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andreyandreev [35.5K]
2 years ago
5

Quadrilateral ABCD ABCD is similar to quadrilateral EFGH EFGH. The lengths of the three longest sides in quadrilateral ABCD ABCD

are 20 feet, 18 feet, and 14 feet long. If the two shortest sides of quadrilateral EFGH EFGH are 6 feet long and 5 feet long, how long is the 4th side on quadrilateral ABCD A:13 ft B:6.4ft C:9.2ft D:11.7

Mathematics
1 answer:
Kay [80]2 years ago
5 0
Sketch the two quadrilaterals and label them as shown in the figure below.

Let x =  length of the 4th side of ABCD.

Because ABCD ~ EFGH, therefore
x/5 = 14/6

That is,
x = (5/6)*14 = 11.67 ft

Answer: D. 11.7 ft

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Barbara drives between Miami, Florida, and West Palm Beach, Florida. She drives 50 mi in clear weather and then encounters a thu
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Distance traveled in clear weather = 50 miles

Distance traveled in thunderstorm = 15 miles

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⇒ Speed in thunderstorm = x-20

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We need to determine average speed in clear weather (i.e. x) and average speed in the thunderstorm (i.e. x-20 ).

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⇒ 1.5 = \frac{50(x-20)+15(x)}{(x)(x-20)}

⇒ 15*x*(x-20) = 10*[50*(x-20)+15*x]

⇒ 15x² - 300x = 500x - 10,000 + 150x

⇒ 15x² - 300x = 650x - 10,000

⇒ 15x² - 950x + 10,000 = 0

⇒ 3x² - 190x + 2,000 = 0

The above equation is in the format of ax² + bx + c = 0

To determine the roots of the equation, we will first determine 'D'

D = b² - 4ac

⇒ D = (-190)² - 4*3*2,000

⇒ D = 36,100 - 24,000

⇒ D = 12,100

Now using the D to determine the two roots of the equation

Roots are: x₁ = \frac{-b+\sqrt{D}}{2a} ; x₂ = \frac{-b-\sqrt{D}}{2a}

⇒ x₁ = \frac{-(-190)+\sqrt{12,100}}{2*3} and x₂ = \frac{-(-190)-\sqrt{12,100}}{2*3}

⇒ x₁ = \frac{190+110}{6} and x₂ = \frac{190-110}{6}

⇒ x₁ = \frac{300}{6} and x₂ = \frac{80}{6}

⇒ x₁ = 50 and x₂ = 13.33

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If speed in clear weather is 13.33 mph then speed in thunderstorm would be negative, which is not possible since speed can't be negative.

Hence, the speed in clear weather would be 50 mph, and in thunderstorm would be 20 mph less, i.e. 30 mph.

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