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Damm [24]
3 years ago
5

In trapezoid ABCD, points M and N are arbitrary points on bases AB and CD respectively. Find the area of the trapezoid, if area

AABN=23 cm2 and area ACDM=18 cm2.
Mathematics
1 answer:
mel-nik [20]3 years ago
6 0

Remark

This is quite a nice little problem. It takes a minute or three to figure out the answer, and when you do, you will be certain that you have been tricked. It is a little like the egg of Columbus.

Solution

The Base of Triangle ABN is AB

The Base of Triangle CDM is CD

The height of both given triangles is h. That is the distance between the two parallel lines.

Area ABN = 1/2*AB * h = 23 cm^2

Area CDM = 1/2*CD * h = 18 cm^2

Now the Area of the trapezoid is

Area_Trapezoid = 1/2 * h (AB + CD)     Using the distributive property Remove the brackets.

Area_Trapezoid = 1/2*AB*h + 1/2*CD*h Did you notice something? Those terms are just the area of the triangles (written above.)

Area Trapezoid = 23 + 18 = 41 cm^2 <<<< Answer

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In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
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Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

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Answer:

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Step-by-step explanation:

Given:

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