Question

construct a triangle of side 4,5&6 and the triangle similar to it whose sides are ⅔ of the corresponding side of Frist ∆le?

explain by images!​

Answer 1

Answer :-

➦ Step 1 : Draw a line segment AB = 4cm. Taking Point A as centre, Draw an arc of 5cm radius. Similarly, taking point B as its centre, Draw an arc of 6cm radius. these arcs with interest each other at point C. Now, AC = 5cm and BC = 6cm and ∆ ABC is the Required Triangle

➦ Step 2 : Draw a ray AX Making an Acute Angle With Line AB on The opposite Side of Vertex C.

➦ Step 3 : Locate 3 points A₁,A₂,A₃ on Line AX Such That AA₁ = A₁A₂ = A₂A₃

➦ Step 4 : Join BA₃ And Draw A line Through A2 parallel to BA₃ to Interest AB at point B'

➦ Step 5 : Draw a line through B' Parallel to the BC to interest AC to C'

∆AB'C' is the Required Answer

═════ Fig ( Refer The Attachment ) ═════

☆ Justification :-

The Construction Can be justified by proving that

• AB' = 2/3 AB
• B'C' = 2/3 BC
• AC' = 2/3 AC

By Construction, We have B'C' || BC

∴ ∠AB'C' = ∠ABC ( corresponding Angles )

In ∆AB'C' & ∆ABC,

⇒∠AB'C' = ∠ABC ( proved Above )

⇒∠B'AC' = ∠BAC ( proved Above )

∴ ∆AB'C' ~ ∆ABC ( AA similarity criteria )

⇒ AB'/AB = B'C'/BC = AC'/AC ___ (1)

In ∆AA₂B' & ∆AA₃B,

⇒ ∠A₂AB' = ∠A₃AB ( common )

⇒ ∠AA₂B' = ∠AA₃B ( corresponding Angles )

∴ ∆AA₂B' ~ ∆AA₃B ( AA similarity criteria )

⇒AB'/AB

⇒AB'/AB = 2/3 ___ (2)

From eq 1 & 2

➙ AB'/AB = B'C'/BC = AC'/AC = 2/3

➙ AB' = 2/3(AB) , B'C' = 2/3(BC)

☆ This Justifies The Construction ☆

$\sf{\red{Angel}}$

Answer 2

Answer:

hi Nithish how r u and how is ur health (◍•ᴗ•◍)

Step-by-step explanation:

please remove ur bio pic , its too bad

allways be happy

keep smiling ( ◜‿◝ )