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2 years ago

Suppose given a trapezoid ABCD with bases BC=5 and AD=8, and leg CD=3. The lines AB and CD meet at a point P. Find DP

Mathematics

2 answers:

LUCKY_DIMON [66]2 years ago

6 0

Answer:

8 units

Step-by-step explanation:

Consider triangles BPC and APD. These two triangles are similar, because

∠DAP≅∠CBP (as corresponding angles);
∠BCP≅∠ADP (as corresponding angles);
∠APD≅∠BPC (by reflexive property of congruence).

Then, by AA theorem, $\triangle BCP\sim \triangle ADP.$

Similar triangles have proportional sides lengths:

$\dfrac{BP}{AP}=\dfrac{CP}{DP}=\dfrac{BC}{AD},\\ \\\dfrac{BP}{AP}=\dfrac{CP}{CP+3}=\dfrac{5}{8}.$

The last proportion gives you

$8CP=5(CP+3),\\ \\8CP=5CP+15,\\ \\3CP=15,\\ \\CP=5\ un.$

Then $DP=CP+CD=5+3=8\ un.$

Luba_88 [7]2 years ago

5 0

Answer:

Step-by-step explanation:

ΔPBC ~ ΔPAD so ...

... BC/AD = CP/DP

... 5/8 = (DP-3)/DP

... 5·DP = 8(DP -3) . . . . multiply by 8·DP

... 24 = 3DP . . . . . . . . . add 24-5·DP

... 8 = DP . . . . . . . . . . . . divide by 3

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A circular platform is to be built in a playground. The center of the structure is required to be equidistant from three support

castortr0y [4]

Answer:

The coordinates for the location of the center of the platform are (0, 1)

Step-by-step explanation:

The equation of the circle of center (h , k) and radius r is:

(x - h)² + (y - k)² = r²

Now,

- The center is equidistant from any point lies on the circumference of the circle

- There are three points equidistant from the center of the circle

- We have three unknowns in the equation of the circle h , k , r

Thus, let's substitute the coordinates of these point in the equation of the circle to find h , k , r.

The equation of the circle is (x - h)² + (y - k)² = r²

∵ Points A(2,−3), B(4,3), and C(−2,5)

- Substitute the values of x and y the coordinates of these points

Point A (2 , -3)

(2 - h)² + (-3 - k)² = r² - - - (1)

Point B (4 , 3)

(4 - h)² + (3 - k)² = r² - - - - (2)

Point C (-2 , 5)

(-2 - h)² + (5 - k)² = r² - - - - (3)

- To find h , k equate equation (1) and (2) and same for equation (2) and (3) because all of them equal r²

Thus;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)² - - - - - (4)

(4 - h)² + (3 - k)² = (-2 - h)² + (5 - k)² - - - - -(5)

- Simplify (5);

h² - 8h + 16 + k² - 6k + 9 = h² + 4h + 4 + k² - 10k + 25

h² and k² will cancel out to give;

-8h - 6k + 25 = 4h - 10k + 29

Rearranging, we have;

12h - 4k = -4 - - - - (6)

Similarly, for equation 4;

(2 - h)² + (-3 - k)² = (4 - h)² + (3 - k)²

h² - 4h + 4 + k² + 6k + 9 = h² - 8h + 16 + k² - 6k + 9

h², k² and 9 will cancel out to give;

4 - 4h + 6k = 16 - 8h - 6k

Rearranging;

4h + 12k = 12 - - - - (7)

Divide by 4 to give;

h + 3k = 3

Making h the subject;

h = 3 - 3k

Put 3 - 3k for h in eq 6;

12(3 - 3k) - 4k = -4

36 - 36k - 4k = -4

40k = 40

k = 40/40

k = 1

h = 3 - 3(1)

h = 0

The coordinates for the location of the center of the platform are (0, 1)

5 0

1 year ago

You just discovered that you have 100 feet of fencing and you have decided to make a rectangular garden. What is the largest are

Rudik [331]

The area is:
A = x * y
The perimeter is:
P = 2x + 2y = 100
We clear y:
2y = 100-2x
y = 50-x
We write the area in terms of x:
A (x) = x * (50-x)
Rewriting:
A (x) = 50x-x ^ 2
Deriving:
A '(x) = 50-2x
We equal zero and clear x:
50-2x = 0
x = 50/2
x = 25
Then, the other dimension is given by:
y = 50-x
y = 50-25
y = 25
Therefore, the largest area is:
A = (25) * (25)
A = 625 feet ^ 2
Answer:
the largest area you can enclose using the materials you have is:
A = 625 feet ^ 2

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Helga [31]

There both types of numbers but a whole number is a number with no exponents or decimals, but decimals are like a broken un half whole number

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10 x 3 tens - unit form and standard form

vfiekz [6]

<span>10X3 tens in unit form is written:
10*3 tens = 30 tens = 300 units

10X3 tens in standard form:
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6 0

2 years ago

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Find all points of intersection of the given curves. (Assume 0 ≤ θ ≤ 2π and r ≥ 0. Order your answers from smallest to largest θ

DerKrebs [107]

Answer:

∅1=15°,∅2=75°,∅3=105°,∅4=165°,∅5=195°,∅6=255°,∅7=285°,

∅8=345°

Step-by-step explanation:

Data

r = 8 sin(2θ), r = 4 and r=4

iqualiting; 8.sin(2∅)=4; sin(2∅)=1/2, 2∅=asin(1/2), 2∅=30°, ∅=15°

according the graph 2, the cut points are:

I quadrant:

0+15° = 15°

90°-15°=75°

II quadrant:

90°+15°=105°

180°-15°=165°

III quadrant:

180°+15°=195°

270°-15°=255°

IV quadrant:

270°+15°=285°

360°-15°=345°

No intersection whit the pole (0)

7 0

2 years ago