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stat1 : QR is a diameter... hence traingle in semicircle is right traingle suff

stat 2 : pythagoras theorem applies here and its true only for right trangles

ans D


mohshu
for option (2)
what if SR =4.5 ???


thanks

rohit8865 got that

only stat is suff

ans A

thanks
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Statement 1 is a property and it is sufficient

Statement 2 : The other side can take any value between 2 and 8; if it is 4 then it is sufficient. Not enough information available.

Option A
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Ans :
st 1 ) sufficient -- QR is diameter . It is property of inscribed triangle in circle . If the one of the sides equal the diameter of the circle . Then the triangle is Right angle triangle .

st 2 ) Not sufficient

two side is given as 3 ,4 but the third side can be 5 or 7 . Apply the property of sum of two side in triangle is greater than the third side . Then you will realize . The triangle can have value as ( 3, 4, 5 --> right angle ) or ( 3,4, 7 --> not a right angle )

Hence A as answer .

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Triangle QSR is inscribed in a circle. Is QSR a right triangle?

(1) QR is a diameter of the circle.
(2) Length QS equals 3 and length QR equals 5.

Attachment:
2017-04-14_1450.png

Triangle QSR is inscribed in a semi-cirlce is QSR a right triangle?

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

(1) QR is a diameter of the circle --> according to the above property QSR must be a right triangle. Sufficient.

(2) Length QS equals 3 and length QR equals to 5 --> it's not necessary QSR to be 3-4-5 right triangle (therefor QR to be diameter/hypotenuse), for example if diameter is more than 5, say 10 than it's possible to inscribe QSR in a semi-circle so that SR would be the largest side and QSR would be obtuse-angled triangle. Not sufficient.

Answer: A.
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A. II statement does not suggest that SR is 4. Hence NS.

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Triangle QSR is inscribed in a circle. Is QSR a right triangle?

(1) QR is a diameter of the circle.
(2) Length QS equals 3 and length QR equals 5.

Attachment:
2017-04-14_1450.png

Triangle QSR is inscribed in a semi-cirlce is QSR a right triangle?

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

(1) QR is a diameter of the circle --> according to the above property QSR must be a right triangle. Sufficient.

(2) Length QS equals 3 and length QR equals to 5 --> it's not necessary QSR to be 3-4-5 right triangle (therefor QR to be diameter/hypotenuse), for example if diameter is more than 5, say 10 than it's possible to inscribe QSR in a semi-circle so that SR would be the largest side and QSR would be obtuse-angled triangle. Not sufficient.

Answer: A.

Hi Bunuel
If it were 3-4-5; 3-4-10; 6-4-5;................any multiple of 3:4:5, then it would be a right angled triangle ?

Also to know if that triangle inscribed in the circle is a right triangle, then diameter would be the longest side of the triangle.

Are there any other factors that can help us understand whether the triangle is a right angled triangle?
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Bunuel
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Triangle QSR is inscribed in a circle. Is QSR a right triangle?

(1) QR is a diameter of the circle.
(2) Length QS equals 3 and length QR equals 5.

Attachment:
2017-04-14_1450.png

Triangle QSR is inscribed in a semi-cirlce is QSR a right triangle?

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

(1) QR is a diameter of the circle --> according to the above property QSR must be a right triangle. Sufficient.

(2) Length QS equals 3 and length QR equals to 5 --> it's not necessary QSR to be 3-4-5 right triangle (therefor QR to be diameter/hypotenuse), for example if diameter is more than 5, say 10 than it's possible to inscribe QSR in a semi-circle so that SR would be the largest side and QSR would be obtuse-angled triangle. Not sufficient.

Answer: A.

Hi Bunuel
If it were 3-4-5; 3-4-10; 6-4-5;................any multiple of 3:4:5, then it would be a right angled triangle ?

Also to know if that triangle inscribed in the circle is a right triangle, then diameter would be the longest side of the triangle.

Are there any other factors that can help us understand whether the triangle is a right angled triangle?

• Any triangle whose sides are in the ratio 3:4:5 is a right triangle. Such triangles that have their sides in the ratio of whole numbers are called Pythagorean Triples. There are an infinite number of them, and this is just the smallest. If you multiply the sides by any number, the result will still be a right triangle whose sides are in the ratio 3:4:5. For example 6, 8, and 10.
• A Pythagorean triple consists of three positive integers \(a\), \(b\), and \(c\), such that \(a^2 + b^2 = c^2\). Such a triple is commonly written \((a, b, c)\), and a well-known example is \((3, 4, 5)\). If \((a, b, c)\) is a Pythagorean triple, then so is \((ka, kb, kc)\) for any positive integer \(k\). There are 16 primitive Pythagorean triples with c ≤ 100:
(3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97).

So, 3:4:10 is not a Pythagorean triple, 3^2 + 4^2 does not equal 10^2. You'll get a Pythagorean triple if you multiply another Pythagorean triple by a positive integer, so multiply the entire ratio, not just one of its numbers.
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