Question

Segment AN is the altitude to side BC in ΔABC. If AB = 3NC and AN = 2NC, prove that AC = BN. (Hint: Use variables in such problems. Let NC = x units and find the other lengths in terms of x.)

Answer 1

Answer :

The proof is as follows :

Step-by-step explanation:

Let NC = x

⇒ AB = 3x and AN = 2x

In Δ ABN, By using Pythagoras theorem,

AB² = BN² + AN²

⇒ BN² = AB² - AN²

⇒ BN² = (3x)² - (2x)²

⇒ BN² = 5x²

⇒ BN = x√5  .......................(1)

Now in ΔANC , Using Pythagoras theorem We have,

AC² = NC² + AN²

⇒ AC² = x² + (2x)²

⇒ AC² = 5x²

⇒ AC = x√5   ....................(2)

From equations (1) and (2) We get,

AC = BN , which is our required result

Answer 2

Answer:

BN=AC=√5 x.

The proof is explained in step-by-step explaination.

Step-by-step explanation:

Let NC=x. It is given that AB=3NC & AN=2NC

⇒ AB=3x & AN=2x

By applying Pythagoras theorem

In triangle ANC,

$AC^{2}=AN^{2}+NC^{2}$

⇒ $AC^{2} = (2x)^{2}+x^{2}$

⇒ $AC^{2}=4x^{2}+x^{2} =5x^{2}$

⇒ $AC=\sqrt{5}x$   →    (1)

Similarly, In triangle ABN,

$AB^{2}=AN^{2}+BN^{2}$

⇒ $(3x)^{2}=BN^{2}+x^{2}$

⇒ $9x^{2} = (BN)^{2}+4x^{2}$

⇒ $BN^{2}=5x^{2}$

⇒ $BN=\sqrt{5}x$   →   (2)

From eq (1) & (2),    AC=BN