Difference between revisions of "2009 Grade 8 CEMC Gauss Problems/Problem 15"
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==Solution 1.1== | ==Solution 1.1== | ||
We can follow the same steps as solution 1. However, when finding <math>QR^{2}</math>, we can also use [[difference of squares]]. | We can follow the same steps as solution 1. However, when finding <math>QR^{2}</math>, we can also use [[difference of squares]]. | ||
+ | |||
<math>QR^{2} = 13^{2} - 12^{2} = (13 - 12) \times (13 + 12) = 1 \times 25 = 25</math> | <math>QR^{2} = 13^{2} - 12^{2} = (13 - 12) \times (13 + 12) = 1 \times 25 = 25</math> | ||
Latest revision as of 11:25, 19 June 2025
Problem
In rectangle ,
and
. The area of rectangle
is
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Solution 1
Because is a rectangle, we can notice that
is a right angle. Thus,
is a right triangle.
This means that we can use the pythagorean theorem to find the length of :
We can now use the formula for the area of a rectangle, which is the rectangle's width multiplied by its height:
~anabel.disher
Solution 1.1
We can follow the same steps as solution 1. However, when finding , we can also use difference of squares.
This eventually leads us to the same answer, or .
~anabel.disher
Solution 1.2
Instead of having to use the pythagorean theorem in solution 1, we can remember that ,
, and
form a pythagorean triple. This gives us
, and eventually leads us to the same answer, or
.
~anabel.disher