Geometrical Concept : Part 1

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.
 AC and BC are
angle bisectors of
BAQ and ABS, respectively. If BAC = 30°,
find
ABC and ACB.

 

A. 60° and 90°
B. 30° and 120°
C. 60° and 30°D. 30° and 90°
 
2.1. If 45° arc of
circle A has the same length as 60° arc of circle B, find the ratio of the
areas of
circle A and
circle B.
A. 16/8
B. 16/9
C. 8/16

 

D.
9/16


3.In the figure given below, lines AB and DE are parallel. What is the value of ∠CDE?
A. 60°
B. 120°
C. 30°D. 150°
 
 
 
 
4.Find the value of a + b in the figure given below:
A. 60°
B. 120°
C. 80°D. 150°
 
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°.
 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
 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,
AreaΔ CFE/Area ΔABC =
(4×1)/(5×2)=2/5,
Therefore,
Area
Δ DEF/Area ΔABC =
1-(3/20+1/8+2/5)=13/40

 

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