# QUESTIONS ON GEOMETRY (PART-6)

#### QUESTIONS ON GEOMETRY (PART-6)

Most of these questions are taken from the previous year’s examinations conducted by the Staff Selection Commission (SSC) on Geometry (Except Circle) of the Quantitative Aptitude (Maths) section of the following exams as well as other similar exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I & Tier-II

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Paper-I & Paper-II

#### QUERY 51

**In a ΔABC, the sides AB and AC have been produced to D and E. Bisectors of ∠CBD and ∠BCE meet at O. If ∠A = 64, then ∠BOC is?**

A) 52

B) 58

C) 26

D) 112

**MAHA GUPTA**

Just remember that

a) External bisectors of two angles of a triangle make angle of 90 – 1/2 the third angle, and

b) Internal bisectors of two angle of a triangle make 90 + 1/2 the third angle.

Here, as that angle is the result of external bisectors, so angle BOC = 90 – (1/2)64 = 58 (option ‘B’)

QUERY 52

**D is a point on the side BC of a ΔABC such that AD ⊥ BC. E is a point on AD for which AE : ED = 5 : 1. If ∠BAD = 30° and tan ∠ACB = 6tan ∠DBE, then ∠ACB =**

A) 30°

B) 45°

C) 60°

D) 15°

**SHIV KISHOR
**

Let ED = x

=> AD = 6x (As ratio of AE : ED = 5 : 1 given)

Therefore, in ΔABD, tan 60 = 6x/BD

In ΔBDE, tan ∠DBE = x/BD

Multiplying both sides by 6

6tan ∠DBE = 6x/BD

=> 6tan ∠DBE = tan 60

=> tan ∠ACB = tan 60

=> ∠ACB = 60 (option ‘C’)

QUERY 53

**In ΔABC the base BC = 10 cm and PQ || BC and cuts the other two sides in 3 : 2. Find PQ in cm.**

A) 6 or 4

B) 7 or 5

C) 8 or 5.5

D) 9 or 6

**MAHA GUPTA
**

PQ cuts the sides in 3 : 2, means either AP : PB = 3 : 2 or 2 : 3 and AQ : QC = 3 : 2 or 2 : 3

In a triangle a parallel line to any of the sides make the triangle into two similar triangles; the original triangle itself and the the other triangle (here APQ) you see in the figure

Here AP and AB are corresponding sides and also PQ and BC are corresponding

Letting AP : PB = 3 : 2

=> AP : AB = 3 : 3+2

=> AP : AB = 3 : 5

Now according to the similarity theorem of triangles

AP/AB = PQ/BC = 3/5 = PQ/10

=> PQ = 6

Similarly, Taking AP : PB = 2 : 3

PQ = 4

Hence 6 or 4 (option ‘A’) is correct.

QUERY 54

**ABC is a right angle triangle, B being the right angle. Mid-points of BC and AC are respectively B’ and A’. The ratio of the area of the quadrilateral AA’ B’B to the area of the ΔABC is?**

A) 1 : 2

B) 2 : 3

C) 3 : 4

D) 4 : 5

**SHIV KISHOR**

If you see AA’B’B is a trapezium in the figure below in which AB is parallel to A’B’

Area of a trapezium = (1/2)*sum of parallel sides*distance between the parallel sides

So area of AA’B’B = (1/2)*(AB + A’B’)*BB’

Now Let the length of AB = a; and BC = b

Now see the following for solution:

So, option ‘C’ is correct

QUERY 55

**Mid points of an equilateral triangle’s sides with 18 cm each side, are joined to form a new triangle and take this process continues to infinite times, then what is the sum of all triangles’ sides?**

A) 81

B) 108

C) 102.5

D) 105

**MAHA GUPTA**

Mid-points of a triangle’s side if joined up to infinite number of times, sum of the perimeters of all the triangles thus formed is equal to double the perimeter of the given triangle.

So the sum of all triangles’ sides here = 2*(18 + 18 + 18) = 108 cm (option ‘B’)

Understand it here

**MEGHA DIXIT
**

Now sum of all the triangles’ sides where P1, P2, P3 … are the perimeters of triangles thus formed

You see series in bracket is a GP where the first term (a) = 1, and the common difference (r) = 1 – 1/2 = 1/2

Sum of an infinite series in a GP = a/(1 – r)

So, the sum here = 1/(1 – 1/2) = 2

Therefore, the sum of all triangles’ sides here = 54*2 = 108 (option ‘B’)

QUERY 56

**The centroid of an equilateral ΔABC is G. If, AB = 9 cm, the length of AG?**

A) 2√3

B) √3

C) 3√3

D) 3√2

**MAHA GUPTA**

Centroid is the point of intersection of all the medians of a triangle. It divides a median by its vertices in 2 : 1. Also a median of an equilateral triangle is its height.

Height of the equilateral triangle = (√3/2)side

So the height of the given triangle = (√3/2)9

As said above height of an equilateral triangle is its median also, and also the centroid of the triangle divides a median by its vertex in 2 : 1

So AG = [(√3/2)9]*(2/3) = 3√3 (option ‘C’)

QUERY 57

**In rectangle ABCD ratio of AB and BC 3 : 2. P is the mid point of AB, then what is sin ∠ CPB?**

A) 1/2

B) 2/3

C) 4/5

D) 3/4

(option ‘C’)

#### QUERY 58

**The perimeter of a triangle is 40 cm and its area is 60 cm2. If the largest side measures 17 cm, then the length (in cm) of the smallest side of the triangle is**

A) 4

B) 6

C) 8

D) 15

**MAHA GUPTA**

Let’s see first whether it qualifies the dimensions that of a right triangle

We see the largest side here is 17, so it must be the hypotenuse if it being a right triangle

Sum of the other sides = perimeter – given side = 40 – 17 = 23

According to the Pythagoras theorem, therefore, 23 must have two such parts whose sum of squares is equal to 17²

We see it has; 15 and 8 such parts. So the the smallest side = 8 cm (option ‘C’)

**SHIV KISHOR**

If criterion of a right triangle is not met we will use Heron’s formula, which is

Area of the triangle = √[s(s – a)(s- b)(s – c)], where s = half parameter, and a, b, c are the sides of the triangle

Now s = 40/2 = 20

a = 17, therefore b + c = 40 – 17 = 23

=> c = 23 – b

Hence, 60 = √[20(20 – 17)(20 – b){20 – (23 – b)}] = 60

Solving, b = 8 (option ‘C’)

**JAYANTCHARAN CHARAN**

The only Pythagoras triplet with 17 as one of the sides is 8, 15, 17, and it’s satisfying all the other things you’ll see, so the smallest side obviously is 8 (option ‘C’)

QUERY 59

**Median of a triangle are 12, 15, 18; then area of the triangle is?**

A) 5√7

B) 15√7

C) 30√7

D) 45√7

**Vijay Bharath Reddy**

Area of a triangle = 4/3(area of triangle formed by its medians)

= 4/3(area of a triangle whose sides are equal to the length of the medians i.e. 12. 15 & 18)

= 4/3(135/4√7) ….(using Heron formula)

= 45√7 (option ‘D’)

QUERY 60

**Find the area of the largest square that can be inscribed in an equilateral triangle whose side is 12 cm.**

A) 22 cm²

B) 27 cm²

C) 31 cm²

D) 48 cm²

**MAHA GUPTA**

Just remember this formula for finding the area of the largest square that can be inscribed in an equilateral triangle

[a(2√3 – 3)]², where ‘a’ is the side of the triangle

So here the area of the largest inscribed circle = [12(2√3 – 3)]² = (24√3 – 36)² = 3024 – 1728√3 = 31 cm² (appox) [option ‘C’]