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

Step-by-step explanation:

using Pythagoras theorem, AB^2+BC^2=AC^2

X^2+(4x-4)^2=(4x-5)^2

X^2+16x^2+16-32x=16x^2+25-40x

x^2+8x-9=0

x^2+9x-x-9=0

x(x+9)-1(x+9)=0

(x+9)(x-1)=0

x=1

the above question is wrong as the sides are coming out to be zero.

Answer 2

$\Large{ \underline{\underline{ \sf{\purple{Correct \: Question:}}}}}$

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

$\Large{ \underline{\underline{ \sf{\purple{Required \: answer:}}}}}$

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