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3 years ago

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the height of the Statue of Liberty is 92 m. This about .27 the height of the Hancock Building in Chicago. What is the approxima

te height of the Hancock Building?

1 answer:

GuDViN [60]3 years ago

4 0

Hancock building * .27 = 92 m

divide each side by .27

Hancock building =92/.27

Hancock building =340.740(repeating)

Answer: the building is approximately 340.74 meters

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1 year ago

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Read 2 more answers

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

7 0

3 years ago