Question

The hypotenous of a right angled triangle is 1 m less than twince the shortest side if the third side is 1 m mire than the shortest side then find the side of the triangle​

Answer 1

Question :

The hypotenous of a right-angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, then find shortest the side of the triangle.

Given :

• Hypotenuse of the triangle = 2(Shortest side of the triangle) - 1.

• Third side of the triangle = Shortest side of the triangle + 1.

To find :

Shortest side of the triangle.

Solution :

Let the shortest side of the triangle be the base of the triangle, denoted by the variable x.

So according to the Question ,

• Hypotenuse of the triangle = (2x - 1) m.

• Third side/height of the triangle = (x + 1) m.

Now we get the triangle as :

$\setlength{\unitlength}{1 cm}\begin{picture}(6,6)\put(2 , 2){\line(1, 1){2}}\put(4 , 2){\line(0 , 1){2}}\put(4 , 2){\line(-1 , 0){2}}\put(1.3,3.){(2x - 1) m}\put(4.1, 3){(x + 1) m}\put(2.6,1.7){x m}\put(1.7,1.8){\bf{A}}\put(4.1,1.8){\bf{B}}\put(4,4){\bf{C}}\end{picture}$

Now by using the Pythagoras theorem , we can find the value of x.

We know the Pythagoras theorem i.e

Pythagoras theorem :-

$\boxed{\bf{H^{2} = P^{2} + B^{2}}}$

Where :

• H = Hypotenuse of the triangle.

• P = Height of the triangle.

• B = Base of the triangle.

By using the Pythagoras theorem and substituting the values in it, we get :

$:\implies \bf{H^{2} = P^{2} + B^{2}}$

$:\implies \bf{(2x - 1)^{2} = (x + 1)^{2} + x^{2}}$

$\boxed{\begin{minipage}{7 cm} \underline{\bf{Identity\:used\::}} \\ \\ \bullet\quad \bf{(a + b)^{2} = a^{2} + b^{2} + 2ab} \\ \\ \bullet\quad \bf{(a - b)^{2} = a^{2} + b^{2} - 2ab} \end{minipage}}$

$:\implies \bf{(2x)^{2} - 2 \times 2x \times 1 + 1^{2} = x^{2} + 2 \times x \times 1 + 1^{2} + x^{2}}$

$:\implies \bf{4x^{2} - 4x + 1 = x^{2} + 2x + 1 + x^{2}}$

$:\implies \bf{4x^{2} - 4x + 1 = 2x^{2} + 2x + 1}$

$:\implies \bf{4x^{2} - 4x + 1 - (2x^{2} + 2x + 1) = 0}$

$:\implies \bf{4x^{2} - 4x + \not{1} - 2x^{2} - 2x - \not{1} = 0}$

$:\implies \bf{2x^{2} - 6x = 0}$

Here the equation formed is quadratic equation so by using the formula for quadratic equation, we can find the value of x.

Formula for quadratic equation :

$\boxed{\bf{\alpha = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}}$

By comparing the above equation by ax² + bx + c , we get that :

• a = 2

• b = (-6)

• c = 0.

By using the Quadratic equation formula and substituting the values in it, we get :

$:\implies \bf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}$

$:\implies \bf{x = \dfrac{-(-6) \pm \sqrt{6^{2} - 4 \times 2 \times 0}}{2 \times 2}}$

$:\implies \bf{x = \dfrac{-(-6) \pm \sqrt{6^{2} - 0}}{4}}$

$:\implies \bf{x = \dfrac{6 \pm \sqrt{6^{2}}}{4}}$

$:\implies \bf{x = \dfrac{6 \pm \sqrt{36}}{4}}$

$:\implies \bf{x = \dfrac{6 \pm 6}{4}}$

$:\implies \bf{x = \dfrac{6 + 6}{4}\:;\:x = \dfrac{6 - 6}{4}}$

$:\implies \bf{x = \dfrac{12}{4}\:;\:x = \dfrac{0}{4}}$

$:\implies \bf{x = 3\:;\:x = 0}$

$\boxed{\therefore \bf{x = 3;0}}$

Hence the value of x is 3.

[Since the value of the side of a figure can't be zero , the value of x as 0 is neglected]

Since we have taken the shortest side of the triangle as x, the measure of shortest side of the triangle is 3 m.

Answer 2

Question

The hypotenus of a right angled triangle is 1 m less than twice the shortest side if the third side 1 m mire then the shortest side then find the side of the triangle.

Given

• The hypotenus of a right angled triangle 1 m less than twice the shortest side
• The third side 1 m mire then the shortest side.

Required to Find

• The side of the triangle.

Solution

Let the length of shortest side = x m

Length of hypotenus = 2x - 1

and The third side = x + 1

Now according to the condition

$\sf \: (2x - {1)}^{2} = {(x)}^{2} + (x + 1 {)}^{2} ..(by \: pythagorus \: theorem)$

$\bf \implies \: 4 {x}^{2} - 4x + 1 = {x}^{2} + {x}^{2} + 2x + 1$

$\bf \implies \: 4 {x}^{2} - 4x + 1 = 2 {x}^{2} - 2x - 1 = 0$

$\bf\implies2 {x}^{2} - 6x = 0$

$\bf\implies \: {x}^{2} - 3x = 0$

$\bf \implies \: x(x - 3) = 0....(dividing \: by \: 2)$

Either x = 0

but it isn't possible. or

$\\ \bf x - 3 = 0$

then x = 3

shortest side = 3m

Hypotenus = 2 × 3 - 1 = 6 - 1 - 5

third side = x + 1 = 3 + 1 = 4

Hence sides are 3, 4, 5 ( in m )