Question

Question 4
Draw an isosceles triangle ABC in which AB=AC = 6 cm and BC=5 cm Construct a triangle PQR similar to $\Delta ABC$ in which PQ = 8 cm, Also justify the construction.

Thinking process
(i)Here, for making two similar triangles with one vertex is base We assume that In $\Delta ABC$ and $\Delta PQR$, vertex B = vertex Q.

(ii) In $\Delta ABC$ and $\Delta PQR$, vertex B = vertex Q.

So, we get the required scale factor.

Now, construct a $\Delta ABC$ and then a $\Delta PBR$, similar to $\Delta ABC$ whose sides are $\frac{PQ}{AB}$ of the corresponding sides of the $\Delta ABC$.

Answer 1

Let $\Delta PQR$ and $\Delta ABC$ are similar triangle, then its scale factor between the corresponding sides is $\frac{PQ}{AB}=\frac{8}{6}=\frac{4}{3}$

Steps of construction
1. Draw a line segment BC = 5 cm
2. Construct OQ the perpendicular bisector of line segment BC meeting BC at $P^{{}^{\prime }}$
3. Taking B and C as centers draw two arcs of equal radius 6cm intersecting each other at A
4. Join BA and CA. So. $\Delta ABC$ is the required isosceles triangle.

5. From B, draw a ray BX making an acute $\angle CBX$
6. Locate four points $B_1,B_2,B_3,B_4$ on BX Such that $B{B}_1=B_1{B}_2=B_3{B}_4$
7. Join $B_{3}C$ and from $B_4$ draw a line $B_{4}R\parallel B_{3}C$ intersecting the extended line segment BC at R.
8. From point R draw $RP\parallel CA$ meeting BA produced at p
Then, $\Delta PBR$ is the required triangle.
Justification
$\because B_{4}R\parallel B_{3}C$
$\therefore \frac{BC}{CR}=\frac{3}{1}$
Now $\frac{BR}{BC}=\frac{BC+CR}{BC}$
$=1+\frac{CR}{BC}=1+\frac{1}{3}=\frac{4}{3}$
Also $RP\parallel CA$
$\therefore \Delta ABC\cong \Delta PBR$
And $\frac{PB}{AB}=\frac{RP}{CA}=\frac{BR}{BC}=\frac{4}{3}$
Hence the new triangle is similar to the given triangle whose sides are $\frac{4}{3}$ times of the corresponding sides of the isosceles $\Delta AC$.