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Yuki888 [10]
3 years ago
10

In quadrilateral ABCD, diagonals AC and BD bisect one another: Quadrilateral ABCD is shown with diagonals AC and BD intersecting

at point P. What statement is used to prove that quadrilateral ABCD is a parallelogram?
A. Angles ABC and BCD are congruent.
B. Sides AB and BC are congruent.
C. Triangles BPA and DPC are congruent.
D. Triangles BCP and CDP are congruent.
Mathematics
2 answers:
EleoNora [17]3 years ago
4 0
B. A parallelogram has congruent sides.

aleksley [76]3 years ago
4 0

Answer with explanation:

It is given that , quadrilateral A BC D, diagonals AC and B D bisect one another at point P.

In ΔAPB and ΔCPD

AP=PC

BP=PD

∠APB =∠CPD→Vertically opposite angles

ΔAPB ≅ ΔCPD→→[SAS]

→AB=CD⇒[CPCT]

→∠A BP=∠C DP⇒[C PCT]

Alternate interior angles are equal , so lines are parallel.

⇒AB║CD, and AB=CD

Similarly, we can prove ΔAPD ≅ ΔBPC, to prove AD║BC, and AD=BC.

⇒A Quadrilateral is a parallelogram , if one pair of opposite sides is equal and parallel.

Option C: The statement which is used to prove that quadrilateral ABCD is a parallelogram→→Triangles B PA and D PC are congruent.

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Answer:

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Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)
ELEN [110]

Question:

Find the point (,) on the curve y = \sqrt x that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})

Step-by-step explanation:

y = \sqrt x can be represented as: (x,y)

Substitute \sqrt x for y

(x,y) = (x,\sqrt x)

So, next:

Calculate the distance between (x,\sqrt x) and (3,0)

Distance is calculated as:

d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}

So:

d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}

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d = \sqrt{x^2 - 6x +9 + x}

Rewrite as:

d = \sqrt{x^2 + x- 6x +9 }

d = \sqrt{x^2 - 5x +9 }

Differentiate using chain rule:

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u = x^2 - 5x +9

\frac{du}{dx} = 2x - 5

So:

d = \sqrt u

d = u^\frac{1}{2}

\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}

Chain Rule:

d' = \frac{du}{dx} * \frac{dd}{du}

d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}

d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}

d' = \frac{2x - 5}{2\sqrt u}

Substitute: u = x^2 - 5x +9

d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}

Next, is to minimize (by equating d' to 0)

\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0

Cross Multiply

2x - 5 = 0

Solve for x

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Substitute x = \frac{5}{2} in y = \sqrt x

y = \sqrt{\frac{5}{2}}

Split

y = \frac{\sqrt 5}{\sqrt 2}

Rationalize

y = \frac{\sqrt 5}{\sqrt 2} *  \frac{\sqrt 2}{\sqrt 2}

y = \frac{\sqrt {10}}{\sqrt 4}

y = \frac{\sqrt {10}}{2}

Hence:

(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})

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