Question

27) In an equilateral triangle ABC, D is a point on side BC such that BD=1/3 BC. Prove that 9AD'=7AB


28) The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find 28 term of this AP.​

Answer 1

Answer: UR WELX

ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .

Now, ∆ABE and ∆AEC

∠AEB = ∠ACE = 90°

AE is common side of both triangles ,

AB = AC [ all sides of equilateral triangle are equal ]

From R - H - S congruence rule ,

∆ABE ≡ ∆ACE

∴ BE = EC = BC/2

Now, from Pythagoras theorem ,

∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)

∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2)

From equation (1) and (2)

AB² - AD² = BE² - DE²

= (BC/2)² - (BE - BD)²

= BC²/4 - {(BC/2) - (BC/3)}²

= BC²/4 - (BC/6)²

= BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9

∵AB = BC = CA

So, AB² = AD² + 2AB²/9

9AB² - 2AB² = 9AD²

Hence, 9AD² = 7AB²

Step-by-step explanation:

Sn = ( n / 2) [ 2a + ( n -1)d

sum of the first 7 terms of an A.P is 63 i. e S7 = 63.

 ( 7 / 2) [ 2a + 6d ] = 63

 2a + 6d = 18 --------(1)

Sum of its next 7 terms = 161.

 Sum of first 14 terms = sum of first 7 terms + sum of next 7 terms.

S14 = 63 + 161 = 224

 ( 14 / 2) [ 2a + 13d ] = 224.

 7 [ 2a + 13d ] = 224.

⇒ [ 2a + 13d ] = 32 -------92)

By Solving equation (1) and (2) we obtain

d = 2 

 a = 3.

t28 = a + ( 28 - 1) d

t28 = 3 + ( 28 - 1) 2

t28 = 57.

28th term= 57.

Answer 2

Answer:

➡ 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 ] .

$\huge \green{ \boxed{ \sf \therefore 9AD^{2} = 7AB^{2}. }}$

Hence, it is proved.