# Geometry - Triangles - Previous Year CAT/MBA Questions

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

**XAT 2020 QADI | Geometry - Triangles**

Two lighthouses, located at points A and B on the earth, are 60 feet and 40 feet tall respectively. Each lighthouse is perfectly vertical and the land connecting A and B is perfectly flat. The topmost point of the lighthouse at A is A’ and of the lighthouse at B is B’. Draw line segments A’B and B’A, and let them intersect at point C’. Drop a perpendicular from C’ to touch the earth at point C. What is the length of CC’ in feet?

- A.
20

- B.
30

- C.
24

- D.
The distance between A and B is also needed to solve this

- E.
25

Answer: Option C

**Explanation** :

∆BAA’ ~ ∆BCC’

∴ $\frac{BC}{BA}=\frac{CC\text{'}}{AA\text{'}}$ …(1)

Also, ∆ABB’ ~ ∆ACC’

∴ $\frac{AC}{AB}=\frac{CC\text{'}}{BB\text{'}}$ …(2)

Adding (1) and (2) we get

$\frac{AC+BC}{AB}=\frac{CC\text{'}}{AA\text{'}}+\frac{CC\text{'}}{BB\text{'}}$

⇒ 1 = $\frac{CC\text{'}}{AA\text{'}}+\frac{CC\text{'}}{BB\text{'}}$

⇒ $\frac{1}{CC\text{'}}=\frac{1}{AA\text{'}}+\frac{1}{BB\text{'}}$

⇒ $\frac{1}{CC\text{'}}=\frac{1}{60\text{'}}+\frac{1}{40}$ = $\frac{1}{24}$

∴ CC’ = 24.

Hence, option (c).

Workspace:

**XAT 2019 QADI | Geometry - Triangles**

The figure below shows two right angled triangles ∆OAB and ∆OQP with right angles at vertex A and P, respectively, having the common vertex O, The lengths of some of the sides are indicated in the figure. (Note that the figure is not drawn to scale.) AB and OP are parallel. What is ∠QOB?

- A.
tan

^{–1}(2/3) - B.
45°

- C.
60°

- D.
30°

- E.
tan

^{–1}(3/2)

Answer: Option B

**Explanation** :

Consider the figure given below.

In ∆POQ, OQ = $\sqrt{{1}^{2}+{2}^{2}}$ = √5

In ∆AOB, OB = $\sqrt{{1}^{2}+{3}^{2}}$ = √10

In ∆DQB, QB = $\sqrt{{2}^{2}+{1}^{2}}$ = √5

Now, in ∆QOB, OB^{2} = OQ^{2} + QB^{2}

⇒ ∆QOB is a right isosceles triangle, right angles at Q.

∴ ∠QOB = 45°

Hence, option (b).

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**XAT 2019 QADI | Geometry - Triangles**

Let ABC be an isosceles triangle. Suppose that the sides AB and AC are equal and let the length of AB be x cm. Let b denote the angle ∠ABC and sin b = 3/5. If the area of the triangle ABC is M sq. cm, then which of the following is true about M?

- A.
x

^{2}/4 ≤ M < x^{2}/2 - B.
3x

^{2}/4 ≤ M < x^{2} - C.
M ≥ x

^{2} - D.
x

^{2}/2 ≤ M < 3x^{2}/4 - E.
M < x

^{2}/4

Answer: Option A

**Explanation** :

Now ∆ABC is an isosceles triangle where AB = AC. Let a perpendicular from A meet BC at D. As ∆ABC is isosceles, AD is a perpendicular bisector and BD = CD.

Given, Sin b = 3/5

⇒ Cos b = 4/5

In ∆ABD,

Sin b = $\frac{3}{5}$ = $\frac{AD}{AB}$

⇒ AD = $\frac{3}{5}AB$ = $\frac{3x}{5}$

Also, Cos b = $\frac{4}{5}$ = $\frac{BD}{AB}$

BD = $\frac{4}{5}AB$ = $\frac{4x}{5}$

∆ABC, BD = CD = 4x/5

∴ BC = 8x/5

⇒ Area of ∆ABC = ½ × BC × AD = ½ × 8x/5 × 3x/5 = 12x/25 = 0.48x.

Looking at the options we can see that M lies between x^{2}/4 and x^{2}/2.

Hence, option (a).

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**IIFT 2019 QA | Geometry - Triangles**

In a triangle, the two longest sides are 13 cm and 12 cm. The angles of the triangle are in arithmetic progression. The radius of the circle inscribed in this triangle is:

- A.
√3 cm

- B.
√3 - 1 cm

- C.
2 cm

- D.
1 cm

Answer: Option D

**Explanation** :

Let the three angles be a − d, a and a + d.

∴ (a − d) + a + (a + d) = 180

∴ a = 60.

Since (a + d) is the largest angle it will be opposite the largest side i.e., 13 and a will be opposite second largest side i.e., 12.

The triangle along with the angles are shown below.

Applying the cosine rule, we get;

Cos 60 = (x^{2} + 13^{2} − 12^{2})/(2 × x × 13)

∴ x = 10.65, 2.34.

Inradius = Area/Semiperimeter.

When x = 10.65, the semiperimeter = (10.65 + 13 + 12)/2 = 17.825.

∴ Inradius = [(1/2) × 10.65 × 13 × Sin60]/17.825 = 3.36.

When x = 2.34, the semiperimeter = (2.34 + 13 + 12)/2 = 13.67.

∴ Inradius = [(1/2) × 2.34 × 13 × Sin60]/13.67 = 0.963 ≈ 1.

Hence option (d).

Workspace:

**IIFT 2019 QA | Geometry - Triangles**

In ∆MNL, line NP bisects the angle MNL. If NP : NL= 2 : 3 and angle MNL = 120°. Then NP : NL: MN is:

- A.
2 : 3 : 4

- B.
2 : 3 : 6

- C.
2 : 3 : 5

- D.
2 : 3 : 9

Answer: Option B

**Explanation** :

Applying cosine rule in ∆NPL, we get;

Cos60 = 1/2 = [NP^{2} + NL^{2} − PL^{2}]/(2 × NP × NL)

∴ 1/2 = [(2x)^{2} + (3x)^{2 }− PL^{2}]/(2 × 2x × 3x)

∴ PL = x√7.

Using angle bisector theorem, MN/NL = MP/PL

So, MN/3x = MP/x√7

∴ MN = MP × (3/√7) ...(I)

Applying cosine rule in ∆MNP, we get;

Cos60 = 1/2 = [MN^{2} + NP^{2} − MP^{2}]/(2 × MN × NP)

Let MN = a, so using (I) MP = (√7/3) a or MP^{2} = (7/9)a^{2}.

∴ 1/2 = [a^{2} + 4x^{2}− (7/9)a^{2} ]/(2 × a × 2x)

∴ a^{2} − 9ax + 18x^{2} = 0

∴ a = 6x or 3x.

If a = 3x, then MN = NL = 3x, so ∆MNL becomes isosceles and hence ∠NPL = 90°, which means that NL^{2} = NP^{2} + PL^{2}.

NL^{2} = 9x^{2} and NP^{2} + PL^{2} = 4x^{2} + (x^{2}/7) = (29/7)x^{2 }≠ 9x^{2}. Hence a ≠ 3x.

∴ a = 6x.

NP : NL: MN = 2x : 3x : 6x = 2 : 3 : 6.

Hence option (b).

Note: We can use the direct formula for the length of the angle bisector L. Consider the triangle drawn below

L = (2abCosQ)/(a + b)

So, 2x = (2 × MN × 3x × Cos60)/(MN + 3x)

Solving, we get; MN = 6x.

Workspace:

**IIFT 2018 QA | Geometry - Triangles**

In the triangle PQR, S is the midpoint of QR. X is any point on PR. T is the point on QR such that PT‖SX. If the area of triangle PQR is 5.8 sq. cm, then the area of triangle RTX is

- A.
2.9 sq. cm

- B.
3.2 sq. cm

- C.
5.8 sq. cm

- D.
2.45 sq. cm

Answer: Option A

**Explanation** :

Since S is the midpoint of QR, A(∆PSR)

= A(∆PQR)/2

= 5.8/2 = 2.9 sq.cm

Now A(∆PSR) = A(∆PSX) + A(∆SXR)

Also, A(∆RTX) = A(∆TSX) + A(∆SXR)

Now, triangles PSX and TSX lie within the same parallel lines – SX and PT – and hence, have the same area.

∴ A(∆PSX) = A(∆TSX)

∴ A(∆PSR) = A(∆RTX) = 2.9 sq.cm

Hence, option (a).

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**XAT 2016 QADI | Geometry - Triangles**

In the figure below, AB = AC = CD. If ∠ADB = 20°, what is the value of ∠BAD?

- A.
40°

- B.
60°

- C.
70°

- D.
120°

- E.
140°

Answer: Option D

**Explanation** :

In ∆ ACD, AC = CD

⇒ ∠CAD = ∠CDA = 20°

In ∆ ABC

∠ACB = ∠CAD + ∠CDA = (20 + 20)° = 40° …(Exterior angle theorem)

Also, AC = AB

⇒ m∠ABC = ∠ACB = 40°

∴ m∠BAD = 180 – 40 – 20 = 120°

Hence, option (d).

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**XAT 2016 QADI | Geometry - Triangles**

The difference between the area of the circumscribed circle and the area of the inscribed circle of an equilateral triangle is 2156 sq. cm. What is the area of the equilateral triangle?

- A.
686√3

- B.
1000

- C.
961√2

- D.
650√3

- E.
None of the above

Answer: Option A

**Explanation** :

Let ‘a’ be the side of the triangle.

Circum-radius of an equilateral triangle = a/√3

∴ Area of the circum-circle = π × (a/√3)^{2} = πa^{2}/3

In-radius of an equilateral triangle = a/2√3

∴ Area of the in-circle = π × (a/2√3)^{2} = πa^{2}/12

According to the question,

πa^{2}/3 - πa^{2}/12 = 2156

⇒ πa^{2}/4 = 2156

⇒ a^{2} = 2156 × 4 × 7/22 = 2744

∴ Area of the equilateral triangle = √3/4 × a^{2} = √3/4 × 2744 = 686√3.

Hence, option (a).

Workspace:

**IIFT 2015 QA | Geometry - Triangles**

A ladder of 7.6 m long is standing against a wall and the difference between the wall and the base of the ladder is 6.4 m. If the top of the aldder now slips by 1.2m, then the foot of the ladder shifts by approximately:

- A.
0.4 m

- B.
0.6 m

- C.
0.8 m

- D.
1.2 m

Answer: Option B

**Explanation** :

AC = ED = 7.6 and AB = 6.4

From the given diagram ; BC^{2 }= 7.6^{2} – 6.4^{2} = 16.8

∴ BC ≈ 4.1 m

When the ladder slips 1.2 m, its top edge would be at D at a height of 4.1 – 1.2 ≈ 2.9 m

∴ BE^{2 }= 7.6^{2} – 2.9^{2} ≈ 49

∴ BE ≈ 7

∴ Ladder shifts approximately 7 – 6.4 = 0.6 m.

Hence, option (b).

Workspace:

**XAT 2015 QA | Geometry - Triangles**

In the diagram below, CD = BF = 10 units and ∠CED = ∠BAF = 30°. What would be the area of triangle AED? (*Note*: Diagram below may not be proportional to scale.)

- A.
100 × (√2 + 3)

- B.
$\frac{100}{(\sqrt{3}+4)}$

- C.
$\frac{50}{(\sqrt{3}+4)}$

- D.
$50\times (\sqrt{3}+4)$

Answer: Option D

**Explanation** :

m∠ECD = m∠BCF = 60°

Also, m∠AFB = 60°, m∠BFC = 30°

∴ m∠AFC = 90°

In a 30° - 60° - 90° triangle, sides are in the ratio $1:\sqrt{3}:2.$

So, in ∆EDC, ED = 10√3 units

Also, in ∆FBC, BF = 10 units

$\Rightarrow \mathrm{FC}=\frac{20}{\sqrt{3}}$ units and BC = $\frac{10}{\sqrt{3}}$

In ∆AFC,

$\mathrm{FC}=\frac{20}{\sqrt{3}}\mathrm{units}\Rightarrow \mathrm{AC}=\frac{40}{\sqrt{3}}\mathrm{units}$

$\therefore \mathrm{AD}=\left(10+\frac{40}{\sqrt{3}}\right)\mathrm{units}$

$\mathrm{A}(\u2206\mathrm{ADE})=\frac{1}{2}\times 10\sqrt{3}\times \left(10+\frac{40}{\sqrt{3}}\right)$

= 50(√3 + 4) sq.units

Hence, option (d).

Workspace:

**XAT 2015 QA | Geometry - Triangles**

The centre of a circle inside a triangle is at a distance of 625 cm. from each of the vertices of the triangle. If the diameter of the circle is 350 cm. and the circle is touching only two sides of the triangle, find the area of the triangle.

- A.
240000

- B.
387072

- C.
480000

- D.
506447

- E.
None of the above

Answer: Option B

**Explanation** :

OA ⊥ PQ, OB ⊥ PR

OP = OQ = OR = 625 cm

In ∆OAQ, OA = 175 cm and OQ = 625 cm ⇒ AQ = 600 cm

Similarly, PA = PB = RB = 600 cm

∆PQR is an isosceles triangle and PQ = PR = 1200 cm

So, PC ⊥ QR

In ∆PBO and ∆PCR,

∠OPB ≅ ∠RPC … (Common angle)

∠PBO ≅ ∠PCR … (Right angle)

∆PBO ~ ∆PCR … (AA test of similarity)

$\therefore \frac{\mathrm{PB}}{\mathrm{PC}}=\frac{\mathrm{BO}}{\mathrm{CR}}=\frac{\mathrm{PO}}{\mathrm{PR}}$

$\therefore \frac{600}{\mathrm{PC}}=\frac{175}{\mathrm{CR}}=\frac{625}{1200}$

∴ PC = 1152 cm and CR = 336 cm

∴ QR = 672 cm

A(△PQR) = $\frac{1}{2}\times 672\times 1152$ = 387072

Hence, option (b).

Workspace:

**IIFT 2014 QA | Geometry - Triangles**

A Boat is being rowed away in still water, from a 210 metres high cliff at the speed of 3 km/hr. What is the approximate time taken for the angle of depression of the cliff at the boat to change from 60 deg. To 45 deg.?

- A.
5 min

- B.
4 min

- C.
1 min

- D.
2 min

Answer: Option D

**Explanation** :

When the boat was at C, the angle of depression = 60° ⇒ m∠ BAC = 30°

When the boat was at D, the angle of depression = 45° ⇒ m∠DAB = 45° ⇒ BD

= AB = 210 m

In ∆ABC, by theorem of 30° – 60° – 90°,

BC = 70√3 ≈ 121.1

DC ≈ 210 – 121.2 = 88.8 m

Speed of the boat = 3 km/ hr = 50 m/min

Time taken = 88.8/ 50 = 1.776 minutes

Hence, option (d).

Workspace:

**IIFT 2013 QA | Geometry - Triangles**

There is a triangular building (ABC) located in the heart of Jaipur, the Pink City. The length of the one wall in east (BC) direction is 397 feet. If the length of south wall (AB) is perfect cube, the length of southwest wall (AC) is a power of three, and the length of wall in southwest (AC) is thrice the length of side AB, determine the perimeter of this triangular building.

- A.
3209 feet

- B.
3213 feet

- C.
3773 feet

- D.
3313 feet

Answer: Option D

**Explanation** :

Let AB = *a*^{3} and AC = 3^{n}

Also

AC = 3 × AB

∴ 3* ^{n}* = 3 ×

*a*

^{3}

∴ 3^{(n – 1)} = *a*^{3}

Now, let the perimeter be equal to *p*

*p *= BC + AC + AB

= 397 + 3* ^{n}* + 3

^{(n – 1)}

(*p* – 397) = 3^{(n – 1)} × (3 + 1)

= 3^{(n – 1)} × 4

Thus the LHS of the above equation should be a multiple of 3 and 4. Substitute the value of perimeter given in the options and verify this. Among the options, only (3313 – 397) is divisible by 3 and 4..

Hence, option (d).

Workspace:

**XAT 2012 QA | Geometry - Triangles**

A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments. Determine the sum of the squares of the lengths of the three paths.

- A.
4000 m

- B.
4800 m

- C.
5600 m

- D.
6400 m

- E.
7200 m

Answer: Option C

**Explanation** :

Here D is midpoint of AC, E is the midpoint of AD and F is the midpoint of CD.

Hence, AE = ED = DF = CF = 20

Let AB = *c* and BC = *a*

Applying Apollonius theorem in ∆ABC, we get,

BD^{2} + 40^{2} = 1/2 × (*c*^{2} + *a*^{2}) = 1/2 × 80^{2}

BD^{2} = 40^{2} … (i)

Now, applying Apollonius theorem in ∆ABD, we get,

BE^{2} + 20^{2} = 1/2 × (*c*^{2} + BD^{2}) … (ii)

Similarly, applying Apollonius in ∆CDB, we get,

BF^{2} + 20^{2} = 1/2 × (*a*^{2} + BD^{2}) … (iii)

Adding (ii) and (iii), and substituting value from (i), we get,

BE^{2} + BF^{2} + 2 × 20^{2} = 1/2 (*a*^{2} + *c*^{2} + 2 × BD^{2}) = 1/2 (80^{2} + 2 × 40^{2}) = 3 × 40^{2}

Hence, BE^{2} + BF^{2} + BD^{2} = 3 × 40^{2} + 40^{2} – 2 × 20^{2} = 5600

Hence, option (c).

Workspace:

**IIFT 2012 QA | Geometry - Triangles**

The perimeter of a right-angled triangle measures 234 m and the hypotenuse measures 97 m. Then the other two sides of the triangle are measured as

- A.
100 m and 37 m

- B.
72 m and 65 m

- C.
80 m and 57 m

- D.
None of the above

Answer: Option B

**Explanation** :

All the three options given satisfy the perimeter criteria. The hypotenuse is the greatest side.

∴ Option 1 is eliminated.

Now, 97^{2} = 9409

**Option 2:**

97^{2} – 72^{2} = 4225 = 65^{2}

**Option 3:**

97^{2} – 80^{2} = 3009 ≠ 57^{2}

Hence, option (b).

Workspace:

**IIFT 2010 QA | Geometry - Triangles**

The area of a triangle is 6, two of its vertices are (1, 1) and (4, –1), the third vertex lies on y = x + 5. Find the third vertex.

- A.
$\left(\frac{2}{5},\frac{27}{5}\right)$

- B.
$\left(-\frac{3}{5},\frac{22}{5}\right)$

- C.
$\left(\frac{3}{5},\frac{28}{3}\right)$

- D.
None of these

Answer: Option A

**Explanation** :

If the coordinates of the vertices of a triangle are (*x*_{1}, *y*_{1}), (*x*_{2}, *y*_{2}) and (*x*_{3}, *y*_{3}) then its area is given by

A = $\frac{1}{2}$$\left|\begin{array}{ccc}{x}_{1}& {x}_{2}& {x}_{3}\\ {y}_{1}& {y}_{2}& {y}_{3}\\ 1& 1& 1\end{array}\right|$

∴ If the third vertex of the given triangle is (x, y)

6 = $\frac{1}{2}$$\left|\begin{array}{ccc}1& 4& x\\ 1& -1& y\\ 1& 1& 1\end{array}\right|$

∴ 12 = 1(−1 − y) − 4(1 − y) + x(1 + 1)

∴ 12 = −1 – y − 4 + 4y + 2x

∴ 2x + 3y = 17 …(i)

But (x, y) also lies on y = x + 5 …(ii)

Solving (i) and (ii)

x = $\frac{2}{5}$, y = $\frac{27}{5}$

Hence, option (a).

Workspace:

**IIFT 2010 QA | Geometry - Triangles**

In a triangle ABC the length of side BC is 295. If the length of side AB is a perfect square, then the length of side AC is a power of 2, and the length of side AC is twice the length of side AB. Determine the perimeter of the triangle.

- A.
343

- B.
487

- C.
1063

- D.
None of these

Answer: Option C

**Explanation** :

2*x*^{2} is a power of 2.

Hence, *x*^{2} is a power of 2.

Hence, *x* is a power of 2.

Also by triangle inequality 3*x*^{2} > 295

∴ *x*^{2} > 98.33

∴ *x* ≥ 10

Hence, *x* is a power of 2 which is greater than 10.

We try with *x* = 16. Then we have,

AB = 256, AC = 512

∴ Perimeter = 256 + 512 +295= 1063, which is there in the options.

Hence, option (c).

Workspace:

**Read the following information carefully and answer the questions**

A warship and a submarine (completely submerged in water) are moving horizontally in a straight line. The Captain of the warship observers that the submarine makes an angle of depression of 30°, and the distance between them from the point of observation is 50 km. After 30 minutes, the angle of depression becomes 60°.

**IIFT 2009 QA | Geometry - Triangles**

Find the distance between them after 30 min from the initial point of reference.

- A.
$\frac{50}{\sqrt{3}}$ km

- B.
25 km

- C.
$\frac{25}{\sqrt{3}}$ km

- D.
25$\sqrt{3}$ km

Answer: Option A

**Explanation** :

Refer to the following figure:

The vertical distance between the warship and the submarine = 25 km

∴ When the angle of depression of the submarine is 60°, the distance between the warship and the submarine is

$\frac{25}{\mathrm{sin}60\xb0}$ = $\frac{50}{\sqrt{3}}$ km

Hence, option (a).

Workspace:

**IIFT 2009 QA | Geometry - Triangles**

If both are moving in same direction and the submarine is ahead of the warship in both the situations, then the speed of the warship, if the ratio of the speed of warship to that of the submarine is 2 : 1, is:

- A.
$\frac{100}{\sqrt{3}}$ km/hr

- B.
100$\sqrt{3}$ km/hr

- C.
200$\sqrt{3}$ km/hr

- D.
$\frac{200}{\sqrt{3}}$ km/hr

Answer: Option D

**Explanation** :

Let the speed of the submarine be x kmph.

Then the speed of the warship is 2x kmph.

Distance travelled by the warship in 30 minutes = x km

Distance travelled by the submarine in 30 minutes = 0.5x km

From the figure,

x = 25$\sqrt{3}$ + 0.5x - $\frac{25}{\sqrt{3}}$

∴ 0.5$\sqrt{3}$x = 50

∴ x =$\frac{100}{\sqrt{3}}$ kmph

∴ The speed of the warship = $\frac{200}{\sqrt{3}}$ kmph.

Hence, option (d).

Workspace:

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