Difference between revisions of "2009 Grade 8 CEMC Gauss Problems/Problem 15"
(Created page with "==Problem== In rectangle <math>PQRS</math>, <math>PQ = 12</math> and <math>PR = 13</math>. The area of rectangle <math>PQRS</math> is {{Image needed}} <math> \text{ (A) }\ 30...") |
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==Solution 1.1== | ==Solution 1.1== | ||
| − | We can follow the same steps as solution 1. However, 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]]. |
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<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> | ||
Revision as of 11:25, 19 June 2025
Problem
In rectangle 
, 
and 
. The area of rectangle is
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
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
