SOLUTION: A banner is in the shape of a right triangle of area 63 in^2. The height of the banner is 4 inches less than twice the width of the banner. Find the height and the width of the ban

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Question 941039: A banner is in the shape of a right triangle of area 63 in^2. The height of the banner is 4 inches less than twice the width of the banner. Find the height and the width of the banner.
Found 3 solutions by Fombitz, macston, MathTherapy:
Answer by Fombitz(32388) About Me  (Show Source):
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A=%281%2F2%29wh=63
h=2w-3
Substituting,
%281%2F2%29w%282w-3%29=63
2w%5E2-3w=126
2%28w%5E2-%283%2F2%29w%29=126
w%5E2-%283%2F2%29w%2B9%2F16=63%2B9%2F16
%28w-3%2F4%29%5E2=1017%2F16
w-3%2F4=0+%2B-+sqrt%281017%29%2F4
Only the positive solution makes sense here.
w=3%2F4+%2B+%283%2F4%29sqrt%28113%29
w=%283%2F4%29%281+%2B+sqrt%28113%29%29
h=2%283%2F4%29%281+%2B+sqrt%28113%29%29-3
h=%283%2F2%29%281+%2B+sqrt%28113%29%29-3
h=-3%2F2+%2B+%283%2F2%29sqrt%28113%29
h=%283%2F2%29%28sqrt%28113%29-1%29


Answer by macston(5194) About Me  (Show Source):
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W=width of banner; H=height of banner=2W-4; A=Area of banner=1/2WH=63 sq in
Substitute for H first equation)in the area equation and solve for W:
1/2WH=63 sq in
1/2W(2W-4)=63 sq in
%281%2F2%29%282W%5E2-4W%29=63+in%5E2
W%5E2-2W=63+in%5E2 Subtract 63in^2 from each side
W%5E2-2W-63in%5E2=63in%5E2-63in%5E2
(((W^2-2W-63in^2=0}}}
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aW%5E2%2BbW%2Bc=0 (in our case 1W%5E2%2B-2W%2B-63+=+0) has the following solutons:

W%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-2%29%5E2-4%2A1%2A-63=256.

Discriminant d=256 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+256+%29%29%2F2%5Ca.

W%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+256+%29%29%2F2%5C1+=+9
W%5B2%5D+=+%28-%28-2%29-sqrt%28+256+%29%29%2F2%5C1+=+-7

Quadratic expression 1W%5E2%2B-2W%2B-63 can be factored:
1W%5E2%2B-2W%2B-63+=+1%28W-9%29%2A%28W--7%29
Again, the answer is: 9, -7. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-63+%29

So W=9,-7
First Let's consider 9
When W=9, H=2W-4=2(9)-4=14
ANSWER Width is 9 in and Height is 14 in, a tall narrow right triangle pointing upward and to the right.
CHECK:
Area=1/2WH=63 in sq
Area=1/2(9in)(14in)=63 in sq
Area=1/2(126in sq)=63 in sq
Area=63 in sq=63 in sq, Check
Next, let's consider -7. At first this seems an unlikely width, but if we consider on an X,Y co-ordinate plane right is the positive direction, it means this banner points left. So solving for height:
H=2W-4=2(-7)-4=-14-4=-18, again negative, so pointing down
So we have a banner of Width=7 in and Height=18 in pointing left and downward
CHECK:
Area=1/2WH=1/2(-7in)(-18 in)=63 in sq
Area=1/2(126 in sq)=63in sq
Area= 63 in sq=63 in sq Check

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

A banner is in the shape of a right triangle of area 63 in^2. The height of the banner is 4 inches less than twice the width of the banner. Find the height and the width of the banner.


Let base of banner be B

Then height = 2B - 4

Since area = 63 in, then it can be said that:

%281%2F2%29B%282B+-+4%29+=+63

%282B%5E2+-+4B%29%2F2+=+63

2%28B%5E2+-+2B%29%2F2+=+63

cross%282%29%28B%5E2+-+2B%29%2Fcross%282%29+=+63

B%5E2+-+2B+=+63

B%5E2+-+2B+-+63+=+0

(B - 9)(B + 7) = 0

B, or base = highlight_green%289%29 inches

Height: 2(9) - 4, or 18 - 4, or highlight_green%2814%29 inches