Before we learn the construction of similar triangle, let us know what similar triangles are

Similar Triangles

If Two triangles ∆ABC and ∆PQR are said to be similar, following two conditions are satisfied:

1. The corresponding angles of the triangles are equal.

i.e.,

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

and

2. Since,∆ABC and ∆PQR are two similar triangles, their corresponding sides are in a ratio or proportion.

That is,

\( \frac {AB}{PQ} \)

We write ∆ABC and ∆PQR are similar as ∆ABC~∆PQR

For example;

Consider the two triangles given in the figure,

If ∆ABC~∆PQR, What is length of the side PR if AB = 6 cm, AC = 8 cm and PQ = 3 cm?

Since, ∆ABC~∆PQR

\( \frac {AB}{PQ} \)

\( \frac {6}{3} \)

PR = \( \frac {8~×~3}{6} \)

Let’s see how to construct similar triangles.

Consider ∆ABC where BC = 6 cm,∠B=40°and ∠C=60°. Draw a triangle similar to ∆ABC with a scale factor 2.

Scale factor is the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle.

Figure 1-Triangle ABC

Here, a scale factor of 2 means that sides of the new triangle which is similar to ∆ABC are twice the sides of ∆ABC.

Let ∆PQR be the new triangle.

QR = 2 × 6 = 12 cm [Scale factor is 2]

∠B = ∠Q = 40°and ∠C = ∠R=60°.

- Draw QR of length 12 cm
- Draw a line through B which makes an angle of 40° from BC.
- Draw a line through C which makes an angle of 60° from BC.
- Mark the intersection point of above two lines as P. ∆PQR is the required triangle (Refer figure).

Figure 2-How to construct similar triangles

Now, suppose the scale factor is a fraction,like \( \frac 54 \)

Then we won’t be able to construct similar triangles precisely.

The method to construct a similar triangle precisely is discussed here.

Problem: Construct a triangle which is similar to ∆ABC with scale factor \( \frac 35 \)

Scale factor \( \frac 35 \)

## Construction of Similar Triangles

The steps of construction of similar triangles are as follows (Refer figure)

- Draw a ray BX which makes acute angle with BC on the opposite side of vertex A.
- Locate 5 points on the ray BX and mark them as B1, B2, B3, B4 and B5 such that B B1 = B1 B2 = B2 B3 = B4 B5.
- Join B5C
- Draw a line parallel to B5C through B3 [Since 3 is the smallest among 3 and 5] and mark C’ where it intersects with BC.
- Draw a line through the point C’ parallel to AC and mark A’ where it intersects AB.
- A’BC’ is the required triangle.

Figure 3-How to construct similar triangles

How can we verify that ∆ABC~∆A’BC’ ?

\( \frac {BC’}{C’C}\)

Therefore,

\( \frac {BC}{BC’}\)

That gives,

\( \frac {BC’}{BC}\)

And, since A’C’ is parallel to AC,

∠ A’C’B=∠ACB [corresponding angles]

∠ABC=∠A’BC’ [common angle]

Therefore,

∆ABC ~ ∆A’BC’

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