Question

I want answer of this questions.. plz slve quickly nd fully ​

Answer 1

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Answer 2

(1)

Okay, here, n is an odd no. except 1.

So let n = 2k + 1 for any positive integer k.

So,

$n^2+(\frac{n^2-1}{2})^2 \\ \\ (2k+1)^2+(\frac{(2k+1)^2-1}{2})^2 \\ \\ 4k^2+4k+1+(\frac{(4k^2+4k+1)-1}{2})^2 \\ \\ 4k^2+4k+1+(\frac{4k^2+4k+1-1}{2})^2 \\ \\ 4k^2+4k+1+(\frac{4k^2+4k}{2})^2 \\ \\ 4k^2+4k+1+(2k^2+2k)^2 \\ \\ 4k^2+4k+1+4k^4+8k^3+4k^2 \\ \\ 4k^4+8k^3+8k^2+4k+1 \\ \\ (2k^2+2k+1)^2 \\ \\ (\frac{4k^2+4k+2}{2})^2 \\ \\ (\frac{4k^2+4k+1+1}{2})^2 \\ \\ (\frac{(2k+1)^2+1}{2})^2 \\ \\ (\frac{n^2+1}{2})^2$

So we get that,

[tex]n^2+(\frac{n^2-1}{2})^2=(\frac{n^2+1}{2})^2 [/tex]

$\therefore\ (n,\ \frac{n^2-1}{2},\ \frac{n^2+1}{2})$ is a Pythagorean triplet.

So this is the actual proof. Hope this may be helpful to you. ^_^

Let me tell something more about this.

 

WHY n ≠ 1?

Because $\frac{n^2-1}{2}$, a member of the Pythagorean triplet, becomes zero.

[tex]\frac{n^2-1}{2}=\frac{1^2-1}{2}=\frac{1-1}{2}=\frac{0}{2}=0 [/tex]

WHY n CAN'T BE EVEN?

Because if n is even, the other two members of the Pythagorean triplet,

$\frac{n^2-1}{2},\ \frac{n^2+1}{2}$

won't be positive integers, as both n² - 1 and n² + 1 becomes odd.

But even it won't be a Pythagoren triplet, the sum of squares of first two is equal to the square of the third.

Let me show you.

Let n = 2k.

$n^2+(\frac{n^2-1}{2})^2 \\ \\ (2k)^2+(\frac{(2k)^2-1}{2})^2 \\ \\ 4k^2+(\frac{4k^2-1}{2})^2 \\ \\ 4k^2+\frac{16k^4-8k^2+1}{4} \\ \\ \frac{16k^2+16k^4-8k^2+1}{4} \\ \\ \frac{16k^4+8k^2+1}{4} \\ \\ (\frac{4k^2+1}{2})^2 \\ \\ (\frac{(2k)^2+1}{2})^2 \\ \\ (\frac{n^2+1}{2})^2$

So here we also get that

[tex] n^2+(\frac{n^2-1}{2})^2=(\frac{n^2+1}{2})^2[/tex]

So we can say that these are also Pythagorean triplets according to this, but the largest two members of this triplet are not integers. So we can't say.

There's a lot more to say about it, but now I'm concluding my words.

But before, let me show you some examples.

If n = 3,

∴ (3, 4, 5) is a Pythagorean triplet.

If n = 5,

∴ (5, 12, 13) is a Pythagorean triplet.

If n=7

∴ (7, 24, 25) is a Pythagorean triplet.

Read more on Brainly.in - https://brainly.in/question/8412312#readmore

(2)

First we've to find the total surface area of the object.

Radius of the cylindrical tent

$=r=105 / 2 = 52.5\ m$

$h=5\ m \\ \\ \\ l=50\ m$

Base area of the cylindrical tent

$=\pi r^2 \\ \\ =(52.5)^2\pi \\ \\ =2756.25\pi\ m^2$

Curved surface area of the cylindrical tent

$=2\pi rh \\ \\ = 2\pi \times 52.5 \times 5 \\ \\ = 525\pi\ m^2$

Curved surface area of the cone

$=\pi rl \\ \\ = \pi \times 52.5 \times 50 \\ \\ =2625\pi\ m^2$

∴ Total surface area = Base area of the tent + CSA of the tent + CSA of the cone

$= 2756.25\pi +525\pi +2625\pi\ m^2 \\ \\ = 5906.25\pi\ m^2 \\ \\ \approx 5906.25 \times 3.14\ m^2 \\ \\ \approx 18545.625\ m^2$

Cost of cloth for 1 square meter = ₹ 15

Cost of cloth for 18545.625 square meter (approx.)

$\approx  18545.625\times 15 \\ \\ \approx 278184.375 \\ \\ \approx \bold{2,78,184}$

Hope this helps you. ^_^

Please mark it as the brainliest if it helps.

Thank you. :-))