SOLUTION: what is the area of a rectangle whose width is 10 in. and whose diagonal measures 26 in

Question 881466: what is the area of a rectangle whose width is 10 in. and whose diagonal measures 26 in
Found 2 solutions by rfer, Chester2020:
Answer by rfer(16322) (Show Source): You can put this solution on YOUR website!
26^2-10^2=576
sqrt 576=24
10*24=240 sq in
Answer by Chester2020(1) (Show Source): You can put this solution on YOUR website!
(REFER's ANSWER IS WRONG!)(24x10) NOT (24x26) (26 is the hypotenuse!!!!)In the pythagorean theorem As + Bs = Cs (s) meaning (square)the diagonal line is hypotenuse and since your using a rectangle all the corners are right angles so this is going to be a 30 45 90 degree right triangle or just call it Leg Leg Hypotenuse. so first you have to find the other side of the rectangle which should also be the last side of your rectangle if i pictured it right.
.
leg1 or As = 10s
Leg2= Bs
hypotenuse or Cs = 26s
10s+Bs=26s (square them) = 100+Bs=676
then simplify as you would do in your normal algebra
100+Bs=676
-100 -100
-------------
Bs=576s
square root of 576=(24)<--Leg2
now you can get the area (10in X 24in = 240in. square)

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