Construct a triangle with sides 5 cm , 6 cm and 7 cm and then another triangle whose sides are 7 5 of the corresponding sides of the first triangle.

Steps of Construction : Step 1: Draw a line segment, A B = 5 cm . Step 2: With A as centre, draw an arc of radius 6 cm . Step 3: With B as centre and radius equal to 7 cm , cut an arc on the previous arc and name it C . Step 4: Join B C and A C . Thus, Δ A B C is obtained. Step 5: Draw a ray A R on the side which is opposite to the point C . Mark points A 1 , A 2 , . . . . A 7 which are equidistant to each other on the same ray. Step 6: Join A 5 B and draw its parallel from A 7 to extension of A B , mark B ' as shown. From B ' draw a parallel to B C to extension of B C and mark C ' . Thus. the triangle A B ' C ' is the required triangle, each of whose sides is 7 5 the corresponding sides of △ A B C .

Construct a ∆ A B C with B C = 7 cm , ∠ B = 60 ° and A B = 6 cm . Construct another triangle whose sides are 3 4 times the corresponding sides of ∆ A B C .

Steps of construction : 1) Draw a line segment B C = 7 cm . 2) At B , construct ∠ X B C = 60 ° . 3) With B as centre and radius 6 cm , draw an arc intersecting X B at A . 4) Join A C to obtain ∆ A B C . 5) Below B C , make an acute ∠ C B Y . 6) Along B Y , mark off 4 points (greater of 3 and 4 in 3 4 ) B 1 , B 2 , B 3 , B 4 such that B B 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 . 7) Join B 4 C . 8) From point B 3 , draw a line parallel to B 4 C intersecting B C at C ' . 9) From point C ' , draw a line parallel to C A intersecting A B at A ' . Thus, ∆ A ' B C ' is the required triangle.