**Dear Readers,**

**Today in this post, we are providing you all the necessary formula and concepts related to Geometry**

**from the Quant section.**

**Keeping in view the recent papers of SSC CGL, we can say that It is an important topic as per SSC CGL, SSC CPO and other Govt. Exam.**

**Fundamental concepts of Geometry:**

**Point:**It is an exact location. It is a fine dot which has neither length nor breadth nor thickness but has position i.e., it has no magnitude.

**Line segment:**The straight path joining two points A and B is called a line segment AB . It has and points and a definite length.

**Ray:**A line segment which can be extended in only one direction is called a ray.

**Intersecting lines:**Two lines having a common point are called intersecting lines. The common point is known as the point of intersection.

**Concurrent lines:**If two or more lines intersect at the same point, then they are known as concurrent lines.

**Angles:**When two straight lines meet at a point they form an angle.

**Right angle:**An angle whose measure is 90° is called a right angle.

**Acute angle:**An angle whose measure is less then one right angle (i.e., less than 90°), is called an acute angle.

**Obtuse angle:**An angle whose measure is more than one right angle and less than two right angles (i.e., less than 180° and more than 90°) is called an obtuse angle.

**Reflex angle:**An angle whose measure is more than 180° and less than 360° is called a reflex angle.

**Complementary angles:**If the sum of the two angles is one right angle (i.e.,90°), they are called complementary angles. Therefore, the complement of an angle θ is equal to 90° – θ.

**Supplementary angles:**Two angles are said to be supplementary, if the sum of their measures is 180°. Example: Angles measuring 130° and 50° are supplementary angles. Two supplementary angles are the supplement of each other. Therefore, the supplement of an angle θ. is equal to 180° – θ..

**Vertically opposite angles:**When two straight lines intersect each other at a point, the pairs of opposite angles so formed are called vertically opposite angles.

**Bisector of an angle:**If a ray or a straight line passing through the vertex of that angle, divides the angle into two angles of equal measurement, then that line is known as the Bisector of that angle.

**Parallel lines:**Two lines are parallel if they are coplanar and they do not intersect each other even if they are extended on either side.

**Transversal:**A transversal is a line that intersects (or cuts) two or more coplanar lines at distinct points.

**1.In the figure**

given below, PQ and RS are two parallel lines and AB is a transversal.

given below, PQ and RS are two parallel lines and AB is a transversal.

**AC and BC are**

angle bisectors of∠

angle bisectors of

**BAQ and**∠

**ABS, respectively. If**∠

**BAC = 30**°

**,**

find∠

find

**ABC and**∠

**ACB.**

**A. 60**° and 9

**0**°

**B. 3**

**0**° and 12

**0**°

**C.**

**60**° and 3

**0**°

**D. 3**

**0**° and 9

**0**°

2.

**1. If 45° arc of**

circle A has the same length as 60° arc of circle B, find the ratio of the

areas ofcircle A has the same length as 60° arc of circle B, find the ratio of the

areas of

**circle A and**

circle B.

circle B.

**A. 16/8**

**B. 16/9**

**C. 8/16**

**D.**

9/169/16

3.In the figure given below, lines AB and DE are parallel. What is the value of ∠CDE?

**A. 60**°

**B.**12

**0**°

**C.**3

**0**°

**D. 15**

**0**°

**4.Find the value of a + b in the figure given below:**

**A. 60**°

**B.**12

**0**°

**C.**8

**0**°

**D. 15**

**0**°

5.Points D, E and F divide the sides of triangle ABC in the ratio 1: 3, 1: 4, and 1: 1, as shown in the figure. What fraction of the area of triangle ABC is the area of triangle DEF?

**A. 16/40**

**B. 13/40**

**C. 14/16**

**D. 12/16**

ANSWERS AND SOLUTION:

**1(A):**∠BAQ

+ ∠ABS = 180°

[Supplementary

angles]

⇒∠BAQ/2 + ∠ABS/2 = 180°/2=90°⇒∠BAC+ ∠ABC= 90°

Therefore, ∠ABC

= 60° and ∠ACB = 90°.

= 60° and ∠ACB = 90°.

**2.(B):**Let the radius of circle A be r1 and that of

circle B be r2.

45/360 x 2π x r1 = 60/360 x 2πx r2 =>

r1/r2= 4/3

r1/r2= 4/3

Ratio of areas =πr1^2/πr2^2 = 16/9

**3(D)**:

We draw a line CF // DE at C, as shown in the figure below.

∠BCF = ∠ABC = 55° ⇒ ∠DCF = 30°.

⇒ CDE = 180° − 30° = 150°.

**4.(C) In the above figure, ∠CED = 180° − 125° = 55°. ∠ACD is the**

exterior angle of ΔABC. Therefore,

∠ACD = a + 45°. In ΔCED, a + 45° + 55° + b = 180° ⇒ a + b = 80°

**5.(B)**AreaΔ ADE/Area ΔABC =

(1×3)/(4×5)=3/20,

AreaΔ BDF/Area ΔABC =

(1×1)/(4×2)=1/8,

(1×1)/(4×2)=1/8,

AreaΔ CFE/Area ΔABC =

(4×1)/(5×2)=2/5,

(4×1)/(5×2)=2/5,

Therefore,

AreaΔ DEF/Area ΔABC =

1-(3/20+1/8+2/5)=13/40

AreaΔ DEF/Area ΔABC =

1-(3/20+1/8+2/5)=13/40

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