# QUESTIONS ON GEOMETRY (PART-4)

#### QUESTIONS ON GEOMETRY (PART-4)

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 31

**In this figure if DE/BC = 1/4 and area ΔADE = 10; find area ΔDEC.**

**MAHA GUPTA**

Considering it a similar triangle case, as it looks, areaADE/areaDEC = AE/EC [According to similar triangle property DE/BC AE/EC]

=> 10/area ΔDEC = AE/(AC – AE) = 1/(4 – 1) = 1/3

=> area ΔDEC = 30 (answer)

QUERY 32

**The bisector of the ∠BAC of ΔABC intersects the side BC at the point D and meets the circumcircle of the ΔABC at E. Then, when it’s always true AB × AC + DE × AE = **

A) AD²

A) AE²

C) CE²

D) CD²

**MAHA GUPTA**

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, ΔABD and ΔAEC are similar triangles (AA property)

hence AB/AD = AE/AC

=> AB.AC = AE.AD

Adding DE.AE to both sides

AB.AC + DE.AE = AE.AD + DE.AE

=> AB.AC + DE.AE = AE(AD + DE)

=> AB.AC + DE.AE = AE. AE

=> AB.AC + DE.AE = AE² (option ‘B’)

QUERY 33

**In a right angle triangle ABC, AD is perpendicular to hypotenuse BC. If AC = 2AB, then the value of BD will be?**

A) BC/2

B) BC/3

C) BC/4

D) BC/5

As triangle BAC is a right angle triangle with BC its hypotenuse

Therefore. BC² = AB² + AC²

=> BC² = x² + (2x)²

=> BC = √5x

Also both triangles ADB and ADC are right triangles in which AB and AC are their hypotenuse respectively;

So in ΔADB, AD = AB² – BD²

And in ΔADC, AD = AC² – CD²

Therefore, AB² – BD² = AC² – CD²

=> x² – BD² = (2x)² – (√5x – BD)²

=> BD = x/√5

=> BD = √5x/5

=> BD = BC/5 (option ‘D’)

#### QUERY 34

**In ΔABC the side BC is extended up to D such that CD = AC. If ∠BAD = 109 and ∠ACB = 72, then the value of ∠ABC = ?**

A) 35°

B) 60°

C) 40°

D) 45°

**MAHA GUPTA
**In ΔACD, AC = CD; therefore ∠CAD = ∠CDA

Also ∠ACB is an exterior angle of this triangle

As an exterior angle of a triangle is equal to the sum of the two opposite (non-adjacent) interior angles; therefore ∠ACB = ∠CAD + ∠CDA

=> ∠CAD + ∠CDA = 72

But both are equal angles; therefore ∠angle CAD = ∠CDA = 72/2 = 36

Now ∠ABD = 180 – (∠BAD + ∠CDA) = 180 – (109 + 36) = 35

Hence ∠ABC = 35 (both ∠ABD and ∠ABC are the same angle) [option ‘A’]

QUERY 35

**ABC is a triangle with AC = BC and ∠ABC = 50°. Side BC is produced to D so that BC = CD. Find ∠BAD.**

A) 50°

B) 45°

C) 75°

D) 90°

In ΔABC, AC = BC, therefore ∠ABC = ∠BAC = 50

As an exterior angle of a triangle is equal to the sum of the two opposite (non-adjacent) interior angles; therefore ∠ACD = ∠ABC + ∠BAC

=> ∠ACD = 50 + 50 = 100

Now, in ΔACD, ∠DAC + ∠CDA = 180 – ∠ACD = 180 – 100 = 80

Also, AC = CD, as AC = BC

Hence, ∠DAC = ∠CDA

=> ∠DAC = 80/2 =40

Now ∠BAD = ∠BAC + ∠DAC = 50 + 40 = 90 (option ‘D’)

QUERY 36

**In ΔABC, P is a point on AB such that ∠ACP = ∠ABC. If AC = 9 cm, CP = 12 cm and BC = 15 cm, find AP.**

A) 11.2 cm

B) 10.2 cm

C) 8.0 cm

D) 7.2 cm

In triangles ACP and ABC

∠A = ∠A (common angles)

and ∠ACP = ∠ABC (given)

Therefore by AA, both the triangles are similar

Hence AP/AC = CP/BC (Corresponding sides of similar triangles are in the same ratio)

Therefore, a/9 = 12/15

=> a = 7.2 cm (option ‘D’)

**SHIV KISHOR
**Let ∠ACP = ∠ABC = α

And ∠CAB = p, also AP = a

in ΔAPC, a/sinα = 12/sinp —(i)

in ΔABC, 9/sinα = 15/sinp —(ii)

Dividing (i) by (ii)

a/9 = 12/15

=> a = 7.2 (option ‘D’)

#### QUERY 37

**In ΔABC, ∠A : ∠B : ∠C = 2 : 3 : 4. A line CD is drawn || to AB, then ∠ACD is?**

A) 40°

B) 60°

C) 80°

D) 20°

∠A : ∠B : ∠C = 2 : 3 : 4

=> ∠A = 40, ∠B = 60, ∠C = 80

Now CD || AB, means ∠BAC = ∠ACD (alternate angles)

Hence ∠ACD = 40 (option ‘A’)

QUERY 38

**Inside a square ABCD, ΔBEC is an equilateral triangle. If CE and BD intersect at O, then ∠BOC = ?**

A) 60

B) 75

C) 90

D) 120

ΔBEC is an equilateral triangle, therefore ∠BCE = 180/3 = 60

As a diagonal of the square cuts an angle of the square into two equal parts, ∠CBD = 90/2 = 45

Hence, ∠BOC = 180 – (45+60) = 75 (option ‘B’)

QUERY 39

**The ratio of the areas of two isosceles triangles having equal vertical angles is 1 : 4. The ratio of their heights will be?**

A) 1 : 2

B) 3 : 4

C) 2 : 3

D) 6 : 7

**MAHA GUPTA**

Height of an isosceles triangle cuts that triangle in two similar triangles, each being similar to to the 0rignal triangle. So half of the smaller triangle will be a similar triangle to the half of the bigger triangle as both the triangles have same vertices in which both the heights will be corresponding sides of those two triangles while the ratio of their areas being the same as of the original triangles.

We also know that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding sides.

Now let the height of the smaller triangle = h

And the height of the bigger triangle = H

Therefore, ^{1}⁄_{4} = ^{h²}⁄_{H²}

=> ^{h}⁄_{H} = √^{1}⁄_{4} = ^{1}⁄_{2 }(option ‘A’)

QUERY 40

**The angles of an acute angle triangle are all distinct integers. If the smallest angle has the largest possible value, find the sum of two other angles.**

A) 60°

B) 61°

C) 121°

D) 125°

**MAHA GUPTA**

If all the angles are distinct integers and the smallest angle has the largest possible value, they must be consecutive integers.

Letting the smallest of them being ‘x’

Therefore, x + (x +1) + (x +2) = 180

=> x = 59

Therefore other angles = 59 +1 = 60, and 59 +2 = 61

Hence their sum = 60 + 61 = 121 (option ‘C’)** **