Question

In triangle abc, angle abc=90°,ab=2a+1,bc=2a^2+2a.find ac in terms of A

Answer 1

S O L U T I O N :

In Δ ABC,we have :

• AB = 2a + 1
• BC = 2a² + 2a
• ABC = 90°

Attachment a figure of right angled Δ according to the given question.

Now,

$\underline{\underline{\tt{Using\:by\:Pythagoras\:theorem\::}}}$

⇒ (hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (AC)² = (BC)² + (AB)²

⇒ (AC)² = (2a² + 2a )²  + (2a + 1)²

⇒ (AC)² = (2a²)² + (2a)² + 2 × 2a² × 2a + (2a)² + (1)² + 2 × 2a × 1

⇒ (AC)² = 4a⁴ + 4a² + 8a³ + 2a² + 1 + 4a

⇒ (AC)² = 4a⁴ + 8a³ + 6a² + 4a + 1

⇒ AC = √4a⁴ + 8a³ + 6a² + 4a + 1

Thus,

The value of AC in terms of A = √4a⁴ + 8a³ + 6a² + 4a + 1 .

Answer 2

Given

In triangle abc, angle abc=90°,ab=2a+1,bc=2a^2+2a.

We Find

AC in terms of A

We Know that

by according to question, we using Pythagoras theorem.

According to the question

⇒ (hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (AC)² = (BC)² + (AB)²

⇒ (AC)² = (2a² + 2a )² + (2a + 1)²

⇒ (AC)² = (2a²)² + (2a)² + 2 × 2a² × 2a + (2a)² + (1)² + 2 × 2a × 1

⇒ (AC)² = 4a⁴ + 4a² + 8a³ + 2a² + 1 + 4a

⇒ (AC)² = 4a⁴ + 8a³ + 6a² + 4a + 1

⇒ AC = √4a⁴ + 8a³ + 6a² + 4a + 1

So,

The value of AC in terms of A = √4a⁴ + 8a³ + 6a² + 4a + 1 .

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