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lys-0071 [83]
3 years ago
9

In triangle ABC, D is a point on line AB and E is a point on line AC such that DE is parallel to BC. If BC = 20 centimeters, and

the area of the trapezoid (trapezium) DBCE is one-fourth the area of triangle ABC, find DE.
Mathematics
1 answer:
sdas [7]3 years ago
8 0

Step-by-step Answer:

If trapezium DBCE is one-fourth of the area of the triangle ABC, that means that the area of triange ADE is (1 - 1/4) = three-fourth of ABC.

Since DE is parallel to BC, we can prove that triangles ADE and ABC are similar.

Similar triangles have corresponding sides proportional, and area is proportional to the square of the linear proportions.

From this we can conclude that

(DE/BC)^2 = 3/4

DE/BC = sqrt(3/4) = sqrt(3)/2

DE = sqrt(3)/2 * 20 = 10 sqrt(3) = 17.320508 cm (to 6 decimal places).

On the subject of similar figures/volumes, if we know the ratio of linear dimensions (such as the side of a cube) as k, then the ratio of AREA of similar squares would be k^2.

Example: A square would have a side of 8 (area=8^2=64), and a (similar) square has a side of 10 (area = 10^2= 100).  The

ratio of areas = 64/100 = (8^2/10^2) = (8/10)^2

Here 8/10 is the ratio of linear dimensions = k, and the ratio of areas is k^2 = (8/10)^2 = 64/100.

The same works for cubes (or similar volumes) where the volumes would be proportional to k^3.

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Solve the given inequality :
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Answer:

x < -5  or  x = 1  or  2 < x < 3  or  x > 3

Step-by-step explanation:

Given <u>rational inequality</u>:

\dfrac{(x-1)^2(x-2)^3}{(x^2-5x+6)^2(x+5)}\geq 0

\textsf{Factor }(x^2-5x+6):

\implies x^2-2x-3x+6

\implies x(x-2)-3(x-2)

\implies (x-3)(x-2)

Therefore:

\dfrac{(x-1)^2(x-2)^3}{(x-3)^2(x-2)^2(x+5)}\geq 0

Find the roots by solving f(x) = 0  (set the numerator to zero):

\implies (x-1)^2(x-2)^3=0

\implies (x-1)^2=0\implies x=1

\implies (x-2)^3=0 \implies x=2

Find the restrictions by solving f(x) = <em>undefined  </em>(set the denominator to zero):

\implies (x-3)^2(x-2)^2(x+5)=0

\implies (x-3)^2=0 \implies x=3

\implies (x-2)^2=0 \implies x=2

\implies (x+5)=0 \implies x=-5

Create a sign chart, using closed dots for the <u>roots</u> and open dots for the <u>restrictions</u> (see attached).

Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.

Test values:  -6, 0, 1.5, 2.5, 4

For each test value, determine if the function is positive or negative:

f(-6)=\dfrac{(-6-1)^2(-6-2)^3}{(-6-3)^2(-6-2)^2(-6+5)}=\dfrac{(+)(-)}{(+)(+)(-)}=+

f(0)=\dfrac{(0-1)^2(0-2)^3}{(0-3)^2(0-2)^2(0+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-

f(1.5)=\dfrac{(1.5-1)^2(1.5-2)^3}{(1.5-3)^2(1.5-2)^2(1.5+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-

f(2.5)=\dfrac{(2.5-1)^2(2.5-2)^3}{(2.5-3)^2(2.5-2)^2(2.5+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+

f(4)=\dfrac{(4-1)^2(4-2)^3}{(4-3)^2(4-2)^2(4+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+

Record the results on the sign chart for each region (see attached).

As we need to find the values for which f(x) ≥ 0, shade the appropriate regions (zero or positive) on the sign chart (see attached).

Therefore, the solution set is:

x < -5  or  x = 1  or  2 < x < 3  or  x > 3

As interval notation:

(- \infty,-5) \cup x=1 \cup (2,3) \cup(3,\infty)

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