please could you help me with this question.

The perimeter of a rectangle is 34cm. given that the diagonal is of length 13cm, and that the width is x cm, derive the equation x^2 - 17x + 60 = 0.

Hence find the dimensions of the rectangle.

Maybe its me being thick.

2. The perimeter is $2x+2y=34--->x+y=17$

Area: $x^{2}+y^{2}=169$

Solve the first one for y and sub into the area:

$x^{2}+(17-x)^{2}-169=0$

Now, expand and solve the resulting quadratic.

3. Thanks galactus

But i don't under stand what u have done.

The answer is 5cm x 12cm.

Do you mean solve the equations as simultaneous equations.

don't understand what you have done.

Originally Posted by galactus
The perimeter is $2x+2y=34--->x+y=17$

Area: $x^{2}+y^{2}=169$

Solve the first one for y and sub into the area:

$x^{2}+(17-x)^{2}-169=0$

Now, expand and solve the resulting quadratic.

don't understand what you have done.

where did u get the formulas.

are of rect = l x h

perimenter = 2l + 2 H

solve it, so i can see how u got the answer

where did u get the 169 from

4. $13^{2}=169$.

I just solved one equation for a variable and subbed into the other. That is all you have to do.

Solve $x+y=17$ for y and sub into the diagonal:

The diagonal is 13. So, by Pythagoras, we have $x^{2}+y^{2}=13^{2}$

We have two equations with two unknowns, solve away. I pretty much showed all the steps in the first post.

5. ## Thank you

Originally Posted by galactus
$13^{2}=169$.

I just solved one equation for a variable and subbed into the other. That is all you have to do.

Solve $x+y=17$ for y and sub into the diagonal:

The diagonal is 13. So, by Pythagoras, we have $x^{2}+y^{2}=13^{2}$

We have two equations with two unknowns, solve away. I pretty much showed all the steps in the first post.
Thank you

Understand what u have done,

cheers