# CRE 1 - Lines & Angles | Geometry - Basics

**CRE 1 - Lines & Angles | Geometry - Basics**

What is the measure of the complementary angle, if the measure of an angle is 57°?

- A.
55°

- B.
33°

- C.
45°

- D.
37°

Answer: Option B

**Explanation** :

Two angles are said to be complementary if the sum of their measures is 90°.

∴ Measure of the complementary angle = 90° − 57° = 33°

Hence, option (b).

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**CRE 1 - Lines & Angles | Geometry - Basics**

What is the measure of the supplementary angle, if the measure of an angle is 57°?

- A.
43°

- B.
133°

- C.
12°

- D.
123°

Answer: Option D

**Explanation** :

Two angles are said to be supplementary if the sum of their measures is 180°.

∴ Measure of the supplementary angle = 180° − 57° = 123°

Hence, option (d).

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**CRE 1 - Lines & Angles | Geometry - Basics**

A line which cuts two or more parallel lines is known as a

- A.
Perpendicular

- B.
Line bisector

- C.
Transversal

- D.
Perpendicular bisector

Answer: Option C

**Explanation** :

A line that cuts two or more parallel lines is known as a transversal.

Hence, option (c).

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**CRE 1 - Lines & Angles | Geometry - Basics**

Which of the following is a reflex angle?

- A.
30°

- B.
120°

- C.
180°

- D.
230°

Answer: Option D

**Explanation** :

A reflex angle is an angle that is greater than 180° but less than 360°.

Hence, 230° is a reflex angle.

Hence, option (d).

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**CRE 1 - Lines & Angles | Geometry - Basics**

Two lines intersect each other and make four angles. Of these, any two adjacent angles are

- A.
Equal

- B.
Complementary

- C.
Supplementary

- D.
None of these

Answer: Option C

**Explanation** :

When two lines intersect each other, any two adjacent angles (of the four angles) make a linear pair and are always supplementary to each other.

Here, ∠1 and ∠2 are adjacent angles. Since they form a linear pair of angles, they are supplementary.

Hence, option (c).

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**CRE 1 - Lines & Angles | Geometry - Basics**

It is given that, m ∠A + m ∠B + m ∠C = 210° and m ∠B is thrice that of m∠C. If ∠B and ∠C are supplementary, the measures of ∠A, ∠B and ∠C respectively are

- A.
30°, 135°, 45°

- B.
30°, 120°, 60°

- C.
45°, 110°, 55°

- D.
30°, 108°, 36°

Answer: Option A

**Explanation** :

m ∠B + m ∠C = 180° (Supplementary angles) … (1) and

m ∠A + m ∠B + m ∠C = 210° …(2)

Solving (1) and (2), we get

∴ m ∠A = 210° − 180° = 30°

Now, m ∠B = 3 × (m ∠C) …(3)

Solving (1) and (3), we get

∴ m ∠B = 135° and m ∠C = 45°

Hence, option (a).

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**CRE 1 - Lines & Angles | Geometry - Basics**

Measure of the supplementary angle of ∠A is 100° more than that of the complementary angle. What is the type of ∠A?

- A.
Acute angle

- B.
Reflex angle

- C.
Obtuse angle

- D.
Not possible

Answer: Option D

**Explanation** :

Let m∠A be x.

Measure of the supplementary angle of ∠A = (180° – x)

Measure of the complementary angle of ∠A = (90° – x)

According to the question,

Measure of the supplementary angle of ∠A = Measure of the complementary angle of ∠A + 100°

∴ (180° – x) = (90° – x) + 100°

∴ 180° – x = 190° − x

⇒ 180° = 190°

This is not possible.

Hence, option (d).

(Note: A complementary angle pair can be formed only by acute angles)

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**CRE 1 - Lines & Angles | Geometry - Basics**

Which of the following is definitely true for two lines in a plane?

- A.
If two lines are not parallel, they will intersect each other at some point

- B.
Two lines perpendicular to a given line, are parallel to each other

- C.
All of the above are true

- D.
None of these

Answer: Option C

**Explanation** :

The statements in options (a) and (b) are true in case both the lines are coplanar i.e. they lie in the same plane.

Hence, both the statements are true.

Hence, option (c).

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**CRE 1 - Lines & Angles | Geometry - Basics**

m∠P is 30° more than the measure of its supplementary angle. Which of the following is the measure of ∠P?

- A.
85°

- B.
105°

- C.
125°

- D.
None of these

Answer: Option B

**Explanation** :

Let m ∠P be x°.

∴ Measure of the supplementary angle of ∠P = x − 30°

The sum of these two angles should be 180°.

∴ x + (x − 30°) = 180°

∴ x = 105°

Hence, option (b).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the given figure, the lines L_{1} and L_{2} are parallel to each other and the line segments PQ and PR are perpendicular to each other. Moreover, m ∠PQR = 55° and lines ST, PQ, and PR intersect at point P. Find m ∠TPR.

- A.
50°

- B.
45°

- C.
35°

- D.
55°

Answer: Option C

**Explanation** :

L_{1} and L_{2} are parallel to each other.

Line PQ is the transversal.

∴ m ∠PQR + m ∠TPQ = 180° ...(∵ ∠PQR and ∠TPQ are interior opposite angles)

⇒ m ∠TPQ = 180° - 55° = 125°

Given, m ∠QPR = 90°

∴ m ∠TPQ = 125° = m ∠QPR + m ∠TPR

⇒ m ∠TPR = 125° - 90° = 35°

Hence, option (c).

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**CRE 1 - Lines & Angles | Geometry - Basics**

A line L_{1} cuts three parallel line segments AB, CD, and EF at points P, Q, and R respectively and another line L_{2} cuts them (AB, CD, and EF) at points S, T, and U respectively. If 3 × PQ = 2 × QR and the length of the segment ST is 4 cm, then what is the length of the segment TU?

- A.
6 cm

- B.
2 cm

- C.
4 cm

- D.
8 cm

Answer: Option A

**Explanation** :

Transversals make intercepts of equal proportions on parallel lines.

Using the Equal Intercept Ratio theorem, we have,

$\frac{\mathrm{PQ}}{\mathrm{QR}}$ = $\frac{\mathrm{ST}}{\mathrm{TU}}$

∴ $\frac{2}{3}$ = $\frac{4}{\mathrm{TU}}$

∴ TU = 6 cm

Hence, option (a).

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**CRE 1 - Lines & Angles | Geometry - Basics**

As shown in the figure, line AC and line FG are parallel to each other and m ∠ABF = 55°. If ray BD and segment EF are parallel and m ∠DBF = 30°, then find m ∠EFG.

- A.
25°

- B.
35°

- C.
45°

- D.
40°

Answer: Option A

**Explanation** :

Ray BD and EF are parallel lines and line BF is the transversal.

∴ m ∠DBF = m ∠EFB = 30° (∠DBF and ∠EFB form a pair of Alternate angles)

m ∠ABF = m ∠BFG = 55°, as they form a pair of alternate angles across lines AC and FG with BF as the transversal.

Now, m ∠BFG = m ∠BFE + m ∠EFG

∴ m ∠EFG = 55° − 30° = 25°

Hence, option (a).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the given figure, lines *l* and *m* are parallel to each other. Transversals AI and BK are also parallel to each other. If m ∠HJI = 50° and m ∠JHI = 45°, then find m ∠CBD.

- A.
105°

- B.
125°

- C.
85°

- D.
120°

Answer: Option C

**Explanation** :

In ∆HJI, m ∠HJI = 50° and m ∠JHI = 45°

Also, m ∠HJI + m ∠JHI + m ∠JIH = 180°

⇒ m ∠JIH = 85° i.e. m ∠JKD = 85° (∠JIH and ∠JKD are corresponding angles)

Now, if we consider lines l and m, BK is the common transversal

∴ ∠JKD = ∠CBD = 85° (alternate interior angles)

Hence, option (c).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the figure, DE and AE bisect ∠ADC and ∠DAB respectively. AB is parallel to CD. Find m ∠AED.

- A.
60°

- B.
75°

- C.
80°

- D.
90°

- E.
100°

Answer: Option D

**Explanation** :

Let m ∠DAB = 2x and m ∠ADC = 2y

∴ m ∠EAB = m ∠EAD = x and m ∠ADE = m ∠EDC = y

∵ AB and CD are parallel, ∠DAB and ∠ADC are interior angles whose sum is 180°.

∴ 2x + 2y = 180°

∴ x + y = 90°

∴ m ∠AED = 180° − (m ∠EAD + m ∠ADE)

= 180° − (x + y)

= 180° − 90°

= 90°

Hence, option (d).

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**CRE 1 - Lines & Angles | Geometry - Basics**

The lines AB, CD, EF and GH are parallel. It is known that AE : CG = 3 : 5 and AC : AG = 1 : 4. If the length of BD is 10 cm, what is the length (in cm) of FH?

- A.
15

- B.
18

- C.
22

- D.
12

Answer: Option C

**Explanation** :

AC : AG = BD : BH = 1 : 4

But, BD = 10 cm

∴ BH = 40 cm

∴ DH = BH − BD = 40 − 10 = 30 cm

Now, AE : CG = BF : DH = 3 : 5

∴ BF/30 = 3/5

∴ BF = 18 cm

∴ FH = BH − BF = 40 − 18 = 22 cm

Hence, option (c).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the given figure, m ∠AEJ = 55°, m ∠BIC = 130°. Also, AE || BF, CG || DH and AD || JH. BI and CI are the angle bisectors of the ∠CBF and ∠BCG respectively. What is the measure of ∠DHG?

- A.
130°

- B.
135°

- C.
140°

- D.
145°

- E.
150°

Answer: Option B

**Explanation** :

∠BFE and ∠AEJ are corresponding angles.

∴ m ∠BFE = m ∠AEJ = 55°

∠CBF and ∠BFE are alternate angles.

∴ m ∠CBF = m ∠BFE = 55°

∵ BI is the bisector of ∠CBF, we have,

∴ m ∠IBC = (m ∠CBF)/2 = 27.5°

∴ m ∠ICB = 180° − (m ∠BIC + m ∠IBC)

= 180° − (130° + 27.5°)= 22.5°

m ∠BCG = 2 × m ∠ICB = 45°

∴ m ∠CGF = 180° − m ∠BCG = 135°

m ∠DHG = m ∠CGF = 135° (∵ m∠DHG and m∠CGF form a pair of Corresponding angles)

Hence, option (b).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the given figure, AB and CD are parallel. m ∠BAE = 45°, m ∠BEC = 55°, m ∠AED = 70° and m ∠ECD = 20°. Find m ∠CED.

- A.
105°

- B.
115°

- C.
130°

- D.
135°

- E.
150°

Answer: Option D

**Explanation** :

Construct a line EF parallel to both AB and CD.

∴ m ∠CEF = m ∠ECD = 20° (∵ ∠CEF and ∠ECD form a pair of Alternate angles)

Also, m ∠AEG = m ∠EAB = 45° (∵ ∠AEG and ∠EAB form a pair of Alternate angles)

∴ m ∠GED = 70° − 45° = 25°

But, m ∠GED + m ∠CED + m ∠CEF = 180°

∴ 25° + m ∠CED + 20° = 180°

∴ m ∠CED = 135°

Hence, option (d).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the given figure, MN and LK are parallel lines. m∠LKO = 65° and m∠KON = 80°. Find m ∠MNO.

- A.
55°

- B.
35°

- C.
45°

- D.
15°

Answer: Option D

**Explanation** :

Draw a line OP parallel to LK.

∵ LK and MN are parallel lines, line OP is parallel to both the lines..

∴ m ∠LKO = m ∠KOP = 65° (∵ ∠LKO and ∠KOP form a pair of Alternate angles)

But, m ∠KON = m ∠KOP + m ∠PON

∴ m ∠PON = 80° − 65° = 15°

∴ m ∠MNO = 15° (∵ ∠MNO and ∠PON form a pair of Alternate angles)

Hence, option (d).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the figure below, y - x = 50°. Find x, y respectively.

- A.
125°, 75°

- B.
75°, 125°

- C.
65°, 115°

- D.
60°, 120°

- E.
None of these

Answer: Option C

**Explanation** :

Given, y - x = 50° ...(1)

Also, since x and y form a linear pair of angles

⇒ y + x = 180° ...(2)

Solving (1) and (2), we get:

x = 65°, y = 115°.

Hence, option (c).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the given figure, find the value of x. Given that AB || DE. (in degree)

Answer: 40

**Explanation** :

Let us draw a line through C parallel to AB.

Now,∠BCL + ∠ABC = 180°

⇒ ∠BCL = 180° - 120° = 60°

Similarly,∠MCD + ∠CDE = 180°

⇒ ∠MCD = 180° - 100° = 80°

Now, ∠BCL + x + ∠MCD = 180°

⇒ x = 180° - 60° - 80° = 40°

Hence, 40.

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the figure below, find the value of x?

- A.
24

- B.
25

- C.
36

- D.
20

- E.
None of these

Answer: Option A

**Explanation** :

A complete circular angle is equal to 360°

⇒ x + 2x + 3x + 4x + 5x = 360

⇒ 15x = 360

∴ x = 24

Hence, option (a).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the figure below, AB || DE and BC || FD.Find ∠FDE, given that ∠ABC = 30°.

- A.
25°

- B.
45°

- C.
30°

- D.
60°

Answer: Option C

**Explanation** :

Let us draw a line CG parallel to DE.

∠ABC = ∠GCB = 30° [alternate interior angles since AB || GC]

Now, since FD || BC and DE || DF,

⇒ ∠FDE = ∠BCG = 30°

Hence, option (c).

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**CRE 1 - Lines & Angles | Geometry - Basics**

In the figure below, find the value of x (in degrees).

- A.
90

- B.
100

- C.
80

- D.
110

Answer: Option C

**Explanation** :

Let us join A and D as shown below.

In ∆CAD, external angle CDE = ∠ACD + ∠CAD

In ∆ADB, external angle BDE = ∠BAD + ∠DBA

Now, x = ∠CDE + ∠BDE = ∠ACD + ∠CAD + ∠BAD + ∠DBA

⇒ x = ∠ACD + ∠CAB + ∠BAD

⇒ x = 20 + 50 + 10 = 80°

Hence, option (c).

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**CRE 1 - Lines & Angles | Geometry - Basics**

If in the given figure PO and QO are bisectors of the ∠AOC and the ∠BOC respectively, then what is the value of ∠POQ?

- A.
100

- B.
90

- C.
60

- D.
120

Answer: Option B

**Explanation** :

Let ∠AOC = 2x and ∠BOC = 2y

Now, 2x + 2y = 180°

⇒ x + y = 90°

Also, ∠POQ = ∠POC + ∠QOC = x + y

∴ ∠POQ = 90°

Hence, option (b).

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