5
$\begingroup$

My friends recently took a Maths GCSE. In the paper, they came across a very difficult question which we spent a full half-hour train journey trying to figure out. We didn't manage it, so I've come here hoping you can help us out.

Here is the question:

enter image description here

The perimeter of the rectangle is 20. Show that $x^2-10x+18=0$

Sorry for my appalling rectangle drawing skills. Drawing with Paint is like trying to staple an apple to your face. Painful and difficult.

$\endgroup$
9
  • $\begingroup$ Is the length of diagonal $8$? $\endgroup$
    – Inceptio
    May 10, 2013 at 16:32
  • 1
    $\begingroup$ Glad you asked, Inceptio...I thought it was "s" at first. $\endgroup$
    – amWhy
    May 10, 2013 at 16:34
  • 1
    $\begingroup$ amWhy, Now it looks more like $\delta$. $\endgroup$
    – Inceptio
    May 10, 2013 at 16:35
  • 1
    $\begingroup$ @Inceptio Sorry, but like I said, Paint happened. By the way, it is indeed 8 $\endgroup$
    – imulsion
    May 10, 2013 at 16:45
  • $\begingroup$ @imulsion: The same paint. $\endgroup$
    – Inceptio
    May 10, 2013 at 16:50

7 Answers 7

Reset to default
5
$\begingroup$

Consider one triangle formed when diagonal is drawn. Let $x$ be one of the sides, other side will be $10-x$

$(10-x)^2+x^2=k^2$

$10^2+2x^2-20x=k^2$ , Looking at the expression $k$ looks like $8$. So, $36+2x^2-10x=0$

$\endgroup$
6
  • $\begingroup$ how does the final equation cancel to $x^2-10x+18$? $\endgroup$
    – imulsion
    May 10, 2013 at 16:53
  • 1
    $\begingroup$ $36+2x^2-10x=2(x^2-10x+18)=0$ You cancel $2$ throughout. $\endgroup$
    – Inceptio
    May 10, 2013 at 16:55
  • $\begingroup$ Thank you! This was such a hard question $\endgroup$
    – imulsion
    May 10, 2013 at 16:56
  • 1
    $\begingroup$ @imulsion: No, it wasn't. You just have to observe things.:) $\endgroup$
    – Inceptio
    May 10, 2013 at 16:57
  • $\begingroup$ i'm not that great at maths or observing things xD $\endgroup$
    – imulsion
    May 10, 2013 at 16:57
3
$\begingroup$

Use the Pythagorean theorem. The width of the rectangle is $w=x$, and the length is $2l=20-2x$, therefore $l=10-x$. Therefore the diagonal $8^2=w^2+l^2=x^2+(10-x)^2$. Now expand this and show that it is equal to the quadratic you stated.

$\endgroup$
2
$\begingroup$

Hint: Let y be the other side. Then $y^2 + x^2 = 64$ and $2(y+x) = 20$.

$\endgroup$
1
$\begingroup$

Let the side not given be $a$. Since the perimeter is 20, we know that $2x+2a=20$ which we can rearrange to obtain $a=10-x$. By the Pythagorean theorem, $x^2+a^2=8^2=64$. Substituting for $a$, we have $x^2+(10-x)^2=64$ which simplifies to $x^2-10x+18=0$; the desired result.

$\endgroup$
0
$\begingroup$

Use the Pythagorean Theorem and simultaneous equations to come up with $x^2 + y^2 = 8^2$ and $2x + 2y = 20$.

Therefore, $x + y = 10$.

x (squared) + (10-x) (squared) = 64

2x(squared) - 20x + 100 = 64

(2x (squared) - 20x + 36)/2 = 0

x - 10x + 18 = 0

$\endgroup$
0
$\begingroup$

Let the height = y.

Therefore we can now create 2 simultaneous equations:-

1) X+y=10 (since perimeter of whole rectangle equals 20, X and y together must equal half ie 10)

2) (8^2)-(x^2)=(y^2) (using pythagrus)

So if we change around the first equation, we can get:- y=10-X, now plugging this into the second equation we get;- (8^2)-(x^2)=(10-X)^2, which is the same as - 64-(x^2)=100-20x+(x^2), which equals (ie, bringing everything on the left hand side over to the right) - 0=36-20x+2(x^2), now divide this whole equation by 2, you get:-

0=18-10x+(x^2),

Which of course when rearranged, is the same as:-

(x^2)-10x+18=0

$\endgroup$
-1
$\begingroup$

Name one side $y$.

$2x+2y=20$ therefore $x+y=10$ and $y=10-x$

By the Pythagorean Theorem,

$(10-x)²+x²=8²$ and therefore

$2x²-20x-100=64$,

$2x²-20x+36=0$,

$x²-10x+18=0$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .