Question

3. Write a Pythagoras triplet whose one member is 18 with step-by-step expanation​

Answer 1

According to Pythagoras Therom, it only applies to right-angled triangles.

The right-angled triangle consists of ⇒

• Base ⇒ $2m$
• Height ⇒ $m^{2}-1$

• Hypotenuse ⇒ $m^{2}+1$

So Pythagorean Therom is ⇒

$\bf Base^{2}+Height^{2}=Hypotenuse^{2}$

$\bf 2m+(m^{2}-1)=(m^{2}+1)$

Now we need to check which one of the above is 18.

Let's first try with the Base.

$\rightarrow 2m=18\\ \\ \rightarrow m=\frac{18}{2} \\ \\ \rightarrow m=9$

∴ 18 could be the base.

Now let's try with Height.

$\rightarrow m^{2}-1=18\\\\ \rightarrow m^{2}=19$

∴ 18 is not the height.

Now let's try with the Hypotenuse.

$m^{2}+1=18$

$m^{2}=17$

∴ 18 is not the hypotenuse.

∴We know that 18 is the base and now we can find out the height and hypotenuse using the formula given for each. 'm' could be substituted for the number 9.

So the base ⇒ $2m$

$\rightarrow 2\times9\\ \\ \rightarrow 18$

∴ The base is 18.

The height ⇒ $m^{2}-1$

$\rightarrow 9^{2}-1\\\\ \rightarrow 81-1\\ \\ \rightarrow 80$

∴ The height is 80.

The hypotenuse ⇒ $m^{2}+1$

$\rightarrow 9^{2}+1\\ \\ \rightarrow 81+1\\\\ \rightarrow 82$

∴ The hypotenuse is 82.

∴ The Pythagoras Triplet is ⇒ $\bf 18,81\ and\ 82$ whose one member is 18.