Triangle Side Determination

$N+1>N-1$ and $N+1 > \dfrac{N}{2}$ . Hence, $N+1$ is the hypotenuse. From the Pythagorean Theorem, we have...

$(N+1)^2=(N-1)^2 + \left(\dfrac{N}{2}\right)^2$

$\Longrightarrow (N+1)^2-(N-1)^2=\dfrac{N^2}{4}$

$\Longrightarrow 4N = \dfrac{N^2}{4}$

$\Longrightarrow 4=\dfrac{N}{4}$

$\therefore ~~~ N = 16$

Notes :

• If $N = 0$ , then $N-1=0-1=-1$ which is impossible since $N-1$ is the side length of a triangle. Hence $N\neq 0$ and so we can divide by $N$ .

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• $(N+1)^2-(N-1)^2=4\cdot N \cdot 1 = 4N$ . This is found from the simple identity $(a+b)^2-(a-b)^2=4ab$ .

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By using Pythagoras Theorem,

(N/2)^2+(N-1)^2=(N+1)^2 By solving we get N^2-16N=0 Hence N=0,16 0 cannot be the length of the side so final answer is 16

As $N+1 > N-1 , \frac{N}{2}$
So $N+1$ is the hypotenus.
Then,
$(N+1)^{2}=(N-1)^{2}+(\frac{N}{2})^{2}$

$(N+1)^2 - (N-1)^2 = \frac{N^{2}}{4}$

$2 × 2N = \frac{N^{2}}{4}$

$4 × N = \frac{N^{2}}{4}$

$4 = \frac{N}{4}$

$N = \boxed{16}$

Note

• I used the difference of square formula in the 3rd line which is $a^{2} -b^{2} = (a+b)(a-b)$ making $a= N+1$ and $b = N-1$

• and i used pythagorean theorem or the Gougu theorem in the 1st line to start with a equation and solve for $N$

This kind of problem (multiple choice) is really a process of elimination, so solve using a middle value. The odds are then 50% that you will know the answer, 50% that you will have to try again. More in my favor is that the creator of the problem chose calculations that will use whole numbers=> N=16; ... If the answer needs to be bigger, then I would try 18 for ease of calc. (Not all problem-solving is using the expected process.)

it must be done by pythagorus theorm.Don't need to go for postulates. Just place N=1-17 and you will find something like (16/2) ^2+ (16-1)^2=?

use pythagorus theorem

It is obvious that we are able to use Pythagorean Theorem. And clearly, the longest side of all is $N+1$ . Hence,

$(N+1)^2=(N-1)^2+(\frac {N}{2})^2$

Solving the above equation gives $N=0$ or $16$ . But side's length is never equals zero. Hence $16$ is the answer

This is Right triangle so we using the Pythagoran theorem

assume a = N/2, b = N - 1, and c = N + 1

Pythagoran theorem = a^2 + b^2 = c^2

subtitute the value we have :

(N/2)^2 + (N - 1)^2 = (N + 1)^2

<=> (N^2 / 4) + N^2 - 2N + 1 = N^2 + 2N + 1

<=> (N^2 / 4) = 4N

<=> N^2 = 16N

<=> N^2 - 16N = 0

<=> N(N - 16) = 0

We have N = 0 or N = 16, because this length we take N = 16

Then value of N = 16

According to Phytogoras" Theorem,

(base)^2 + (height)^2 = (hypotenuse)^2

(N-1)^2 +(N/2)^2 = (N+1)^2

N^2-2N+1+(N^2/4)=N^2+2N+1

-2N+(N^2/4) =2N

N^2/4 =4N

N^2 =16N

N=16

• We note that if N is a non-zero real positive number then $N+1 \geq \frac{N}{2} \geq {N-1}$
• This fact follows the next $N+1$ is the hypothenuse and $\frac{N}{2}$ and $N-1$ are legs $(N+1)^2=(\frac{N}{2})^2+(N-1)^2$
• Simplyfing $N^2-16N=0 \Rightarrow N=16$