Geometry - Quadrilaterals & Polygons - Previous Year CAT/MBA Questions

Answer: Option A

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Text Explanation :

Given that, CPD is an equilateral triangle

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∠CPD = ∠PDC = ∠DCP = 60°

AQ || PC ⇒ ∠CPD = ∠QAP = 60°
BC || AD ⇒ ∠QAP = ∠AQB = 60°

⇒ AQCP is a parallelogram, hence AQ = PC = a

∆AQB is a 30°-60°-90° triangle.
⇒ BQ = 1/2 × AQ = a/2

Comapring ∆AQB and ∆CPD
Base AQ = 1/2 × PD while the heights of both triangles is same.
⇒ Area of ∆AQB = 1/2 × Area of ∆CPD = 2√3

Hence, option (a).

Answer: Option A

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Text Explanation :

Answer: Option B

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Text Explanation :

Answer: Option C

Video Explanation

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Text Explanation :

Length of the diagonal = √(30²+40² ) = 50 m.

Each circle on the end of the diagonal will touch sides of the rectangular field

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Using Pythagoras' theorem, the distance between the vertex of the rectangle and centre of the first circle drawn on the diagonal (OC) = 1.25√2

Distance between the vertex of the rectangle and circumference of the first circle drawn on the diagonal (OD) = 1.25√2 - 1.25 = 0.51 meters

Space that cannot be used to draw circle otherwise they will go outside rectangle on every diagonal = 0.51 × 2 = 1.02 meters

Space that can be used to draw circles = length of diagonal - unused space = 50 - 1.02 = 48.98 meters

On every diagonal, maximum number of such circles = usable length/diameter of each circle = 48.98/2.5 = 19.6

∴ Maximum 19 circles can be drawn on a diagonal.

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Now, on every diagonal, one circle will be at the centre (intersection of diagonals) and 9 circles will be on each half of the diagonal

⇒ The circle in centre will be common for both diagonals. 

∴ Total circles = 19 + 19 – 1 = 37.

Hence, option (c).

Answer: Option D

Video Explanation

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Text Explanation :

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Length of the diagonal AC = 2√2
⇒ AX = ½ of diagonal = √2

Length of AC₁ = 2/3 × √2 = 2√2/3

⇒ C₁X = AX – AC₁ = √2 - 2√2/3 = √2/3

Using Pythagoras theorem:
XM = $\sqrt{{\left(\frac{2}{3}\right)}^2-{\left(\frac{\sqrt{2}}{3}\right)}^2}$ = $\frac{\sqrt{2}}{3}$

In ∆C₁XM,
C₁X = XM = √2/3 and ∠C₁XM = 90°
⇒ ∠XC₁M = 45°

And ∠LC₁M = 90°

∴ Area of ∆LC₁M = ½ × 2/3 × 2/3 = 2/9

Area of minor arc LM = Area of sector LC₁M - Area of ∆LC₁M

= $\frac{90}{360}\times \mathrm{\pi }\times {\left(\frac{2}{3}\right)}^2-\frac{2}{9}$ = $\frac{\mathrm{\pi }-2}{9}$

∴ Area of overlapping region =  $2\left(\frac{\mathrm{\pi }-2}{9}\right)$

Hence, option (d).

Answer: Option D

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Text Explanation :

The slope of any line parallel to 2y - x - 5 = 0 would be 1/2
The slope of any line parallel to y + 2x - 7 = 0 would be -2
Product of slopes = -1 which shows that these 2 lines would be perpendicular i.e. we have a parallelogram
whose diagonals are perpendicular to each other. This means that the quadrilateral is atleast a rhombus
and it would be a square if we can get info that all sides are equal to each other.

Answer: Option A

Video Explanation

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Text Explanation :

Refer the diagram below:

In ∆ACL,
Sin∠BAC = Sin∠LAC = perpendicular/hypotenuse = CL/AC   …(1)

Similarly, in ∆ADM,
Sin∠BAD = Sin∠MAD = DM/AD   …(2)

(1) ÷ (2)

⇒ Sin∠BAC : Sin∠BAD = CL/AC : DM/AD

Now, CL = DM

Hence, Sin∠BAC : Sin∠BAD = 1/AC : 1/AD = AD : AC

Hence, option (a).

Answer: Option D

Video Explanation

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Text Explanation :

In □EFGH, EG is the diagonal. 

Also, EI and GJ are the perpendicular bisectors of the equilateral triangles AEB and GCD. 

Let us suppose AB = ’a’ units.

BC will also be ‘a’ units since AB and BC are sides of the same square.
 
In ∆ABE, EI = a√3/2 (perpendicular of an equilateral triangle is √3/2 times the side)

Similarly, GJ = a√3/2
 
Also, IJ = BC = a

∴ EG = EI + IJ + GJ = a√3/2 + a + a√3/2 = a(√3 + 1)

∴ Side of square EFGH = a(√3 + 1)/√2

⇒ Area of square EFGH = [a(√3 + 1)/√2]² = (2 + √3)a²

Also, area of square ABCD = a²

Area of square ABCD = a²

Ratio of area of EFGH to area of ABCD = (2 + √3)a² : a² = (2 + √3) : 1

Hence, option (d).

Answer: Option E

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Text Explanation :

We know that the diagonals of a rhombus bisect each other at right angles.

So, let’s assume that the two diagonals have lengths 6x and 8x.

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⇒ (3x)² + (4x)² = 152

⇒ 25x² = 225

⇒ x² = 9

⇒ x = 3

∴ The length of the diagonals are 18 and 24.

∴ Area of the rhombus = ½ × product of the diagonals = ½ × 18 × 24 = 216

Hence, option (e).

Answer: Option E

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Text Explanation :

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Let d₁ = x and d₂ = 2x.

Now, SR would be shortest when SR = d1 = x

Area of PQRS = 30 square units.

∴ Area of PQRS = ½ × QS × 2x + ½ × QS × x

= (3/2) × QS × x = 30 

⇒ QS × x = 20

Smallest value of x = 1 units.

⇒ QS = 20

∴ ∆QSR is right angled at S

⇒ QR² = QS² + SR²

⇒ QR² = 400 + 1

⇒ QR = slightly more than 20.

Hence, option (e).

Answer: Option A

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Text Explanation :

Let the length of the rectangle be ‘l’ and breadth be ‘b’.

The given figure is.

Let us draw a line MS parallel to DC

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∆RMS ~ ∆RDQ

∴ RM : MD = RS : SQ = 1 : 3

Let the length of the rectangle (i.e., AB = CD) ABCD be 8l and width be 8b (i.e., AD = BC).

∴ AR = 4b and RM = b

Also, MS = l and SL = 7l.

Also, PL : LC = 1 : 3 [∵ RS : SQ = 1 : 3]

Area (∆ARS) = ½ × AR × SM = 2lb

Area (∆RDQ) = ½ × RD × DQ = 8lb

Area (∆ABP) = ½ × AB × BP = 16lb

Area (PSQC) = Area (∆PSL) + Area(∆SLQC) 
                         = ½ × SL × PL + ½ × (SL + QC) × LC 
                         = 3.5lb + 16.5lb
                         = 20lb

Area of ∆ASP = Area of rectangle ABCD – [Area (∆ARS) + Area (∆RDQ) + Area (∆ABP) + Area (PSQC)]
                          = 64lb – [2lb + 8lb + 16lb + 20lb]
                          = 64lb - 46lb = 18lb

∴ $\frac{Area(∆ASP)}{Area\left(ABCD\right)}$ = $\frac{18lb}{64lb}$ = $\frac{36}{128}$

Hence, option (a).

Answer: Option A

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Text Explanation :

Since all triangles are equilateral triangles

This is possible when YM = MN = NZ = 1/3 of YZ = 54/3 = 18 cm.

∴ Side of the regular hexagon MNOPQRM is 18 cm.

∴ Area of the hexagon = 6 × √3/4 × (18)² = 486√3.

Hence, option (b).

Answer: Option D

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Text Explanation :

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Let DQ = x and PB = 2x.

From the diagram A(△BDP) = ½ × AD × PB = ½ × 9 × 2x.

Also A(△BDQ) = ½ × BC × DQ = ½ × 13 × x

A(ꐎBPDQ) = A(△BDQ) + A(△BDP) = ½ × 31 × x = 150

∴ x = 300 / 31

∴ x can have integer values from 1 to 9.

Hence , there are 9 values of DQ.

Hence, option (d).

Answer: Option B

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Text Explanation :

A regular hexagon comprises six equilateral triangles.
∴ Area of hexagon = 6 × area of equilateral triangle = area of six equilateral triangles

P, R and T are alternate vertices of the hexagon. The triangle formed by joining them splits each equilateral triangle into half. Six half-equilateral triangles is equivalent to three equivalent triangles.

∴ Area of triangle PRT = area of three equilateral triangles.

∴ Required ratio = (area of three equilateral triangles) / (area of six equilateral triangles) = 0.5

Hence, option (b).

Answer: Option A

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Text Explanation :

Let m∠DAB = θ ⇒ m∠BCD = 2θ
□PBCD is a parallelogram.
∴ m∠DPB = 2θ
m∠PBC = m∠PDC = (180 – 2θ)
∠DPB is an exterior angle of ∆PAB.
∴ By exterior angle theorem, m∠PBA = θ
as, m∠DAB = θ
Thus, in ∆PAB, PA = PB

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∴ m∠ABC = θ + (180 – 2θ) = 180 – θ
According to the given conditions, 10x + y = 1120 and 10x = 1000
Solving the two equations, we get x = 100 and y = 120
sin (180 – θ) = sin θ
Applying sine rule to ∆PAB,

$\frac{100}{\sin \mathrm{\theta }}=\frac{120}{\sin (180-2\mathrm{\theta })}$

$$\begin{gathered} \frac{}{} \\ \frac{100}{\sin \mathrm{\theta }}=\frac{120}{\sin 2\mathrm{\theta }} .....\left(\mathrm{i}\right) \end{gathered}$$

sin 2θ = 2 × sin θ × cos θ

∴ (i) becomes

$\frac{100}{\sin \theta }=\frac{120}{2\times \sin \theta \times \cos \theta }$

$$\begin{gathered} \frac{}{} \\ \therefore \cos \theta =\frac{3}{5}\Rightarrow \sin \theta =\frac{4}{5} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{}}{} \\ \therefore \sin \angle ABC=\sin (180-\theta )=\sin \theta =\frac{4}{5} \end{gathered}$$

Hence, option (a).

Answer: Option D

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Text Explanation :

A square is a cyclic quadrilateral. So we can find the answer using a square.

The area of a square with side x is x²

The area of a square with side 3x is 9x²

∴ The percentage increase in area when all sides increase three fold = 800%.

Hence, option (d).