pythagoras

**simonmann** · May 28th 2008, 05:51 AM

the lengths, in centimetres, of the sides of a right-angled triangle are:
x, x+2 and x+5

i) use pythagoras' theorem to write down an equation in x. show that it simplifies to
x^2-6x-21=0
ii) solve the equation

iii) write down the length of hypotenuse of the triangle.

**Soroban** · May 28th 2008, 06:16 AM

Hello, simonmann!

The problem is straight-foward.
Exactly where is your difficulty?

The lengths of the sides (in cm) of a right triangle are: $<br /> x,\:x+2,\:x+5$

a) Use Pythagoras' Theorem to write an equation in $x.$

$a^2 + b^2 \:=\:c^2 \quad\Rightarrow\quad x^2 + (x+2)^2 \:=\:(x+5)^2$

Show that it simplifies to : . $x^2-6x-21\:=\:0$

We have: . $x^2 + x^2 + 4x + 4 \;=\;x^2 + 10x + 25\quad\Rightarrow\quad x^2-6x-21 \:=\:0$

b) Solve the equation.

Quadratic Formula: . $x \;=\;\frac{-(\text{-}6) \pm\sqrt{(\text{-}6)^2 - 4(1)(\text{-}21)}}{2(1)} \;=\;\frac{6\pm\sqrt{120}}{2} \;=\;3 \pm\sqrt{30}$

And the positive root is: . $x \:=\:3 + \sqrt{30}$ cm.

c) Find the length of hypotenuse of the triangle.

The hypotenuse is: . $x + 5 \;=\;(3 + \sqrt{30}) + 5 \;=\;8 + \sqrt{30}$ cm.

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