Solution:
We are going to find the maximum value of Area of Triangle.
Given,
The legs of a right triangle have lengths A and B
A + B = 10 A + B = 10 A + B = 10
Here Base = A and Height = B
We have a formula for area of the triangle, that is
Area of the Right triangle =1 2 × B a s e × h e i g h t \frac {1} {2} \times Base \times height 2 1 × B a se × h e i g h t
= 1 2 × A × B \frac {1} {2} \times A \times B 2 1 × A × B
Now we can solvve for A and B using the concept A.M (Arithmetic mean) and G.M (Geometric mean)
A . M o f A a n d B = A + B 2 A.M \space of \space A \space and \space B = \frac {A+B} {2} A . M o f A an d B = 2 A + B
G . M o f A a n d B = A B G.M \space of \space A \space and \space B = \sqrt {AB} G . M o f A an d B = A B
We know,
G . M ≤ A . M G.M \le A.M G . M ≤ A . M
A B ≤ A + B 2 \sqrt {AB}\le \frac {A+B} {2} A B ≤ 2 A + B
A B ≤ 10 2 A B ≤ 5 \sqrt {AB}\le \frac {10} {2} \\\sqrt {AB}\le 5 A B ≤ 2 10 A B ≤ 5
A B ≤ 25 AB \le 25 A B ≤ 25 Maximum Area would be = 1 2 A B = 12.5 \frac {1}{2} AB = 12.5 2 1 A B = 12.5 , if the A and B are equal
So, A = B = 5.
Answer: A = B = 5
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