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Vedmedyk [2.9K]
3 years ago
12

In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th

e ratio of BE:EC?
Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
8 0

Answer:

\frac{BE}{EC} =\frac{1}{3}

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

\frac{BK}{CK} =\frac{1}{4}

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE                ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE     ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK  and Δ EBK are similar

so we have

\frac{AD}{BE} =\frac{DK}{BK}

\frac{AD}{BE} =\frac{4}{1}

\frac{BC}{BE}=\frac{4}{1}     ( ABCD is parallelogram so AD=BC)

\frac{BE+EC}{BE}=\frac{4}{1}         ( BC= BE+EC)

\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}

1+\frac{EC}{BE}=4

\frac{EC}{BE}=3  ( subtracting 1 from both side )

\frac{EC}{BE}=\frac{3}{1}

taking reciprocal both side

\frac{BE}{EC} =\frac{1}{3}


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Natali5045456 [20]

Answer:

P(A given B) = 3/4

Step-by-step explanation:

As we know that it is conditional probability, where the probability of an event depends on the event that has certain probability of occurrence.

The formula for conditional probability is:

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Where a and B are events. The probability of event B is known and we also know probability of A∩B

So, putting the values in the formula:

P(A\ given\ B) = \frac{P(A and B)}{P(B)}\\= \frac{\frac{3}{10} }{\frac{2}{5} } \\=\frac{3}{10} *\frac{5}{2}\\=\frac{3}{4}

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3 years ago
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Answer:

y=2x +3

Step-by-step explanation:

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21/ 21 What is the equivalent​ decimal?
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an isosceles triangle has congruent sides of 20cm. the base is 10cm. find the height of the triangle.​
dolphi86 [110]

Answer:

\large\boxed{A_\triangle=25\sqrt{15}\ cm^2}

Step-by-step explanation:

Look at the picture.

The formula of an area of a triangle:

A_\triangle=\dfrac{bh}{2}

<em>b</em><em> - base</em>

<em>h</em><em> - height</em>

<em />

We need a length of a height.

Use the Pythagorean theorem:

leg^2+leg^2=hypotenuse^2

We have:

leg=5,\ leg=h,\ hypotenuse=20

Substitute:

5^2+h^2=20^2

25+h^2=400             <em>subtract 25 from both sides</em>

h^2=375\to h=\sqrt{375}\\\\h=\sqrt{(25)(15)}\\\\h=\sqrt{25}\cdot\sqrt{15}\\\\h=5\sqrt{15}\ cm

Calculate the area:

A_\triangle=\dfrac{(10)(5\sqrt{15})}{2}=\dfrac{50\sqrt{15}}{2}=25\sqrt{15}\ cm^2

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Perpendicular lines meet at a 90 degree angle.

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