Question

If the area of a right angled triangle with hypotenuse 5 is maximum, then its perimeter is

• A. 12
• B. \( 2\sqrt{3}+\sqrt{13}+5 \)
• C. \( 7+\sqrt{21} \)
• D. \( 5(\sqrt{2}+1) \)

Hint:

For a right-angled triangle with fixed hypotenuse \(h\), the area \( A = \frac{1}{2}ab \) is maximized when the triangle is isosceles, i.e., \( a=b \). In this case, \( a^2+a^2 = h^2 \implies 2a^2=h^2 \implies a = h/\sqrt{2} \). The sides are \( h/\sqrt{2}, h/\sqrt{2}, h \). Perimeter \( = h/\sqrt{2} + h/\sqrt{2} + h = \sqrt{2}h + h = h(\sqrt{2}+1) \).

Answer 1

The Correct Option is D

Let the sides of the right-angled triangle containing the right angle be \(a\) and \(b\).
Let the hypotenuse be \(h\).
Given \(h=5\).
By Pythagoras theorem, \( a^2+b^2 = h^2 = 5^2 = 25 \).
Area of the triangle \( A = \frac{1}{2}ab \).
We want to maximize area A.
Let \( a = 5\cos\theta \) and \( b = 5\sin\theta \), where \( \theta \) is one of the acute angles.
Then \( A(\theta) = \frac{1}{2} (5\cos\theta)(5\sin\theta) = \frac{25}{2} \sin\theta\cos\theta = \frac{25}{4} (2\sin\theta\cos\theta) = \frac{25}{4}\sin(2\theta) \).
Area A is maximum when \( \sin(2\theta) \) is maximum, which is 1.
This occurs when \( 2\theta = \pi/2 \implies \theta = \pi/4 \).
When \( \theta = \pi/4 \), the triangle is an isosceles right-angled triangle.
Sides are \( a = 5\cos(\pi/4) = 5 \cdot \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \).
And \( b = 5\sin(\pi/4) = 5 \cdot \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} \).
Check: \( a^2+b^2 = (\frac{5}{\sqrt{2}})^2 + (\frac{5}{\sqrt{2}})^2 = \frac{25}{2} + \frac{25}{2} = \frac{50}{2} = 25 = 5^2 \).
Correct.
The perimeter P is \( a+b+h \).
\[ P = \frac{5}{\sqrt{2}} + \frac{5}{\sqrt{2}} + 5 = \frac{10}{\sqrt{2}} + 5 = 5\sqrt{2} + 5 \] \[ P = 5(\sqrt{2}+1) \] This matches option (4).