Question

In ∆ABC ,angle B = 90° find the sides of the triangle if AB=xcm , BC= (4x-4) cm and AC= (4x-5 ) cm​

Answer 1

In ∆ABC, Angle B is 90°. The largest angle in the triangle is Angle B, hence the longest side 'hypotenuse' will be AC according to rule, largest angle is opposite to the longest side in a triangle.

So, AB and BC are the leg sides and AC is the hypotenuse. Using Pythagoras theoram,

➛ AB² + BC² = AC²

We have,

• AB = x cm

1. BC = 4x + 4 cm

• AC = 4x + 5 cm

Plugging the given values,

➛ x² + (4x + 4)² = (4x + 5)²

➛ x² + 16x² + 32x + 16 = 16x² + 40x + 25

➛ 17x² + 32x + 16 = 16x² + 40x + 25

➛ 17x² + 32x + 16 - 16x² - 40x - 25 = 0

➛ x² - 8x - 9 = 0

Finding the zeroes,

➛ x² - 9x + x - 9 = 0

➛ x(x - 9) + 1(x - 9) = 0

➛ (x + 1)(x - 9) = 0

Then, x = -1 or 9 but sides cannot be negative hence, x = 9.

Now:-

• AB = 9 cm

• BC = 4(9) + 4 cm = 40 cm

• AC = 4(9) + 5 cm = 41 cm

Answer 2

*Correct Question :-

• In ∆ABC ,angle B = 90° find the sides of the triangle if AB=xcm , BC= (4x+4) cm and AC= (4x+5 ) cm

*Required Answer :-

In ∆ABC, Angle B is 90°. The largest angle in the triangle is Angle B, hence the longest side 'hypotenuse' will be AC according to rule, largest angle is opposite to the longest side in a triangle.

So, AB and BC are the leg sides and AC is the hypotenuse. Using Pythagoras theoram,

➛ AB² + BC² = AC²

We have,

• AB = x cm
• BC = 4x + 4 cm
• AC = 4x + 5 cm

Plugging the given values,

➛ x² + (4x + 4)² = (4x + 5)²

➛ x² + 16x² + 32x + 16 = 16x² + 40x + 25

➛ 17x² + 32x + 16 = 16x² + 40x + 25

➛ 17x² + 32x + 16 - 16x² - 40x - 25 = 0

➛ x² - 8x - 9 = 0

Finding the zeroes,

➛ x² - 9x + x - 9 = 0

➛ x(x - 9) + 1(x - 9) = 0

➛ (x + 1)(x - 9) = 0

Then, x = -1 or 9 but sides cannot be negative hence, x = 9.

Finding all the sides:

• AB = 9 cm
• BC = 4(9) + 4 cm = 40 cm
• AC = 4(9) + 5 cm = 41 cm

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