CONCEPTS TO BE COVERED:-
Introduction,Basic constructions and Some constructions of triangles.
INTRODUCTION:-
In this chapter we are going to learn the constructions of angles.
ANGLE:-
Angles are geometric figures formed by two rays meeting at a common endpoint.
1.Angles are measured in degrees.
2.The common point of contact is called the "vertex" of an angle.
*Let us know the types of "Angles".
BASIC CONSTRUCTIONS:
We have already learnt how to construct a circle, the perpendicular bisector of a line segment, angles of 30°,60°,90° and 120°, and the bisector of a given angle,without giving nay justification for these constructions.In this section,we will construct some of these with reasoning behind,why these constructions are held.
Given an angle ABC, we want to construct its bisector.
STEPS OF CONSTRUCTION:-
1.Taking B as center and any radius, draw an arc to intersect the rays BA and BC, say at E and D respectively (see fig.I)
2.Next,taking D and E as centers and with the radius more than 1/2DE, draw an arcs to intersect each other, sat at F.
3.Draw the ray BF [see fig.II]. This ray BF is the required bisector of the angle ABC.
Let us see how this method gives us the required angle bisector.
Join DE and EF.
In triangles BEF and BDF,
BE=BD (Radii of the same arc)
EF=DF (ARcs of equal radii)
BF=BF (common)
Therefore,△BEF≅△BDF (SSS rule)
∠EBF=∠DBF (CPCT)
CONSTRUCTION-2:To construct the perpendicular bisector of a given line segment.
Given a line segment AB,we want to construct its perpendicular bisector.
STEPS OF CONSTRUCTION:-
1.Taking A and B as centers and radius more than 1/2AB, draw an arc on both sides of the line segment AB (to intersect each other).
2.Let these arcs intersect each other at P and Q.Join PQ (see fig)
3.Let PQ intersect AB at the point M.Then line PMQ is the required perpendicular bisector of AB.
This method gives us the perpendicular bisector of AB.
Join A and B to both P and Q to form AP, AQ, BP and BQ.
In triangles, PAQ and PBQ.
AP=BP (Arcs of equal radii)
AQ=BQ (Arcs of equal radii)
PQ=PQ (common)
Therefore,△PAQ≅△PBQ (SSS rule)
so, ∠APM=∠BPM (CPCT)
In triangles PMA and PMB
AP=BP (Arcs of equal radii)
PM=PM (common)
∠APM=∠BPM (CPCT)
△PMA≅△PMB (SAS Rule)
AM=BM and ∠PMA=∠PMB (CPCT)
∠PMA+∠PMB=180° (linear pair)
we get,
∠PMA=∠PMB=90°
Therefore PM, is PMQ is the perpendicular bisector of AB.
CONSTRUCTON-3:To construct an angle of 60° at the initial point of a given ray.
Let us take a ray AB with initial point A (see fig.I). We want to construct a ray AC such that ∠CAB=60°, one way of doing so is given below.
STEPS OF CONSTRUCTION:-
1.Taking A as center and some radius, draw an arc of a circle, which intersects AB, say at a point D.
2.Taking D as center and with the same raidus as before, draw an arc intersecting the previously drawn arc, say at a point E.
3.Draw the ray AC passing through E (see fig).Then ∠CAB is the required angle of 60°.
Join DE
Then AE=AD=DE (By construction)
Therefore,△EAD is an equilateral triangle and the ∠EAD, which is same as ∠CAB i.e equal to 60°.
EXAMPLES:
1.Construct the angle of 90° at the initial point of a given ray and justify the construction.
2.Construct the angle of 45° at the initial point of a given ray and justify the construction.
3.Construct the angles of the following measurements:
(i)30° (ii)22.5° (iii)15°
4.Construct the following angles and verify by measuring them by a protractor.
(i)75° (ii)105° (iii)135°
5.Construct an equilateral triangle, given its side and justify the construction.
SOME CONSTRUCTIONS OF TRIANGLES:
Some basic constructions have been discussed above .Now,some constructions of triangles will be done by using the SAS,SSS,ASA and RHS rules to give the congruency of two triangles.
SAS(Side Angle Side):-If any two side and the angle included between the sides of one triangle are equivalent to the corresponding two sides and the angle between the sides of a second triangle then the two triangles are said to be congruent.
SSS(Side Side Side):-If the three sides of one triangle are equal to the corresponding three sides of the other triangle,then the two triangles are said to be congruent.
ASA(Angle Side Angle):-If any two angles and the side included between the angles of one triangle is equal to the corresponding two angles with side included between the angles of other triangle,then the two triangles are said to be congruent.
RHS RULE:-In two right angled triangles, if the length of the hypotenuse and one side of triangle, is equal to the length of the hypotenuse and corresponding side of the other triangle, then two triangles are congruent.
CONSTRUCTION-4:To construct a triangle, given its base, a base angle and sum of other two sides.
Given the base BC, a base angle, say ∠B and the sum AB+AC of the other two sides of a triangle ABC.
STEPS OF CONSTRUCTION:-
1.Draw the base BC and at the point B make an angle, say XBC equal to the given angle.
2.Cut a line segment BD equal to AB+AC from the ray BX.
3.Join DC and make an angle DCY equal to ∠BDC.
4.Let CY intersect BX at A (see fig).
Then, ABC is the required triangle.
Base BC and ∠B are drawn as given.Next in triangle ACD.
∠ACD=∠ADC
Therefore,AC=AD and
AB=BD-AD=BD-AC
AB+AC=BD.
ALTERNATIVE METHOD:-Follow the first two steps as above.Then draw perpendicular bisector PQ of CD to intersect BD at a point A (see fig). Join AC. Then ABC Is the required triangle.Note that A lies on the perpendicular of CD, therefore AD=AC.
The construction of the triangles is not possible if the sum AB+AC≤BC.
CONSTRUCTION-5:To construct a triangle given its base , a base angle and the difference of the other two sides.
Given the base BC, a base angle say ∠B and the difference of other two sides AB-BC or AC-AB, we have to construct the triangle ABC.Clearly there are following two cases.
Case-(i):-Let AB>AC that is AB-AC is given.
STEPS OF CONSTRUCTION:-
1.Draw the base BC and at point B make an angle say XBC equal to the given angle.
2.Cut the line segment BD equal to AB-AC from ray BX.
3.Join DC and draw the perpendicular bisector, say PQ Of DC.
4.Let it intersect BX at a point A. Join AC (see fig).
Then ABC is required triangle.
Base BC and ∠B are drawn as given.
The point A lies on the perpendicular bisector of DC.
Therefore,AD=AC
BD=AB-AD
BD=AB-AC.
Case-ii:-Let AB<AC that is AC-AB is given.
STEPS OF CONSTRUCTION:-
1.Same as in case(i).
2.Cut line segment BD equal to AC-AB from the line BX extended on opposite side of line segment BC.
3.Join DC and draw the perpendicular bisector, PQ of DC.
4.Let PQ intersect BX at A. Join AC (see fig). Then, ABC is the required triangle.
CONSTRUCTION-6:-To construct a triangle given its perimeter and its two base angles.
Given the base angles,∠B and ∠C and BC+CA+AB. We have to construct
STEPS OF CONSTRUCTION:-
1.Draw a line segment XY which is equal to the BC+CA+AB.
2.∠LXY is equal to the ∠B and ∠MYX is equal to the ∠C.
3.Bisect the ∠LXY and ∠MYX and these bisectors meet at point A.
4.Draw perpendicular bisectors of PQ of AX and RS of AY.
5.Now, PQ intersect XY at B and RS intersect XY at C. Then join AB and AC.
6.ABC is the required triangle. And B lies on the peroendicular bisector PQ of AX.
Therefore,XB=AB and CY-AC
BC+CA+AB=BC+XB+CY=XY
∠BAX= ∠AXB (In △AXB, AB=XB)
∠ABC=∠BAX+∠AXB
∠ABC=2∠AXB
∠ABC=∠LXY
Similarly,∠ACB=∠MYX.
EXAMPLE:-Construct a triangle ABC, in which ∠B=60° and ∠C=45° and AB+BC+CA=11cm
STEPS OF CONSTRUCTION:-
1.Draw a line segment PQ=11cm (=AB+BC+CA).
2.At P construct an angle 60° at Q construct an angle 45°.
3.Bisect these angles. Let the bisectors of these angles intersect at a point A.
4.Draw perpendicular bisectors DE Of AP to intersect PQ at B and FG of AQ to intersect PQ at C.
5.Join AB and AC.
6.ABC is the requires triangle.
EXAMPLES:-
1.Construct a triangle ABC in which BC=7cm ∠B=75° and AB+AC=13cm.
2.Construct a triangle ABC in which BC=8cm ∠B=45° and AB-AC=3.5cm.
3.Construct a triangle PQR in which QR=6cm, ∠Q=60° and PR-PQ=2cm.
4.Construct a triangle XYZ in which ∠Y=30°, ∠Z=90 ° and XY+YZ+ZX=11cm.
5.Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18cm.
MULTIPLE CHOICE QUESTIONS:-
1.To divide a line segment AB in the ratio p:q (p,q are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum numbers of these points is
(a)greater of p and q (b)p+q
(c)p+q-1 (d)pq
Answer:-(b)p+q
2.To draw a pair of tangents to a circle which are incident to each other at an angle of 35°. It is required to draw tangents at the end points of these two radii of the circle.The angle between which is
(a)105° (b)70°
(c)140° (d)145°
Answer-(d)145°
3.To divide a line segment AB in the ratio 5:7, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is
(a)8 (b)10
(c)11 (d)12
Answer-(d)12
4.To divide a line segment AB in the ratio 4:7, ray AX is drawn first such that ∠BAX is an acute angle and then points A1,A2,A3......... are located at equal distances on the ray AX and the point B is joined to
(a)A12 (b)A11
(c)A10 (d)A9
Answer:-(b)A11
5.To divide a line segment AB in the ratio 5:6, draw a ray AX such that ∠BAX is an acute angle, then draw a ray B4 parallel to AX and the points A1,A2,A3..... and B1,B2,B3...... are located at equal distances on ray AX and B4, respectively. Then the points joined are
(a)A5 and B6 (b)A6 and B5
(c)A4 and B5 (d)A5 and B4
Answer:-(a)A5 and B6
6.To construct a triangle similar to a given △ABC with its sides 3/7 of the corresponding sides of △ABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B1,B2,B3 on BX at equal distances and next step is to join
(a)B10 to C (b)B3 to C
(c)B7 to C (d)B4 to C
Answer:-(c)B7 to C
7.To draw a pair of tangent to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle,the angle between them should be (a)135° (b)90°
(c)60° (d)120°
Answer:-(d)120°