Question

PQRS is a square. If $$PQ=10cm$$ then $$PR=........cm$$.
A). $$10$$
B). $$20$$
C). $$10\sqrt{2}$$
D). $$2\sqrt{10}$$

Answer 1

Hint: The length of one side of a square is given. We know that all four sides of a square will have the same length. Here we have to find the diagonal of the square. A diagonal of a square can split the square into two right-angled triangles where the diagonal will be the hypotenuse so using the Pythagoras theorem, we can find the length of the diagonal.
Formula:
Pythagoras theorem: $$A{C}^2=A{B}^2+B{C}^2$$ where $$ABC$$ is a right-angled triangle $$\left(\mathrm{\angle }B\right)={90}^o$$.
And some other formulas that we need to know:
1.$$\sqrt{x^2}=x$$
2.$$\sqrt{ab}=\sqrt{a}\times \sqrt{b}$$
3.$$\sqrt{a}\times \sqrt{a}=a$$

Complete step-by-step solution:
It is given that $$PQRS$$ is a square with $$PQ=10cm$$. We know that all the sides of the square will have the same length.

Therefore $$QR=10cm,RS=10cm$$ & $$SP=10cm$$.
We aim to find the length of $$PR$$ which is the diagonal of a square$$PQRS$$.
By Pythagoras theorem, we know that for a right-angled triangle $$\mathrm{\Delta }ABC$$,$$A{C}^2=A{B}^2+B{C}^2$$ that is hypotenuse square is equal to the sum of an opposite square and adjacent square. From the diagram, let us take the right-angled triangle$$\mathrm{\Delta }PSR$$.
So, for the right-angled triangle $$\mathrm{\Delta }PSR$$ we have $$P{R}^2=P{S}^2+S{R}^2$$.
We know that $$\left(\mathrm{\angle }S\right)={90}^o,SR=10cm$$ & $$PS=10cm$$.
Substituting these values in $$P{R}^2=P{S}^2+S{R}^2$$we get
$$P{R}^2={10}^2+{10}^{{}^2}$$
Simplifying this we get
$$P{R}^2=100+100$$
On simplifying this we get
$$P{R}^2=200$$
But we need the value for $$PR$$ so let us take the square root
$$\sqrt{P{R}^2}=\sqrt{200}$$
Now let us write $$200$$ in terms of its factors
$$\sqrt{P{R}^2}=\sqrt{20\times 10}$$
Now let us write $$20$$ in terms of its factors
$$\sqrt{P{R}^2}=\sqrt{2}\times \sqrt{10}\times \sqrt{10}$$
Let us split the square root using the formula, $$\sqrt{ab}=\sqrt{a}\times \sqrt{b}$$
$$\sqrt{P{R}^2}=\sqrt{2}\times \sqrt{10}\times \sqrt{10}$$
Noe using the formula $$\sqrt{a}\times \sqrt{a}=a$$ we get
$$\sqrt{P{R}^2}=\sqrt{2}\times 10$$
$$\sqrt{P{R}^2}=10\sqrt{2}$$
Again, using the formula $$\sqrt{x^2}=x$$ we get
$$PR=10\sqrt{2}$$
Therefore, the length of the diagonal of the square $$PQRS$$ is $$PR=10\sqrt{2}cm$$.
Now let us see the options, option (a) $$10$$ cannot be the right answer since we got the length of $$PR$$ as $$10\sqrt{2}cm$$.
Option (b) $$20$$ cannot be the right answer since we got the length of $$PR$$ as $$10\sqrt{2}cm$$ from the above calculation.
Option (c) $$10\sqrt{2}$$ is the right answer since we got the length of $$PR$$ as $$10\sqrt{2}cm$$ from the above calculation.
Option (d) $$2\sqrt{10}$$ cannot be the right answer since we got the length of $$PR$$ as $$10\sqrt{2}cm$$ from the above calculation.
Thus, option (c) $$10\sqrt{2}$$ is the correct answer.

Note: We can also take another right-angled triangle that is $$\mathrm{\Delta }PQR$$and use the Pythagoras theorem to find the length of the diagonal $$PR$$, we will get the same answer as we got above. This is because all the sides of the square take the same length.