lengths of side of a triangle ?

Question

Which of the following groups of numbers could be the lengths of the sides of a right triangle?

Ⅰ. 1, 4, sqrt (17)
Ⅱ. 4, 7, sqrt (11)
Ⅲ. 4, 9, 6

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

Answer 1

The answer is (A).

The Pythagorean Theorum shows it. Take the two smallest sides of any right triangle and square each and their sum has to equal the square of the longest side. (a^2 + b^2 = c^2, with "c" being the longest side)

I: 1^2 + 4^2 = 1 + 16 = 17. Since sqrt(17) squared is 17, this is one.
II: {sqrt(11)}^2 + 4^2 = 11 + 16 = 27. Since 7^2 = 49, this is not one.
III: 4^2 + 6^2 = 16 + 36. Since 9^2 = 81, this is not one.

So only "I" is a right triangle making (A) the answer.

Answer 2

Its only A because:

to find the sides you need to find the hypothoneus. Here is the formula. a^2 + b^2 = c^2

so 1^2 + 4^2 = sqrt 17

1 * 1 = 1 + 4 * 4 = 16 so 16 + 1 = 17 then take the square root.

Answer 3

A amd only a becaus a^2 + b^2 = c^2

1^2 = 1 4^2 = 16 1+16 = 17

Answer 4

Hello Amuthan A,
The anser to the questin is " (A) I only"
Because the Pathagorean Therom says, " a^2+b^2=c^2"
and since C is always the longest side then 17^(1/2) or Sqrt of 17 is 17 when squared. and so that is C and if A and B added squared added to gether gets you 17 then it is correct.

sqrt(1^2+4^2)=4.123105626
and
sqrt of (17)=4.123105626

Answer 5

In an isosceles precise triangle both equivalent aspects must be the legs, because the hypotenuse is continually the longest aspect in an exact triangle. enable x be the dimensions of a leg. Then, by using Pythagoras, the dimensions of the hypotenuse is: c = ?(x² + x²) = ?(2x²) = (?2) x yet, you're informed that c is 4 cm. longer than x: c = x + 4 (?2) x = x + 4 ... substituted for c (?2)x - x = 4 ... upload -x to each aspect (?2 - a million)x = 4 .... ingredient on the left x = 4/(?2 - a million) .... divide each aspect by using (?2 - a million) So, the reply is 4/(?2 - a million) centimeters.

Answer 6

I only, because acording to pitagoras

c^2 = a^2+b^2

and let's say that a = 1 and b = 4 then

c = sqrt (1^2+4^2) = sqrt (1+16) = sqrt(17)

hope this helps.

Answer 7

Pythagorean theorem : a^2 + b^2 = c^2

1*1 + 4*4 = 17 c = sqrt 17

the first one is the only one that works

Answer 8

the answer is A.
because
by using pythagoras theorem,

side a^2+side b^2=side ^2c

and in triangle a 1^2+4^2=17
hypotenuse=sqrt(17).
it satisfy the condition

but the remaing 2 don't satisify the condition

Answer 9

Awnser is (A)

In any traingle H<(P+B) So Ans. II is not possible.

For a right angle traingle H=Sqrt.((B*B)+(P*P))
For Case I Equation will be H = Sqrt. ((1*1)+(4*))
= Sqrt. (1+16)
= Sqrt. (17)

In case III Eq. will be H = Sqrt. ((4*4) + (6*6))
= Sqrt. (16 + 36)
= Sqrt. (52)

So. Awnser could be (A) i.e. Case I only

Answer 10

it's A.
want to know the reason....
using pythagoras theorem,
1^2+4^2=17
hypotenuse=sqrt(17).
but the other 2 don't satisify the condition.