Common Core Math

Properties of a Rectangle Rhombus and Square is always a confusing concept for students. Learning every individual topic is important to score good marks in the exam. So, we have explained every individual topic clearly in a detailed manner in this article. Therefore, those who wish to learn the concepts of Parallelogram and its properties, problems, can completely learn the Parallelogram concepts on our website.

Rectangle

A rectangle is said to be a parallelogram when it has all 4 angles having equal measure.

Properties of Rectangle

• The Opposite sides of a rectangle are parallel.
• Also, the Opposite sides of a rectangle are equal in length.
• Diagonals are equal in length.
• The interior angles are 90 degrees each.
• Diagonals bisect each other.
• It has horizontal and vertical lines of symmetry.
• Each of the diagonal bisects the rectangle into 2 congruent triangles.
• If you combine the 4 sides of a rectangle, then the mid-points of it form a rhombus.

Rectangle Formulas

If l is the length of the rectangle and b is the breadth of the rectangle, then
Area = lb square units
Perimeter = 2 (l+b) units.

Diagonal Properties of a Rectangle

Prove that the diagonals of a rectangle are equal and bisect each other.

Proof:
Let PQRS be a rectangle that has diagonals PQ and QS intersect at the point O.

From ∆ PQR and ∆ QPS,
PQ = QP (common)
∠PQR = ∠QPS (each equal to 90º)
QR = PS (opposite sides of a rectangle).
Therefore, ∆ PQR ≅ ∆ QPS (by SAS congruence)
⇒ PR = QS.
Hence, the diagonals of a rectangle are equal.
From ∆ OPQ and ∆ ORS,
∠OPQ = ∠ORS (alternate angles)
∠OQP = ∠OSR (alternate angles)
PQ = RS (opposite sides of a rectangle)
Therefore, ∆OPQ ≅ ∆ ORS. (by ASA congruence)
⇒ OP = OR and OQ = OS.
This shows that the diagonals of a rectangle bisect each other.

Hence, the diagonals of a rectangle are equal and bisect each other.

Rhombus

The rhombus is a quadrilateral that consists of four sides with equal lengths.

Properties of Rhombus

• The Rhombus consists of parallel and equal opposite sides. As it consists of parallel and equal opposite sides, it is said to be a parallelogram.
• All available sides (4 sides) are equal.
• Also, opposite angles in a rhombus are equal.
• Diagonals bisect each other.
• Diagonals of a rhombus intersect each other at right angles.
• Furthermore, Diagonals bisect opposite vertex angles.
• Every diagonal divides the rhombus into 2 congruent triangles.

Rhombus Formula

If b is the side, a and b are the two diagonals of the rhombus, then
Area = ab/2 Square units.
Perimeter = 4b units

Diagonal Properties of a Rhombus

Prove that the diagonals of a rhombus bisect each other at right angles.

Proof:
Let PQRS be a rhombus whose diagonals AC and BD intersect at point O.

The diagonals of a parallelogram bisect each other. Also, we know that every rhombus is a parallelogram.
So, the diagonals of a rhombus bisect each other.
Therefore, OP = OR and OQ = OS
From ∆ ROQ and ∆ ROS,
RQ = RS (sides of a rhombus)
RO = RO (common).
OQ = OS (proved)
Therefore, ∆ ROQ ≅ ∆ ROS (by SSS congruence)
⇒ ∠ROQ = ∠ROS
But, ∠ROQ + ∠ROS = 2 right angles (linear pair)
Therefore, ∠ROQ = ∠ROS = 1 right angle.

Hence, the diagonals of a rhombus bisect each other at right angles.

Square

A square is a rectangle that has all equal sides.

Properties of Square

• The opposite sides of a square are parallel.
• All 4 sides are equal in length.
• Diagonals are equal in length.
• Diagonals bisect opposite vertex angles.
• The interior angles of a square measure 90 degrees each.
• Diagonals bisect each other at right angles.
• It has 4 lines of symmetry – a horizontal, a vertical, and 2 diagonals.
• Each diagonal bisects the square into 2 congruent triangles.

Square Formula

If b is the side of the square, then
Area = b² square units
Perimeter = 4b units.

Diagonal Properties of a Square

Prove that the diagonals of a square are equal and bisect each other at right angles.

Proof:
We know that the diagonals of a rectangle are equal.
Also, every square is a rectangle.
Therefore, the diagonals of a square are equal.
Again, the diagonals of a rhombus bisect each other at right angles. But, every square is a rhombus.
So, the diagonals of a square bisect each other at right angles.

Hence, the diagonals of a square are equal and also bisect each other at right angles.

Note 1: If the diagonals of a quadrilateral are equal but it is not necessary to be a rectangle.
Note 2: If the diagonals of a quadrilateral interest at a point with right angles then also it is not necessary to become a rhombus.

Construction Of Quadrilaterals is easy if you have a complete grip on the concept. A quadrilateral is a polygon with 4 sides, 4 angles, and also 4 vertices. When you add the interior angles of a quadrilateral, then you can get 360 degrees. The quadrilateral side lengths and angles may different. Depending on the lengths and angles of the sides, you can easily know what is the name of the quadrilateral. You can easily construct a quadrilateral by considering the following criteria.

(i) 4 sides and 1 diagonal is given.
(ii) 3 sides and including 2 angles are given
(iii) 2 sides and three angles are given
(iv) 3 sided and 2 diagonals are given
(v) 4 sides and 1 angle is given

1. Construct the Quadrilateral when 4 sides and 1 diagonal is given
PQ = 5 cm
QR = 4.5 cm
RS = 3.8 cm
PS = 4.4 cm
Diagonal PQ = 6 cm

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.

1. Draw a line segment of length 5 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 6 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 4.5 cm. Mark the point as R where the two arcs cross each other. Join the points Q and R as well as P and R.
4. By taking the point P as a center, draw an arc with a radius of 4.4 cm.
5. By taking the point R as a center, draw an arc with a radius of 3.8 cm.
6. Mark the point as S where the two arcs cross each other. Join the points R and S as well as P and S.

The final result is the required quadrilateral.

2. Construct the Quadrilateral when 3 sides and including 2 angles are given
PQ = 3.8 cm
QR = 4.2 cm
PS = 5.2 cm
∠QPS = 180º
∠PQR = 80º

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.

1. Draw a line segment of length 3.8 cm and mark the ends as P and Q.
2. Take point P as a center and make a point by taking 180º using a protector.
3. Next, take point P as a center and draw an arc by taking the radius 5.2 cm. Mark the point as S where the point and arc cross each other. Join the points P and S.
4. By taking the point Q as a center, make a point by taking 80º using a protector.
5. By taking the point Q as a center, draw an arc with a radius of 4.2 cm.
6. Mark the point as R where the point and arc cross each other. Join the points Q and R.
7. Finally, join the points R and S and draw a line segment.

The final result is the required quadrilateral.

3. Construct the Quadrilateral when 3 sides and including 2 angles are given
AB = 4.7 cm
∠ABC = 120°,
BC = 4 cm,
∠BCD = 100°, and ∠BAD = 60°.

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.

1. Draw a line segment of length 4.7 cm and mark the ends as A and B.
2. Take point A as a center and make a point by taking 60º using a protector.
3. Next, take point B as a center and make a point by taking 120º using a protector. Mark the point as D where the two points are meet at a point. Join the points A and D.
4. By taking the point B as a center, draw an arc with a radius of 4 cm.
5. Mark the point as C where the point and arc cross each other. Join the points B and C.
6. Finally, join the points C and D and draw a line segment.

The final result is the required quadrilateral.

4. Construct the Quadrilateral when 3 sided and 2 diagonals are given
PQ = 4.2 cm
QR = 4 cm
PS = 3.2 cm
Diagonal PR = 5 cm
Diagonal QS = 4.6 cm

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.

1. Draw a line segment of length 4.2 cm and mark the ends as P and Q.
2. Take the point P as a center and draw an arc by taking the radius 5 cm.
3. Next, take point Q as a center and draw an arc by taking the radius 4 cm. Mark the point as R where the two arcs cross each other. Join the points Q and R as well as P and R.
4. By taking the point P as a center, draw an arc with a radius of 3.2 cm.
5. By taking the point Q as a center, draw an arc with a radius of 4.6 cm.
6. Mark the point as S where the two arcs cross each other. Join the points R and S as well as P and S.

The final result is the required quadrilateral.

5. Construct the Quadrilateral when 4 sides and 1 angle is given.
PQ = 4 cm, QR = 3.6 cm, RS = 4.7 cm, PS = 5.2 cm and ∠B = 80°.

Steps of Construction:
Firstly, draw a rough figure of the quadrilateral with the given dimensions.

1. Draw a line segment of length 4 cm and mark the ends as P and Q.
2. Take point Q as a center and make a point by taking 80º using a protector.
3. Next, take point Q as a center and draw an arc with a radius of 3.6 cm. Mark the point as R where the two points are meet at a point. Join the points Q and R.
4. By taking the point P as a center, draw an arc with a radius of 5.2 cm.
5. By taking the point R as a center, draw an arc with a radius of 4.7 cm.
6. Mark the point as S where the two arcs cross each other. Join the points P and S, R and S.

The final result is the required quadrilateral.

A Transversal Line is a line that crosses at least two lines (Parallel Lines) and intersects with them. Find Problems on Transversal Lines with Solutions for a better understanding of the concept. Check concept and examples of Lines and Angles on our website for free.

From the above figure, A and B are parallel lines and line C is intersecting both lines at points D and E. The line C is called as Transversal Line.

From the above figure, The line M intersecting Lines N and O at a point c. But the Lines N and O are not parallel lines. Therefore, Line M is not a Transversal Line.

Angles made by the Transversal with Two Lines

From the above figure, Line C intersecting Line A and B (parallel lines). The angles formed around the Transversal line are called Transversal angles. From the given figure, ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7, ∠8 are the angles with special names.

Interior Angles

The angles that present in the area enclosed between two parallel lines and that are intersected by a transversal are also called interior angles. From figure, ∠3, ∠4, ∠5, ∠6 are interior angles.

Exterior Angles

The angles that present outside two parallel lines and that are intersected by a transversal are called Exterior angles. From the figure, ∠1, ∠2, ∠7, ∠8 are Exterior angles.

Corresponding Pair of Angles

The corresponding pair of angles present on the same side of the transversal. In the Pair of corresponding angles, if one is an interior angle, the other will be an exterior angle. Also, they do not form a linear pair. From the figure, (∠2, ∠6); (∠1, ∠5); (∠3, ∠7); (∠4, ∠8) are the Corresponding pair of angles.

Pair of Alternate Angles

Pair of alternate angles present on the opposite sided of the transversal. Both Pair of Alternate Angles either interior angles or both are exterior angles. They will not form a linear pair. From the figure, (∠4, ∠6) and (∠3, ∠5) are interior alternate angles. Also, (∠1, ∠7) and (∠2, ∠8) are exterior alternate angles.

Pair of Co-interior or Conjoined or Allied Angles

Pair of co-interior or Conjoined or Allied Angles are pairs of interior angles that present on the same side on the transversal. From the figure, (∠3, ∠6); (∠4, ∠5) are co-interior angles.

Two Parallel Lines are cut by the Transversal

When two parallel lines A and B are intersected by a Transversal line C, then

• the pairs of alternate angles are equal ∠4 = ∠6, ∠ 1 = ∠7, ∠3 =∠5, ∠2 = ∠8.
• the pairs of corresponding angles are equal ∠2 = ∠6, ∠1 = ∠5, ∠3 = ∠7, ∠4 = ∠8.
• Interior angles on the same side of transversal are supplementary ∠4 + ∠ 5 = 180°, ∠6 = 180°.

Converse

When two parallel lines are intersected by a Transversal line

• pairs of alternate angles are equal
• pairs of corresponding angles are equal
• interior angles on the same side of the transversal are supplementary.

Parallel Lines are two lines in a plane, they do not intersect and parallel to each other. The distance between the parallel lines is always the same. The symbol to denote parallel lines is ||. Learn and practice all Lines and Angles concepts and problems on our website.

From the above figure, Line A and Line B are parallel lines. We can write it as A || B and we can read it as A is parallel to B.

Properties Of Angles Associated with Parallel Lines

From the above figure, A and B are two parallel lines and C is passing through two parallel lines by intersecting them at a point. C is a transversal line. If a line intersects two or more lines at distinct points then it is known as a transversal line. When two lines meet at a point in a plane, they are known as intersecting lines.

• The pair of corresponding angles is equal (∠4 = ∠8); (∠3 = ∠7); (∠2 = ∠6); (∠1 = ∠5).
• The pair of exterior alternate angles is equal (∠2 = ∠8); (∠1 = ∠7).
• The pair of interior alternate angles is equal (∠3 = ∠5); (∠4 = ∠6).
• Interior angles present on the same side of the transversal are supplementary, i.e., ∠4 + ∠5 = 180° and ∠3 + ∠6 = 180°.

Example:

Look at the above figure and check out the parallel lines, transversal line, and etc.,
From the above figure, AB and CD are parallel lines and MN is a transversal line that cuts the parallel lines AB and CD.
(i) Interior and exterior alternate angles are equal.
i.e. ∠4 = ∠5 and ∠3 = ∠6 [Interior alternate angles]
∠2 = ∠7 and ∠1 = ∠8 [Exterior alternate angles]
(ii) Corresponding angles are equal.
i.e. ∠2 = ∠6; ∠1 = ∠5; ∠3 = ∠7 and ∠4 = ∠8
(iii) Co-interior or allied angles are supplementary.
i.e. ∠4 + ∠6 = 180° and ∠3 + ∠5 = 180°

Conditions of Parallelism

Two straight lines are cut by a transversal, and if

• the two pairs of corresponding angles are equal and also the lines are parallel to each other.
• if the pair of interior angles on the same side of the transversal are supplementary, then the two straight lines are parallel.
• the pair of alternate angles are equal, then the straight lines are parallel to each other.

Parallel Rays

Parallel Rays are two rays that do not intersect each other and have the same distance every time. A ray consists of one endpoint. If AB and CD are two rays and parallel rays, then it is represented as ray AB ∥ ray CD.

Parallel Segments

If two line segments are parallel then those two line segments are called Parallel Segments. The line segment is a line that consists of two endpoints. If AB and CD are two line segments and parallel, then it is represented as line segment AB ∥ line segment CD.

Learn what are Vertically Opposite Angles and how to calculate them in this article. Vertically Opposite Angles are the angles formed when two lines intersect each other. The vertically opposite angles are always equal to each other. The adjacent angles of vertically opposite angles are always supplementary angles and they form 180º when we add them. Find Lines and Angles concepts and problems on our website.

If one angle is 35º, then the other angle is 180º – 35º.

From the above figure, the line segment PQ and RS meet at the point O and these two lines are intersecting lines.

In a pair of intersecting lines, the angles which are opposite to each other form a pair of vertically opposite angles. From the figure, ∠SOQ and ∠POR form a pair of vertically opposite angles.

∠SOQ = ∠POR
∠SOP = ∠ROQ

Note: A vertical angle and its adjacent angle is supplementary to each other. It means they add up to 180 degrees.

Vertical Angles Theorem and Proof

Theorem: In a pair of intersecting lines the vertically opposite angles are equal.

Proof: Consider two lines PQ and RS which intersect each other at O. The two pairs of vertical angles are:
i) ∠SOQ = ∠POR
ii) ∠SOP = ∠ROQ

It can be seen that ray OS stands on the line PQ and according to Linear Pair Axiom, if a ray stands on a line, then the adjacent angles form a linear pair of angles.
Therefore, ∠SOQ + ∠SOP = 180° —(1) (Linear pair of angles)
Similarly, Ray OP stands on the line SR.
Therefore, ∠SOP + ∠POR = 180° —(2) (Linear pair of angles)
From (1) and (2),
∠SOQ + ∠SOP = ∠SOP + ∠POR
⇒ ∠SOQ = ∠POR —(3)
Also, OQ stands on the line SR.
Therefore, ∠SOQ + ∠QOR = 180° —(4) (Linear pair of angles)
From (1) and (4),
∠SOQ + ∠SOP = ∠SOQ + ∠QOR
⇒ ∠SOP = ∠QOR —(5)
Thus, the pair of opposite angles are equal.

Hence, proved.

Vertically Opposite Angles Examples

1. In the given figure, find the measure of unknown angles.

Solution:
From the given figure
(i) ∠C = 60° vertically opposite angles
(ii) ∠B = 90° vertically opposite angles
(iii) ∠B + ∠A + 60° = 180° (straight angle)
90° + ∠ A + 60° = 180°
150° + ∠ A = 180°
Therefore, ∠A = 180° – 150° = 30°
(iv) ∠A = ∠D vertically opposite angles

Therefore, ∠D = 30°

2. In the given figure, lines AB, CD, EF intersects at O. If M : N : P = 1 : 2 : 3, then find the values of M, N, P.

Solution:
We know that the sum of all the angles at a point is 360°.
∠AOC = ∠DOB = M° (Pair of vertically opposite angles are equal.)
∠FOB = ∠AOE = N° (Pair of vertically opposite angles are equal.)
∠EOD = ∠COF = P° (Pair of vertically opposite angles are equal.)
Therefore, ∠AOE + ∠AOC + ∠COF + ∠FOB + ∠BOD + ∠DOE = 360°
N + M + P + N + M + P = 360°
⟹ 2M + 2N + 2P = 360°
⟹ 2(M + N + P) = 360°
⟹ M + N + P = 3̶6̶0̶°/2̶
⟹ M + N + P = 180° ——— (i)
Let the common ratio be a.
Therefore, M = a, N = 2a, P = 3a
Therefore, from the equation (i) we get;
a + 2a + 3a = 180°
⟹ 6a = 180°
⟹ a = 1̶8̶0̶°/6̶
⟹ a = 30°
Therefore, M = a, means M = 30°
N = 2a, means N = 2 × 30 = 60°
P = 3a, means P = 3 × 30 = 90°

Therefore, the measures of the angles are 30°, 60°, 90°.

Do you want to know What is linear pair of angles? Linear Pair of Angles are similar to adjacent angles but only the difference is adding two angles we get 180º in linear angles. Linear Pair of Angles will also have a common vertex, common arm like adjacent angles. Their interiors will also do not overlap. Remember that all adjacent angles may not form a linear pair. Learn all the Lines and Angles concepts in one place on our website.

Properties of Linear Pair of Angles

Check the below Linear Pair of Angles properties listed below.

• They have a common vertex.
• They have a common arm.
• Also, they do not overlap.
• The sum of the angles is 180°.

From the above figure, AO and OC are two opposite rays and ∠AOB and ∠BOC are the adjacent angles. Therefore, ∠AOB and ∠BOC form a linear pair.

Explanation for Linear Pair of Angles

The angle between the two straight lines is 180° and they form a straight angle. The line segment is the portion of a line that consists of two endpoints. Also, a line with one endpoint is known as a ray. If you consider a line mn and the middle point is o. A ray OP dividing the two lines MO and ON. The complete angle MOP and angle PON becomes 180º to form a linear pair of angles.

∠MOP + ∠PON = ∠MON = 180°

We have also given Lines and Angles concepts and problems for free of cost on our website.

Axioms

Axiom 1: If a ray stands on a line then the adjacent angles form a linear pair of angles.

From the above figure, there are different line segments available that are passing through point O. From the figure, we can say that Angle BOD and Angle BOC are Linear Pair of Angles.

Linear Pair of Angles Examples

In the given figure, ∠AOC and ∠ BOC form a linear pair if x – y = 50°, find the value of x and y.

Solution:
Given x – y = 50° ………… (i)
We know that, x + y = 180° ………… (ii)
Adding (i) and (ii)
x – y + x + y = 180° + 50°
2x = 230°
x = 240°/2
Therefore, x = 115°

Since, x – y = 50°
or, 115° – y = 50°
or, 115° – 115° – y = 50° – 115°
or, -y = -65°
Therefore, y = 65°

The final answer is x = 115° and y = 65°.

Adjacent Angles are the angles that have a common vertex and common arm. The other arms of the two adjacent angles present on the opposite sides of the common arm. The adjacent angles can be a supplementary angle or complementary angle when they share the common vertex and side. Find other concepts of Lines and Angles along with the adjacent angles on our website with a clear explanation.

From the above figure, YO is the common arm and Y is the common vertex. The above figure shows a pair of adjacent angles. The Ray XY and ray YZ are present on the opposite sides of the common arm YO. Therefore, ∠XYO and ∠OYZ are adjacent angles.

∠XYZ and ∠XYO are not adjacent angles, because their other arms YZ and YO are not on the opposite sides of the common arm XY.

Adjacent Angle Example

If you take a wall clock, you can see a minute hand, second hand, and also an hour hand. The minute hand and second hand of the clock form one angle represented as ∠XYO and the hour hand form another angle with the second hand represented as∠OYZ. Both the angles i.e.∠XYO and ∠OYZ are present on the opposite sides of the common hand. Therefore, these two angles are adjacent angles.

Properties of Adjacent Angles

Check out the Properties of Adjacent Angles mentioned below as a part of your preparation. They are as such

• Adjacent Angles share the common vertex
• Angles do not overlap
• They share the common arm
• There is no common interior-point in Adjacent Angles.
• It may be complementary or supplementary angles.
• There should be a non-common arm on both sides of the common arm.

Adjacent Angles Solved Problems

1. Are the following angles adjacent? Give reasons.

Solution:

(a) From figure a, ∠1, and ∠2 don’t have a common arm. Therefore, ∠1 and ∠2 are not adjacent angles.
(b) Adjacent angles will not overlap each other. From figure b, ∠1 and ∠2 interiors are overlapped. Therefore, ∠1 and ∠2 are not adjacent angles.
(c) Adjacent angles must have a common vertex. from figure c, ∠1, and ∠2 don’t have a common vertex. Therefore, ∠1 and ∠2 are not adjacent angles.
(d) From figure (d), ∠1 and ∠2 are adjacent angles. Because they have a common vertex, a common arm, and their interiors do not overlap.

Complementary and Supplementary Angles are the angles formed by adding two angles. If the sum of the two angles is 90º, they are called Complementary angles. Or else, if the sum of the two angles is 180º, they are called Supplementary angles. Find different problems on Supplementary and Complementary Angles in this article. Also, we have given all the concepts available in Lines and Angles on our website.

Complementary Angles

The two angles are said to be complementary angles if their sum is one right angle i.e. 90°. Each angle is called the complement of the other. If x° is one angle then (90 – x)° is the other angle in Complementary Angles.

For example, if 70° is one angle in Complementary Angles, then the other angle will be 20°. 70° and 20° complement each other.

Example 1: To find the complement of 3y + 52°, subtract the given angle from 90 degrees.
90º – (3y + 52º) = 90º – 3y – 52º = -3y + 38º

The complement of 3y + 52° is 38º – 3y.

Facts of complementary angles

• Two complementary angles are acute but vice versa is not possible
• Two right angles cannot complement each other
• Also, two obtuse angles cannot complement each other

Supplementary Angles

The two angles are said to be Supplementary angles if their sum is 180°. Two angles are added and formed a straight line in Supplementary Angles. If x is one angle then (180 – x)° is the other angle in Supplementary Angles.

For example, if 130° is one angle in Supplementary Angles, then the other angle will be 50°. By adding 130 and 50 degrees, it forms a straight line. In Supplementary Angles, both angles are said to be a supplement to each other.

Solved Examples on Supplementary and Complementary Angles

1. Find the complement of 50 degrees?

Solution:
The given angle is 50 degrees, then, the Complement is 40 degrees.
We know that Sum of Complementary angles = 90 degrees.
By adding 40 degrees to 50 degrees, the total angle becomes 90 degrees.
So, 50° + 40° = 90°

The final answer is 90°

2. Find the Supplement of the angle 1/4 of 200°.

Solution:
Given that the angle 1/4 of 200°
Convert 1/4 of 200°
That is, 1/4 x 200° = 50°
Supplement of 50° = 180° – 50° = 130°
By adding 130 degrees to 50 degrees, the total angle becomes 180 degrees.

Therefore, Supplement of the angle 1/4 of 200° is 130°

3. The measures of two angles are (x + 15)° and (3x + 25)°. Find the value of x if angles are supplementary angles.

Solution:
We know that, Sum of Supplementary angles = 180 degrees.
So, (x + 15)° + (3x + 25)° = 180°
Add the x terms and numbers.
4x + 40° = 180°
Move 40° to the right side and subtract it from 180°.
4x = 140°
Move 4 to the right side and divide it from 140°.
x = 35°

The value of x is 35 degrees.

4. The difference between two complementary angles is 54°. Find both the angles.

Solution:
Given that the difference between two complementary angles is 54°.
Let, the first angle = x degrees, then, Second angle = (90 – x)degrees {as per the definition of complementary angles}
Difference between angles = 54°
Now, (90° – x) – x = 54°
90° – 2x = 54°
– 2x = 54° – 90°
-2x = -36°
x = 36°/2°
x = 18°
Again, Second angle = 90° – 18° = 72°

Therefore, the required angles are 18°, 72°.

5. Find the complement of the angle 4/5 of 90°.

Solution:
Convert 4/5 of 90°
Multiply 4/5 with 90°
4/5 × 90° = 72°
Complement of 72° = 90° – 72° = 18°

Therefore, a complement of the angle 4/5 of 90° = 18°

6. Find the supplement of the angle 2/3 of 90°.

Solution:
Convert 2/3 of 90°
2/3 × 90° = 60°
Supplement of 60° = 180° – 60° = 120°

Therefore, a supplement of the angle 2/3 of 90° = 120°

7. The measure of two complementary angles are (3x – 6)° and (x – 4)°. Find the value of x.

Solution:
According to the problem, (3x – 6)° and (x – 4)°, are complementary angles’ so we get;
(3x – 6)° + (x – 4)° = 90°
or, 3x – 6° + x – 4° = 90°
or, 3x + x – 6° – 4° = 90°
or, 4x – 10° = 90°
Add 10° on both sides.
or, 4x – 10° + 10° = 90° + 10°
or, 4x = 100°
or, x = 100°/4°
or, x = 25°

Therefore, the value of x = 25°.

Supplementary angles are the angles that are added up to 180 degrees. If one angle is 120 degrees then the other angle is 60 degrees in supplementary angles because by adding 120 and 60, we get 180 degrees. By adding two supplementary angles, they form a straight line and a straight angle. One important thing in supplementary angles is the two angles need not be next to each other.

The important geometry concepts Lines and Angles are explained on our website with detailed explanations and solved examples. Learn all the concepts and improve your preparation level easily.

Supplementary Angle Definition

Supplementary Angle form 180º by adding two angles.

From the given figure, Angle AOC is one angle and angle BOC is another angle. By adding the two angles AOC and BOC, we get 180º. ∠AOC and ∠BOC are Supplementary Angles.
∠AOC + ∠BOC = 180°

Properties of Supplementary Angles

Have a look at the important properties of supplementary angles explained in the below modules. They are as such

• The two angles are said to be supplementary angles when they add up to form 180°.
• The two supplementary angles together make a straight line, but the angles need not be together.
• Two acute angles cannot be a supplement by each other.
• “S” of supplementary angles stands for the “Straight” line. This means they form 180°.
• The two right angles are always supplementary.

Adjacent and Non-Adjacent Supplementary Angles

The supplementary angles may be classified as either adjacent or non-adjacent. The adjacent supplementary angles have the common line segment or arm with each other. Also, the non-adjacent supplementary angles do not have the line segment or arm.

Supplementary Angles Theorem

The supplementary angle theorem states that if two angles are supplementary to the same angle, then the two angles are said to be congruent.

Proof:

If ∠a and ∠b are two different angles that are supplementary to a third angle ∠c, such that,
∠a + ∠c = 180 ……. (1)
∠b + ∠c = 180 ……. (2)
Then, from the above two equations, we can say,
∠a = ∠b

Hence proved.

Supplementary Angles Examples

1. Verify if 125°, 55° are a pair of supplementary angles?

Solution:
Add the given two angles and check if the resultant angle is 180° or not.
125° + 55° = 180°

The resultant angle is 180°. Hence, they are a pair of supplementary angles.

2. Find the supplement of the angle (30 + x)°.

Solution:
To find the supplement of (30 + x)°, subtract it from 180°
Supplement of the angle (30 + x)° = 180° – (30 + x)°
= 180° – 30° – x°
= (150 – x) °

3. If angles of measures (y – 2)° and (3y + 7)° are a pair of supplementary angles. Find the measures.

Solution:
Since (y – 2)° and (3y + 7)° represent a pair of supplementary angles, then their sum must be equal to 180°.
Therefore, (y – 2) + (3y + 7) = 180
y – 2 + 3y + 7 = 180
y + 3y – 2 + 7 = 180
4y + 5 = 180
Subtract 5 from both sides
4y + 5 – 5 = 180 – 5
4y = 180 – 5
4y = 175
y = 175/4
y = 43.75°

Therefore, we know the value of y = 43.75°, put the value in place of y
y – 2
= 43.75 – 2
= 41.75°

And again, 3y + 7
= 3 × 43.75° + 7
= 118 + 5
= 138.25°
Therefore, the two supplementary angles are 41.75° and 138.25°.

4. Two supplementary angles are in the ratio 5 : 4. Find the measure of the angles.

Solution:
Let the common ratio be m.
If one angle is 5m, then the other angle is 4m.
Therefore, 5m + 4m = 180
9m = 180
m = 180/9
m = 20
Put the value of m = 20

One angle is 5m
= 5 × 20
= 100°

And the other angle is 4m
= 4 × 20
= 80°

Therefore, the two supplementary angles are 100° and 80°.

5. In the given figure find the measure of the unknown angle.

Solution:
x + 45° + 25° = 180°
The sum of angles at a point on a line on one side of it is 180°
Therefore, x + 70° = 180°
Subtract 70° from both sides
x + 70° – 70° = 180° – 70°
x = 110°

Complementary Angles are the angles with 90°. When we add two angles then the resultant angle should be 90° to call those angles complementary angles. In complementary angles, each angle complements another angle. If an angle is 90 degrees, then it can’t be called a complementary because it doesn’t have any pairs. It consists of only one angle with 90 degrees.

The angles become complementary when the sum of the two angles becomes 90 degrees. Therefore, an angle with 90 degrees cannot be called complementary angles. Find all the Lines and Angles concepts on our website with solved examples.

From the above figure, the ∠AOB and ∠BOC are complementary as ∠AOB + ∠BOC = 35° + 55° = 90°. Therefore, they both are Complementary Angles. Also, angles 35 and angle 55 complements each other.

Example 1: If you take a triangle with one angle of 50 degrees and another angle of 40 degrees, the combined angle will become 90 degrees. The two 40 and 50 angles are called complementary angles.

Example 2: If you take a triangle with one angle of 15 degrees and another angle of 75 degrees, the combined angle will become 90 degrees. The two 15 and 75 angles are called complementary angles.

Example 3: If you take a triangle with one angle of 10 degrees and another angle of 80 degrees, the combined angle will become 90 degrees. The two 10 and 80 angles are called complementary angles.

Observations of Complementary Angles

• If two angles complement each other, then each angle must be an acute angle. But any two acute angles need not be complementary.
  For example, angles of measures 30° and 40° are not a complement to each other.
• Two right angles cannot complement each other.
• Also, two obtuse angles cannot complement each other.

Complementary Angles Examples

1. Find the complement of below angles

(a) 58°

Solution:
To find the complement of 58°, subtract it from 90°
90° – 58° = 32°

Therefore, the complement of 58° is 32°

(b) 37°40′

Solution:
To find the complement of 37°40′, subtract it from 90°
90° – 37°40′
90° = 89°60′
= 89°60′ – 37°40′
= 52°20′
Therefore, the complement of 37°40′ is 52°20′

(c) y + 42°

Solution:
To find the complement of y + 42°, subtract it from 90°
90° – (y + 42°)
= 90° – y – 42°
= 56° – y
Therefore, the complement of y + 42° is 56° – y

2. Find the complement of the angle (20 + x)°

Solution:
To find the complement of (20 + x)°, subtract it from 90°
90° – (20 + x)
= 90° – 20 – x°
= 70° – x° = (70 – x)°
Therefore, the complement of (20 + x)° is (70 – x)°

3. Find the measure of an angle that is 36° less than its complement.

Solution:
Let the unknown angle be y, then measure of its complement = (90° – y)
According to the given question,
(90° – y) – y = 36°
90° – y – y = 36°
90° – 2y = 36°
Subtract 90° from both sides
90° – 90° – 2y = 36° – 90°
-2y = -54°
y = 54/2 = 27°
y = 27°
Therefore, 90 – y (Put the value of y = 27°)
= 90 – 27°
= 63°
Therefore, the pair of complementary angles are 63° and 27°

Have a look at Some Geometric Terms and Results. We have given a useful reference of geometric terms and their definitions along with solved examples. Most of the students feel difficult to understand Some Geometric Terms. So, to help such students, we explained all Geometric Terms in a clear and understandable way. We can get strong and exact results using Certain Geometric Statements. So, make use of the given formulas and get the easy process to solve problems.

We have also given Lines and Angles concepts and problems for free of cost on our website.

Some Geometric Terms and Results

• The sum of all the angles at a point is 360°.

i.e., ∠7 + ∠8 + ∠9 + ∠10 + ∠11 + ∠12 = 360°

• The sum of all the angles about a point on a straight line on one side of if it is 180°.

i.e., ∠5 + ∠6 + ∠7 + ∠8 = 360°

Some Important Geometric Terms

1. Equal Angles

The two angles are considered to be equal if they have the same degree measure.

Given that two angles. The angles are ∠ABC and ∠PQR.
∠ABC = 90°; ∠PQR = 90°.
∠ABC and ∠PQR are equal angles of measure 90°.

2. Bisector of an Angle

A ray that divides the given angle into two equal angles is called an angle bisector.

Given that ∠XYZ along with a ray YO that divides ∠XYZ. In the adjoining figure, the ray YO divides ∠XYZ into two equal angles ∠XYO and ∠OYZ
i.e., ∠XYO = ∠OYZ.

3. Perpendicular Lines

Perpendicular Lines are the lines that intersect each other to form right tangles between them. In the adjoining e, lines AB and CD intersect at 0 such that ∠COB = ∠ COA = ∠AOD = ∠BOD = 90°.
Therefore, we say that AB is perpendicular to CD, i.e., (AB ⊥ CD).

4. Perpendicular Bisector

Perpendicular Bisector is the line that passes through the midpoint of the given line segment and also it is perpendicular to it. Here, the CD is the line segment. AB is the perpendicular bisector as ∠AOB = ∠AOD = 90° and CO = OD.

Related angles are nothing but the pairs of angles and assigned with specific names that we come across. Related angles have some conditions to mention. Learn the detailed concept of Related angles with images and examples in this article. Improve your preparation level by reading the entire concept without missing any subtopic. We have given complete information about Lines and Angles on our website for free of cost.

Different Types of Related Angles

Check different types of Related Angles along with examples to clearly understand the concept. We have explained each of them with definitions, solved examples, etc. Refer to the following modules and get a grip on it.

Complementary Angles

If the sum of the measures of two angles is about 90°, those angles are called complementary angles.

Facts of Complementary Angles

• The two right angles never complement each other.
• Also, the two obtuse angles never complement each other.
• Two complementary angles are always acute but there is no possibility of vice versa.

Example:

Let us take two angles which are complementary angles. If one angle is a, then the other angle is 90° – a.
An angle of 40° and another angle of 50° are complementary angles of each other.

Also, complement of 40° is 90° – 40° = 50°.
And complement of 50° is 90° – 50° = 40°

The ∠XOY is 40° and ∠MON is 50°. By adding two angles ∠XOY and ∠MON, we get 90° as they are the Complementary Angles.

Therefore, the sum of the ∠XOY and ∠MON is 90°
∠XOY + ∠MON = 90°.

Supplementary Angles

Supplementary Angles are the angles when the sum of the measures of two angles is 180°. If the sum of the two angles forms a straight angle, then those angles are called Supplementary Angles. If one angle is a, then the other angle is 180° – a.

Example:

Let us take two angles which are Supplementary angles. If one angle is x, then the other angle is 180° – x.
An angle of 110° and another angle of 70° are supplementary angles of each other. Also, a supplement of 110° is 180° – 110° = 70°.
And the supplement of 60° is 180° – 70° = 110°.

From the above figure, the ∠XOY is 110° and ∠MON is 70°. By adding two angles ∠XOY and ∠MON, we get 180° as they are the Supplementary Angles.

Therefore, the sum of the ∠XOY and ∠MON is 180°
∠XOY + ∠MON = 180°.

Adjacent Angles

The two angles are called to be Adjacent angles when they have a common arm, a common vertex, and also the non-common arms present on the opposite side of the common arm.

From the above figure, ∠ABD and ∠CBD are adjacent angles with the common arm BD. The B is the common vertex and BA, BC is opposite sides of BD.

Linear Pair

When two adjacent angles form a linear pair of angles with the non-common arms are two opposite rays. In other words, the sum of two adjacent angles is 180°.

From the above figure, the ∠XOY and ∠XOZ are two adjacent angles. By adding two angles ∠XOY and ∠XOZ, we get 180°.

Therefore, the sum of the ∠XOY and ∠XOZ is 180°
∠XOY + ∠XOZ = 180°.

Vertically Opposite Angles

The arms of the lines are opposite in direction and both lines are interesting to each other in Vertically opposite angles. The pair of vertically opposite angles are equal.

From the above figure, the ∠MOR and ∠SON and ∠MOS and ∠RON are pairs of vertically opposite angles.

Theorems on Related Angles

1. If a ray stands on a line, then the sum of adjacent angles formed is 180°.

Given: A ray BE standing on (AC) ⃡ such that ∠ABE and ∠CBE are formed.

Construction: Draw BD ⊥ AC.

Proof: Take the angle ABE.
Now ∠ABE = ∠ABD + ∠DBE ……………. (1)
Take the angle CBE.
Also ∠CBE = ∠CBD – ∠DBE ……………. (2)
Now, add equation 1 and equation 2.
Adding (1) and (2),
∠ABE + ∠CBE = ∠ABD + ∠CBD + ∠DBE – ∠DBE
∠DBE – ∠DBE = 0
= ∠ABD + ∠CBD
∠ABD = 90°; ∠CBD = 90°
Substitute ∠ABD and ∠CBD values in ∠ABD + ∠CBD
= 90° + 90°
= 180°

2. The sum of all the angles around a point is equal to 360°.

Given: A point O and rays OA, OB, OC, OD, OE which make angles around O.

Construction: Draw OD opposite to ray OA

Proof: Since, OB stands on DA therefore
∠AOB + ∠BOD = 180°
∠BOD = ∠BOC + ∠COD
Substitute ∠BOD = ∠BOC + ∠COD in ∠AOB + ∠BOD = 180°
∠AOB + (∠BOC + ∠COD) = 180°
∠AOB + ∠BOC + ∠COD = 180° ……………. (i)
Again OE stands on DA, therefore
∠DOE + ∠EOA = 180°
∠EOA = ∠EOF + ∠FOA
Substitute ∠EOA = ∠EOF + ∠FOA in ∠DOE + ∠EOA = 180°
∠DOE + (∠EOF + ∠FOA) = 180°
∠DOE + ∠EOF + ∠FOA = 180° ……………. (ii)
Now, add euqtion (i) and equation (ii)
Adding (i) and (ii),
∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF + ∠FOA
= 180° + 180°
= 360°

3. If two lines intersect, then vertically opposite angles are equal.

Given: MN and AB intersect at point O.

Proof: OB stands on MN.
Therefore, ∠MOB + ∠BON = 180° ……………. (i)
MO stands on AB
∠MOB + ∠MOA = 180° ……………. (ii)
From (i) and (ii),
∠MOB + ∠BON = ∠MOB + ∠MOA
∠BON + ∠MOA
Similarly, ∠MOB = ∠AON can be proved.

Comparison of Integers is a bit confusing topic and it requires proper understanding to solve correctly. We have provided the necessary information here when comparing or ordering the integer numbers. Know various methods like comparing integers using absolute value and using number line etc. Follow all the tips given and solve the problems related to it easily. Check the below sections to know the detailed information regarding the comparison and ordering of integers.

Comparison of Integers – Introduction

Are you confused while comparing the integers? We are sure you will get clarity on things by the end of this article. Most people feel confused while comparing negative integers, but if you follow an exact procedure, then it is easy for you to solve all the problems. Two Integers are compared mostly using conditional or arithmetic methods. It is also done using the number line method. When the integers are represented on a number line, we observe that, as we are moving towards the right, the value of the number increases. In the same way, when we move towards the left, the value decreases.

The right side consists of whole numbers and the left side consists of a negative number and zero is present in the middle of the number line. Zero is less than all the positive numbers and greater than negative numbers. It is neither a positive nor a negative number. If you have to order the integer numbers from least to highest, then you would order the numbers from the right of the number line to the left.

Symbolic Representation for comparing Integers

Integers can be compared using the symbols ‘<‘, ‘>’ or ‘=’

Procedure for Comparing or Ordering of Integer Numbers

Follow the below-listed steps while comparing or ordering integer numbers. They are along the lines

1. Consider, there are 2 numbers to compare. In that case, one of the numbers is greater than or less than or equal to other numbers.
2. If the count of the first number is higher than the second number, then that first number is greater than the second number. The symbol ‘>’ is used to represent greater than and vice versa. The symbol ‘<‘ is used to represent less than the symbol.
3. For example, consider 15 and 10 as the two integer numbers to be compared. 15 is greater than 10, as 15 has a higher count when compared to the number 10.
4. Using a number line, point all the integer numbers on a number line and thus the comparison can be done easily.

Key Points to Remember while comparing Integers

Do remember the following points while ordering integers or comparing them. These help you to solve your problems easily.

1. As we know, the integer is the set of whole numbers and also their opposites. In other words, Integer numbers are positive or negative whole numbers.
2. Integers do not include fractions and decimals which are known as rational numbers. Integers only include whole numbers.
3. Look at the integers and compare the values and then plot the integers on a number line.
4. Compare the integer values and use the symbols of ‘>’,'<‘,’=’. Drawing the number line will help you to visualize the integers.
5. Then, order the integer values from least to greatest.
6. Arrange in order from ‘Least’ to ‘Greatest’ i.e. left to right. ‘Greatest’ to ‘Least’ is right to left.
7. Consider the absolute value of the number which is the distance from the number line. Absolute values are always positive because it signifies the distance.

Comparing Integers Examples

Example 1:

Josh has $23, his friend Krista has $25 and another friend Mark has $8, his friend Luis has -$5, another friend Tessa has -$10. Josh wants to order and compare the amount of money they have?

Solution:

First, draw the number line and make sure you include all the numbers you have to plot. Plot all the above money values on the number line.

Then, find out the number which is furthest to the left of the number line. -10 is the smallest number which means Tessa is with the least amount of money.

After -10, -5 is the second least number. Therefore, Luis has the second least amount of money. Next comes the Mark with the next least amount of money of $8. Thus the pattern follows with Josh and Krista.

This analysis will give you the numbers that are ordered from least to greatest.

Therefore, the final solution is in the order of -$10,-$5,$8,$23,$25. Thus, Tessa has less amount of money and Krista has the highest amount of money.

Example 2:

Choose the inequality symbol which goes in the blank to make the given statement true: -22 _____ -32

Solution:

First, visualize or draw a number line. Then, plot the points of -22 and -32.

Then, determine where -22 is on the drawn number line when compared to -32. -22 will be to the right of -32.

As it is on the right side, it determines -22 is greater than -32.

The final solution is -22>-32

The inequality symbol which goes in the blank is “>”

Example 3:

Arrange the numbers −4, 2, 8, 9, −11, −5 in order using the number line?

Solution:

First of all, draw a number line and include all the numbers in such a way that the least number -11 is included and the highest number 9 is included.

Once, you draw the number line, start plotting the numbers as per the ascending order. First, start by -11, then -5 followed by -4 and so on.

Continue by reading the points on the plotted number line from left to right which gives you the order of the numbers from least to highest.

The final solution is 11, -5, -4, 2, 8, 9.

Example 4: 

Arrange the numbers 5, −16, 6, -12, 1, 4 in order using the number line?

Solution:

First of all, draw a number line that includes all the numbers given in the above question. The least number must be greater than -16 and the highest number must be greater than 6.

Once, you start drawing the number line, make sure you arrange the numbers in ascending order. The first least number comes as -16, then -12 followed by 1 and so on.

Now, plot the numbers from left to right which you have previously arranged in order. Once you plot the numbers, you will get the list of numbers from least to highest values.

The solution is -16, -12, 1, 4, 5, 6.

Example 5: 

Arrange the numbers −3, −2, −7, −12, −1 in order using the number line?

Solution:

The first step is to draw the number line by including all the numbers that are given in the question. Make sure you draw a number line by including a few extra numbers.

Now, that you are ready with the number line, the next step is to plot the numbers on the graph. Analyze the order of the numbers from least to highest. The least number from the given series is -12. -7 is the second least number, then comes -3, and so on.

Once you are ready with the order of numbers, you can easily plot them on the number line. Plot the numbers from left to right in the order you arranged previously. After plotting the numbers, you will find the final result of numbers from least to highest values.

The final solution is -12,-7,-3,-2,-1.

Here is the detailed information about Comparison of Integers and also how to order the integers. This concept will help you throughout your integers problem-solving methods. I hope, you got clarity on how to order, compare the integers, and also how to include inequality symbols for various integer numbers. For more details regarding other mathematical topics, stay tuned to our site.

The polygon area is the region occupied by the polygon. The basic types of polygons are regular polygon and irregular polygon. Learn about the area of a polygon, polygon definition, central point of a polygon, radius of the inscribed circle, circumscribed circle, and polygon of n sides.

Types of Polygons

Polygon is a closed shape in a two dimensional plane with straight lines. It has an infinite number of sides but all sides are straight lines. The line segments of a polygon are called the sides or edges. The point of intersection of two line segments is called the vertex.

Based on the sides and angles, the polygons are classified into different types. They are regular polygon, irregular polygon, convex polygon, and concave polygon. The regular polygon measures all sides and angles equal. Some of the polygons are triangle, square, pentagon, hexagon, and others.

Central Point of a Polygon:

The inscribed and the circumscribed circles of a polygon have the same center, called the central point of the polygon.

The radius of the Inscribed Circle of a Polygon:

The length of the perpendicular from the central point of a polygon upon any one of its sides is called the radius of the inscribed circle of the polygon. The radius of the inscribed circle of a polygon is denoted by r.

The radius of the Circumscribed Circle of a Polygon:

The line segment joining the central point of a polygon to any vertex is the radius of the circumscribed circle of the polygon. The radius of the circumscribed circle of a polygon is denoted by R.

Properties of Polygon

The polygon’s properties are based on the sides and angles. Refer to the below properties and learn entirely about Polygons. They are as follows

• The sum of all interior angles of an n-sided polygon is (n – 2) x 180°.
• The number of triangles formed by joining the diagonals from one corner of a polygon = n – 2.
• The number of diagonals in a polygon with n sides = n(n – 3)/2.
• The measure of each interior angle of n-sided regular polygon = [(n – 2) × 180°]/n
• The measure of each exterior angle of n sided regular polygon = 360°/n

Area of Regular Polygons Formulas

• If a regular polygon has three sides, then it is called a triangle and its area formula is ½ x base x height
• If the regular polygon has 4 sides, then it called square, and its area formula is side².
• If the regular polygon has 5 sides, then it is called the pentagon and its area formula is /2 x side length × distance from the center of sides to the center of the pentagon.
• If the regular polygon has 6 sides, then it is called the hexagon and its area formula is (3√3)/2 × distance from the center of sides to the center of the hexagon.
• The area of the rectangle formula is length x breadth.
• The area of the rhombus formula is 1/2 x product of diagonals.
• The area of the hexagon is [3(√3)a²/2] square units.
• The area of the octagon is 2a² (1 + √2) square units.
• The area of a polygon is n/2 × a × √(R² – a²/4) square units.

How to Find the Area of N- Sided Polygons?

Follow the below mentioned simple steps and instructions to calculate the area of a polygon having N-Sides easily.

• If the given is a regular polygon, then substitute the values in the formula.
• If the given figure is an irregular polygon or regular polygon having n sides can use the below process.
• Divide the geometric figure into the combination of regular polygons like triangle, square, rectangle, etc.
• Find the area of each shape.
• Add up those areas to get the polygon area in square units.

Area of a Polygon Word Problems

Example 1.

Find the area of a regular pentagon whose perimeter is 40 units and whose apothem is 5 units?

Solution:

Given that,

The perimeter of the regular pentagon = 40 units

Apothem = 5 units

Area of the regular pentagon = ½ (perimeter) (apothem)

= ½ (40) (5)

= 20 x 5 = 100 sq units

Therefore, the area of regular pentagon = 100 sq units.

Example 2.

Find the area of a regular octagon each of whose sides measures 25 cm?

Solution:

Given that,

Side of a regular octagon a = 25 cm

Area of octagon A = 2a² (1 + √2)

Substitute a = 25 in above equation.

A = 2 x 25² (1 + √2)

= 2 x 625 (1 + √2) = 1250 (1 + √2)

= 1250 x 2.414 = 3,017.76 cm²

Therefore, regular octagon area is 3017.76 cm².

Example 3.

The area of a regular pentagon is 215 cm². Find the length of the side of the pentagon?

Solution:

Given that,

Regular pentagon area A = 215 cm²

Pentagon area formula is 1/4 ((√(5 (5 + 2 √5) s²))

Pentagon area = 215 cm²

1/4 ((√(5 (5 + 2 √5) s²)) = 215

(√(5 (5 + 2 √5) s²)) = 215 x 4 = 860

s² = 860/(√(5 (5 + 2 √5))

s² = 860/6.88 = 125

s = √125

s = 11.18

Therefore, side of regular pentagon is 11.18 cm.

Example 4.

Find the area of a regular polygon having 12 sides and each side length is 5 m?

Solution:

Given that,

Number of sides of a polygon n = 12

Side length a= 5 cm

Area of regular polygon A = [n * a² * cot(π/n)]/4

A = [12 x 5² x cot(π/12)]/4

= [12 x 25 x (2 + √3)]/4

= (300 x (2 + √3))/4

= 75 (2 + √3)

= 75 (3.73) = 279.9 cm²

Therefore, the area of a regular polygon is 279.9 cm².

The trapezium is a quadrilateral that has four sides and one set of parallel sides. The area of trapezium is the region covered by the solid in a two-dimensional plane. Other types of quadrilaterals are parallelogram, rectangle, square, and rhombus. Find the definition, formula, and solved examples on the trapezium area.

Trapezium Definition

The trapezium is a flat closed shape having 4 straight sides with one pair of parallel sides. The parallel sides are called the bases and the non-parallel sides are called the legs. The different types of the trapezium are isosceles trapezium, scalene trapezium, and right trapezium. Some of the properties of the trapezium are along the lines:

• Every trapezium has exactly one pair of parallel and opposite sides.
• The diagonals intersect each other.
• The non-parallel sides in trapezium are unequal except in isosceles trapezium.
• The line that joins the mid-points of the non-parallel sides is always parallel to the bases or parallel sides which is equal to half of the sum of parallel sides.
• In isosceles trapezium, the non parallel sides are congruent
• The sum of interior angles is equal to 360°.
• The sum of two adjacent angles is 180°.

Area of a Trapezium

The trapezium area is the region covered by the trapezium in a two-dimensional plane. It can be measured by taking the average of two bases and multiplied the result by altitude. Therefore, the area of the trapezium formula is given as,

Trapezium Area A = h(a + b)/2 square units

Where,

a, b are the bases

h is the height or altitude

Area of an isosceles Trapezium = 1/2 [√(c² – (a – b)²) (a + b)]

Where,

a = AB, b = CD, c = AD

The perimeter of trapezium P = a + b + c + d units

Perimeter of the Isosceles Trapezium = a + b + 2c units

Prove that:

Area of trapezium ABCD = 1/2[AD + BC] h

Area of trapezium ABCD = area (∆DFA) + area (rectangle DFEC) + area (∆CEB)

= (¹/₂ × AF × DF) + (FE × DF) + (¹/₂ × EB × CE)

= (¹/₂ × AF × h) + (FE × h) + (¹/₂ × EB × h)

= h/2 x (AF + FE + EB)

= h/2 × (AB + FE)

= h/2 x (AB + DC)

= ¹/₂ × (sum of parallel sides) × (distance between them)

Therefore, The area of a trapezium = ¹/₂ × (sum of parallel sides) × (distance between them)

How to find the Trapezium Area?

Follow the steps to calculate the Area of the Trapezium by referring to the below steps. They are as such

• Observe the question carefully and make a note of known parameters.
• Find either height, base sides, or sum of parallel sides, the distance between the parallel sides.
• Substitute these values in the formula to get the area.

Area of a Trapezium Questions and Answers

Example 1.

In a trapezium, the measurement of one parallel side two more than the other parallel side and the height is 6 cm. The area of the trapezium is 216 cm². Find the lengths of the two parallel sides?

Solution:

Let a, b are the parallel sides of the trapezium.

According to the question,

One parallel side is two more than the other parallel side

a = b + 2

Area of the trapezium = 216 cm²

½(a + b)h = 216

Substitute a = b + 2, h = 6 cm in above equation.

½(b + 2 + b)4 = 216

2(2b + 2) = 216

2b + 2 = 216/2

2b + 2 = 108

2b = 108 – 2

2b = 106

b = 106/2

b = 53

Put b = 53 in a = b + 2

a = 53 + 2 = 55

So, the lengths of the two parallel sides are 55 cm, 53 cm.

Example 2.

The area of a trapezium is 148 cm² and its height is 12 cm. If one of the parallel sides is double of the other, find the two parallel sides?

Solution:

Given that,

Area of the trapezium = 148 cm²

Height = 12 cm

Let one side of the trapezium is a, then another parallel side is b.

As per the question,

One of the parallel sides is double of the other.

a = 2b

Area of the trapezium = 148 cm²

½(a + b)h = 148

Substitute a = 2b, h = 12 in above equation.

½(2b + b) x 12 = 148

3b x 6 = 148

18b = 148

b = 148/18

b = 8.2

Put b = 8.2 in a = 2b

a = 8.2 x 2

a = 16.4

Therefore, the two parallel sides of length are 16.4 cm, 8.2 cm.

Example 3.

Find the area of a trapezium whose parallel sides are AB = 42 cm, CD = 16 cm, and non parallel side AD = 15 cm, BC = 15 cm?

Solution:

In trapezium ABCD,

CE = 15 cm

CD = 16 cm, AE = 16 cm

EB = AB – AE = 42 – 16 = 26 cm

Now, in ∆ EBC

S = (15 + 15 + 26)/2

S = 56/2 = 28

Area of triangle = √(s(s – a) (s – b) (s – c))

= √(28 (28 – 15) (28 – 15) (28 – 26))

= √(28 (13) (13) (2))

= √(9464) = 97.2 cm²

Area of ∆EBC = 1/2 × EB × CP

97.2 = 1/2 × 26 × CP

CP x 13 = 97.2

CP = 97.2/13 = 7.48 cm

Therefore, h = 7.48 cm

Area of trapezium = 1/2(p₁ + p₂) × h

= 1/2(58) × 7.48

= 29 x 7.48 = 216.92 cm²

Therefore, area of a trapezium is 216.92 cm².

Example 4.

The length of the parallel sides of a trapezium is in the ratio 5: 8 and the distance between them is 18 cm. If the area of the trapezium is 256 cm², find the length of the parallel sides?

Solution:

Given that,

The ratio of parallel sides of the trapezium is 5: 8.

The distance between parallel sides = 18 cm

Area of the trapezium = 256 cm²

Let the common ratio be x,

Then, parallel sides are 5x, 8x.

Area of trapezium = 1/2 (p₁ + p₂) h

256 = 1/2 (5x + 8x) x 18

256 = 9(13x)

256 = 117x

x = 256/117

x = 2.188

5x = 5 x 2.188 = 10.94, 8x = 8 x 2.188 = 17.5

Therefore, the length of parallel sides area are 10.94 cm and 17.5 cm.

FAQs on Area of a Trapezium

1. What is the formula for the Area of a trapezium?

Area of trapezium = ½ x Sum of parallel sides x Distance between the parallel sides

Trapezium area = ½ x (a+b) x h

2. What are the applications of a trapezium?

It is used in various physics computations and mathematical calculations.

3. What are the properties of Trapezium?
The trapezium has 4 unequal sides, 2 are parallel and 2 are nonparallel. The sum of the interior angles is 360 degrees and diagonals bisect each other.