# Geometry - Triangles - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Geometry - Triangles. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2022 QA Slot 2 | Geometry - Triangles**

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If ∠BAC = 45° and ∠ABC = θ, then AD/BE equals

- A.
1

- B.
√2 cosθ

- C.
√2 sinθ

- D.
(sinθ + cosθ)/√2

Answer: Option C

**Explanation** :

Area of ∆ABC = $\frac{1}{2}$ × AB × AC × sin 45° = $\frac{AB\times AC}{\sqrt{2}}$ ...(1)

Area of ∆ABC = $\frac{1}{2}$ × BA × BC × sin θ° = $\frac{BA\times BC\times \mathrm{sin}\theta}{2}$ ...(2)

(1) = (2)

⇒ $\frac{AB\times AC}{\sqrt{2}}$ = $\frac{BA\times BC\times \mathrm{sin}\theta}{2}$

⇒ AC = $\frac{BC\times \mathrm{sin}\theta}{\sqrt{2}}$ ...(3)

Area of ∆ABC = $\frac{1}{2}$ × AD × BC = $\frac{1}{2}$ × AC × BE

⇒ $\frac{AD}{BE}$ = $\frac{AC}{BC}$

⇒ $\frac{AD}{BE}$ = $\frac{BC\times \mathrm{sin}\theta}{\sqrt{2}\times BC}$ = $\frac{\mathrm{sin}\theta}{\sqrt{2}}$

Hence, option (b).

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**CAT 2022 QA Slot 2 | Geometry - Triangles**

The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD, in cm, is

- A.
√7

- B.
√6

- C.
√5

- D.
√8

Answer: Option A

**Explanation** :

Area of ∆ACD is half of area of ∆ABD.

Since their height is same, ratio of their areas will be same as the ratio of their bases.

⇒ BD = 2CD

⇒ BD = 2 cm and CD = 1 cm.

Also, BE = 3/2 = 1.3

⇒ ED = 2 – 1.5 = 0.5 cm.

Also, AE = height of the equilateral triangle = $\frac{\sqrt{3}}{2}$ × 3 = $\frac{3\sqrt{3}}{2}$

In ∆AED

⇒ AD^{2} = AE^{2} + ED^{2}

⇒ AD^{2} = ${\left(\frac{3\sqrt{3}}{2}\right)}^{2}$ + (0.5)^{2}

⇒ AD^{2} = $\frac{27}{4}$ + $\frac{1}{4}$ = 7

⇒ AD = $\sqrt{7}$

Hence, option (a).

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**CAT 2022 QA Slot 3 | Geometry - Triangles**

Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be

- A.
6

- B.
8

- C.
12

- D.
4

Answer: Option C

**Explanation** :

Initially both ships are at a distance of 24 kms from the port (A).

Let the distance travelled by faster ship be F kms when slower ship travels 8 kms.

Now, ∆ABC is a right triangle when ∠BAC = 60°

⇒ AC = ½ × AB = 8 kms.

∴ 24 – F = 8

⇒ F = 16 kms

Ratio of speeds of faster and slower ships = 16 : 8 = 2 : 1

Now, when the faster ships reaches port it would have travelled 24 kms and the slower ship would have travelled 12 kms.

∴ Slower ship will be 24 – 12 = 12 kms from the port.

Hence, option (c).

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**CAT 2022 QA Slot 3 | Geometry - Triangles**

Suppose the medians BD and CE of a triangle ABC intersect at a point O. If area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is

Answer: 9

**Explanation** :

BD, CE and AF are medians of the triangle ABC.

We know centroid (O) divides the triangle in 6 smaller triangles of equal area.

⇒ Area(∆EOB) = Area(∆BOF) = Area(∆FOC) = Area(∆COD) = 1/6 × 108 = 18.

∆AED ~ ∆ABC

⇒ $\frac{\mathrm{Area}(\u2206\mathrm{AED})}{\mathrm{Area}(\u2206\mathrm{ABC})}$ = ${\left(\frac{\mathrm{AE}}{\mathrm{AB}}\right)}^{2}$ = $\frac{1}{4}$

⇒ Area(EBCD) = ¾ × 108 = 81

Now,

Area(∆EOD) = Area(EBCD) – [Area(∆EOB) + Area(∆BOF) + Area(∆FOC) + Area(∆COD)]

= 81 – 72 = 9

Hence, 9.

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**CAT 2021 QA Slot 2 | Geometry - Triangles**

Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is

Answer: 30

**Explanation** :

Consider the figure below.

Consider ∆AED and ∆BED.

Height of both triangles is same, hence ratio of area will be same as ratio of their base.

∴ Area(∆AED)/Area(∆BED) = AD/BD = 2/1

Area (∆BED) = 8/2 = 4

∴ Area (∆ABE) = 8 + 4 = 12

Now, consider ∆AED and ∆BED.

Height of both triangles is same, hence ratio of area will be same as ratio of their base.

∴ Area(∆ABE)/Area(∆CBE) = AE/CE = 2/3

Area (∆CBE) = 3/2 × 12 = 18

∴ Area (∆ABC) = 12 + 18 = 30 sq. cm.

Hence, 30.

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**CAT 2021 QA Slot 3 | Geometry - Triangles**

In a triangle ABC, ∠BCA = 50°. D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then ∠FDE, in degrees is equal to

- A.
96

- B.
100

- C.
80

- D.
72

Answer: Option C

**Explanation** :

In the figure above:

In ∆ABC,

∠A + ∠B + ∠C = 180°

⇒ x + y = 130°

Since DA = DE

⇒ ∠DAE = ∠DEA = x

⇒ ∠ADE = 180° – 2x

Also, DB = DF

⇒ ∠DBF = ∠DFB = y

⇒ ∠BDC = 180° – 2y

∠ADE + ∠EDF + ∠FDB = 180°

⇒ 180 – 2x + ∠EDF + 180 – 2y = 180

⇒ ∠EDF = 2x + 2y – 180

⇒ ∠EDF = 260° – 180° = 80°

Hence, option (c).

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**CAT 2020 QA Slot 2 | Geometry - Triangles**

From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is s. Then the area of the triangle is

- A.
S

^{2}/√3 - B.
√3S

^{2}/2 - C.
S

^{2}/2√3 - D.
2S

^{2}/√3

Answer: Option A

**Explanation** :

The figure can be drawn as below.

Sum of areas of three smaller ∆s = Area of ∆ABC

⇒ $\frac{1}{2}\times {h}_{1}\times a$ + $\frac{1}{2}\times {h}_{2}\times a$ + $\frac{1}{2}\times {h}_{3}\times a$ = $\frac{\sqrt{3}}{4}{a}^{2}$

⇒ $\frac{1}{2}\times s\times a$ = $\frac{\sqrt{3}}{4}{a}^{2}$

⇒ a = $\frac{2s}{\sqrt{3}}$

∴ Area of ∆ABC = $\frac{\sqrt{3}}{4}{a}^{2}$ = $\frac{\sqrt{3}}{4}\times \frac{4}{3}\times {s}^{2}$ = $\frac{{s}^{2}}{\sqrt{3}}$

Hence, option (a).

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**CAT 2019 QA Slot 1 | Geometry - Triangles**

Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is

- A.
5 : 6

- B.
4 : 5

- C.
3 : 4

- D.
2 : 3

Answer: Option D

**Explanation** :

The hexagon will be regular only if the hexagon is symmetrical with resepect to the original triangle.

This is only possible when the corners cut are all equilateral (of let's say side x) and the side length of the hexagon is equal to side length of the equilateral corner (again x).

H = 6 × $\frac{\sqrt{3}}{4}$x^{2} (∵ A regular hexagon consists of six equilateral triangles of side length equal to the side length of the hexagon)

T = $\frac{\sqrt{3}}{4}$(3x)^{2} = $\frac{9\sqrt{3}}{4}$x^{2} (∵ Side length of the triangle = x + x + x = 3x)

∴ $\frac{\mathrm{H}}{\mathrm{T}}$ = $\frac{6}{9}$ = $\frac{2}{3}$.

Hence, option (d).

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**CAT 2019 QA Slot 2 | Geometry - Triangles**

In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is

- A.
80

- B.
68

- C.
72

- D.
78

Answer: Option C

**Explanation** :

As AD and BE are medians, so G is the centroid of ∆ABC.

So, AG : GD = BG : GE = 2 : 1.

∴ AG = 8 cm, GD = 4 cm, BG = 6 cm and GE = 3 cm.

If we join centroid with all 3 vertices, we get 3 triangles of equal areas.

∴ Area of ∆ABC = 3 × Area of ∆ABG

⇒ Area of ∆ABC = 3 × (1/2 × 8 × 6) = 72 sq. cm.

Hence, option (a).

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**CAT 2019 QA Slot 2 | Geometry - Triangles**

Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

- A.
10

- B.
8√2

- C.
6√2

- D.
5

Answer: Option A

**Explanation** :

∠BAC = 90°, so BC can be considered diameter of a circle with center anywhere on BC.

AP is maximum when it is the perpendicular bisector from the vertex to the hypotenuse. So, BP = PC = 20/2 = 10. This implies that P is the center of the circle.

∴ AP = BP = PC = 10 cm.

Hence, option (b).

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**CAT 2018 QA Slot 1 | Geometry - Triangles**

Given an equilateral triangle T_{1} with side 24 cm, a second triangle T_{2} is formed by joining the midpoints of the sides of T_{1}. Then a third triangle T_{3} is formed by joining the midpoints of the sides of T_{2}. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T_{1}, T_{2}, T_{3},... will be

- A.
248√3

- B.
164√3

- C.
188√3

- D.
192√3

Answer: Option D

**Explanation** :

Consider ∆ABC (i.e. T_{1}) and ∆DEF (i.e., T_{2}).

D and E are midpoints of AB and AC respectively. Therefore, BC = 2 × DE

Side of T_{2} = 1/2 × Side of T_{1}

Area of T_{1} = A(T_{1}) = √3/4 × 24^{2}

Similarly, A(T_{2}) = √3/4 × 12^{2}

A(T_{3}) = √3/4 × 6^{2} ... and so on

Sum of areas of infinitely many T_{i}’s

= √3/4 × 24^{2 }+ √3/4 × 12^{2 }+ √3/4 × 6^{2} + ...

= √3/4 (24^{2 }+ 12^{2 }+ 6^{2} + ...)

Here, (24^{2 }+ 12^{2 }+ 6^{2} + ...) is an infinite series with r = 1/4

Hence, = √3/4 (24^{2 }+ 12^{2 }+ 6^{2} + ...) = $\frac{\sqrt{3}}{4}\left(\frac{{24}^{2}}{\left(1-{\displaystyle \frac{1}{4}}\right)}\right)$ = 192√3

Hence, option (d).

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**CAT 2018 QA Slot 2 | Geometry - Triangles**

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

**Explanation** :

Since points P and Q lie on the circumference of the circle drawn with BC as diameter, angles BPC and BQC are right angled triangles.

Now, area of triangle ABC = $\frac{1}{2}\times AB\times PC$ = $\frac{1}{2}\times AC\times BQ$

Therefore, BQ = $\frac{\mathrm{AB}\times \mathrm{PC}}{\mathrm{AC}}$ = $\frac{30\times 20}{25}$ = 24

Hence, 24.

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**CAT 2017 QA Slot 1 | Geometry - Triangles**

From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is:

- A.
225√3

- B.
500/√3

- C.
275√3

- D.
250√3

Answer: Option B

**Explanation** :

The medians of a triangle divide the triangle into six parts of equal area.

Area of the triangle) = $\sqrt{s(s-a)(s-b)(s-c)}$ = $250\sqrt{3}$

Area of GBC = $\frac{1}{3}$(Area of the triangle)

∴ Area of the remaining portion = 2/3 × 250√3 = 500/√3

Hence, option (b).

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**CAT 2017 QA Slot 1 | Geometry - Triangles**

Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq. cm, of the region enclosed by BPC and BQC is:

- A.
9π - 18

- B.
18

- C.
9π

- D.
9

Answer: Option B

**Explanation** :

AB = a (a = 6)

BC = a√2

BCQB is a semicircle of radius = half of BC = $\frac{\mathrm{a}}{\sqrt{2}}$

Area of semicircle BCQB = ½ × π × ${\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}^{2}$ = ¼ × π × a^{2}

ACPBA is a quarter circle (quadrant) of radius a.

Area of BCPB = Area of sector ABPCA - Area of triangle ABC

= ¼ × π × a^{2} - Area of triangle ABC

Now, Area of CPBQC = Area of semicircle CQB - Area of CPBC

= ¼ × π × a^{2} - ( ¼ × π × a^{2} - Area of triangle ABC)

= Area of triangle ABC

∴ Area of region enclosed by CPBQC = Area of ∆ABC = ½ × 6 × 6 = 18.

Hence, option (b).

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**CAT 2017 QA Slot 1 | Geometry - Triangles**

Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is:

Answer: 24

**Explanation** :

Given triangle is right triangle with smaller sides as 15 and 20, hence the hypotenuse will be 25.

Altitude from right vertex to hypotenuse = (15 × 20)/25 = 12

This is the shortest distance from A to BC. At 30 km/hr = 12/30 hrs = 12/30 × 60 mins = 24 mins.

Hence, 24.

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**CAT 2017 QA Slot 2 | Geometry - Triangles**

Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular distance of P from each of AB, BC, and CA is 4(√2 - 1) m, then the area, in sq cm, of the triangle ABC is

Answer: 16

**Explanation** :

Now since the perpendicular distance of P from each 3 sides is the same, it is the incenter of the triangle ABC.

Let ‘x’ be one of the equal sides of the isosceles right angled triangle ABC.

So the length of the hypotenuse will be √2x.

Area of right triangle ABC = $\frac{1}{2}{\mathrm{x}}^{2}$

So the Area of triangle ABC can also be expressed as a product of the in radius ‘r’ and semi perimeter ‘s’.

s = $\frac{x+x+\sqrt{2}x}{2}$ = x + $\frac{x}{\sqrt{2}}$

∴ Area = $\left(x+\frac{x}{\sqrt{2}}\right)\left[4\right(\sqrt{2}-1\left)\right]$ = $\frac{{x}^{2}}{2}$

⇒ 4√2x - 4x + 4x - 2√2x = $\frac{{x}^{2}}{2}$

⇒ 2√2x = $\frac{{x}^{2}}{2}$

⇒ x = 4√2

Area of triangle ABC = ½ × 4√2 × 4√2 = 16 sq.units.

Hence, 16.

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**CAT 2008 QA | Geometry - Triangles**

In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?

- A.
17.05

- B.
27.85

- C.
22.45

- D.
32.25

- E.
26.25

Answer: Option E

**Explanation** :

We know that the area (A) of triangle (ABC) is related to the circum radius (R) and sides of the triangle as follows:

R = $\frac{AB\times BC\times AC}{4A}$

Where, Area (A) = ½ × AD × BC

∴ R = $\frac{\mathrm{AB}\times \mathrm{BC}\times \mathrm{AC}}{4\times {\displaystyle \frac{1}{2}}\times \mathrm{AD}\times \mathrm{BC}}$ = $\frac{\mathrm{AB}\times \mathrm{AC}}{2\mathrm{AD}}$ = $\frac{17.5\times 9}{2\times 3}$ = 26.25 cm

Hence, option (e).

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**CAT 2008 QA | Geometry - Triangles**

Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer, then how many such triangles exist?

- A.
5

- B.
21

- C.
10

- D.
15

- E.
14

Answer: Option C

**Explanation** :

We know that for an obtuse triangle of sides a, b and c (where c is the largest side),

a^{2} + b^{2} < c^{2}

We also know that for a triangle, a + b > c

Let the third side be x.

These present us with two limiting cases.

**Case 1**: Let 8 cm and 15 cm be the shorter sides. The value of the largest side (x) must be greater than

$\sqrt{{8}^{2}+{15}^{2}}=17cm$

Also, x < 8 + 15 = 23.

The possible integer values of x are 18, 19, 20, 21 and 22 cm.

We cannot consider values from 23 onwards because 8 + 15 = 23 and this violates the second condition.

**Case 2:** Let 8 and x be the shorter sides and 15 cm is the largest side.

The value of the remaining side (x) must be less than

$\sqrt{{15}^{2}-{8}^{2}}=12.69cm$

Also, x > 15 - 8 = 7

The possible integer values are 12, 11, 10, 9 and 8 cm.

We cannot consider values less than 8 because 7 + 8 = 15 and this violates the second condition.

Thus, we have 10 possible values for x.

Hence, option (c).

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**CAT 2005 QA | Geometry - Triangles**

Consider the triangle ABC shown in the following figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and ∠BCD = ∠BAC. What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC?

- A.
7/9

- B.
8/9

- C.
6/9

- D.
5/9

Answer: Option A

**Explanation** :

m ∠BCD = m ∠BAC and B is common to triangles ABC and CBD.

∆ABC is similar to ∆CBD.

AB/CB = BC/BD = AC/CD

AB/12 = 12/9 = AC/6

AB = 16 cm and AC = 8 cm

AD = AB – BD = 16 – 9 = 7 cm

∴ Perimeter of ∆ADC = 7 + 6 + 8 = 21 cm

∴ Perimeter of ∆BDC = 9 + 6 + 12 = 27 cm

∴ Required ratio = 21/27 = 7/9

Hence, option (a).

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**Each question is followed by two statements, A and B. Answer each question using the following instructions**

**Choose 1 **if the question can be answered by using one of the statements alone but not by using the other statement alone.

**Choose 2 **if the question can be answered by using either of the statements alone.

**Choose 3 **if the question can be answered by using both statements together but not by either statement alone.

**Choose 4 **if the question cannot be answered on the basis of the two statements.

**CAT 2003 QA - Leaked | Geometry - Triangles**

D, E, F are the mid-points of the sides AB, BC and CA of triangle ABC respectively. What is the area of DEF in square centimetres?

A. AD = 1 cm, DF = 1 cm and perimeter of DEF = 3 cm

B. Perimeter of ABC = 6 cm, AB = 2 cm, and AC = 2 cm

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option B

**Explanation** :

D, E and F are the midpoints of AB, BC and CA respectively.

**Using statement A alone:**

∵ AD = 1 cm

∴ BD = 1 cm …(Since D is the mid-point)

∵ DF = 1 cm

∴ BC = 2 cm …(Since D and F are the mid-points and DF is parallel to BC)

∴ BE = 1 cm and EC = 1 cm

∵ Perimeter of DEF = 3 cm

∴ EF = 1 cm

∴ Area of ∆DEF can be obtained.

So, statement A alone is sufficient to obtain the answer.

**Using statement B alone:**

∵ Perimeter of ∆ABC = 6 cm, AB = 2 cm, AC = 2 cm

∴ BC = 2 cm

∴ DE = 1, EF = 1 and DF = 1

So, statement B alone is also sufficient to obtain the answer.

Hence, option (b).

Workspace:

**CAT 2003 QA - Leaked | Geometry - Triangles**

In a triangle ABC, AB = 6, BC = 8 and AC = 10. A perpendicular dropped from B, meets the side AC at D. A circle of radius BD (with centre B) is drawn. If the circle cuts AB and BC at P and Q respectively, then AP : QC is equal to

- A.
1 : 1

- B.
3 : 2

- C.
4 : 1

- D.
3 : 8

Answer: Option D

**Explanation** :

In ∆ABC, AB = 6 cm, BC = 8 and AC = 10

∵ 6, 8, 10 are Pythagorean triplets, ∆ABC is a right triangle.

BD is the perpendicular drawn from B to AC.

∴ A(ΔABC) = 1/2 × 6 × 8 = 1/2 × BD × 10

∴ BD = 4.8 cm

A circle with centre B is drawn which will intersect AB at P and BC at Q.

∴ AP = AB − PB = 1.2 and QC = BC − BQ = 3.2

∴ AP : QC = 3 : 8

Hence, option (d).

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**CAT 2003 QA - Leaked | Geometry - Triangles**

In the diagram given below, ∠ABD = ∠CDB = ∠PQD = 90°. If AB : CD = 3 : 1, the ratio of CD : PQ is

- A.
1 : 0.69

- B.
1 : 0.75

- C.
1 : 0.72

- D.
None of the above

Answer: Option B

**Explanation** :

Consider the figure below,

As AB || CD and ∠ABD = ∠CDB = ∠PQD = 90°

∴ ∠BAP = ∠CDP and ∠ABP = ∠DCP

∆CPD ~ ∆BPA …(AAA test)

$\therefore \frac{CP}{PB}=\frac{x}{3x}=\frac{1}{3}$

If CP = y,

PB = 3y

Now, ∆CBD ~ ∆PBQ …(AA test)

$\frac{CD}{PQ}=\frac{CB}{PB}=\frac{y+3y}{3y}=\frac{4}{3}$= 1 : 0.75

Hence, option (b).

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**CAT 2003 QA - Retake | Geometry - Triangles**

A piece of paper is in the shape of a right angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. There was a 35% reduction in the length of the hypotenuse of the triangle. If the area of the original triangle was 34 square inches before the cut, what is the area (in square inches) of the smaller triangle?

- A.
16.665

- B.
16.565

- C.
15.465

- D.
14.365

Answer: Option D

**Explanation** :

Since DE is parallel to AC, ∆ABC is similar to ∆DBE by AAA rule of similarity,

i.e. ΔABC ~ ΔDBE

When two triangles are similar, the ratio of their areas is equal to the ratio of squares of their corresponding sides.

$\therefore \frac{Area(\u2206ABC)}{Area(\u2206DBE)}={\left(\frac{AC}{DE}\right)}^{2}={\left(\frac{1}{0.65}\right)}^{2}$

∴ Area (∆DBE) = (0.65)^{2} × Area (∆ABC)

∴ Area (∆DBE) = 0.4225 × 34 = 14.365

Hence, option (d).

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**CAT 2003 QA - Retake | Geometry - Triangles**

Two straight roads R_{1} and R_{2} diverge from a point A at an angle of 120°. Ram starts walking from point A along R_{1} at a uniform speed of 3 km/hr. Shyam starts walking at the same time from A along R_{2} at a uniform speed of 2 km/hr. They continue walking for 4 hours along their respective roads and reach points B and C on R_{1} and R_{2}, respectively. There is a straight line path connecting B and C. Then Ram returns to point A after walking along the line segments BC and CA. Shyam also returns to A after walking along line segments CB and BA. Their speeds remain unchanged. The time interval (in hours) between Ram’s and Shyam’s return to the point A is:

- A.
$\frac{10\sqrt{19}+26}{3}$

- B.
$\frac{2\sqrt{19}+10}{3}$

- C.
$\frac{\sqrt{19}+26}{3}$

- D.
$\frac{\sqrt{19}+10}{3}$

Answer: Option B

**Explanation** :

After 4 hours,

Distance travelled by Ram, AB = 4 × 3 = 12 km

Distance travelled by Shyam, AC = 4 × 2 = 8 km

Ram’s return path = BC + CA = (x + 8) km

Shyam’s return path = CB + BA = (x + 12) km

Now dropping perpendiculars from C and A to form the right triangle AOC,

In ΔAOC,

OA = AC sin30˚ = 8 × 1/2 = 4

OC = AC sin60° = 8 × $\frac{\sqrt{3}}{2}=4\sqrt{3}$

∴ BC = $\sqrt{\left({\left(4\sqrt{3}\right)}^{2}+{16}^{2}\right)}$

BC = $\sqrt{304}$

BC = $4\sqrt{19}$

Time taken by Ram to travel = $\frac{BC+CA}{3}=\frac{4\sqrt{19}+8}{3}$ ...(i)

Time taken by Shyam to travel = $\frac{BC+BA}{2}=\frac{4\sqrt{19}+12}{2}\phantom{\rule{0ex}{0ex}}$ ...(ii)

∴ Time interval between the two reaching point A is (i) subtracted from (ii)

= $\frac{2\sqrt{19}+10}{3}$

Hence, option (b).

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**CAT 2003 QA - Retake | Geometry - Triangles**

In the figure (not drawn to scale) given below, P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD is parallel to CP. In ∆ARC, ∠ARC = 90°, and in ΔPQS, ∠PSQ = 90°. The length of QS is 6 cm. What is the ratio AP : PD?

- A.
10 : 3

- B.
2 : 1

- C.
7 : 3

- D.
8 : 3

Answer: Option C

**Explanation** :

In ∆ABC, PQ is parallel to AC,

BP : AP = BQ : QC = 3 : 4

In ∆PBC, QD is parallel to CP,

BD : DP = BQ : QC = 3 : 4

The given figure may be depicted as follows:

Let AP = 4x and PB = 3x

Since DB : PD = 3 : 4,

∴ PD = $\frac{4}{7}\times 3x=\frac{12x}{7}$

∴ AP : PD = 4x : $\frac{12x}{7}$ = 7 : 3.

Hence, option (c).

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