Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.

Name: Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.
Uploaded: Wed, 2021-01-20 18:30
Description: Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.

Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.

Answers (1)

Solution

Steps of construction
1. Draw line BC = 5 cm
2.   Taking B and C as centres, draw two arcs of equal radius 6 cm intersecting each other at point A.
3.   Join AB and AC DABC is required isosceles triangle
4    From B draw ray   $B_{X}$ with an acute angle $CB{B}'$
6.   draw $B_{1},B_{2},B_{3},B_{4}$ at $BX$ with equal distance
7.   Join $B_{3}C$ and from $B_{4}$ draw line $B_{4}D\parallel B_{3}C,$ , intersect extended segment BC at point D.
8.   From point D draw $DE\parallel CA$ meting BA produced at E.
Then EBD is required triangle. We can name it PQR.
Justification
$\because B_{4}D\parallel B_{3}C$
$\therefore \frac{BC}{CD}= \frac{3}{1}\Rightarrow \frac{CD}{BC}= \frac{1}{3}$
$Now\, \therefore \frac{BD}{BC}= \frac{BC+CD}{BC}= 1+\frac{CD}{BC}= 1+\frac{1}{3}= \frac{4}{3}$
$Also\, DE\parallel CA$
$\therefore \bigtriangleup ABC\sim \bigtriangleup DBE$
$\frac{EB}{AB}= \frac{DE}{CA}= \frac{BD}{BC}= \frac{4}{3}$