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# In the figure, ABC and ADC are right triangles.

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Joined: 01 Sep 2010
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17 Mar 2019, 04:14
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Difficulty:

55% (hard)

Question Stats:

50% (01:59) correct 50% (02:39) wrong based on 20 sessions

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In the figure, ABC and ADC are right triangle..jpg [ 16.2 KiB | Viewed 357 times ]

In the figure, ABC and ADC are right triangles. Which of the following could be the lengths of AD and DC, respectively?

(I) $$\sqrt{3}$$ and $$\sqrt{4}$$
(II) 4 and 6
(III) 1 and $$\sqrt{24}$$
(IV) 1 and $$\sqrt{26}$$

(A) I and II only
(B) II and III only
(C) III and IV only
(D) IV and I only
(E) I, II and III only

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17 Mar 2019, 06:03
carcass wrote:
Attachment:
In the figure, ABC and ADC are right triangle..jpg

In the figure, ABC and ADC are right triangles. Which of the following could be the lengths of AD and DC, respectively?

(I) $$\sqrt{3}$$ and $$\sqrt{4}$$
(II) 4 and 6
(III) 1 and $$\sqrt{24}$$
(IV) 1 and $$\sqrt{26}$$

(A) I and II only
(B) II and III only
(C) III and IV only
(D) IV and I only
(E) I, II and III only

Property: For a triangle to exist the sum of any two sides must be GREATER than the third side

With two sides of triangle ABC known as 3 and 4, the third side AC becomes 5 units long (Right angle triangle property)

Now checking options

(I) $$\sqrt{3}$$ and $$\sqrt{4}$$ but $$√3^2 + √4^2 ≠ 5^2$$ hence not a possible case
(II) 4 and 6 but $$4^2 + 5^2 ≠ 6^2$$ hence not a possible case
(III) 1 and $$\sqrt{24}$$ but $$1^2 + √24^2 = 5^2$$ hence a possible case
(IV) 1 and $$\sqrt{26}$$ but $$1^2 + 5^2 = √26^2$$ hence a possible case

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17 Mar 2019, 08:49
IMO C ...

We can see that AC = 5
As ADC is a right angle triangle and only options 3rd and 4th give the pythagoras triplets

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Re: In the figure, ABC and ADC are right triangles.   [#permalink] 17 Mar 2019, 08:49
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