Geometry - Triangles - Previous Year CAT/MBA Questions
The best way to prepare for Geometry - Triangles is by going through the previous year Geometry - Triangles XAT questions. Here we bring you all previous year Geometry - Triangles XAT questions along with detailed solutions.
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Consider a right-angled triangle ABC, right angled at B. Two circles, each of radius r, are drawn inside the triangle in such a way that one of them touches AB and BC, while the other one touches AC and BC. The two circles also touch each other (see the image below).
If AB = 18 cm and BC = 24 cm, then find the value of r.
- (a)
3 cm
- (b)
4 cm
- (c)
3.5 cm
- (d)
4.5 cm
- (e)
None of the remaining options is correct.
Answer: Option B
Text Explanation :
Workspace:
Ramesh and Reena are playing with triangle ∠ABC. Ramesh draws a line that bisects ; this line cuts BC at D. Reena then extends AD to a point P. In response, Ramesh joins B and P. Reena then announces that BD bisects ∠PBA, hat a surprise! Together, Ramesh and Reena find that BD = 6 cm, AC = 9 cm, DC = 5 cm, BP = 8 cm, and DP = 5 cm.
How long is AP?
- (a)
11.5 cm
- (b)
11.75 cm
- (c)
10.5 cm
- (d)
11 cm
- (e)
10.75 cm
Answer: Option B
Text Explanation :
Given:
BD = 6 cm, AC = 9 cm, DC = 5 cm, BP = 8 cm, and DP = 5 cm.
Since AD is the angular bisector applying the angular bisector theorem we have:
=
Hence : Considering AB = x cm.
=
x = 10.8 cm.
Now since BD is the angular bisector for angle PBA we have :
Applyinh the internal angle bisector theorem:
=
Considering AD = y cm.
=
y = 6.75 cm.
AP = AD + DP.
= 6.75 + 5 = 11.75 cm
Workspace:
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
Text Explanation :
∆BAA’ ~ ∆BCC’
∴ …(1)
Also, ∆ABB’ ~ ∆ACC’
∴ …(2)
Adding (1) and (2) we get
⇒ 1 =
⇒
⇒ =
∴ CC’ = 24.
Hence, option (c).
Workspace:
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
Text Explanation :
Consider the figure given below.
In ∆POQ, OQ = = √5
In ∆AOB, OB = = √10
In ∆DQB, QB = = √5
Now, in ∆QOB, OB2 = OQ2 + QB2
⇒ ∆QOB is a right isosceles triangle, right angles at Q.
∴ ∠QOB = 45°
Hence, option (b).
Workspace:
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)
x2/4 ≤ M < x2/2
- (b)
3x2/4 ≤ M < x2
- (c)
M ≥ x2
- (d)
x2/2 ≤ M < 3x2/4
- (e)
M < x2/4
Answer: Option A
Text 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 = =
⇒ AD = =
Also, Cos b = =
BD = =
∆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 x2/4 and x2/2.
Hence, option (a).
Workspace:
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
Text 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).
Workspace:
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
Text 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 = πa2/3
In-radius of an equilateral triangle = a/2√3
∴ Area of the in-circle = π × (a/2√3)2 = πa2/12
According to the question,
πa2/3 - πa2/12 = 2156
⇒ πa2/4 = 2156
⇒ a2 = 2156 × 4 × 7/22 = 2744
∴ Area of the equilateral triangle = √3/4 × a2 = √3/4 × 2744 = 686√3.
Hence, option (a).
Workspace:
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)
- (c)
- (d)
Answer: Option D
Text 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
So, in ∆EDC, ED = 10√3 units
Also, in ∆FBC, BF = 10 units
units and BC =
In ∆AFC,
= 50(√3 + 4) sq.units
Hence, option (d).
Workspace:
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
Text 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)
∴ PC = 1152 cm and CR = 336 cm
∴ QR = 672 cm
A(△PQR) = = 387072
Hence, option (b).
Workspace:
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
Text 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,
BD2 + 402 = 1/2 × (c2 + a2) = 1/2 × 802
BD2 = 402 … (i)
Now, applying Apollonius theorem in ∆ABD, we get,
BE2 + 202 = 1/2 × (c2 + BD2) … (ii)
Similarly, applying Apollonius in ∆CDB, we get,
BF2 + 202 = 1/2 × (a2 + BD2) … (iii)
Adding (ii) and (iii), and substituting value from (i), we get,
BE2 + BF2 + 2 × 202 = 1/2 (a2 + c2 + 2 × BD2) = 1/2 (802 + 2 × 402) = 3 × 402
Hence, BE2 + BF2 + BD2 = 3 × 402 + 402 – 2 × 202 = 5600
Hence, option (c).
Workspace:
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