# QUESTIONS ON GEOMETRY (PART-5)

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#### QUESTIONS ON GEOMETRY (PART-5)

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 41

**How many sides does a regular polygon have whose interior and exterior angle are in the ratio of 2:1.**

A) 3

B) 5

C) 6

D) 12

**MAHA GUPTA**

Interior and exterior angles are always supplementary i.e. INTERIOR ANGLE + EXTERIOR ANGLE = 180

Their ratio is given 2 : 1

=> EXTERIOR ANGLE of the polygon = 180*(1/3) = 60

But sum of exterior angles of a polygon is always 360°

Therefore the number of sides = 360/60 = 6 (option ‘C’)

QUERY 42

**Taking any three of the line segments out of segments of length 2, 3, 5, 6 cm; the number of triangles formed is?**

A) 4

B) 3

C) 2

D) 1

**MAHA GUPTA**

We just have to keep in mind here that at each given condition the sum of two sides of a triangle is always greater than the third

So combinations of 2, 5, 6 and 3, 5, 6 only are possible

Hence 2 (option C’) is the answer.

QUERY 43

**In ΔABC, X and Y are points on sides AB and BC respectively such that XY || AC and XY divides the triangular region ABC into two parts equal in area; then AX/AB is equal to?**

A) (2 + √2)/2

B) (√2 + 3)/2

C) (2 – √2)/2

D) (3 – √2)/2

**MAHA GUPTA**

Draw the figure

XY || AC, therefore ΔABC is similar to ΔXBY

so area ΔABC/area ΔXBY = (AB/XB)² [The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides] —-(i)

Also area ΔABC = 2 area ΔXBY [given]

So, Area ΔABC/area ΔXBY = 2/1 —-(ii)

From (i) and (ii)

(AB/XB)² = 2/1

=> AB/XB = √2/1

=> XB/AB = 1/√2

=> 1 – XB/AB = 1 – 1/√2

=> (AB – XB)/AB = (√2 – 1)/√2

=> AX/AB = (√2 – 1)/√2 = (2 – √2)/2 (option ‘C’)

QUERY 44

**POLYGON and THINGS RELATED**

**MAHA GUPTA**

1. Polygon is a closed plane figure bounded by straight lines; e.g.

Triangle (3 lines)

Quadrilateral (4 lines)

Pentagon (5 lines)

Hexagon (6 lines

Heptagon (7 lines)

Octagon (8 sides)

Nonagon (9 lines)

Decagon (10 sides)

2. REGULAR POLYGON

If a polygon is regular its all sides/interior angles are equal.

3. EACH EXTERIOR ANGLE OF A REGULAR POLYGON of ‘n’ sides

(360/n)°

4. EACH INTERIOR ANGLE OF REGULAR POLYGON OF ‘n’ SIDES

180 – exterior angle; if exterior angle is not given it’s [180(n – 2)]/2

5. SUM OF ALL EXTERIOR ANGLES OF A POLYGON

360°

6. SUM OF ALL INTERIOR ANGLES OF A POLYGON of ‘n’ sides

(n – 2)180

7. NUMBER OF DIAGONALS OF A POLYGON OF ‘n’ SIDES

n(n – 3)/2

QUERY 45

**ΔPQR circumscribes a circle with centre O and radius x cm such that ∠PQR is 90. If PQ = 3 and QR=4; then value of x?**

A) 2

B) 1

C) 1.5

D) 2.5

**MAHA GUPTA**

As one of the angles (here ∠PQR) is 90 of a triangle and circumscribing a circle, means the side opposite the 90° angle is the diameter of the circle. And obviously it’s the hypotenuse of that triangle also.

By Pythagoras

3² + 4² = diameter²

=> diameter = √25 = 5

hence x (radius) = 5/2 = 2.5 (option ‘D’)

QUERY 46

**ΔABC is right angled at B. AB = 7, AC = 25 and D is a point on BC such that AD is the bisector of ∠A, what is the length of AD?**

A) 9

B) 8.75

C) 12.5

D) 13

**MAHA GUPTA**

In Triangle AC is the hypotenuse; therefore BC = √(AC² – AB²) = √(25² – 7²) = 24

Now let BD = x; hence DC = 24 – x

According to the bisector angle theorem

7/x = 25/(24 – x)

=> x = 21/4

Now in ΔABD; AD is the hypotenuse, other sides being 7 and 21/4

Hence AD = √[7² + (21/4)²] = 35/4 = 8.75 (option ‘B’)

QUERY 47

QUERY 47

**The sides of a triangle are 5, 13 and 10 cm. The radius of its in-circle is?**

A) (3√14)/7 cm

B) (2√14)/7 cm

C) (3√14)/8 cm

D) √14/7 cm

**MAHA GUPTA**

Radius of the in-circle of a triangle = (2*area of the triangle)/Perimeter

Now the perimeter of the triangle = 5+13+10 = 28 cm

Using Heron’s formula the area of the triangle = √[s(s -a)(s – b)(s -c)]; where s = semi-perimeter and a, b, c are sides of the triangle

= √[14(14 – 5)(14 – 13)(14 – 10)]

= 6√14

Hence the radius of the in-circle = (2*6√14)/28 = (3√14)/7 (option ‘A’)

QUERY 48

**The sides of a triangle are 6 cm, 8 cm and 10 cm. The area of greatest square that can b inscribed in it is?**

A) 576/49 cm²

B) 15 cm²

C) 24/7 cm²

D) 500/49 cm²

**MAHA GUPTA**

Sides 6, 8 and 10 means it’s a right angle triangle as 10² = 6² + 8²

The side of the square inscribed in a right triangle = product of the smaller sides/sum of the smaller sides

so side of the square = (6*8)/(6+8) = 24/7

Therefore area of the square = (24/7)² = 576/49 (option ‘A’)

QUERY 49

**If the altitudes of a triangle are 20, 28 and 35. What will be the area of the triangle?**

A) 1231√6/6

B) 1225√6/6

C) 1231√5/5

D) 1231√3/6

**MAHA GUPTA**

Area of the triangle when altitudes are given = [(LCM of altitudes/4)²](√6/6)

Now LCM of 20, 28, 35 = 140

So, the area of the triangle = [(140/4)²](√6/6) = 1225√6/6 (option ‘B’)

QUERY 50

**There is a ΔABC in which AD, BE and CF are the perpendiculars drawn on opposite sides and they intersect each other at O. If ∠BOC=65, find ∠BAC.**

A) 100

B) 95

C) 115

D) 105

**MAHA GUPTA**

The intersection of the three lines which passes through the 3 vertices of a triangle and perpendicular to opposite 3 sides is called orthocenter.

Therefore, this way we’ll have a quadrilateral AFOE in which angle AFO and angle AEO will be 90° each because of perpendiculars.

Now ∠BOC = 65 given. But it’s vertically opposite angle of ∠FOE of quadrilateral AFOE; therefore ∠FOE = ∠BOC = 65

Hence, the fourth angle of this quadrilateral, FAE = 360 – (90+90+65) = 115.

But ∠FAE and ∠BAC are same, so ∠BAC = 115 (option ‘C’)