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Anuta_ua [19.1K]
4 years ago
9

The length of a side of a triangle is 36. A line parallel to that side divides the triangle into two parts of equal area. Find t

he length of the segment determined by the points of intersection between the line and the other two sides of the triangle.

Mathematics
1 answer:
Hitman42 [59]4 years ago
5 0

Answer:

Length of BE is \frac{36}{\sqrt2}units

Step-by-step explanation:

The length of a side of a triangle is 36. A line parallel to that side divides the triangle into two parts of equal area. we have to find the length of the segment determined by the points of intersection between the line and the other two sides of the triangle.

Now, the two parts of triangle have equal area ∴ ar(ADE)=ar(BDEC)

⇒ ar(ADE)=\frac{1}{2}ar(ABC)

⇒ \frac{ar(ADE)}{ar(ABC)}=\frac{1}{2}

In ΔABE and ΔABC,

∠A=∠A     (Common)

∠ABE=∠ABC   (Corresponding angles)

By AA similarity, ΔABE is similar to ΔABC.

Hence by area side proportionality theorem

\frac{ar(ADE)}{ar(ABC)}=(\frac{BE}{BC})^2

⇒ \frac{1}{2}=\frac{BE^2}{36^2}

⇒ BE^2=\frac{36^2}{2}

⇒ BE=\frac{36}{\sqrt2}units

Hence, length of BE is \frac{36}{\sqrt2}units

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\mathbf{S(t)=200(\frac{105}{100})^{x}}

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Step-by-step explanation:

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S(2) means the number of students in school after two year when the number increased by 5% than previous year which is S(1)  

S(2) = S(1) + 5% of S(1) = \textrm{S}(1)(\frac{105}{100}) = 200(\frac{105}{100})(\frac{105}{100}) = 200(\frac{105}{100})^{2}  

.  

.  

.  

.  

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Similarly \mathbf{S(x)=200(\frac{105}{100})^{x}}  

<h3>The predicted amount spent per student over time, A(t) </h3>

Rate of decrements is 2% per year.  

A function 'A(t)' which gives the amount spend on each student in school after 't' years.  

A(0) means the initial year when the number of students is 40.  

A(0) = 40  

A(1) means the amount spend on each student in school after one year when the amount decreased by 2% than previous year which is 40.  

A(1) = 40 + 2% of 40 = 40-\frac{2}{100}\time40 = 40(1-\frac{2}{100}) = 40(\frac{98}{100})  

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A(2) = A(1) + 2% of A(1) = \textrm{A}(1)(\frac{98}{100}) = 40(\frac{98}{100})(\frac{98}{100}) = 40(\frac{98}{100})^{2}  

.  

.  

.  

.  

.  

Similarly \mathbf{A(x)=40(\frac{98}{100})^{x}}  

<h3>The predicted total expense for supplies each year over time, E(t)</h3>

Total expense = (number of students) ×  (amount spend on each student)

E(t) = S(t) × A(t)

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\mathbf{E(t)=8000(\frac{10290}{10000})^{x}}

(NOTE : The value of x in all the above equation is between zero(0) to ten(10).)

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