Room Addition

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phillygurl

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Mr. Lon Green is hiring your architectural firm to build a rectangular studio on the south side of his house. The studio is attached so that north side of the studio will be a portion of the current south side of his house. For energy efficiency, Mr. Green wants to make the walls of the studio out of rammed earth, 2 feet thick. The inside south wall of the studio will be twice as long as the inside west wall.
Mr. Green would also like to build a semicircular patio around the studio. The studio will be circumscribed by the patio. A large tree ( circumference = 20”) is 20 feet from south side of the house. The external area of the room addition is to be 392 square feet.

1. Determine the inside dimensions of the room addition, having an external area of 392 square feet and walls that are 2 feet thick.

12L x 24W
• Define the exterior dimensions with binomial expressions
14L x 28w

• Simplify the resulting quadratic equation

ax^2 + bx + c =0
14x^2 + 28 = 0

• Solve for the unknown

14(x^2 + 2) = 0
14(x + 1)^2

2. Determine how far the patio extends past the south wall of the room addition. ( Use the Pythagorean Theorem)
a^2 + b^2 = c^2
x^2 + (x + 1)^2 = c^2
19x^2 + (38x + 1)^2 = 0
19 x 1^2 + (38(1) + 1)^2 = 0
19 x 1 + (38 + 1)^2 = 0
19 + 1444 + 1 = 0
1464 = 0
*** doesn't work out - what am I doing wrong? ***

3. What accommodations must be made for the tree?
None - I extended the patio an additional 5ft which makes it shorter than the 6ft difference

4. How could the interior dimensions be changed to eliminate the tree from consideration? ( Reverse engineer the solution.) - This doesn't apply because the patio came up shorter and the tree was never a factor.

Is this the best possible answer for this equation?
 
Try this...

LW = Area

the inner length is twice the inner width, so, set W = x and L = 2x for the inside dimensions. The outer dimensions would be W = x + 2 and L = 2x + 4, for 3 walls 2 feet thick.

then,

(2x + 4)(x + 2) = 392 this will give you the info for part 1.

for part 2,

The diameter of the circle circumscribing the rectangle would be the hypotenuse of a triangle with legs of (x+2) and (2x + 4). The outer L and W dimensions of the rectangle. (draw it out)

Let me know if this helps
 
phillygurl said:
14(x^2 + 2) = 0
14(x + 1)^2
Click to expand...
Philly, (x + 1)^2 = x^2 + 2x + 1, NOT x^2 + 2 ; that's quite basic stuff.
Also, you seem unaware of how to get the area of a rectangle.

I think it's time (and fair to you) to tell you that you are evidently not ready
for this type of problem, and need serious classroom help.
 
Hi Phillygurl!
A have an alternate approximation (I don't think tutor_joel is right)

1.- Find exterior and interior dimensions. Remember that inside long is 2 feet less than exterior and inside wide is 4 feet shorter. Try to not express it as a sum (like x+b) because this will make things messy.
2.- To find the distance ("how far the patio extends past" of the room addition) using the Pythagoream Theorem, you need to find a radius (an hypotenuse). The right triangle you need to know is the one whose legs are: the exterior long of the room L and half of exterior width W:
L[sup:1o96widp]2[/sup:1o96widp]+ (W/2)[sup:1o96widp]2[/sup:1o96widp]=R[sup:1o96widp]2[/sup:1o96widp]
I got R=28.86, check the work in case of mistakes from me. I changed inches to feet.
3.- Regards to the tree, it depends on where it is located and from where you are considering the 20 feet of distance. And you have to decide (or consider both cases) if the 20 feet distance is from the center of the tree or from the circunference. In the first case you have to find the radius of the tree (that is why you need the circumference of the tree), in the second case no. I will take the second case:
Lets suppose that the dimensions of the propperty allow build the room in such a way that the tree is outside of the patio desired.

Since the patio radius is 28.86 feet, and the distance from the tree to the house is 20 feet, considering that the tree has to be outside of the patio, then you can build a right triangle with one leg 20 feet and hypotenuse 28.86. Using Pythagorean Theorem find the other leg (namely b) and substract half the outside width of your room (b-W/2). That is the distance you have to consider to choose the place where the room will be located.
4.- Work backwards thinking the patio radius is less or equal to 20 feet, and the external sides of your room will change. May be you know a technique that stands for "Reverse engineer the solution". The tree is a factor
I hope this help, let me know if you have any question.
GARF
 
I assumed that all 4 walls were 2 feet thick, and that was wrong. For only 3 walls 2 feet thick (not the attached wall to the house), making the correction was simple.

Outer W = x+2; L = 2x+4

Mine is correct now. Post your answers and I'll let you know.

The tree does interfere with the patio, assuming the tree is 20 feet south of the original house and not 20 ft south of the addition.
 
garf said:
2.- To find the distance ("how far the patio extends past" of the room addition) using the Pythagoream Theorem, you need to find a radius (an hypotenuse). The right triangle you need to know is the one whose legs are: the exterior long of the room L and half of exterior width W:
L[sup:3d31u8oe]2[/sup:3d31u8oe]+ (W/2)[sup:3d31u8oe]2[/sup:3d31u8oe]=R[sup:3d31u8oe]2[/sup:3d31u8oe]
I got R=28.86, check the work in case of mistakes from me. I changed inches to feet.
Click to expand...

I'm not sure what you're doing here garf. The room is "circumscribed", so the circular diameter of a circumscribed rectangle is simply the diagonal distance from corner to corner of the rectangle (Draw it). Oh, looks like perhaps you're putting a half circle on the south side of the room.
 
Thank you all for responding - I am trying to figure out how you both tutor joel and garth came up with 28.86 as the radius because I didn't come up with the same thing and I am questinoing it because I read the equation as to be no more than 20ft. The patio is being circumscribed by the patio (joel you were correct) and (garth, you also) because it is 20ft from south side of house so the tree would be a factor.

After I take a break from this - I will work it out again because it is due tomorrow and I will post my answer.

Thanks again!!
 
okay this is what I came up with - I hope I am on the right page.

) Determine the inside dimensions of the room addition, having an external area of 392 square feet and walls that are 2 feet thick.
W = X L = 2X
• Define the exterior dimensions with binomial expressions
W = x + 2
L = 2x + 4
For 3 walls @ 2ft thick

• Simplify the resulting quadratic equation
(2x + 4)(x + 2) = 392

• Solve for the unknown
(2 x 12 + 4)(12 + 2) = 392
(24 + 4)(14) = 392
(28)(14) = 392
392 = 392

2. Determine how far the patio extends past the south wall of the room addition. ( Use the Pythagorean Theorem)
A^2 + b^2 = c^2
14^2 + 14^2 = c^2
196 + 196 = c^2 ?
392 = c^2 141. ) Determine the inside dimensions of the room addition, having an external area of 392 square feet and walls that are 2 feet thick.
W = X L = 2X
• Define the exterior dimensions with binomial expressions
W = x + 2
L = 2x + 4
For 3 walls @ 2ft thick

• Simplify the resulting quadratic equation
(2x + 4)(x + 2) = 392

• Solve for the unknown
(2 x 12 + 4)(12 + 2) = 392
(24 + 4)(14) = 392
(28)(14) = 392
392 = 392

2. ( 30 points) Determine how far the patio extends past the south wall of the room addition. ( Use the Pythagorean Theorem)
A^2 + b^2 = c^2
14^2 + 14^2 = c^2
196 + 196 = c^2
392 = c^2
C = ?392
C = 19.798



3. What accommodations must be made for the tree?
In this scenario no further accommodations needs to be made because the patio is under the 20ft and would not interfere with the tree.

4. How could the interior dimensions be changed to eliminate the tree from consideration? ( Reverse engineer the solution.)
The dimensions could be changed by making it smaller which would also result in the elimination of the tree.
 
ok, is that tree number 20" circumference or diameter? I only ask because it says "large tree". So, a 20" circ is approx 6" Dia, not really a large tree.

ok, ok, wait a minute. in part 1, you're fine up until

(2x+4)(x+2) = 392

now you need to do F.O.I.L method.

F first
O outside
I inside
L last

First: 2x^2
Out: (2x)(2) = 4x
Inside: 4x
Last: 8

2x^2 + 4x + 4x + 8 = 392

2x^2 +8x + 8 = 392 (combine like terms)

x^2 +4x + 4 = 196 (divide by 2)

x^2 + 4x -192 = 0 (set to zero)

Solve for x

Show all your steps, you can't skip things, that's where mistakes creep in. This problem is advanced; you need to work on some more foil practice problems.
Circumscribed
 
phillygurl said:
okay this is what I came up with - I hope I am on the right page.

) Determine the inside dimensions of the room addition, having an external area of 392 square feet and walls that are 2 feet thick.
W = X L = 2X
• Define the exterior dimensions with binomial expressions
W = x + 2
L = 2x + 4
For 3 walls @ 2ft thick

• Simplify the resulting quadratic equation
(2x + 4)(x + 2) = 392

• Solve for the unknown
(2 x 12 + 4)(12 + 2) = 392
(24 + 4)(14) = 392
(28)(14) = 392
392 = 392

2. Determine how far the patio extends past the south wall of the room addition. ( Use the Pythagorean Theorem)
A^2 + b^2 = c^2
14^2 + 14^2 = c^2 WATCH OUT! ARE YOU CONSIDERING THE DIMENSIONS OF THE RIGHT TRIANGLE? IT SEEMS LIKE YOU ARE TAKING THE DIMENSIONS OF A SQUARE ROOM, NOT A RECTANGLE. IF THE ROOM HAS TO BE SURROUNDED BY A SEMICIRCLE, THE CENTER OF THE CIRCLE HAS TO BE AT THE MIDPOINT OF WIDTH OF RECTANGLE, IN THE WALL THAT BELONGS BOTH TO THE HOUSE AND TO THE STUDIO.
196 + 196 = c^2 ?
392 = c^2 141. ) Determine the inside dimensions of the room addition, having an external area of 392 square feet and walls that are 2 feet thick.
W = X L = 2X
• Define the exterior dimensions with binomial expressions
W = x + 2
L = 2x + 4
For 3 walls @ 2ft thick

• Simplify the resulting quadratic equation
(2x + 4)(x + 2) = 392

• Solve for the unknown
(2 x 12 + 4)(12 + 2) = 392
(24 + 4)(14) = 392
(28)(14) = 392
392 = 392

2. ( 30 points) Determine how far the patio extends past the south wall of the room addition. ( Use the Pythagorean Theorem)
A^2 + b^2 = c^2
14^2 + 14^2 = c^2
196 + 196 = c^2
392 = c^2
C = ?392
C = 19.798



3. What accommodations must be made for the tree?
In this scenario no further accommodations needs to be made because the patio is under the 20ft and would not interfere with the tree.

4. How could the interior dimensions be changed to eliminate the tree from consideration? ( Reverse engineer the solution.)
The dimensions could be changed by making it smaller which would also result in the elimination of the tree.
Click to expand...
 
Thanks Tutor Joel and Garth - I too thought that I wasn't ready for this class however the assetment and grade average in the prior class said differently. I have two weeks to go and I currently have a B. I am hoping to make it and pass the final exam. Thanks Again for your patience and help.
 
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