Question

the side of a triangle are three consecutive integers . a prependicular is drawn on the second largest side . if it divides the second largest side into two parts of lengths p and q respectively , then find the value of ( p-q )² . no spam please I mark you brianlist​

Answer 1

Answer:

Using sine law, n-1

sina

n+1 sin2a

→ 2cosa=

(n-1) n+1

cosa= 2(n-1)

n+1

2n(n+1)

n 2

+(n+1)

2 -(n-1)

2

2(n-1) (n+1)

(using cosine law)

2n+(n+1)

n

2

+4n

2(n-1) (n+1)

2(n+1) n+4

2(n-1) n+1

(n+1)

2 =(n+4)(n-1)

Answer 2

The value of $(p-q)^{2}$ = 16

Given:

• A triangle with three consecutive integers as sides.
• A perpendicular drawn on second largest side.

To find:

Value of $(p-q)^{2}$

Basic algebraic identities used: -

• $(x-y)^{2} = x^{2} + y^{2} - 2xy$
• $(x+y)^{2} = x^{2} + y^{2} + 2xy$
• $(x^{2} - y^{2}) = (x+y) (x-y)$
  

Step by step explanation:

As we have been given three consecutive sides of a triangle,
Let (x-1), x, and (x+1) be the lengths.
Let our triangle be ΔABC.
Therefore, AB = (x-1), BC = (x), AC = (x+1)
Second largest side is BC. Let AD be the altitude drawn perpendicular to BC. (D is the midpoint of BC)
BD = p and CD = q        ................... (given)
⇒ (p + q) = x

In ΔABD,
$AB^{2} = AD^{2} + BD^{2}$          ................. (Pythagoras theorem)
⇒ $AD^{2} = AB^{2} - BD^{2}$                              -----------------------(1)

In ΔACD,
$AC^{2} = AD^{2} + CD^{2}$           ................. (Pythagoras theorem)
⇒ $AD^{2} = AC^{2} - CD^{2}$                             -----------------------(2)

From (1) and (2) Equating AD on both sides
$AB^{2} - BD^{2} = AC^{2} - CD^{2}$
⇒ $(x-1)^{2} - p^{2} = (x+1)^{2} - q^{2}$        
⇒ $x^{2} +1 -2x -p^{2} = x^{2} + 1+ 2x -q^{2}$              
⇒ $-p^{2} + q^{2} = 2x + 2x$
⇒ $p^{2} - q^{2} = -4x$
⇒ (p - q) (p + q) = -4x
⇒ (p - q) x = -4x
⇒ (p - q) = -4
Squaring both sides
⇒ $(p-q)^{2} = 16$

∴ The value of $(p-q)^{2}$ = 16.

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