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Bunuel
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shashaankbhat
How can we assume that the triangle is right angled triangle. Neither the angle is given in the figure nor it is said "Drawn to Scale"
Experts please help

AB = 8, BC = 6, AC = 10 satisfy AB^2 + BC^2 = AC^2, so the triangle is indeed right-angled at B.
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Ah missed it. Thank you very much for the explanation Bunuel
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Bunuel

In the figure, if AB = 8, BC = 6, AC = 10 and CD = 9, then AD =

(A) 12
(B) 13
(C) 15
(D) 17
(E) 24

Attachment:
triangle.jpg
\(AB = 8\) & \(BD = 15\) in the right Angled Triangle..

SO, Hypotenuse \(AD = 17\) , following Pythagorean triple (8, 15, 17) , Answer must be (D)
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Bunuel

In the figure, if AB = 8, BC = 6, AC = 10 and CD = 9, then AD =

(A) 12
(B) 13
(C) 15
(D) 17
(E) 24

Attachment:
triangle.jpg

\(AB = 8, BC = 6, AC = 10\) tell us that it is a right angle triangle. \(AC = \sqrt{15^2 + 8^2} = 17\)

Another way (probably easier).

Any side of the triangle cannot be more than that of the sum of the other two sides.
Sum of the other two sides = 23

Hypotenuse is the largest side of a right angle triangle. We already have a base of 15.

Only one option remains.

Answer is D.
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Bunuel

In the figure, if AB = 8, BC = 6, AC = 10 and CD = 9, then AD =

(A) 12
(B) 13
(C) 15
(D) 17
(E) 24

Attachment:
triangle.jpg

Since triangle ABC is a 6-8-10 triangle, it’s a right triangle with right angle at B. We can find AD, the hypotenuse of right triangle ABD, using the Pythagorean theorem,

AB^2 + BD^2 = AD^2

8^2 + 15^2 = AD^2

64 + 225 = AD^2

289 = AD^2

17 = AD

Answer: D
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