Question

Construct a triangle of sides 4 cm, 5 cm. and 6 cm. and then a triangle similar to it whose sides are 2÷3 of the corresponding sides of the first triangle
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Answer 1

Step-by-step explanation:

1) draw a segment AB=6cm

2) with a as centre draw radius of AC=4cm, draw an arc

3) with B as centre BC =5cm

4) join AC to BC to obtain ∆abc

5) below AB make acute angle BAX

6) along AX mark off three points ( greater of 2 and 3 in 2/3)

7) draw A2B' || A3B

8) construct B'C' || BC

9) hence the triangle is divided into the ratio 2:3

Answer 2

$\huge\star{\purple{\underline{\underline{\mathfrak{Explanation:}}}}}$

______________________________

$\bold{\pink{STEPS-OF-CONSTRUCTION:}}$

• $\large{\pink\leadsto}$Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.

• $\large{\pink\leadsto}$Take the point A as centre, and draw an arc of radius 5 cm.

• $\large{\pink\leadsto}$Similarly, take the point B as its centre, and draw an arc of radius 6 cm .

• $\large{\pink\leadsto}$The arcs drawn will intersect each other at point C.

• $\large{\pink\leadsto}$Now, we obtained AC = 5 cm and BC = 6 cm and therefore ΔABC is the required triangle.

• $\large{\pink\leadsto}$Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

• $\large{\pink\leadsto}$Locate 3 points such as A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A2 = A2A3.

• $\large{\pink\leadsto}$Join the point BA3 and draw a line through A2 which is parallel to the line BA3 that intersect AB at point B’.

• $\large{\pink\leadsto}$Through the point B’, draw a line parallel to the line BC that intersect the line AC at C’.

• Therefore, ΔAB’C’ is the required triangle.

$\huge\star{\purple{\underline{\mathfrak{Justification:}}}}$

• AB’  =(2/3)AB
• B’C’ = (2/3)BC
• AC’ = (2/3)AC

From the construction, we get B’C’ || BC

∴ ∠ AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’

∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)

In ΔAAB’ and ΔAAB, ∠A2AB’=∠A3AB (Common)

From the corresponding angles, we get,

∠AA2B’=∠AA3B

Therefore, from the AA similarity criterion, we obtain

Δ AA2B’ and AA3B

So, AB’/AB = AA2/AA3

Therefore, AB’/AB = 2/3 ……. (2)

From the equations (1) and (2),

we get AB’/AB= B’C’/BC = AC’/ AC = 2/3

This can be written as AB’  = (2/3)AB

B’C’ = (2/3) BC

AC’= (2/3) AC

$\large\star{\purple{\underline{\mathfrak{Hence,justified!}}}}$