Suppose ABCD is a rectangle. Find AB and AD if point M is the midpoint of
1 answer:
We know that
Perimeter=2AB+2AD=34 in-----> AB+AD=17-----> AB=17-AD-----> equation 1
MA=MD
MA²=AB²+(AD/2)²
for the triangle AMD
AD²=[MA²+MD²]----> AD²=2*[MA²]----> AD²=2*[AB²+(AD/2)²]---> equation 2
Perimeter=2AB+2AD=34 in-----> AB+AD=17-----> AB=17-AD-----> equation 1
MA=MD
MA²=AB²+(AD/2)²
for the triangle AMD
AD²=[MA²+MD²]----> AD²=2*[MA²]----> AD²=2*[AB²+(AD/2)²]---> equation 2
I substitute 1 in 2
AD²=2*[(17-AD)²+(AD/2)²]----> AD²=2*[289-34AD+AD²+0.25AD²]
AD²=578-68AD+2.50AD²--------> 1.50AD²-68AD+578
1.50AD²-68AD+578=0
using a graph tool to solve the quadratic equation
see the attached figure
AD1=11.33 in
AD2=34 in----------is not solution because (AB+AD=17)
Solution is AD=11.33 in
AB=17-11.33--------> 17-11.33-----> AB=5.67 in
the answer is
AD=11.33 in
<span>AB=5.67 in</span>
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Answer:
(x, y) = (1/2, -1)
Step-by-step explanation:
Subtracting twice the first equation from the second gives ...
(2/x +1/y) -2(1/x -5/y) = (3) -2(7)
11/y = -11 . . . . simplify
y = -1 . . . . . . . multiply by y/-11
Using the second equation, we can find x:
2/x +1/-1 = 3
2/x = 4 . . . . . . . add 1
x = 1/2 . . . . . . . multiply by x/4
The solution is (x, y) = (1/2, -1).
_____
<em>Additional comment</em>
If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.
A graph of the original equations gives you an extraneous solution of (x, y) = (0, 0) along with the real solution (x, y) = (0.5, -1).
