Round 48,382 to thenearest100:1,000:10,000:
1 answer:
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<h3>Answer: 16 ft by 25 ft</h3>
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Explanation:
Refer to the diagram below.
The outer dimensions 24 ft and 33 ft shrink down to 24-2x ft and 33-2x ft respectively. This subtraction of 2x is due to subtracting two copies of x per side.
The carpet has area of (24-2x)(33-2x)
Cynthia can afford to buy 400 sq ft of carpet
So we set 400 equal to that previous expression and solve for x
(24-2x)(33-2x) = 400
24(33-2x) - 2x(33-2x) = 400
792-48x - 66x + 4x^2 = 400
4x^2 - 114x + 792 = 400
4x^2 - 114x + 792-400 = 0
4x^2 - 114x + 392 = 0
From here, use the quadratic formula to isolate x.
Plug in a = 4, b = -114, c = 392
The two possible solutions are x = 24.5 and x = 4
But if x = 24.5, then 24 - 2x = 24 - 2*24.5 = -25 which isn't possible. We cannot have a negative width or negative length for the carpet.
Luckily x = 4 does work since
24 - 2x = 24 - 2*4 = 16
33 - 2x = 33 - 2*4 = 25
Both results are positive.
Therefore, the carpet has dimensions of <u>16 ft by 25 ft</u>
Check: 16*25 = 400, so the answer is confirmed
Answer:
∠EFD ≅ ∠EGD ⇒ A
Arc ED ≅ arc FD ⇒ C
m arc FD = 120° ⇒ E
Step-by-step explanation:
Let us revise some facts
- Equal chords subtended equal arcs
- The measure of an inscribed angle is one-half the measure of the central angle which subtended by the same arc
- The measure of a central angle is equal to the measure of its subtended arc
- If one angle of an isosceles triangle measure 60° then the triangle is equilateral
- The sum of the measures of the interior angles of any quadrilateral is 360°
In the quadrilateral CDGE
∵ m∠G = 60°
∵ m∠GDC = m∠GEC = 90°
- By using the 5th rule above
∴ m∠G + m∠GDC + m∠DCE + m∠GEC = 360°
∴ 60 + 90 + m∠DCE + 90 = 360
∴ 240 + m∠DCE = 360
- Subtract 240 from both sides
∵ m∠DCE = 120°
In circle C
∵ ∠DCE is a central angle subtended by arc DE
∵ ∠DFE is an inscribed angle subtended by arc DE
- By using the 2nd rule above
∴ m∠DFE = m∠∠DCE
∵ m∠DCE = 120°
∴ m∠DFE = (120)
∴ m∠DFE = 60°
- That means ∠EFD ≅ ∠EGD because their measure is 60°
∴ ∠EFD ≅ ∠EGD
In Δ EFD
∵ EF = FD
∵ m∠DFE = 60°
- By using the 4th rule above
∴ Δ EFD is an equilateral triangle
∴ ED = FD = FE
In circle C
∵ Side ED subtended by arc ED
∵ Side FD subtended by FD
∵ Side ED ≅ side FD ⇒ proved
- By using the 1st rule above
∴ Arc ED ≅ arc FD
∵ m∠ECD = 120°
∵ ∠ECD is a central angle subtended by arc ED
- By using the 3rd rule above
∴ m∠ECD = m arc ED
∴ m of arc ED = 120°
∵ Arc ED ≅ arc FD
∴ m arc ED = m arc FD
∴ m arc FD = 120°