Question

In a right anfled triangle the hypotenuse is 6 m more than the shorter side and third side is 3m less than the hypotenuse then find the angles of a rigjt angled triangle​

Answer 1

Question :- In a right angled triangle the hypotenuse is 6 m more than the shorter side and third side is 3m less than the hypotenuse then find the all three sides of the ∆ ?

Answer :-

→ Let Hypotenuse = x m.

→ Than, shorter side = (x - 6)m.

→ Third side = (x - 3) = (x - 3)m .

So, By Pythagoras Theoram ,

→ (x - 3)² + (x - 6)² = x²

→ x² - 6x + 9 + x² - 12x + 36 = x²

→ x² - 18x + 45 = 0

Splitting The Middle Term now,

→ x² - 15x - 3x + 45 = 0

→ x(x - 15) -3(x - 15) = 0

→ (x - 15) (x - 3) = 0

Putting Both Equal to Zero,

→ x = 15 or 3.

since Hypotenuse is Not Less than 6 here.

So,

→ Hypotenuse = x m. = 15m.

→ Than, shorter side = (x - 6)m. = 15 - 6 = 9m .

→ Third side = (x - 3) = (x - 3)m . = 15 - 3 = 12m .

So , Sides of ∆ are 9, 12 & 15m.

______________________

Now, Lets Try to Solve Similar Correct Question :-

"The hypotenuse of a right triangle is 6m more than the twice of the shortest side. If the third side is 2m less than the hypotenuse, find the sides of the triangle." ?

Solution :-

→ Let Shorter side = x m.

→ Than, Hypotenuse = (2x + 6)m.

→ Third side = (2x + 6) - 2 = (2x + 4)m .

So, By Pythagoras Theoram ,

→ x² + (2x + 4)² = (2x + 6)²

→ x² + 4x² + 16 + 16x = 4x² + 36 + 24x

→ x² + 16x - 24x + 16 - 36 = 0

→ x² - 8x - 20 = 0

Splitting The Middle Term now,

→ x² - 10x + 2x - 20 = 0

→ x(x - 10) + 2(x - 10) = 0

→ (x - 10)(x + 2) = 0

Putting Both Equal to Zero now,

→ x - 10 = 0

→ x = 10

Or,

→ x + 2 = 0

→ x = (-2) { Side in Negative ≠ }.

So,

→ Shorter side = x m. = 10m.

→ Than, Hypotenuse = (2x + 6)m. = (2*10+6) = 26m .

→ Third side = (2x + 6) - 2 = (2x + 4)m . = (2*10+4) = 24m.

Hence, The Three sides of Right Angled ∆ are 10,24 & 26m.

Answer 2

$\underline{\underline{\large{\bf{\green{QuEsTion}}}}}$

In a right angled triangle the hypotenuse is 6 m more than the shorter side and third side is 3 m less than the hypotenuse. then, find the sides of  right angled triangle .

$\underline{\underline{\large{\bf{\green{SoluTion}}}}}$

Let, the shorter side of triangle = x

then, hypotenuse will be = x + 6

third side will be = ( x+6) - 3 = x+3

By Pythagoras theorem

(hypotenuse)²= (shorter side)²+(third side)²

(x+6)² = x² + (x+3)²

x²+36+12 x = x²+ x² + 9 +6 x

x² - 6 x - 27 = 0

(splitting the middle term )

x² - 9 x+3 x - 27=0

x (x-9) + 3 (x-9) = 0

(x+3)(x-9)=0

x = - 3 or x = 9

∵ Length cannot be negative

hence, we will take x = 9

so,

sides of triangle will be

x = 9 cm

x + 6 (hypotenuse) = 9+6 =15 cm

x + 3 = 9+3 = 12 cm.