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#1 2006-03-02 07:56:42

nebraskanative
Member
Registered: 2006-03-02
Posts: 1

Help Please

What is the area of a rhombus that has one side length 10 and diagonals that differ by 4

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#2 2006-03-02 08:55:41

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,582

Re: Help Please

Well a rhombus has all four sides the same length and is a special parallelogram,
so all sides are 10.  Hope that helps you.


igloo myrtilles fourmis

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#3 2006-03-02 09:05:48

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,582

Re: Help Please

Draw the slanted square (rhombus), draw the diagonals so now you have a
picture of kite without the tail flying in the wind.
Notice the right triangles.
Assign values like x and x-2 and 10 and apply pythagoreans theorm
Then solve the quadratic equation using quadratic formula if it's hard.
Then the area is four right triangles, or two rectangles, if you pair up
the right triangles by sliding them all around.


igloo myrtilles fourmis

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#4 2006-03-02 09:21:33

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,555

Re: Help Please

Yes! The diagonals of a rhombus are at right angles.

So imagine a triangle whose base is the base of the rhombus, but whose apex is the middle, then use Pythagoras Theorem

a[sup]2[/sup] + b[sup]2[/sup] = c[sup]2[/sup]

Remember, these are going to be half-diagonals, so the difference will only be 2, and lets call the short diagonal "x":

x[sup]2[/sup] + (x+2)[sup]2[/sup] = 10[sup]2[/sup]

x[sup]2[/sup] + x[sup]2[/sup] + 4x + 2[sup]2[/sup] = 10[sup]2[/sup]

2x[sup]2[/sup]  + 4x + 4 = 100

2x[sup]2[/sup]  + 4x -96 = 0

This is a Quadratic Equation and the solutions are 6 and -8. The lengths won't be negative so "6" is the answer.

That is the short "half-diagonal", so the full diagonal will be 12, and the longer diagonal 16.

The area will be ½ × 12 × 16 = 96 (because you can calculate the area of a rhombus by multiplying the diagonals and then halving)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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