Constructions

 Mind Maps

Class X - maths: Constructions

Q) How to construct different types of quadrilaterals?

Q) How many diagonal are there in a quadrilateral?

Q) Construction of Pie Chart { The Value Of The Component /________?

Q) An angle whose measure is 90° is called ___________?

Q) An angle whose measure is 120° is called ___________?

Q) Use your protractor to draw these angles: 45, 60.

Q) The four line segments AB, BC, CD and DA are called ________ of a quadrilateral.

Q) The mid-point of the diameter is the _________ of the circle.

Q) A diameter of a circle has the extreme points (8, 9) and (-7, -5). What would be the co-ordinates of the centre?

Q) A point divides internally the line-segment joining the points (7, 9) and (-8, 4) in the ratio 4 : 3. Find the co-ordinates of the point?

Q) A (8, 5) and B (7, -6) are two given points and the point C divides the line-segment AB externally in the ratio 7 : 5. Find the co-ordinates of C?

Q) Find the ratio in which the line-segment joining the points (7, - 9) and (7,2) is divided by the Y-axis?

Q) Find the ratio in which the point (- 33, 26) divides the line segment joining the points (- 6, 2) and (8, - 9)?

Q) To divide a given line segment AB at a point P such that AP?PB = 2 ? 5 ,the line is to be divided in the ratio

Q) Is the construction of a triangle possible in which the lengths of sides are 25 cm, 23 cm and 9 cm?

Q) To draw two tangents to a circle of radius 3 cm and center O, from a point P at a distance of 5 cm from O,a circle is drawn with center at mid-point of OP and radius

Q) Is it possible to have a triangle whose sides are 5 cm, 6 cm and 4 cm?

Q) Construct a quadrilateral ABCD in which AB = 9.8cm,BC = 3.3cm, CD = 6.6cm, AD = 4.2cm and diagonal AC = 10cm?

Q) Construct a quadrilateral ABCD in which AB = 8cm BC = 2.8 cm, AD = 8cm,diagonal AC = 2cm and diagonal BD = 2.8cm?

Q) Construct a quadrilateral ABCD in which AB = 3.6 cm, ∠ABC = 80°,BC = 8cm,∠BAD = 120° and AD = 5 cm?

Q) Construct a quadrilateral PQRS in which PQ = 8.5 cm ∠PQR = 120°, QR = 6.8 cm, ∠QRS = 100° and ∠QPS = 60.

Q) Construct a quadrilateral ABCD in which AB = 9.8 cm, BC = 3.5cm, CD = 7.5 cm, AD = 9cm and ∠B = 70°.

Q) To divide a line segment internally in the ratio 3 ? 5 , the numbers of arcs to be drawn on a ray inclined to the line segment is

Q) To draw two tangents to a circle,which are inclined at an angle of 60 ° , the perpendiculars are to be drawn at the ends of two radii which are at an angle of?

Q) A point divides internally the line-segment joining the points (8, 9) and (-7, 4) in the ratio 2 : 3. Find the co-ordinates of the point.

Q) A (4, 5) and B (7, -1) are two given points and the point C divides the line-segment AB externally in the ratio 4 : 3. Find the co-ordinates of C.

Q) To a circle ,two tangents are to be drawn which are inclined at an angle of 70 °.For that we have to draw perpendiculars to two radii which are at a an angle of?

Q) To drawn two tangents to a circle of radius 3 cm and center O from a point P,8 cm away from O ,we draw a circle with center at the mid point of OP and radius equal to

Contents
• Constructions
• Division of a Line Segment
• To divide a line segment in a given ratio
• Construction Of Triangle
• Construction Of Tangents To A Circle
• To Construct a Right Triangle when its Hypotenuse and One Side are given
• To Construct A Triangle Whose Three Sides Are Given
• To Construct a Triangle when Two of its Angles and the included Side are given
• To Construct A Triangle When Two Of Its Sides And The Included Angles Are Given
• Construction of Angles by using Compass
• Interior and Exterior of an Angle
• Construction of quadrilaterals when four sides and one diagonal are given
• Construction of quadrilaterals when three sides and two diagonals are given
• Construction of quadrilaterals when three sides and two included angles are given
• Construction of quadrilaterals when two adjacent sides and three angles are given
• Construction of quadrilaterals when four sides and one angle are given
• Construct Different Types of Quadrilaterals
• Examples of Properties of Triangle
• Construction of Pie Chart
• example on construction of pie chart/pie graph
• Line Graph
• Steps of construction of line graph
• Division of Line Segment
• External Division of line segment
• Example on Division of Line Segment
Formulae

• Division of a Line Segment
Suppose a line segment is given and you have to divide it in a given ratio say 3:2
• Measuring the length and then marking a point on it that divides it in the given ratio

Construction Of Triangle

• To construct a triangle similar to a given triangle as per given scale factor.
• This construction involves two different situations. In one, the triangle to be constructed is smaller and in the other it is larger than the given triangle.

Construction Of Tangents To A Circle

• If centre of the circle is not given, you may locate its centre first by taking any two non-parallel chords and then finding the point of intersection of their perpendicular bisectors

A simple closed figure formed by joining four line segments is called a quadrilateral. It has 4 sides, 4 angles, 4 vertices and two diagonals.

How to construct different types of quadrilaterals?
The different types of quadrilaterals are constructed and classified by relationships of their sides, angles and diagonals.

Construction of Pie Chart
{ The Value Of The Component / Total Value } x 360°

If the values of observation/components are expressed in percentage, then the centre angle corresponding to particular observation/component is given by
{ Percentage Value Of Component / 100 } x 360°

Constructions

Division of a Line Segment
• Suppose a line segment is given and you have to divide it in a given ratio say 3:2.
• Measuring the length and then marking a point on it that divides it in the given ratio. But suppose you do not have any way of measuring it precisely, how would you find the point?
• We give below two ways for finding such a point

To divide a line segment in a given ratio.

Given a line segment AB,we want to divide it in the ratio m:n,where both m and n are positive integers.To help you to understand it,we shall take m = 3 and n = 2.

Steps of Construction
1. Draw any ray AX, making an acute angle with AB.
2. Locate (5 = m + n) points A1,A2,A3,A4 and A5 on AX so that
AA1 = A1A2 = A2A3 =A3A4= A4A5.
3. Join BA5.
4. Through the point A3(m = 3),draw a line parallel to A5B (by making an angle equal to∠AA5B) at intersecting AB at the point C. Then,AC:CB=3:2.
Alternative Method

Steps of Construction :
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray BY parallel to AX by making /_ABY equal to /_BAX.
3. Locate the points A1,A2,A3(m = 3) on AX and B1,B2(n = 2) on BY such that
AA1 = A1A2 = A2A3= BB1 = B1B2.
4. Join A3B2.Let it intersect AB at a point C (see Fig.11.2).

Then AC:CB = 3:2.
Here △AA3C is similar to △BB2C
Then AA3/BB2=AC/BC
BY construction,A A_3/BB_2=3/2 therefore AC/BC=3/2

Construction Of Triangle
• To construct a triangle similar to a given triangle as per given scale factor.
• This construction involves two different situations. In one, the triangle to be constructed is smaller and in the other it is larger than the given triangle.
• Here, the scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle.

Let us take the following examples for understanding the constructions involved. The same methods would apply for the general case also.

Example 1: Construct a triangle similar to a given triangle ABC with its sides equal to 3/4 of the corresponding sides of the triangle ABC.

Steps of Construction :
1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
2. Locate 4(the greater of 3 and 4 in 3/4) points B1, B2, B3 and B4 on BX so that
BB1= B1B2= B2B3= B3B4.
3. Join B4C and draw a line through B3(the 3rd point,3 being smaller of 3 and 4 in 3/4) parallel to B4C to intersect BC at C'.
4. Draw a line through C' parallel to the line CA to intersect BA at A'(see Fig.11.3). Then, △A'BC'is the required triangle. Let us now see how this construction gives the required triangle.). BC'/C'C=3/1

Therefore,BC/BC'=(BC'+C'C)/BC'=1+1/3=3/4

Example 2 : Construct a triangle similar to a given triangle ABC with its sides equal to 5/3 of the corresponding sides of the triangle ABC (i.e.,of scale factor 5/3).

Steps of Construction:
• Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
• Locate 5 points (the greater of 5 and 3 in 5/3) B1, B2, B3, B4and B5 on BX so thatBB1= B1B2= B2B3= B3B4= B4B5.
• Join B3(the 3rd point,3 being smaller of 3 and 5 in 5/3)to C and draw a line through B5 parallel to B3C,intersecting the extended line segment BC at C'.
• Draw a line through C' parallel to CA intersecting the extended line segment BA at A'(see Fig.11.4).

Then A'BC'is the required triangle.
For justification of the construction, note that △ABC ~ △A'B'C'
Therefore(AB )/(A'B)= (AC)/(A'C') = (BC)/(BC')
But(BC)/(BC') = (BB_3)/(BB_5) = 3/5
(BC')/ (BC)=5/3
therefore (A'B)/(AB) = (A'C)/(AC)= (BC')/(BC)

Construction Of Tangents To A Circle

Construction

To construct the tangents to a circle from a point outside it. We are given a circle with centre O and a point P outside it. We have to construct the two tangents from P to the circle.

Steps of Construction
1. Join PO and bisect it. Let M be the midpoint of PO.
2. Taking M as centre and MO as radius, draw a circle.Let it intersect the given circle at the points Q and R.
3. Join PQ and PR. Then PQ and PR are the required two tangents (see Fig. 11.5).
• Now let us see how this construction works. Join OQ.
Then /_PQO is an angle in the semicircle and therefore /_PQO = 90°,PQ _|_ OQ
• OQ is a radius of the given circle,PQ has to be a tangent to the circle. Similarly,PR is also a tangent to the circle.

Note : If centre of the circle is not given, you may locate its centre first by taking any two non-parallel chords and then finding the point of intersection of their perpendicular bisectors

To Construct a Right Triangle when its Hypotenuse and One Side are given
• To construct a right triangle when its hypotenuse and one side are given, let's follow the examples.

1.Draw a right angled triangle at C in which AB = 7 cm and BC = 5 cm.

Steps Of Construction
1. Draw BC = 5 cm.
2. Make /_BCX = 90°.
3. With B as centre and radius 7 cm,cut off BA = 7 cm.
4. Join AB.

∴ DeltaABC is the required Delta.

2.Draw a right angled triangle at C in which XY = 7.5 cm and YZ = 5.5 cm.

Steps of Construction
1. Draw YZ = 5.5cm.
2. Make/_YZA = 90°.
3. With Y as centre and radius 7.5cm,cut off XY = 7.5 cm.
4. Join XY.

∴ DeltaXYZ is the required triangle.

To Construct A Triangle Whose Three Sides Are Given

1. Draw a triangle ABC such that BC = 4 cm, AB = 3.5 cm, AC = 4.5 cm.
[Let us draw a rough sketch of the triangle ABC. Name the vertices and show the given measures.]

Steps of Construction
1. Draw a line segment BC = 4 cm.
2. With B as Centre and radius 3.5 cm,draw an arc.
3. With C as centre and radius 4.5 cm,draw another arc cutting the previous arc at A.
4. Join AB and AC.

∴ DeltaABC is the required triangle.

2.Draw a triangle ABC such that BC = 4.5 cm, AB = 3.5 cm, AC = 5 cm.

Steps of Construction

Now let's follow these steps one by one

1. Draw a line segment BC = 4.5 cm.
2. With B as Centre and radius 3.5 cm, draw an arc.
3. With C as centre and radius 5 cm, draw another arc cutting the previous arc at A.
4. Join AB and AC.

∴ DeltaABC is the required triangle.

To Construct a Triangle when Two of its Angles and the included Side are given

To construct a triangle when two of its angles and the included side are given, let's follow the examples.
1. Draw a DeltaLMN in which/_ M = 50° and /_N = 60° and side MN = 5 cm.

Steps of Construction
1. Draw a line segment MN = 5 cm.
2. Make/_XMN = 50°
3. Make /_YNM = 60°
4. Let XM and YN intersect at L.

2. Draw a DeltaMNO in which /_M = 70° and /_N = 60° and side MN = 7 cm.

Steps of Construction
1. Draw a line segment MN = 7 cm.
2. Make/_AMN = 70°
3. Make /_BNM = 60°
4. Let AM and BN intersect at O.

∴ DeltaOMN is the required triangle.

To Construct A Triangle When Two Of Its Sides And The Included Angles Are Given

To construct a triangle when two of its sides and the included angles are given,let's follow the examples.

1. Draw a triangle ABC in which AB = 5cm, BC= 7cm, and/_B = 75°.

Steps of Construction
1. Draw a line segment AB = 5cm.
2. Make/_ABC = 75°.
3. Cut of BC = 7cm.
4. Join AC.

∴ DeltaABC is the required triangle.

2.Draw a triangle ABC in which AB = 4.5cm, BC = 6cm, and/_B = 80°.

Steps of Construction
1. Draw a line segment AB = 4.5cm.
2. Make/_ABC = 80°.
3. Cut of BC = 6cm.
4. Join AC.

∴ DeltaABC is the required triangle.

Construction of Angles by using Compass

1. Construction of an Angle of 60° by using Compass

Step Of Construction
1. Draw a ray OA.
2. With O as centre and any suitable radius draw an arc above OA cutting it at a point B.
3. With B as centre and the same radius as before,draw another arc to cut the previous arc at C.
4. Join OC and produce it to D.

Then /_AOD = 60°.

2. Construction of an Angle of 120° by using Compass

Step of Construction
1. Draw a ray OA.
2. With O as centre and any suitable radius draw an arc cutting OA at B.
3. With B as centre and the same radius cut the arc at C, then with C as centre and same radius cut the arc at D. Join OD and produce it to E.

Then, /_AOE = 120°.

3. Construction of an Angle of 30° by using Compass

Step of Construction
1. Construction an angle/_AOD = 60° as shown.
2. Draw the bisector OE of/_AOD.

Then, /_AOD = 30°.

4. Construction of an Angle of 90° by using Compass

Step of Construction
1. Take any ray OA.
2. With O as centre and any convenient radius, draw an arc cutting OA at B.
3. With B as centre and the same radius, draw an cutting the first arc at C.
4. With C as centre and the same radius, cut off an arc cutting again the first arc at D.
5. With C and D as centre and radius of more than half of CD, draw two arcs cutting each other at E, join OE.

Then, /_EOA = 90°.

5. Construction of an Angle of 75° by using Compass

Step of Construction
1. Take a ray OA.
2. With O as centre and any convenient radius, draw an arc cutting OA at C.
3. With C as centre and the same radius, draw an cutting the first arc at M.
4. With M as centre and the same radius, cut off an arc cutting again the first arc at L.
5. With L and M as centre and radius of more than half of LM, draw two arcs cutting each other at B, join OB which is making 90°.
6. Now with N and M as centres again draw two arcs cutting each other at P.
7. Join OP.

Then,/_POA = 75°.

6. Construction of an Angle of 105° by using Compass

Step of Construction
1. After making 90°angle take L and N as centre and draw two arcs cutting each other at S.
2. Join SO.

Then, /_SOA = 105°.

7. Construction of an Angle of 135° by using Compass

Step of Construction
1. Construct /_AOD = 90°
2. Produce /_AO to B.
3. Draw OE to bisect /_DOB.
4. /_DOE = 45°
5. /_EOA = 45° + 90° = 135°

Then, /_EOA = 135°.

8. Construction of an Angle of 150° by using Compass

Step of Construction
1. Construct /_AOC = 120°
2. Produce /_AO to B.
3. Draw OD to bisect /_COB.
4. Now /_COD = 30°
5. Therefore, /_AOD = 120°+ 30° = 150°

Then, /_AOD = 150°

Interior and Exterior of an Angle

• The shaded portion between the arms BA and BC of the angle ABC can be extended indefinitely.
• This portion is called the interior of the angle. X is a point in the interior of the angle. The point Y lies in the exterior of the angle. The point Z lies on the angle.
• In the above figure, here /_1 is called the interior angle because it lies inside the two arms. /_2 is called the exterior angle.
• Whenever two rays meet, two angle are formed - one an interior angle and other an exterior angle. The size of an angle is measured by the amount of turn or rotation of two arms and not by how long the arms appear to be.
• Here /_a is greater than /_b,/_b is greater than /_c.

• A simple closed figure formed by joining four line segments is called a quadrilateral. It has 4 sides, 4 angles, 4 vertices and two diagonals.
• In a quadrilateral if A, B, C, D are four points in a plane such that no three of them are collinear and the line segments AB, BC, CD and DA do not intersect except at their end points. Then, the figure formed by these four line segments is called the quadrilateral ABCD.

• the four points A, B, C, D are called its vertices.
• the four line segments AB, BC, CD and DA are called its sides
• /_DAB, /_ABC, /_BCD and /_CDA are called its angles, to be denoted by /_A, /_B, /_C and /_D respectively, and
• the line segments AC and BD are called its diagonals.

Construction of quadrilaterals we will learn how to construct a quadrilateral. We know that a quadrilateral has ten parts in all: four sides, four angles and two diagonals. To construct a quadrilateral, we shall need data about five specified parts of it.

We consider the following five cases and explain the construction in each case by an example. We divide the required quadrilateral into two triangles which can be easily constructed. These two triangles together will form a quadrilateral.

I.Construction of quadrilaterals when four sides and one diagonal are given

1. Construct a quadrilateral ABCD in which AB =4.8cm,BC = 4.3cm, CD = 3.6cm, AD = 4.2cm and diagonal AC=6cm.
Solution:

First we draw a rough sketch of quadrilateral ABCD and write down its dimensions, as shown. We may divide it into two triangles,namely DEltaABC and DeltaACD.(Rough Sketch)

Steps of Construction

Step 1: Draw AB = 4.8 cm.
Step 2: With A as center and radius equal to 6 cm,draw an arc.
Step 3: With B as center and radius equal to 4.3 cm, draw another arc, cutting the previous arc at C.
Step4: Join BC.
Step 5: With A as center and radius equal to 4.2 cm, draw an arc.
Step 6: With C as center and radius equal to 3.6 cm, draw another arc, cutting the previous arc at D.
Step 7: Join AD and CD.
Then, ABCD is the required quadrilateral.

II. Construction of quadrilaterals when three sides and two diagonals are given

2. Construct a quadrilateral ABCD in which AB = 4cm BC = 3.8 cm, AD = 3cm,diagonal AC = 5 cm and diagonal BD = 4.6cm.
Solution:

Steps of Construction

Step 1: Draw AB = 4 cm.
Step 2: With A as center and radius equal to 5 cm, draw an arc.
Step 3: With B as center and radius equal to 3.8 cm, draw another arc, cutting the previous arc at C.
Step 4: Join BC.
Step 5: With A as center and radius equal to 3 cm, draw an arc.
Step 6: With B as center and radius equal to 4.6 cm draw another arc,cutting the previous arc at D.
Step 7: Join AD and CD.
Then, ABCD is the required quadrilateral.

III. Construction of quadrilaterals when three sides and two included angles are given:

3. Construct a quadrilateral ABCD in which AB = 3.6 cm, /_ABC = 80°,BC = 4cm,/_BAD = 120° and AD = 5 cm.
Solution:

Steps of Construction:

Step 1: Draw AB = 3.6 cm.
Step 2: Make /_ABX = 80°.
Step 3: With B as center and radius equal to 4 cm, draw an arc, cutting BX at C.
Step 4: Make /_BAY = 120°.
Step 5: With A as center and 5 cm as radius, draw an arc, cutting AY at D.
Step 6: Join CD.
Then, ABCD is the required quadrilateral.

IV. Construction of quadrilaterals when two adjacent sides and three angles are given

4. Construct a quadrilateral PQRS in which PQ = 4.5 cm /_PQR = 120°, QR = 3.8 cm, /_QRS = 100° and /_QPS = 60°.

Steps of Construction

Step 1: Draw PQ = 4.5 cm.
Step 2: Make /_PQX = 120°.
Step 3: With Q as center and radius 3.8 cm, draw an arc, cutting QX at R. Join QR.
Step 4: Make /_QRY = 100°.
Step 5: /_QPZ = 60° so that PZ and RY intersect each other at the point S.
Then, PQRS is the required quadrilateral.

V.Construction of quadrilaterals when four sides and one angle are given:

5. Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.4cm, CD = 4.5 cm, AD = 5cm and /_B = 80°.
Solution:

Steps of Construction

Step 1: Draw AB = 3.8 cm.
Step 2: Make /_ABX = 80°.
Step 3: From B, set off BC = 3.4 cm.
Step 4:With A as center and radius equal to 5 cm draw an arc.
Step 5: With C as center and radius equal to 4.5 cm, draw another arc, cutting the previous arc at D.
Step 5: Join AD and CD.
Then ABCD is the required quadrilateral.

How to construct different types of quadrilaterals?
The different types of quadrilaterals are constructed and classified by relationships of their sides, angles and diagonals.
Some of the constructions of different types of quadrilaterals are given below along with the steps-by-step explanation.

1.Construct a parallelogram ABCD in which AB = 6cm, BC = 4.5cm and diagonal AC = 6.8cm.
Solution:

Steps of Construction
1. Draw AB = 6 cm.
2. With A as center and radius 6.8 cm, draw an arc.
3. With B as center and radius 4.5 cm draw another arc,cutting the previous arc at C.
4. Join BC and AC.
5. With A as center and radius 4.5 cm, draw an arc.
6. With C as center and radius 6 cm draw another arc, cutting the previously drawn arc at D.
7. Join DA and DC.

Then, ABCD is the required parallelogram.

2. Construct a parallelogram,one of whose sides is 5.2cm and whose diagonals are 6cm and 6.4cm. Solution:

We know that the diagonals of a parallelogram bisect each other. Make a rough sketch of the required parallelogram, as shown

Steps of Construction
1. Draw AB = 5.2 cm.
2. With A as center and radius 3.2 cm, draw an arc.
3. With B as center and radius 3 cm draw another arc, cutting the previous arc at O.
4. Join OA and OB.
5. Produce AO to C such that OC = AO and produce BO to D such that OD = OB.
6. Join AD, BC and CD.

Then, ABCD is the required parallelogram.

3. Construct a parallelogram whose diagonals are 5.4 cm and 6.2 cm and an angle between them is 70°.

Solution:
We know that the diagonals of a parallelogram bisect each other. So, we may proceed according to the steps given below.

Steps of Construction
1. Draw AC = 5.4 cm.
2. Bisect AC at O.
3. Make /_COX = 70° and produce XO to Y.
4. Set off OB = 1/2 (6.2)= 3.1cm and OD =1/2 (6.2)=3.1 cm as shown.
5. Join AB, BC, CD and DA.

Then, ABCD is the required parallelogram.

4.Construct a rectangle ABCD in which side BC = 5 cm and diagonal BD = 6.2 cm.

Solution:
First draw a rough sketch of the required rectangle and write down its dimensions. Now, we may construct it by following the steps given below

Steps of Construction
1. Draw BC = 5 cm.
2. Draw CX _|_ BC.
3. With B as center and radius 6.2 cm draw an arc,cutting CX at D.
4. Join BD.
5. With D as center and radius 5 cm, draw an arc.
6. With B as center and radius equal to CD draw another arc, cutting the previous arc at A.

Then, ABCD is the required rectangle.

5. Construct a square ABCD, each of whose diagonals is 5.2 cm.

Solution:
We know that the diagonals of a square bisect each other at right angles. So, we proceed according to the following steps.

Steps of Construction
1. Draw AC = 5.2 cm.
2. Draw the right bisector XY of AC, meeting AC at O.
3. From O set off OB = 1/2 (5.2) = 2.6 cm along OY and OD = 2.6 cm along OX.
4. Join AB, BC, CD and DA.

Then, ABCD is the required square.

6. Construct a rhombus with side 4.2 cm and one of its angles equal to 65°.

Solution:
Clearly, the adjacent angle = (180° -65°) = 115°. So, we may proceed according to the steps given below.

Steps of Construction
1. Draw BC = 4.2 cm.
2. Make /_CBX = 115° and /_BCY = 65°.
3. Set off BA = 4.2 cm along BX and CD = 4.2 cm along CY.

Then, ABCD is the required rhombus.

For Example

1. In a right triangle, if one angle is 50°, find its third angle.
Solution:

Delta PQR is a right triangle, that is, one angle is right angle. Given, /_PQR = 90°
/_QPR=50°
Therefore, /_QRP = 180°-(/_Q + /_P)
= 180°-(90°+50°)
= 180°-140°
/_R =40°

Property 2: Relation between lengths of the side

Draw a DeltaABC. Measure the length of its three sides. Let the lengths of the three sides be AB = 5 cm, BC = 7 cm, AC = 8 cm. Now add the lengths of any two sides compare this sum with the lengths of the third side.

1. AB + BC = 5 cm + 7 cm = 12 cm
Since 12 cm > 8 cm
Therefore, (AB + BC) > AC
2. BC + CA = 7 cm + 8 cm = 15 cm
Since 15 cm > 5 cm
Therefore, (BC + CA) > AB
3. CA + AB = 8 cm + 5 cm = 13 cm
Since 13 cm >cm
Therefore, (CA + AB) > BC
4. Thus, we conclude that the sum of the length of any two sides of a triangle is greater than the length of the third side.

For Example

1. Is it possible to have a triangle whose sides are 5 cm, 6 cm and 4 cm?
Solution:

The lengths of the sides are 5 cm, 6 cm, 4 cm,
(a) 5 cm + 6 cm > 4 cm.
(b) 6 cm + 4 cm > 5 cm.
(c) 5 cm + 4 cm > 6cm.
Hence, a triangle with these sides is possible.

Examples of Properties of Triangle

We will solve some of the examples of properties of triangle.

#### 1. In the triangle, given write the names of its three sides, three angles and three vertices.

Solution:
Three sides of DeltaPQR are: PQ, QR and RP
Three angles of DeltaPQR are: /_PQR, /_QRP and R/_PQ
Three vertices of deltaPQR are: P, Q and R

#### 2. Measures of two angles of a triangle are 65° and 40°. Find the measure of its third angle.

Solution:
Measures of two angles of a delta are 65° and 40°
Sum of the measures of two angles = 65° + 40° = 105°
Sum of all three angles of delta = 180°
Therefore, measure of the third angle = 180° - 105° = 75°

#### 3. Is the construction of a triangle possible in which the lengths of sides are 5 cm, 4 cm and 9 cm?

Solution:
The lengths of the sides are 5 cm, 4 cm, 9 cm.
5 cm + 4 cm = 9 cm.
Then the sum of two smaller sides is equal to the third side. But in a triangle, the sum of any two sides should be greater than the third side.
Hence, no triangle possible with sides 5 cm, 4 cm and 9 cm.

Construction of Pie Chart

It is a circular graph which is used to represent data. In this :

• Various observations of the data are represented by the sectors of the circle.
• The total angle formed at the centre is 360°.
• The whole circle represents the sum of the values of all the components.
• The angle at the centre corresponding to the particular observation component is given by

{ The Value Of The Component / Total Value } x 360°

• If the values of observation/components are expressed in percentage, then the centre angle corresponding to particular observation/component is given by
• { Percentage Value Of Component / 100 } x 360°

How to construct a pie chart?
Steps of construction of pie chart for a given data:

• Find the central angle for each component using the formula given on the previous page.
• Draw a circle of any radius.
• Starting with the horizontal radius, draw radii, making central angles corresponding to the values of respective components.
• Repeat the process for all the components of the given data.
• These radii divide the whole circle into various sectors.
• Now, shade the sectors with different colours to denote various components.
• Thus, we obtain the required pie chart.

• Solved example on construction of pie chart/pie graph:

1. The following table shows the numbers of hours spent by a child on different events on a working day.

Represent the adjoining information on a pie chart

Activity no.of hours
School 6
Sleep 8
playing 2
Study 4
T.V 1
Others 3

The central angles for various observations can be calculated as:

Activity No.of Hours Measure of central angle
School 6 (6/[24] **360)^° = 90°
Sleep 8 (8/[24] **360)^° = 120°
Playing 2 (2/[24] ** 360)^° = 30°
Study 4 (4/[24]**360)^° = 60°
T.V 1 (1/[24] **360)^° = 15°
Others 3 (3/[24] **360)^° = 45°

Represent these angles within the circle as different sectors. Then we now make the pie chart:

#### 2. The favourite flavours of ice-cream for the children in a locality are given in percentage as follow. Draw the pie chart to represent the given information

Flavours % of Students Prefer the Flavours
Vanilla 25 %
Strawberry 15 %
Chocolate 10 %
Kesar-Pista 30 %
Mango Zap 20 %
The central angles for various observations can be calculated as:
Flavours % of Students Prefer the Flavours Measure of Central Angles
Vanilla 25 % ([25]/[100] ** 360)^° = 90°
Strawberry 15 % ([15]/[100] ** 360)^° = 54°
Chocolate 10 % ([10]/[100] ** 360)^° = 36°
Kesar-Pista 30 % ([30]/[100] ** 360)^° = 108°
Mango Zap 20 % ([20]/[100] ** 360)^° = 72°

Now, we shall represent these angles within a circle to obtain the required pie graph.

Line Graph

The data which changes over a period of time can be displayed through a line graph.
In line graph:

• Points are plotted on the graph related to two variables
• Points are joined by the line segments.

Steps of construction of line graph:

• On a graph, draw two lines perpendicular to each other intersecting at O. The horizontal line is x-axis and vertical line is y-axis.
• Mark points at equal intervals along x-axis and write the names of the data items whose values are to be marked.
• Along the y-axis, choose an appropriate scale considering the given values. Now, make the points
• Join each point with the successive point using a ruler. Thus, a line graph is obtained

1. The vehicular traffic at a busy road crossing in a particular place was recorded on a particular day from 6am to 2 pm and the data was rounded off to the nearest tens.

 Number of Vehicles Time in Hours Number of Vehicles 6- 7 7- 8 8 - 9 9 - 10 10 - 11 11 -12 12 - 1 1 - 2 100 450 1250 1050 750 600 550 200

Division of Line Segment

(i) Internal Division of line segment:
Let (x1, y1) and (x2, y2) be the Cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY and the point R divides the line-segment PQ internally in a given ratio m : n (say), i.e., PR : RQ = m : n. We are to find the co-ordinates of R.

• Let, (x, y) be the required co-ordinate of R . From P, Q and R, draw PL, QM and RN perpendiculars on OX. Again, draw PT parallel to OX to cut RN at S and QM at T. Then,
• PS = LN = ON - OL = x - x_1;
PT = LM = OM - OL = x_2 - x_1;
RS = RN - SN = RN - PL = y - y_1;
and QT = QM - TM = QM - PL = y_2 - y_1
Again, [PR]/[RQ] = m/n
or, [RQ]/[PR]= n/m
or, [RQ]/[PR] + 1 = n/m + 1
or, [RQ + PR]/PR = (m + n)/m
o, [PQ]/[PR] = (m + n)/m

• Now, by construction, the triangles PRS and PQT are similar; hence,
• [PS]/[PT] = [RS]/[QT] = [PR]/[PQ]
Taking, [PS]/[PT] = [PR]/[PQ] we get,
(x - x_1)/(x_2 - x_1) = m/(m + n)
or, x (m + n) - x_1 (m + n) = mx_2 - mx_1
or, x ( m + n) = mx_2 - mx_1 + m x_1 + nx_1 = mx_2 + nx_1
Therefore, x = (mx_2 + n x_1)/(m + n)
Again, taking [RS]/[QT] = [PR]/[PQ] we get,
(y - y_1)/(y_2 - y_1) = m/(m + n)
or, ( m + n) y - ( m + n) y_1 = my_2 - my_1
or, ( m+ n)y = my_2 - my_1 + my_1 + ny_1 = my_2 + ny_1
Therefore, y = (my_2 + ny_1)/(m + n)
Therefore, the required co-ordinates of the point R are
((mx_2 + nx_1)/(m + n), (my_2 + ny_1)/(m + n))

(ii) External Division of line segment:

• Let (x1, y1) and (x2, y2) be the Cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY and the point R divides the line-segment PQ externally in a given ratio m : n (say) i.e., PR : RQ = m : n. We are to find the co-ordinates of R.
• Let, (x, y) be the required co-ordinates of R. Draw PL, QM and RN perpendiculars on OX. Again, draw PT parallel to OX to cut RN at S and QM and RN at S and T respectively, Then,
• PS = LM = OM - OL = x_2 - x_1
PT = LN = ON - OL = x - x_1
QT = QM - SM = QM - PL = y_2 - y_1
and RT = RN - TN = RN - PL = y - y_1
Again, [PR]/[RQ] = m/n
or, [QR]/[PR] = n/m
or, 1 - [QR]/[PR]= 1 - n/m
or, [PR - RQ]/[PR] = (m - n)/m
or, [PQ]/[PR] = (m -n)/m
Now, by construction, the triangles PQS and PRT are similar; hence,
[PS]/[PT] = [QS]/[RT] = [PQ]/[PR]
Taking, PS/PT = PQ/PR we get,
(x_2 -x_1)/(x - x_1) = (m - n)/m
or, (m - n)x - x_1(m - n) = m (x_2 - x_1)
or, (m - n)x = mx_2 - mx_1 + mx_1 - nx_1 = mx_2 - nx_1.
Therefore, x = (mx_2 - nx_1)/(m -n)

Again, taking [QS]/[RT] = [PQ]/[PR] we get,
(y_2 - y_1)/(y - y_1) = (m - n)/m
or, (m - n)y - (m - n)y_1 = m(y_2 - y_1)
or, (m - n)y = my_2 - my_1 + my_1 - ny_1 = my_2 - ny_1
Therefore, x = (my_2 - ny_1)/(m - n)
Therefore, the co-ordinates of the point R are
((mx_2 - nx_1)/(m - n), (my_2 - ny_1)/(m - n))

Corollary:

To find the co-ordinates of the middle point of a given line segment:

Example on Division of Line Segment

#### 1. A diameter of a circle has the extreme points (7, 9) and (-1, -3). What would be the co-ordinates of the centre?

Solution:
Clearly, the mid-point of the given diameter is the centre of the circle.
Therefore, the required co-ordinates of the centre of the circle = the co-ordinates of the mid-point of the line-segment joining the points (7, 9) and (-1, -3)
= ((7 - 1)/2, (9 - 3)/2 ) = (3, 3).

#### 2. A point divides internally the line-segment joining the points (8, 9) and (-7, 4) in the ratio 2 : 3. Find the co-ordinates of the point.

Solution:
Let (x, y) be the co-ordinates of the point which divides internally the line-segment joining the given points. Then,
x = (2**(- 7) + 3**8)/(2 + 3) = (-14 + 24)/5 = 10/5 = 2
And y = (2 **4 + 3 ** 9)/(2 + 3) = (8 + 27)/5 = 35/5 = 5
Therefore, the co-ordinates of the required point are (2, 7).

Note:
To get the co-ordinates of the point in question we have used formula, x = (mx_2 + n x_1)/(m + n) and y = (my_2 + ny_1)/(m + n). For the given problem, x1 = 8, y1 = 9, x2 = -7, y2 = 4, m = 2 and n = 3.)

#### 3. A (4, 5) and B (7, -1) are two given points and the point C divides the line-segment AB externally in the ratio 4 : 3. Find the co-ordinates of C.

Solution:
Let (x, y) be the required co-ordinates of C. Since C divides the line-segment AB externally in the ratio 4 : 3 hence,
x = (4 * 7 - 3 * 4)//(4 - 3) = (28 - 12)//1 = 16
And y = (4 * (-1) - 3 * 5)//(4 - 3) = (-4 - 15)//1 = -19
Therefore, the required co-ordinates of C are (16, - 19).

Note: To get the co-ordinate of C we have used formula, x = (mx_1 + n x_1)/(m + n) and y = (my_2 + ny_1)/(m + n). In the given problem, x2 = 4, y1 = 5, x2 = 7, y2 = - 1, m = 4 and n = 3).

#### 4. Find the ratio in which the line-segment joining the points (5, - 4) and (2, 3) is divided by the x-axis.

Solution:
Let the given points be A (5, - 4) and B (2, 3) and x-axis. intersects the line-segment bar(AB)at P such that AP : PB = m : n.
Then the co-ordinates of P are ((m * 2 + n * 5)/(m + n), (m * 3 + n *(-4))/(m + n)). Clearly,
the point P lies on the x-axis; hence, y co-ordinate of P must be zero. Therefore, (m ** 3 + n ** (-4))/(m + n) = 0
or, 3m - 4n = 0
or, 3m = 4n
or, m/n = 4/3
Therefore, the x-axis divides the line-segment joining the given points internally in 4 : 3.

#### 5. Find the ratio in which the point (- 11, 16) divides the line segment joining the points (- 1, 2) and (4, - 5).

Solution:
Let the given points be A (- 1, 2) and B (4, - 5) and the line-segment AB is divided in the ratio m : n at (- 11, 16). Then we must have,
-11 = [m** 4 + n ** (-1)]/(m + n)
or, -11m - 11n = 4m - n
or, -15m = 10n
or, m/n = (10)/(-15) = - 2/3
Therefore, the point (- 11, 16) divides the line-segment AB externally in the ratio 3 : 2.

Note: (i) A point divides a given line-segment internally or externally in a definite ratio according as the value of m:n is positive or negative. (ii) See that we can obtain the same ratio m : n = - 2 : 3 using the condition 16 = (m ** (-5) +n ** 2)/(m + n))

Practice Test on Angles

Geometry practice test on angles, the questions we practised and discussed under worksheets on angles are given here in geometry practice test.
1.Classify the following angles:

1. 30°
2. 66°
3. 75°
4. 111°
5. 180°

2. Use your protractor to draw these angles:

1. 90°
2. 180°
3. 105°

3.Identify which of the following pairs of angles are complementary or supplementary?

1. 160°,30°
2. 45°,45°
3. 30°,60°
4. 135°,45°
5. 10°,170°
6. 20°,70°

4. Find the complement of each of the following angles:

1. 85°
2. 38°
3. 90°
4. 15°
5. 49°

5. Find the supplement of each of the following angles?

1. 40°
2. 127°
3. 35°

6. Draw a pair of supplementary angles such that one of them measures:

1. 35°
2. 135°
3. 105°

7.Construct the angles of the following measures with the help of a compass:

1. 30°
2. 105°
3. 75°

8.An angle whose measure is 90° is called __________

9. An angle whose measure is greater than 90° but less than 180° is called an __________

In maths practice test on quadrilateral worksheet we will practice different types of questions in quadrilateral. Students can practice the questions of quadrilateral worksheet before the examinations to get more confident.

1. Fill in the blanks:

3. A quadrilateral has ________vertices, no three of which are_______.
5. A diagonal of a quadrilateral is a line segment that joins two _______ vertices of the quadrilateral.
6. The sum of the angles of a quadrilateral is _________.

1. How many pairs of adjacent sides are there? Name them.
2. How many pairs of opposite sides are there? Name them.
3. How many pairs of adjacent angles are there? Name them.
4. How many pairs of opposite angles are there? Name them.
5. How many diagonals are there? Name them.
1. Prove that the sum of the angles of a quadrilateral is 360°.
2. The three angles of a quadrilateral are 76°, 54° and 108°. Find the measure fourth angle.
3. The angles of a quadrilateral are in the ratio 3 : 5 : 7 : 9. Find the measure of each of these angles.
4. A quadrilateral has three acute angles, each measuring 75°. Find the measure of the fourth angle.
5. Three angles of a quadrilateral are equal and the measure of the fourth angle is 120°. Find the measure of each of the equal angles.
6. Two angles of a quadrilateral measure 85° and 75° respectively. The other two angles equal. Find the measure of each of these equal angles.
7. Answers for quadrilateral worksheet are given below to check the exact answers of the above questions.

1. four
2. four
3. four, collinear
4. two
5. opposite
6. 360°
1. four; (AB,BC),(BC, CD),(CD, DA),(DA, AB)
3. four; (/_A,/_B),(/_B,/_C),(/_C,/_D), (/_D, /_A)
4. two; (/_A,/_C),(/_B,/_D)