Geometry Questions and answers

Topics

General Knowledge
General Science
General English
Aptitude
General Computer Science
General Intellingence and Reasoning
Current Affairs
Exams
Elementary Mathematics
English Literature

Elementary Mathematics

11). The side AC of a \( \Large \triangle ABC \) is extended to D such that BC = CD. If \( \Large \angle ACB \) is \( \Large 70 ^{\circ} \), then what is \( \Large \angle ADB \) equal to?

A). \( \Large 35 ^{\circ} \)

B). \( \Large 45 ^{\circ} \)

C). \( \Large 70 ^{\circ} \)

D). \( \Large 110 ^{\circ} \)

View Answer

Correct Answer: \( \Large 35 ^{\circ} \)

\( \Large \angle ACB + \angle BCD = 180 ^{\circ} [linear\ pair] \)

\( \Large \angle BCD = 180 ^{\circ} - 70 ^{\circ} = 110 ^{\circ} \)

In \( \Large \triangle BCD, \)

\( \Large BC = CD \)

\( \Large \angle CBD = \angle CDB \) ...(i)

[angles opposite to equal sides]

Also, \( \Large \angle BCD + \angle CBD + \angle CDB = 180 ^{\circ} \)

\( \Large 2 \angle CDB = 180 ^{\circ} - \angle BCD \)

= \( \Large 180 ^{\circ} - 110 ^{\circ} = 70 ^{\circ} \)

\( \Large \therefore \angle CDB = \angle ADB \)

= \( \Large \frac{70 ^{\circ} }{2} = 35 ^{\circ} \)

12). E is the mid-point of the median AD of a \( \Large \triangle ABC \). If BE is extended it meets the side AC at F, then CF is equal to

A). \( \frac{AC}{3} \)

B). \( \frac{2 AC}{3} \)

C). \( \frac{AC}{2} \)

D). None of these

View Answer

Correct Answer: \( \frac{2 AC}{3} \)

Draw line segment DG parallel to BF

Then in, \( \Large \triangle ADG, \)

\( \Large EF \parallel DG \)

and AE = ED

\( \Large \therefore AF = GC \) ...(i)

Similarly, in \( \Large \triangle BCF, \)

\( \Large DG \parallel BF \)

and BD = DC

\( \Large \therefore FG = GC \) ...(ii)

From Eqs. (i) and (ii),

\( \Large CF = \frac{2}{3}AC \)

13). Consider the following statements
I. If G is the centroid of \( \Large \triangle ABC \), then GA = GB = GC.
II. If H is the orthocentre of \( \Large \triangle ABC \), then HA = HB = HC.
Which of the statements given above is/are correct?

A). Only I

B). Only II

C). Both I and Il

D). Neither I nor ll

View Answer

Correct Answer: Neither I nor ll

GA = GB = GC is true only and only for equilateral triangle and here it is not given that ABC is an equilateral triangle. So, statement I is not correct. Similarly, statement II will also hold only for equilateral triangle. Hence, it is also not correct.

14). The three sides of a triangle are 15, 25 and x units. Which one of the following is correct?

A). 10 < x < 40

B). \( \Large 10 \le x \le 40 \)

C). \( \Large 10 \le x < 40 \)

D). \( \Large 10 < x \le 40 \)

View Answer

15). If AD is the internal angle bisector of \( \Large \triangle ABC \) with AB = 3 cm and AC = 1 cm, then what is BD : BC equal to?

A). 1: 3

B). 1: 4

C). 2 : 3

D). 3: 4

View Answer

Correct Answer: 3: 4

\( \Large \triangle ABC, AD is\ the\ internal\ angle\ bisector\ of\ \angle A.\\ Using\ property\ of\ internal\ angle\ bisector. \)

\( \Large \frac{BD}{CD} = \frac{AB}{AC} \)

=> \( \Large \frac{CD}{BD} = \frac{AC}{AB} \)

=> \( \Large \frac{CD}{BD}+1 = \frac{AC}{AB}+1 \)

=> \( \Large \frac{CD + BD}{BD} = \frac{AC + AB}{AB} \)

=> \( \Large \frac{BC}{BD} = \frac{3+1}{3} => \frac{BD}{BC} = \frac{3}{4} \)

\( \Large \therefore BD : BC = 3 : 4 \)

16). In a \( \Large \triangle ABC \), if \( \Large \angle A \) = 115 degree, \( \Large \angle C \) = 20 degree and D is a point on BC such that \( \Large AD \perp BC \) and BD = 7 cm, then AD is of length

A). 15 cm

B). 5 cm

C). 7 cm

D). 10 cm

View Answer

Correct Answer: 7 cm

\( \Large \triangle ABC \)

\( \Large \angle A = 115 ^{\circ}, \angle C = 20 ^{\circ} \)

\( \Large \therefore \angle B = 180 ^{\circ} - \left(115 ^{\circ} +20 ^{\circ} \right) = 45 ^{\circ} \)

Now, in \( \Large \triangle ABD \)

\( \Large \frac{AD}{BD} = \tan 45 ^{\circ} \)

=> AD = BD = 7 cm

17). In \( \Large \triangle ABC \), \( \Large DE \parallel BC \), where DE intersects AB and AC at the points D and E, respectively. If AD = 6 cm, DB = 12x - 6 cm, AE = 2x cm and CE = 16 - 2x cm, then the value of x is

A). 6 cm

B). 4 cm

C). 2 cm

D). 8 cm

View Answer

Correct Answer: 2 cm

Given. \( \Large DE \parallel BC \)

Since, \( \Large \triangle ADE - \triangle ABC \)

If any line parallel to one side of triangle divides the other two sides proportionally

Then, \( \Large \frac{AB}{AD} = \frac{AC}{AE} \)

=> \( \Large \frac{12x}{6} = \frac{16}{2x} \)

=> \( \Large x^{2} = \frac{16 \times 6}{12 \times 2} = 4 \)

\( \Large x = 2 cm \)

18). In \( \Large \triangle ABC \), D and E are points on sides AB and AC, such that \( \Large DE \parallel BC \). If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, then the value of x is

A). 4

B). 2

C). 1

D). 8

View Answer

19). In \( \Large \triangle ABC \), AB = AC and D is a point on AB, such that AD = DC = BC. Then \( \Large \angle BAC \), is

A). \( \Large 40 ^{\circ} \)

B). \( \Large 45 ^{\circ} \)

C). \( \Large 30 ^{\circ} \)

D). \( \Large 36 ^{\circ} \)

View Answer

Correct Answer: \( \Large 36 ^{\circ} \)

Given that AB = AC and AD = CD = BC

\( \Large \angle ABC = \theta \)

Then, \( \Large \angle ACB = \theta \) [\( \Large \because AB = AC \)]

=> \( \Large \angle BAC = 180 ^{\circ} - 20 \)

=> \( \Large \angle ACD = 180 - 20 \) [\( \Large \because AD = CD \)]

\( \Large \angle BCD = \angle AC B - \angle ACD \)

=> \( \Large \angle BCD = \theta - \left(180 ^{\circ} - 20 \right) = 30 - 180 ^{\circ} \)

and \( \Large \angle BDC = \theta \) [\( \Large \because CD = BC \)]

Now, in \( \Large \triangle BCD \)

\( \Large \angle CBD + \angle BDC + \angle BCD = 180 ^{\circ} \)

=> \( \Large \theta + \theta + 30 - 180 ^{\circ} = 180 ^{\circ} \)

=> \( \Large 50 = 360 ^{\circ} => \theta = 72 ^{\circ} \)

\( \Large \therefore \angle BAC = 180 ^{\circ} - 20 \)

= \( \Large 180 ^{\circ} - 144 = 36 ^{\circ} \)

20). In a \( \Large \triangle ABC \), \( \Large \angle A \) : \( \Large \angle B \) : \( \Large \angle C \) = 2 : 3 : 4. A line CD drawn parallel to AB, then \( \Large \angle ACD \) is

A). \( \Large 80 ^{\circ} \)

B). \( \Large 20 ^{\circ} \)

C). \( \Large 40 ^{\circ} \)

D). \( \Large 60 ^{\circ} \)

View Answer

Correct Answer: \( \Large 40 ^{\circ} \)

Let the angles be 2x, 3x and 4x.

Then. \( \Large 2x + 3x + 4x = 180 ^{\circ} \)

\( \Large 9x = 180 ^{\circ} => x = 20 ^{\circ} \)

\( \Large \therefore \angle A = 2x = 2 \times 20 = 40 ^{\circ} \)

\( \Large \angle B = 3x = 3 \times 20 = 60 ^{\circ} \)

\( \Large \angle C = 4x = 4 \times 20 = 80 ^{\circ} \)

Now, \( \Large AB \parallel CD\ and\ AC\

be\ the\ transversal. \)

Then, \( \Large \angle BAC = \angle ACD \)

[alternate interior angles]

\( \Large \therefore \angle ACD = 40 ^{\circ} \)

1	2	3	4	5	6	7

Go to :

Total Pages : 53