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frozen [14]
3 years ago
7

The trapezoid ABCD has bases AB = 15 and CD = 5, thigh AD = 9 and diagonal BD = 12. Find:

Mathematics
1 answer:
Fed [463]3 years ago
3 0

Answer:

The height of the trapezoid is 6.63 units

The perimeter of the trapezoid is 38 units

Step-by-step explanation:

Whenever a geometry problem is given, it is often useful if it is sketched out. A sketch of this problem can be found in the image attached.

A)

We can see that a right-angled triangle is formed between points BED, with line BE being the height, h.

To get the dimensions of the line EB, we subtract the dimensions of DC from AB. This will give 15 -5 = 10

hence the dimensions of the righ angled triangle are

DE= h

DB = 12 (diagonal)

EB = 10

From Pythagoras' theorem,

h =\sqrt{12^2 -10^2}=6.63

The height of the trapezoid is 6.63

B)

We can get the perimeter of the trapezoid by adding the dimensions of all four sides together.

This will be

AD + DC + CB + AB

However we can assume for this case that it is a symmetrical trapezoid, and hence AD = CB

Thus, perimeter =

2 (AD) + DC +AB

2(9) +5 +15 = 38.

The perimeter of the trapezoid is 38 units

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Answer: x=65

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Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein
ivanzaharov [21]

Answer:

1508527.582 cm³/min

Step-by-step explanation:

The net rate of flow dV/dt = flow rate in - flow rate out

Let flow rate in = k. Since flow rate out = 6800 cm³/min,

dV/dt = k - 6800

Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone

dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)

So, dV/dt = (πr²dh/dt)/3

Let h = height of tank = 12 m, r = radius of tank = diameter/2 = 3/2 = 1.5 m, h' = height when water level is rising at a rate of 21 cm/min = 3.5 m and r' = radius when water level is rising at a rate of 21 cm/min

Now, by similar triangles, h/r = h'/r'

r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm

Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.

The net rate of increase of water is dV/dt = (πr'²dh/dt)/3

dV/dt = (π(262.5 cm)² × 21 cm/min)/3

dV/dt = (π(68906.25 cm²) × 21 cm/min)/3

dV/dt = 1447031.25π/3 cm³

dV/dt = 4545982.745/3 cm³

dV/dt = 1515327.582 cm³/min

Since dV/dt = k - 6800 cm³/min

k = dV/dt - 6800 cm³/min

k = 1515327.582 cm³/min - 6800 cm³/min

k = 1508527.582 cm³/min

So, the rate at which water is pumped in is 1508527.582 cm³/min

5 0
3 years ago
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