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# In the figure above, Δ POQ and Δ AOB are both right-angled at point O.

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In the figure above, Δ POQ and Δ AOB are both right-angled at point O.  [#permalink]

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Updated on: 29 May 2017, 10:43
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Difficulty:

75% (hard)

Question Stats:

63% (02:58) correct 38% (02:52) wrong based on 56 sessions

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In the figure above, Δ POQ and Δ AOB are both right-angled at point O. If BQ = OB and OA = AP, and the length of side AB is 5 units, what is the area of the shaded region?

(1) OA = OB + 1
(2) If a rectangle is drawn with sides of length 2QB and $$\frac{PA}{4}$$, the area of the rectangle will be 6 square units

Attachment:

oQ6sNCK.png [ 33.11 KiB | Viewed 1249 times ]

Originally posted by niteshwaghray on 29 May 2017, 09:11.
Last edited by Bunuel on 29 May 2017, 10:43, edited 1 time in total.
Edited the question.
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Re: In the figure above, Δ POQ and Δ AOB are both right-angled at point O.  [#permalink]

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29 May 2017, 11:01
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1

In the figure above, Δ POQ and Δ AOB are both right-angled at point O. If BQ = OB and OA = AP, and the length of side AB is 5 units, what is the area of the shaded region?

(1) OA = OB + 1

OA^2 + OB^2 = AB^2

(OB + 1)^2 + OB^2 = 5^2

We can find OB, then OA. We also know that BQ = OB and OA = AP. So, we know everything about the figure and will be able to answer the question. Sufficient.

(2) If a rectangle is drawn with sides of length 2QB and $$\frac{PA}{4}$$, the area of the rectangle will be 6 square units

2QB*PA/4 = 6;

Since BQ = OB and OA = AP, then 2OB*OA/4 = 6 --> OB*OA = 12 --> OA = 12/OB.

OA^2 + OB^2 = AB^2

(12/OB)^2 + OB^2 = 5^2

The same here: we can find OB, then OA. We also know that BQ = OB and OA = AP. So, we know everything about the figure and will be able to answer the question. Sufficient.

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Re: In the figure above, Δ POQ and Δ AOB are both right-angled at point O.  [#permalink]

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05 May 2020, 04:42
niteshwaghray wrote:

In the figure above, Δ POQ and Δ AOB are both right-angled at point O. If BQ = OB and OA = AP, and the length of side AB is 5 units, what is the area of the shaded region?

(1) OA = OB + 1
(2) If a rectangle is drawn with sides of length 2QB and $$\frac{PA}{4}$$, the area of the rectangle will be 6 square units

(1) sufic
ob=y, oa=y+1
$$(y^2)+(y+1)^2=5^2,y^2+y-12=0,y=3$$

(2) sufic
2qb*pa/4=6, qb*pa=12, qb=ob=x, pa=oa=y, xy=12
$$x^2+y^2=25,x^2+(12/x)^2=25, x^4+144=25x^2$$
$$x^4-25x^2+144=0,(x^2-9)(x^2-16)=0,x=(4,3),y=(3,4)$$

ans (D)
Re: In the figure above, Δ POQ and Δ AOB are both right-angled at point O.   [#permalink] 05 May 2020, 04:42