Question 16774: Can someone help me with this problem? The base of a parallelogram is 10 cm long, the height is 5cm, and the longer diagonal is 13cm long. How long is the shorter diagonal? First I used the pythagorean theorem c2=a2+b2....c=13cm, a=?, b= 5cm.....from there I am still confused please help...Thank you!
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! If you will draw the parallelogram with lower left vertex at A, lower right at B, upper right at C, and upper left at D, then drop perpendicular lines from upper vertices C and D down to the base line AB. In making this drawing of parallelogram ABCD, draw it so that AC is the longest diagonal, which is 13 cm. Let E be the point on extended line AB that is directly below point C, and let F be the point on line AB that is directly below point D. By the Theorem of Pythagoras, it is possible to find AE (since triangle AEC is a right triangle):  . AE=12. Since AB=10, this means that BE = 12-10=2 cm.
Next, find the length of line segment FB (in the middle of line AE). Since BE = 2, by symmetry, AF = 2, so FB= 8. There is another right triangle FBD, where the legs are FD= 5cm and FB=8, so the hypotenuse of this right triangle is the other diagonal:
There are probably easier ways to do this, and I apologize for the lengthy description. They say a picture is worth a thousand words, but I don't know how to draw a figure in algebra.com.
R^2 at SCC
|
|
|
