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11. Constructions
1. Introduction
In Class IX, you have done certain constructions using a straight edge (ruler) and a compass, e.g., bisecting an angle, drawing the perpendicular bisector of a line segment, some constructions of triangles etc. and also gave their justifications. In this chapter, we shall study some more constructions by using the knowledge of the earlier constructions. You would also be expected to give the mathematical reasoning behind why such constructions work.
2 Division of a Line Segment
subpose a line segment is given and you have to divide it in a given ratio, say 3 : 2. You may do it by measuring the length and then marking a point on it that divides it in the given ratio. But subpose 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.
Let us see how this method gives us the required division.
Since A_{3}C is parallel to A_{5}B, therefore,
AA_{3}/A_{3}A_{5} = AC/CB (By the Basic Proportionality Theorem)
By construction, AA_{3}/A_{3}A_{5} = 3/2, Therefore, AC/CB = 3/2
This shows that C divides AB in the ratio 3 : 2.
Then AC : CB = 3 : 2.
Why does this method work? Let us see.
Here △ AA_{3}C is similar to △ BB_{2}C. (Why ?)
Then AA_{3}/BB_{2} = AC/BC.
Since by construction, AA_{3}/BB_{2} = 3/2 ,therefore, AC/BC = 3/2
In fact, the methods given above work for dividing the line segment in any ratio.
We now use the idea of the construction above for constructing a triangle similar to a given triangle whose sides are in a given ratio with the corresponding sides of the given triangle.
Construct a triangle similar to a given triangle ABC with its sides equal to 3/4 of the corresponding sides of the triangle ABC (i.e., of scale factor 3/4).
Given a triangle ABC, we are required to construct another triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Steps of Construction :
- Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
- Locate 4 (the greater of 3 and 4 in 3/4 ) points B_{1}, B_{2}, B_{3} and B_{4} on BX so that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.
- Join B_{4}C and draw a line through B_{3} (the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to B_{4}C to intersect BC at C'.
- Draw a line through C' parallel to the line CA to intersect BA at A' (see Fig. 3).
Fig 3
Then, △A'BC' is the required triangle.
Let us now see how this construction gives the required triangle.
By Construction 1, BC'/C'C = 3/1.
Therefore, BC/BC' = (BC'+C'C)/BC' = 1+C'C/BC' = 1+(1/3) = 4/3 , i.e., BC' / BC = 3/4
Also C'A' is parallel to CA. Therefore, △ A'BC' ~ △ABC. (Why ?)
So, A'B = AB = A'C'/AC = BC'/BC = 3/4.
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).
Given a triangle ABC, we are required to construct a triangle whose sides are of the corresponding sides of △ABC.
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) B_{1}, B_{2}, B_{3}, B_{4} and B5 on BX so that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}.
- Join B_{3} (the 3rd point, 3 being smaller of 3 and 5 in ) to C and draw a line through B_{5} parallel to B_{3}C, 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. 4).
Fig 4
Then A'BC' is the required triangle.
For justification of the construction, note that △ ABC ~ △ A'BC'. (Why ?)
Therefore, AB/A'B = AC/A'C' = BC/BC'
But BC/BC' = BB_{3}/BC_{5} = 3/5
So, BC/BC' = 5/3 and, therefore, A'B/AB = A'C'/AC = BC'/BC = 5/3
Remark :
In Examples 1 and 2, you could take a ray making an acute angle with AB or AC and proceed similarly.3 Construction of Tangents to a Circle
You have already studied in the previous chapter that if a point lies inside a circle, there cannot be a tangent to the circle through this point. However, if a point lies on the circle, then there is only one tangent to the circle at this point and it is perpendicular to the radius through this point. Therefore, if you want to draw a tangent at a point of a circle, simply draw the radius through this point and draw a line perpendicular to this radius through this point and this will be the required tangent at the point.
You have also seen that if the point lies outside the circle, there will be two tangents to the circle from this point.
We shall now see how to draw these tangents.
Now let us see how this construction works.
Join OQ. Then ∠PQO is an angle in the semicircle and, therefore,
∠PQO = 90°
Can we say that PQ ⊥ OQ?
Since, 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. Then you could proceed as above.