just three more?

Question

The side of a large tent is in the shape of an isosceles triangle whose area is 54 ft² and whose base is 6 feet shorter than twice its height. Find the height and the base of the tent (height by base).


A farmer plans to use 21 m of fencing to enclose a rectangular pen having an area of 55 m². Only three sides of the pen need fencing because of an existing wall. Find the dimensions of the pen.


A decorator plans to place a rug in a room 9 m by 12 m in such a way that a uniform strip of flooring around the rug will remain uncovered. If the rug is to cover half the floor space, what should the dimensions of the rug be?

Answer 1

1)
((b x h):2 = 54
(b = 2h - 6

Substitute:
(2h - 6)h = 54 * 2
2h² - 6h - 108 = 0 (:2)
h² - 3h - 54 = 0
-3² - 4.1.-54 = 9 + 216 = 225

h = (3 +/- \/225) : 2
h = (3 + 15) : 2
h = 9 ft

b = 2h - 6
b = 2(9) - 6
b = 18 - 6
b = 12 ft
Answer: The base is 12ft and the height is 9 ft.
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2)
A = 55m²
(B*h = 55
(b + 2h = 21
b = 21 - 2h
(21 - 2h)*h = 55
21h - 2h² - 55 = 0
21² - 4.-2.-55 = 1
441 - 440 = 1

h = (-21 +/- 1):2.-2
h = (-21 + 1): -4 = 5

b = 21 - 2h
b = 21 - 2(5)
b = 21 - 10
b = 11m
Answer: The dimensions of the pen are probably 5 x 11.
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3)
A = B * h =
A = 9*12 = 108 m²
A (rug) = 108/2 = 54m²
Answer: The dimensions of the rug should be 54m². In that case, only two sides of rug will be uncovered because the difference between the wall and rug was not determined.
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Answer 2

1) Let h = height. Then the base is 6 feet shorter than twice it's height, or b = 2h - 6

The area of a triangle is given by the formula
A = (1/2)bh, but we know A and we know b, so

54 = (1/2)[2h - 6]h
108 = [2h - 6]h
108 = 2h^2 - 6h
54 = h^2 - 3h
h^2 - 3h - 54 = 0
(h - 9) (h + 6) = 0

Therefore, h = 9, -6. We can discard -6 though, since height cannot be negative.

The height of the tent is 9, the base of the tent is 2h - 6 = 2(9) - 6, or 18 - 6 = 12.

2) This is a rectangular pen, whose area would be defined as
A = length times width.
Let L = length and W = width. Then

A = LW, and since we're given the area as 55,
55 = LW

The perimeter of the pen would *normally* be
P = L + W + L + W, but since there are only three sides,
P = L + W + L, or
P = 2L + W

Since P = 21,

21 = 2L + W

Two equations, two unknowns:

55 = LW
21 = 2L + W

55 = LW means that W = 55/L, therefore, plugging this into the second equation gives us

21 = 2L + [55/L]. Multiply L both sides to give us

21L = 2L^2 + 55. Bring everything over to the right hand side to give us

0 = 2L^2 - 21L + 55

Factor

0 = (2L - 11) (L - 5), giving us
L = 11/2, 5

Therefore, the length of the pen is 11/2 OR 5, and the width of the pen is 55/L, 10 or 11

Two dimensions possible are: (11/2 x 10) or (5 x 11)

Answer 3

let the base be x
x=2h-6
area=(1/2)(h)(2h-6)=54
2h^2-6h=108
h^2-3h=54
h^2-3h-54=0
(h-9)(h+6)=0
h=9
base=18-6=12

2.2l+w=21
lw=55
l=55/w
sub in (1)
2(55/w)+w=21
110+w^2=21w
w^2-21w+110=0
(w-11)(w-10)=0
w=10 or 11
so l=5 or 5.5

3.area of the room=108 m^2
area of the rug=54
let the width of the emptyspace all round be x
area of that space=54
2*12*x+2(9-2x)*x=54
24x+18x-4x^2-54=0
-4x^2+42x-54=0
2x^2-21x+27=0
2x^2-18x-3x+27=0
(2x-3)(x-9)=0
x=3/2 or 9
9 is absurd and so unacceptable
so it is 3/2m

no more nathalie

Answer 4

I agree with zero, phooey on those problems. I'm going to kick back and watch Zoolander for the 5th time; might even make some popcorn.

Answer 5

who the hell wants to read such a long question. forget it.