Question

If one side of a right-angled triangle (with all sides integers) is 15 cm, then
what is the maximum perimeter of the triangle?
यदि एक समकोण त्रिभुज (जिसकी सभी भुजाएं पूर्णाक में है) की एक भुजा 15 cm है, तो इस त्रिभुज की
अधिकतम परिमाप क्या होगी?​

Answer 1

$\bold{\sf{Smile\:please :D}}$

$\bold{\sf{Smile\: because \:it\: is\: the\:key\:that\: fits\: the\: lock\:of\:everybody 's\:heart.}}$

$\bold{\huge{\underline{\underline{\sf{AnsWer:}}}}}$

$\bold{\large{\mathtt{\blue{Maximum\:Perimeter\:of\:the\:triangle\:is\:240\:cm}}}}$

$\bold{\huge{\underline{\underline{\sf{Explanation:}}}}}$

The sides of the right angled triangle :

1. a
2. b
3. c

We have one of the sides given as 15 cm.

Let,

1. a = $\tt{Side_1}$
2. b = $\tt{Side_2}$
3. c = $\tt{Hypotenuse}$

Property :

• In a right angled triangle, sum of the squres of the two sides is equal to the square of the hypotenuse.

From the property we infer,

• a² + b² = c²

Assuming 15 as the hypotenuse of the right angled triangle, we would get the following pythagorean triplets,

• (12,9,15)

Perimeter = 12 + 9 + 15 = 36

We know that there are better possibilities of getting maximum perimeter for the triangle if we neglect assuming 15 as the hypotenuse.

Let's move on.

Assume 15 as one of the two sides of the right angled triangle.

Now, this creates possibility of 2 pythagorean triplet.

• (17,15,8)

Perimeter :

★ $\mathtt{a\:+\:b\:+\:c}$

★ $\mathtt{17\:+\:15\:+\:8}$

★ $\mathtt{40}$

The pythagorean triplet (17,15,8) gives us perimeter equals to 40 cm.

• (15,25,20)

Perimeter :

★ $\mathtt{a+b+c}$

★ $\mathtt{15+25+20}$

★ $\mathtt{60}$

The pythagorean triplet (15,25,20) results in perimeter equals to 60 cm.

From the above three cases (12,9,15) , ( 17,15,8) and (15,25,20) the maximum perimeter is obtained from the triplet (15,25,20)

Does that mean we have solved out of the question?

No. Actually to solve this question you should know the following technique/concept.

Concept :

• Difference between the consecutive square of numbers is an odd number.

For better explanation, let's take an example.

Two consecutive numbers :

1. 6
2. 5

Difference between squares :

• 6² - 5² = 36 - 25 = 11

The difference obtained in the above example between the consecutive squares is 11 which is an odd number.

Now let's use this same technique for our question and let's see if we get any better triplet.

But how to apply? Let me tell you the main concept.

We know the square of 15 is 225.

To find two consecutive squres, you just simply have to divide the difference obtained between two consecutive squres by 2. To obtain the larger number of the two, add 0.5 to the result obtained after dividing and for obtaining the smaller number just subtract 0.5 from the result obtained.

In the example of 6 and 5, difference between the square of the numbers is 11. So here's what the main thing approaches.

★ $\mathtt{\dfrac{11}{2}\:+\:0.5}$

★ $\mathtt{5.5\:+\:0.5}$

★ $\mathtt{6}$

Similary doing it for 5 too,

★ $\mathtt{\dfrac{11}{2}\:-\:0.5}$

★ $\mathtt{5.5\:-\:0.5}$

★ $\mathtt{5}$

And here we are! We obtained 6 and 5 using the difference between the squres.

Let's now give final touch!;)

Finding two consecutive squres whose difference is 225.

★ $\tt{\dfrac{225}{2}\:+\:0.5}$

★ $\tt{112.5+0.5}$

★ $\tt{113}$

For the remaining side :

★ $\mathtt{\dfrac{225}{2}\:-\:0.5}$

★ $\mathtt{112.5-0.5}$

★ $\mathtt{112}$

Difference between squares :

• 113² - 112² = 12769 - 12544 = 225

Now, we have the following set of sides :

• 113
• 112
• 15

The hypotenuse of the right angled triangle would be 113 and $\tt{Side_1}$ and $\tt{Side_2}$ would be 112 and 15 respectively.

Let's verify if the triplet forms a pythagorean triplet.

★ $\mathtt{113^2\:=\:112^2\:+\:15^2}$

★ $\mathtt{12769\:=\:12544+225}$

★ $\mathtt{12769=12769}$

The square of larger side (113) is equal to the sum of the squres of the other two sides.

$\tt{\therefore{(113,112,15)\:forms\:a\:pythagorean\:triplet}}$

Perimeter :

★ $\mathtt{a+b+c}$

★ $\mathtt{113+112+15}$

★ $\mathtt{240}$

$\bold{\large{\boxed{\sf{\red{Maximum\:perimeter\:of\:the\:triangle\:is\:240\:cm}}}}}$