Question

9adsquare = 7absqaure . chapter - similarity of triangles

Answer 1

Let us assume that each side of triangle ABC is ‘a’

Then, BD = a/3, MC = a/2 and

DM =2a/3 -a/2 =a/6

also, AM = a√3/2

According to Pythagoras theorem, in triangle ADM;

AD² =AM²+DM²

OR  ,AD² =(a√3/2)² +(a/6)²

=3a²/4+a²/36 

=27a²+a²/36

AD² =28a²/36 = 7a²/9

=7AD² =9a²

=7AD² =9AB² ......PROVED

Answer 2

Step-by-step explanation:

Given :-

A ∆ABC in which AB = BC = CA and D is a point on BC such that BD = ⅓BC.

To prove :-

9AD² = 7AB² .

Construction :-

Draw AL ⊥ BC .

Proof :-

In right triangles ALB and ALC, we have

AB = AC ( given ) and AL = AL ( common )

∴ ∆ALB ≅ ∆ ALC [ By RHS axiom ] .

So, BL = CL .

Thus, BD = ⅓BC and BL = ½BC .

In ∆ALB, ∠ALB = 90° .

∴ AB² = AL² + BL² .......(1) [ by Pythagoras' theorem ] .

In ∆ALD , ∠ALD = 90° .

∴ AD² = AL² + DL² . [ by Pythagoras' theorem ] .

⇒ AD² = AL² + ( BL - BD )² .

⇒ AD² = AL² + BL² + BD² - 2BL.BD .

⇒ AD² = ( AL² + BL² ) + BD² - 2BL.BD .

⇒ AD² = AB² + BD² - 2BL.BD. [ using (1) ]

⇒ AD² = BC² + ( ⅓BC )² - 2( ½BC ). ⅓BC .

[ ∵ AB = BC, BD = ⅓BC and BL = ½BC ] .

⇒ AD² = BC² + 1/9BC² - ⅓BC² .

⇒ AD² = 7/9BC² .

⇒ AD² = 7/9AB² [ ∵ BC = AB ] .

⇒ 9 AD² = 7 AB²

Hence proved