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Mathematics is one of the most important that is asked in the government recruitment exams. Generally, there are questions asked related to basic concepts and formulas of trigonometry. To let you make the most of the Mathematics section, we are providing important facts related to Geometry. Also, RRB NTPC, CGL, and State exams are nearby with bunches of posts for the interested candidates in which Mathematics is a major part. We have covered important notes and questions focusing on these prestigious exams. We wish you all the best of luck in come over the fear of the Mathematics section.
- Time and Work Notes: Short Tricks To Solve Questions
- Simple Interest Formula, Concept, and Study Notes
Fundamental concepts of Geometry
- Point: It is an exact location. It is a fine dot that 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 points and a definite length.
- Ray: A line segment that 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 than 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 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.
Triangles: Part-1
- A centroid divides a median in a 2:1 ratio. A centroid is a point of intersection of three medians.
- The ratio of two adjacent sides of a triangle is equal to the two parts of the third side which make by the internal angle bisector.
- In an equilateral triangle, the internal angle bisector and the median are the same.
- Any two of the four triangles formed by joining the mid-point of the sides of a given triangle are congruent.
- Two triangles have the same area if they have the same base and lie between two parallel lines.
- In a triangle side opposite a smaller angle is smaller in comparison to the side which is opposite a greater angle.
- When the corresponding sides of two triangles are in proportion then the corresponding angles are also in proportion.
- If the two triangles are similar, then we have the following results –
The ratio of area of two triangles = Ratio of squares of corresponding sidesThe ratio of sides of two triangles= Ratio of height (Altitudes)= Ratio of medians= Ratio of angle bisector= Ratio of in radii/circum radii= Ratio of Perimeter
- Some important terms:
- Orthocenter → Point of intersection of three Altitudes.
- Incentre → Point of the intersection of the angle bisectors of a triangle
- Circumcenter → Point of intersection of the perpendicular bisectors of the sides.
- Median → Line joining the mid-point of a side to the vertex opposite to the sides
- In a right-angled triangle, the triangle on each side of the altitude drawn from the vertex of the right angle to the hypotenuse is similar to the original triangle and to each other too.
Some questions are based on lines & angle
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. If the 45° arc of circle A has the same length as the 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?
ANSWERS AND SOLUTIONS:
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
an exterior angle of ΔABC. Therefore, ∠ACD = a + 45°. In ΔCED, a + 45° + 55° + b = 180° ⇒ a + b = 80°
an 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, AreaΔ DEF/Area ΔABC = 1-(3/20+1/8+2/5)=13/40
