# QUESTIONS ON GEOMETRY (PART-1)

#### QUESTIONS ON GEOMETRY (PART-1)

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 1

**Two medians AD and BE of triangle ABC intersect at G at right angle. If AD = 9 cm and BE = 6 cm, then BD is ?**

A) 2

B) 3

C) 4

D) 5

**Libnitz
**Medians of a triangle always intersect in the ratio of 2 : 1

So AG : DG = 2 : 1

But AD = 9 cm (given)

Therefore DG = 9*/(1/3) = 3 cm

Same way, BG = 4 cm

Now as ΔBDG is a right triangle (medians cut at right angle)

BD= √(4² + 3²) = 5 (Using pythagoras) (option ‘D’)

QUERY 2

**G is the centroid of Triangle ABC, and AG=BC. Find angle BGC.
**

A) 45

B) 30

C) 90

D) 80

**RONNIE BANSAL**

We know the centroid of a triangle is the intersecting points of all medians of that triangle; and it divides a median in the ratio of 2 : 1, greater side of the median being attached to its vertex.

Now draw the figure accordingly in which median from A meets the sides BC say at D. Concentrate on the ΔBGC in the figure as we have to find the ∠BGC. In this triangle again we have two triangles namely BGD & CGD. Now we have to just give our all attention to the sides BC and GD as AG is given equal to BC.

As AD is the median, GD=half of AG (since the centroid cuts a median in the ratio of 2 : 1. Hence GD will be half of BC also as AG=BC (given). But point D is the midpoint of BC (since AD is the median). Therefore GD=BD=CD. Means both triangles BGD & CGD are isosceles triangles in which BD=GD & CD=GD respectively..

Therefore ∠DBG=∠BGD and ∠DCG=∠CGD. But ∠BGD+∠CGD=∠BGC; hence

∠BGC=∠DBG+∠DCG. Now if you’ll think over it a bit you will find ∠BGC = 90 (option ‘C’)

QUERY 3

**If a regular polygon has each of its angles equal to 3/5 times of two right angles, then the number of sides is ?**

A) 5

B) 6

C) 7

D) 8

**Lokesh Shravan**

Here each angle (internal) of the polygon is (3/5)*2*90 = 108

So external angle is 180 – 108 = 72

So the number of sides = 360/72 = 5 (option ‘A’) —– [Number of sides of a polygon = 360/external angle]

QUERY 4

**The two diagonals of a rhombus are of lengths 55 cm and 48 cm. If P is the perpendicular height of the rhombus then which is correct?**

A) 36 < P < 37

B) 35 < P < 36

C) 34 < P < 35

D) 33 < P < 34

**RONNIE BANSAL**

Let ABCD be the rohombus with B and D as obtuse angles and perpendicular from D meets AB at E (internally). Let diagonals AC and BD meet at O.

Both triangles AOB and DEB are right triangles (as perpendiculars of a rhombus bisect each other at right angles; and DE is perpendicular on the side AB. Also angle B is common to both. Therefore triangles AOB and DEB are similar.

Being such AO/DE = AB/DB

therefore DE = (AO*DB)/AB

As triangle AOB is right angled so AB = √(AO² + BO²) = 73/2

Hence DE i.e. P = (55/2)*48/(73/2) = 36.164 that gives option ‘A’ (36 < P < 37) (0ption ‘A’)

QUERY 5

**D is a point on the side AC of triangle ABC. P, Q, X, Y are mid-points of AB, BC, AD, DC. What will be the ratio of PX and QY?**

A) 1:1

B) 1:2

C) 2:1

D) 2:3

**RONNIE BANSAL**

In ΔABD, P and X are mid points of AB and AD. So, PX || BD and PX = BD/2

Similarly in ΔCBD, QY || BD and QY = BD/2

Therefore, PX = QY

Hence option (1 : 1) is correct (option ‘A’)

QUERY 6

**The ratio of the areas of the two isosceles triangles having the same vertical angle is 1:4. The ratio of their heights is ?**

A) 1:4

B) 1:2

C) 4:1

D ) 2:1

**KANW@LJEET
**Let their heights and bases be h1, h2 and b1, respectively

Therefore [(1/2)*b1*h1]/[(1/2)*b2*h2] = 1/4

or (b1/b2)*(h1/h2) = 1/4………………………eq1

but as the height of triangle increases, its base will also increase in same proportion

ie, (b1/b2) = (h1/h2)

putting this value in eq1 (h1/h2)^{2} = 1/4

or h1/h2 = 1/2 (option ‘B’)

**Amul Baby
**Vertical angles of both the triangles are equal (given)

As both the triangles are isosceles, therefore other two angles of each triangle are also equal to the angles of the 2nd triangle.

Hence according to AAA property both the triangles are similar.

We know that when triangles are similar then the ratio of their heights = ratio of the square root of areas

As the ratio of the areas of the triangles is 1 : 4; letting areas 1 and 4 units respectively Ratio of heights = √(1/4) = 1/2 = 1: 2 (option ‘B’)

#### QUERY 7

A) 5

B) 7

C) 6

D) 6.5

**MAHA GUPTA**

∠A = ∠CED (given)

∠C = ∠C (common)

Therefore by AA similarity ΔABC is similar to ΔEDC, in which sides AB and ED are corresponding, sides BC and DC are corresponding and sides CA and CE are corresponding.

Hence CA/CE = AB/ED

=> 15/10 = 9/x

=> x = 6 (option ‘C’)

QUERY 8

**The difference between the radii of the circular ends of a bucket is 20 cm. If height of the bucket is 15 cm, find its slant height. **

A) 30 cm

B) 25 cm

C) 15 cm

D) 20 cm

**JAYANTCHARAN CHARAN**

If observed carefully we’ll get a right triangle in the end, with 20 cm as base and 15 cm as height.

The smallest pythagoras triplet is 3, 4, 5

The question is simply an extension to the same.

We can also write this triplet like this:

3 x 5, 4 x 5, 5 x 5 (multiplying each side by 5)

= 15, 20, 25

Now its clear from above that the longest side will be the desired result; so answer is 25 (option ‘B’)

QUERY 9

**In ΔABC, P & Q are mid points of sides AB & AC respectively. R is a point on the segment PQ such that PR : QR = 1 : 2. If PR = 2 cm then BC = ?**

A) 12 cm

B) 6 cm

C) 10 cm

D) 8 cm

**Amit jha**

Given PR : QR = 1 : 2 and PR = 2 cm

=> QR = 4 cm; and PR + RQ = PQ = 6 cm

Remember if P and Q are the mid points of two sides (here AB and AC) of a triangle then the side joining them i.e. PQ must be half of the third side (here BC)

Hence in this question BC = 2PQ = 2*6 = 12 cm (option ‘A’)

QUERY 10

**The lengths of two adjacent sides of a parallelogram are 5 cm and 3.5 cm respectively. One of its diagonals is 6.5 cm long, the area of the parallelogram is?**

A) 8√3 cm²

B) 12√3 cm²

C) 15 cm²

D) 10√3 cm²

**MAHA GUPTA**

The two adjacent sides of the parallelogram and its one diagonal will make a triangle; and that triangle will be half the area of the parallelogram, as a diagonal of a parallelogram cuts it into two equal triangles.

Now length of the three sides of the triangle thus got = 5 cm, 5.3 cm and 6 cm

Now its area = √[s(s – a)(s- b)(s – c)]; where s = (1/2)(a+b+c) and a, b, and c are the three sides of the triangle.

s = (1/2)(5+3.5+6.5) = 7.5 cm

Hence the area of that triangle = √[7.5(7.5 – 5)(7.5 – 3.5)( 7.5 – 6.5) = √75 = 5√3

But the area of the parallelogram is twice the area of that triangle ………explained above

So, its area = 2*5√3 = 10√3 cm² (option ‘D’)

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