∆ABC ~ ∆PBQ. In ∆ABC, AB = 3 cm, ∠B = 90°, BC = 4 cm. Ratio of the corresponding sides of two triangles is 7: 4. Then construct ∆ABC and ∆PBQ

#### Solution

**Analysis:** As shown in the figure,

Let B–A–P and B–C–Q.

∆PBQ ∼ ∆ABC

∴ ∠PQB ≅ ∠ACB .....[Corresponding angles of similar triangles]

`"PB"/"AB" = "BQ"/"BC" = "PQ"/"AC"` .....(i) [Corresponding sides of similar triangles]

∴ `"PB"/"AB" = "BQ"/"BC" = "PQ"/"AC" = 7/4` ......[Given]

∴ Sides of ∆PBQ are longer than corresponding sides of ∆ABC.

∴ If seg BC is divided into 4 equal parts, then seg BQ will be 7 times each part of seg BC.

So, if we construct ∆ABC point Q will be on side BC, at a distance equal to 7 parts from B.

Now, point P is the point of intersection of ray AB and a line through Q, parallel to AC.

∴ ∆PBQ is the required triangle similar to ∆ABC.**Steps of construction:**

- Draw seg BC of length 4 cm.
- Take ∠B as 90° and draw an arc of 3 cm on it. Name the point as A.
- Join seg AC to obtain ∆ABC.
- Draw ray BX such that ∠CBX is an acute angle.
- Locate points B
_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}, B_{7}on ray BX such that, BB_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}= B_{4}B_{5}= B_{5}B_{6}= B_{6}B_{7}. - Join point C and B
_{4}. - Through point, B
_{7}draw a line parallel to seg CB_{4}which intersects seg BC at point Q. - Draw a line parallel to AC through Q to intersect line AB at point P.

∆PBQ is the required triangle similar to ∆ABC.