Maharashtra Board Class 9 Maths Part II Chapter 4 Constructions of Triangles PDF Download

Constructions Of Triangles

To construct a triangle, if following information is given.

Base, an angle adjacent to the base and sum of lengths of two remaining sides.

Base, an angle adjacent to the base and difference of lengths of remaining two sides.

Perimeter and angles adjacent to the base.

Let's Recall

In previous standard we have learnt the following triangle constructions.

To construct a triangle when its three sides are given.

To construct a triangle when its base and two adjacent angles are given.

To construct a triangle when two sides and the included angle are given.

To construct a right angled triangle when its hypotenuse and one side is given.

Perpendicular Bisector Theorem

Every point on the perpendicular bisector of a segment is equidistant from its end points.

Every point equidistant from the end points of a segment is on the perpendicular bisector of the segment.

Let's Learn

Constructions Of Triangles

To construct a triangle, three conditions are required. Out of three sides and three angles of a triangle two parts and some additional information about them is given, then we can construct a triangle using them.

We frequently use the following property in constructions.

If a point is on two different lines then it is the intersection of the two lines.

Teacher's Note

Triangle constructions are very useful in real life. For example, builders use these methods to build strong triangle-shaped roofs for houses.

Exam Trick

Remember: Perpendicular bisector means a line that cuts a line into two equal parts at a 90-degree angle. Just like dividing a pizza into two equal slices.

Points To Remember

A triangle needs three pieces of information to be drawn.

The perpendicular bisector is a line that cuts another line into two equal parts.

If a point is on the perpendicular bisector, it is the same distance from both ends of the line.

We use rays and angles to help us construct triangles.

Construction I

To Construct A Triangle When Its Base, An Angle Adjacent To The Base And The Sum Of The Lengths Of Remaining Sides Is Given

Ex. Construct Triangle ABC in which BC = 6.3 cm, angle B = 75° and AB + AC = 9 cm.

Solution

Let us first draw a rough figure of expected triangle.

Explanation

As shown in the rough figure, first we draw seg BC = 6.3 cm of length. On the ray making an angle of 75° with seg BC, mark point D such that BD = AB + AC = 9 cm.

Now we have to locate point A on ray BD.

BA + AD = BA + AC = 9

Therefore AD = AC

Therefore point A is on the perpendicular bisector of seg CD.

Therefore the point of intersection of ray BD and the perpendicular bisector of seg CD is point A.

Steps Of Construction

Draw seg BC of length 6.3 cm.

Draw ray BP such that m angle PBC = 75°.

Mark point D on ray BP such that d(B,D) = 9 cm.

Draw seg DC.

Construct the perpendicular bisector of seg DC.

Name the point of intersection of ray BP and the perpendicular bisector of CD as A.

Draw seg AC.

Triangle ABC is the required triangle.

Teacher's Note

This method helps us make triangles when we know the base, one angle, and the sum of the other two sides. Think of it like knowing the bottom of a tent, one corner angle, and the total length of two poles.

Exam Trick

Remember: The perpendicular bisector of CD helps us find point A. It is like finding the middle point of a seesaw where both sides are equal distance.

Points To Remember

We need to draw a ray at the given angle from the base.

Mark point D at a distance equal to the sum of the two sides.

Find the perpendicular bisector of the line from D to the end of the base.

The point where the ray meets the perpendicular bisector is our third corner of the triangle.

Practice Set 4.2

Construct triangle PQR, in which QR = 4.2 cm, m angle Q = 40° and PQ + PR = 8.5 cm.

Construct triangle XYZ, in which YZ = 6 cm, XY + XZ = 9 cm. angle XYZ = 50°.

Construct triangle ABC, in which BC = 6.2 cm, angle ACB = 50°, AB + AC = 9.8 cm.

Construct triangle ABC, in which BC = 5.2 cm, angle ACB = 45° and perimeter of triangle ABC is 10 cm.

Construction II

To Construct A Triangle When Its Base, Angle Adjacent To The Base And Difference Between The Remaining Sides Is Given

Ex (1) Construct triangle ABC, such that BC = 7.5 cm, angle ABC = 40°, AB - AC = 3 cm.

Solution

Let us draw a rough figure.

Explanation

AB - AC = 3 cm, therefore AB > AC.

Draw seg BC. We can draw the ray BL such that angle LBC = 40°. We have to locate point A on ray BL. Take point D on ray BL such that BD = 3 cm.

Now, B-D-A and BD = AB - AD = 3.

It is given that AB - AC = 3.

Therefore AD = AC.

Therefore point A is on the perpendicular bisector of seg DC.

Therefore point A is the intersection of ray BL and the perpendicular bisector of seg DC.

Steps Of Construction

Draw seg BC of length 7.5 cm.

Draw ray BL such that angle LBC = 40°.

Take point D on ray BL such that BD = 3 cm.

Construct the perpendicular bisector of seg CD.

Name the point of intersection of ray BL and the perpendicular bisector of seg CD as A.

Draw seg AC.

Triangle ABC is required triangle.

Teacher's Note

This method works when we know the base, one angle, and the difference between two sides. For example, if one side of a triangle frame is 3 cm longer than another side, we can use this method to build it.

Exam Trick

Remember: The difference means one side is bigger than the other. Mark point D at the difference distance, then use the perpendicular bisector to find the third corner.

Points To Remember

Mark point D on the ray at a distance equal to the difference of the two sides.

The difference means how much longer one side is compared to the other side.

Use the perpendicular bisector of seg CD to find where the third corner should be.

The intersection point of the ray and the perpendicular bisector is our third corner.

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