# QUESTIONS ON GEOMETRY (PART-2)

#### QUESTIONS ON GEOMETRY (PART-2)

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 11

**If O is the circumcentre of ΔABC and ΔOBC= 35°; then ∠BAC = ?**

A) 55

B) 110

C) 70

D) 35

**Amit jha**

The circumcentre is at the intersection of the perpendicular bisectors of the triangle’s sides. It is a point in a triangle from which all vertices of that triangle are equidistant.

So in the given triangle OA = OB = OC

Now, as OB and OC are equal in ΔOBC, hence it is an isosceles triangle.

As in ΔOBC, ∠OBC= 35, therefore ∠OCB = 35° (both are equal angles of ΔOBC)

Therefore the third angle of the ΔOBC; ∠BOC = 180 – (35 +35) = 110°

Also remember that an angle subtended by a side on the circumcentre is double of angle subtended by that side on opposite vertex of the triangle..

Therefore ∠A = half of ∠BOC = 110/2 = 55° (option ‘A’)

QUERY 12

**ABC is a triangle and its medians CD and BE intersect each other at M; then find the ratio of area of ∠MDE to the area of ∠ABC**

A) 1:3

B) 1:12

C) 1:6

D) 1:9

**Vijay Bharath Reddy
**As CD and BE are medians, D and E are the mid-points of AB and AC respectively

Therefore DE = BC/2

Let height of ∠ABC be ‘h’, then height of ∠ADE = h/2 (by similarity of triangles)

and height of ∠MBC = h/3 (since median divides heights in the ratio 2:1)

Therefore, height of ∠MDE = (h – (h/2 + h/3) = h/6

Now, area of ∠MDE = (1/2)*DE*h/6 = (1/2)*(BC/2)*h/6 = (1/12)*(1/2*BC*h) = (1/12)*area of ∠ABC

So the required ratio is 1 : 12 (option ‘B’)

QUERY 13

**The bisector of angle A of ΔABC cuts BC at D and the circumcircle of the triangle at E; then**

A) AB:AC = BD: DC

B) AD:AC = AE: AB

C) AB:AD = AC: AE

D) AB:AD = AE: AC

**Vijay Bharath Reddy
**Consider ΔABD and ΔAEC

∠BAD = ∠EAC (since AE is bisector of ∠A)

∠ABD = ∠AEC (angles made by chord AC in the same segment of a circle)

therefore, ABD and AEC are similar triangles (AA property)

hence AB/AE = AD/AC

=> AB:AD = AE:AC (option ‘D’)

QUERY 14

**In a 8 cm equilateral triangle two other equilateral triangles are made from its centroid so that they divide the main equilateral triangle in three equal areas; then find the length of side of the smaller triangle.**

A) 4

B) 2√2

C) 8/√3

D) 8/3

**Vijay Bharath Reddy
**Area of the smaller triangle = (1/3)area of the bigger triangle

= ⅓[(√3/4)a²]; where ‘a’ is the side of the bigger triangle

= ⅓[(√3/4)*8²]

= (16/3)√3

Side of an equilateral triangle = √(4area/√3)

Now letting the side of smaller triangle as ‘t’

t = √[4{(16/3)√3}/√3]

=>√[{(64/3)√3}/√3]

=> t = √64/3

=> 8/√3 (option ‘C’)

QUERY 15

**What is the difference between the areas of the circumcircle and incircle of a regular polygon of length ‘2a’ as its side.**

A) πa²

B) a²

C) 2a²

D) 4a²

**Vijay Bharath Reddy
**Let ‘R’ be the radius of circumcircle (line from centre to vertex) and ‘r’ be the radius of incircle (perpendicular line from centre to the side)

When you draw the figure, you’ll, thus, find a right angle triangle with ‘R’ as the hypotenuse and ‘r’ & ‘a’ as two of its sides.

Therefore according to Pythagoras, R² = r² + a²

=> R² – r² = a²

As ‘R’ is the radius of the circumcircle and ‘r’ being the radius of the incircle, their areas respectively = πR² and πr²

Thus the difference in areas = πR² – πr²

=π(R² – r²)

= πa² (option ‘A’)

QUERY 16

**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) 26.25

**Vijay Bharath Reddy
**Area of a triangle circumscribed by a circle = abc/(4r), where a, b, c are sides of the triangle & r the radius of the circle.

Also area of a triangle = base*height/2

Therefore according to the question

17.5*9*BC/(4r) = BC*3/2

=> r = 26.25 (option ‘D’)

QUERY 17

**Medians of a triangle are 18 cm, 24 cm and 30 cm. Find its area.**

A) 288 cm²

B) 480 cm²

C) 216 cm²

D) 360 cm²

**SANDEEP SINGH SONI
**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. 18, 24 & 30)

Here we see that these sides are making a right triangle

Therefore the area of the given triangle = (4/3)×[½×Product of the smaller sides]

= (4/3)×½×18×24

= 288 cm² (option ‘A’)

#### QUERY 18

**In the given figure AD is the bisector of angle A, find AC.
**

A) 5 cm

B) 5.5 cm

C) 4.5 cm

D) 5.25 cm

**MAHA GUPTA
**Here the ANGLE BISECTOR OF TRIANGLE theorem will apply. According to this theorem an angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

Therefore, 3/4 = AC/6

=> AC = 18/4 = 4.5 cm (option ‘C’)

#### QUERY 19

**In ΔABC, ∠A=90, AD is perpendicular to BC whereas D lies on BC. If BC = 8, AC= 6, then area of ΔABC : area of ΔDAC ?**

A) 16:9

B) 25:9

C) 4:3

D) 25:16

**RONNIE BANSAL
**In triangles ABC and DAC

∠A = ∠D (both given right angles)

∠C = ∠C (Common angle to both the triangles)

So by AA both the triangles ABC and DAC are similar

If carefully seen side BC of ΔABC and side AC are corresponding ones

We know in two similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides.

Hence area of ΔABC : area of ΔDAC = BC² : AC²

= 8² : 6²

= 64 : 36

= 16 : 9 (option ‘A’)

#### QUERY 20

**If O is the orthocenter of the ΔABC and ∠BAC = 80, then measure of ∠BOC is?**

A) 120

B) 90

C) 80

D) 100

**Atul Roy**

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

Now let the line CD through O meets AB at D and and BE through O meets AC at E This way we’ll have a quadrilateral ADOE in which ∠ADO and ∠AEO will be 90 each; and ∠DAE = ∠BAC = 80 (given)

Therefore ∠DOE = 180 – 80 = 100

But ∠DOE and ∠BOC are opposite angles

Hence ∠BOC = ∠DOE = 100 (option ‘D’)