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# In the figure above, ABCD is a rectangle, and each of AP and CQ is per

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In the figure above, ABCD is a rectangle, and each of AP and CQ is per [#permalink]
Bunuel wrote:

In the figure above, ABCD is a rectangle, and each of AP and CQ is perpendicular to BD. If DP = PQ = QB, what is the ratio of AB to AD?

(A) $$\sqrt{2}$$ to 1

(B) $$\sqrt{3}$$ to 1

(C) $$\sqrt{3}$$ to $$\sqrt{2}$$

(D) 2 to 1

(E) 2 to $$\sqrt{3}$$

Are You Up For the Challenge: 700 Level Questions

Attachment:
1.png

Let the length be 'L' & breadth be 'B'

Note that, Rectangle ABCD is divided into 4 equal right angled triangles APB, BQC, CQD & DPA
--> Area of each triangle = 1/4(LB) = LB/4

Also, area of triangle ABD = BCD = $$\frac{1}{2}(LB) = \frac{LB}{2}$$

Length of the diagonal BD = $$\sqrt{L^2 + B^2}$$
So, Length of DP = PQ = QB = $$\frac{1}{3}*\sqrt{L^2 + B^2}$$

In triangle ABD, Area = 1/2*AP*BD = 1/2*AD*AB
--> AP*BD = AD*AB
--> AP*$$\sqrt{L^2 + B^2}$$ = B*L
--> $$AP = \frac{LB}{\sqrt{L^2 + B^2}}$$
--> $$AP^2 = \frac{L^2B^2}{(L^2 + B^2)}$$ ....... (1)

In triangle APD, $$AD^2 = AP^2 + PD^2$$
--> $$B^2 = AP^2 + (\frac{1}{3}*\sqrt{L^2 + B^2})^2$$
--> $$B^2 = AP^2 + (\frac{1}{9}*(L^2 + B^2)$$
--> $$AP^2 = \frac{8B^2 - L^2}{9}$$ ....... (2)

From (1) & (2),
$$\frac{L^2B^2}{(L^2 + B^2)} = \frac{8B^2 - L^2}{9}$$
--> $$9L^2B^2 = 7L^2B^2 - L^4 + 8B^4$$
--> $$L^4 + 2L^2B^2 - 8B^4 = 0$$
--> $$(L^2 + 4B^2)(L^2 - 2B^2) = 0$$
--> $$L^2 - 2B^2 = 0$$
or
$$L^2 = 2B^2$$
$$L/B = \sqrt{2}$$

$$\frac{AB}{AD} = \frac{L}{B} = \frac{\sqrt{2}}{1}$$

Option A
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Re: In the figure above, ABCD is a rectangle, and each of AP and CQ is per [#permalink]
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Re: In the figure above, ABCD is a rectangle, and each of AP and CQ is per [#permalink]
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