for ΔABC,
a² + b² = 20² = 400 ...(i)
In ΔABC and ΔPAC,
∠BAC = ∠APC (= 90°)
∠C = ∠C (common)
∴ ΔABC ~ ΔPAC (by AA)
[Ratio of corresponding sides of two similar triangles are equal]

...(ii)
For maximum value of AP, we have to maximize the product ab.
Applying AM ≥ GM inequality, we get


⇒ ab ≤ 200
Hence, the maximum value of ab = 200
From eq. (ii), it is clear that value of AP will be maximum when value of ab will be maximum
∴ maximum value of AP = 200 / 20 = 10

Draw the third median CF. We know the following facts.
(i) The intersection point of medians i.e. centroid (G) divides each median into 2:1

(ii) All three medians divide the triangle into 6 parts of equal area.
Now, Area of triangle
∴ area of triangle ABC = 6 × (Area of ΔBGD)
= 6 × 12 = 72

From the figure, OC = AD – r = 4 – r
Applying Pythagoras theorem in right triangle AOC, we get
AC² = AO² + OC²
⇒ (4 + r)² = 4² + (4 – r)²
⇒ (4 + r)² – (4 – r)² = 16
⇒ 8 × 2r = 16
∴ r = 1

From the diagram, it is clear that AB is the slant height of the equilateral triangle and is also the slant height of the pyramid.

On applying Pythagoras theorem in triangle AOB,
OB² = AB² – OA² = (10√3 )² − 10² = 200
∴ Height of the pyramid OB = 10√2 cm

To find the minimum number of cylinders, the volume of each of the cylinder must be HCF of 405, 783 and 351
HCF (405, 783, 351) = 27
Therefore, volume of each cylinder = 27cc
Hence number of cylinders of iron = 405 / 27 = 15
Number of cylinders of aluminium = 783 / 27 = 29
and number of cylinders of copper = 351 / 27 = 13
Therefore, the total number of a cylinders
= 15 + 29 + 13 = 57
Since, volume of each cylinder = 27cc
∴ πr²h = 27 ⇒ π × 3² × h = 27
⇒ h = 3 / π
Now, total surface area of each cylinder = 2πr (r + h)

∴ Total surface area of 57 cylinders
= 57 × 18(π + 1) sq. cm. = 1026 (π + 1)

The formula for each interior angle =where ‘n’ 
is the side of the regular polygon.
Now, (an interior angle of B) =


Hence, polygon with each side = a + b = 4 + 8 = 12, will have each interior angle

Let the area of ABCD be 100.
∴ Length of each side of ABCD = 10
Area of EFGH is 62.5,
Therefore length of each side of EFGH = √62.5
Triangles AEH, BFE, CGF and DHG are congruent by ASA.
Let AE = BF = CG = DH = x, then EB = FC = DG = AH = 10 – x
In right triangle EAH, AE² + AH² = EH²
x² + (10 − x)² = (√62.5)² ⇒ 4x² – 40x + 75 = 0
⇒ (2x – 5) (2x – 15) = 0
∴ x = 2.5 or 7.5
Since it’s given that CG is longer than EB, therefore CG = 7.5 and EB = 2.5.
∴ EB : CG = 1 : 3

_{Area of ∆ OCD =}

Area of the parallelogram ABCD = (base) (height)
= CD × AP = 72 sq.cm.
⇒ 72 = 9 × (AP) ⇒ AP = 8cm

Let the 6 cm long chord be x cm away from the centre of the circle. Let the radius of the circle be r cm. The perpendiculars from the centre of the circle to the chords bisect the chords.
r² = x² + 3² = (x + 1)² + 2²
On solving, x = 2 and r = √13

shortest distance according graph = 1
y = |x – 1| + | x + 1 |
let x, – 1 ≤ x ≤ 1
y = – x + 1 + x + 1 ⇒ y = 2

Enclosed Area = Area of semi circle BQC with centre O – Area of Arc BPC with O
 ...(i)
Now, Area of triangle ABC + BPCO (Arc) = Area of Quarter circle (ABPC)

Area of BPCO = 9π –18
Put the value in equation (i)
Enclosed Area =  = 18 sq cm

Let G be a point on AC such that DG is parallel to BF.

Here, BD = 8 cm
DC = 6 cm and
AE : ED = 3 : 4
AF = 12 cm
AC = ?

In ΔQTR,

Let the radius of the bigger circle be R and the smaller circle be r and the side of the square is 2a.
∴ OE = R – EF
= R – [2R – (2r + 2a)]
OE² + EB² = OB²
i.e  [2a +2r – R]² +a²  = R²
a = 9 (∵ 2a = 18); R = 15
∴ (18 + 2r – 15)² + 9² =15²
∴ 2r + 3 = 12

Radius of smaller circle = 4.5 cm.

∴  AB = 20
BC = 21

and (20 – r) + (21 – r) = 29
or, 2r = 41 – 29
⇒ r = 6 cm

Area of the shaded region

PQ is perpendicular to line y = 

∴ Co-ordinates of point Q =  and Co-ordinates of point P = (0, 4).
Hence, 

Here, ∠EOQ = 85° and
∠BOD = 15°
∴ ∠EOD = 180° – (85° + 15°) = 80°
∴ In ∠OED
OE = OD (radius), ∠OED = ∠ODE = θ° (let)
∴ θ + θ + 80° = 180
⇒ 2θ° = 180 – 80 = 100°

∴ ∠OED = ∠ODE = 50°
In ∠EOC,
∠EOC = 80° + 15° = 95°
and ∠OEC = 50°
∴ ∠ECA =180° – (95° + 50°) = 35°

The figure would be as shown above.
Join EC.
Let ∠BAD = x°
and ∠ABD = y°
⇒ ∠DAC = x°
and ∠AEC = y°
In ΔABD and ΔAEC,
∠ABD = ∠AEC
∠BAD = ∠EAC
∴ ΔABD ~ ∠AEC

Given that PT||SR|| QU
PT = a units, QU = b units.
ΔPTQ and ΔSRQ are similar.
∴ We have

 ...(i)
ΔUQP and ΔRSP are similar.

Combining (i) and (ii)

 

Option (a) – The length of AE is always equal to the length of CF because E and F is midpoint of side AB and CD.
So, statement – I is always true.
Option (b) : the Length of BC is equal to length of EF so statement II is always False.
Option (c) : Length of AE is equal to the length of DF because of EF is parallel to BC
Option (d) : ∠AEF = ∠EFC = 90°
So statement III is also true.

Here, Two cases are possible
Case I : When B lies in greater arc AC.
Then,

  ∠OAC = ∠OCA = 20°
⇒ ∠AOC = 140°

Case II : When B lies in smaller arc AC.
Then, 

Let D be a point O is the greater arc AC.
∠OAC =∠OCA = 20°
⇒ ∠AOC = 140°

ABCD is a cyclic quadrilateral
∴ ∠ABC = 180° – 70° = 110°

Let the two sides of the rectangle be x and 2x and ∠AED be θ.
∴ ∠DEC = θ

∴ ∠CEB = 180 – 2θ
Also, ∠EDC = ∠AED = θ (AB is parallel to CD.) in ΔDEC,
∠DEC = ∠EDC ⇒ CD = EC = 2x
In ΔBCE,

Now, in ΔBCE
2θ = 90° + 60° (Exterior angle is equal to the sum of opposite interior angles.)
⇒ θ = 75°