﻿ Class 10 NCERT Math Solution
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TOPICS
Unit-11(Examples)

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 (i.e., of scale factor 3/4).

Solution :-
```  Given that :
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 :
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 Figure).
Then, Δ A′BC′ is the required triangle.

Let us now see how this construction gives the required triangle.
By construction, BC'/C'C = 3/1
Therefore, BC/BC' = (BC' + CC')/BC' = 1 + CC'/BC' = 1 + 1/3 = 4/3
i.e., BC'/BC = 3/4
Also C'A' is parallel to CA.
Therefore, Δ A'BC' ~ Δ ABC.
So, A'B/AB = A'C'/AC = BC'/BC = 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).

Solution :-
```  Given that :
A triangle ABC, we are required to construct another triangle whose
sides are 5/3 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 5 points (the greater of 5 and 3 in 5/3) B₁, B₂, B₃, B₄ and B₅ on BX so that
BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
3. 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′.
4. Draw a line through C′ parallel to CA intersecting the extended line segment BA at A′ (see Figure).
Then, Δ A′BC′ is the required triangle.

For justification of the construction, note that Δ ABC ~ Δ A′BC′
Therefore, AB/A'B = AC/A'C' = BC/BC'
But, BC/BC' = BB₃/BB₅ = 3/5
So, BC'/BC = 5/3, and
Therefore, A'B/AB = A'C'/AC = BC'/BC = 5/3
```
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