Question

A folded napkin has a triangular cross-section of sides x cm, (x+1) cm and (x+2) cm. If one of the angles of the triangle is 90 degrees, find the value of x.

Answer 1

Answer:

this is the answer of your question.

Answer 2

The value of x is 3 cm.

Given:

• A folded napkin has a triangular cross-section of sides x cm, (x+1) cm and (x+2) cm.
• If one of the angles of the triangle is 90 degree.

To find:

• find the value of x.

Solution:

Concept to be used:

Pythagoras theorem can be used.

Hypotenuse²= Base²+ Perpendicular².

Step 1:

Draw the right triangle and mention sides.

As hypotenuse is the longest side,

So,

Hypotenuse = (x+2) cm

Base = (x+1) cm

and Perpendicular= x cm

Note*: Base and Perpendicular can be interchanged.

Step 2:

Apply Pythagoras theorem.

$( {x + 2)}^{2} = ( {x + 1)}^{2} + {x}^{2}$

open identity

$\bf ( {a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

So,

${x}^{2} + 4x + 4 = {x}^{2} + 2x + 1 + {x}^{2}$

or

${x}^{2} - 2 {x}^{2} + 4x - 2x + 4 - 1 = 0$

or

$- {x}^{2} + 2x + 3 = 0$

or

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

Factorise the quadratic polynomial by splitting middle term.

${x}^{2} - 3x + x - 3 = 0$

or

$x(x - 3) + 1(x - 3) = 0$

or

$(x - 3)(x + 1) = 0$

or

$\bf \red{x = 3}$

or

$x = - 1$

Discard the negative value.

Thus,

Value of x is 3 cm.

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