2 thoughts on “In ∆ABC ,angle B = 90° find the sides of the triangle if AB=xcm , BC= (4x-4) cm and AC= (4x-5 ) cm”
*CorrectQuestion:–
In ∆ABC ,angle B = 90° find the sides of the triangle if AB=xcm , BC= (4x+4) cm and AC= (4x+5 ) cm
*RequiredAnswer:–
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.
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.
*Correct Question :–
*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,
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:
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HOPE \: IT \: HELPS \: YOU.
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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,
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:–