Let's consider the diagram below to find out the maximum possible area of a triangle with two of the sides given
We are given length of two sides BC = b and AC = a. We know that area of triangle = \(\frac{1}{2}\) * base * height = \(\frac{1}{2}\) * b * AD
Since we know the lengths of two sides as a and b, we need to express AD in terms of 'a' as 'b' is already present in the expression of the area.
We see that triangle ADC is a right angled triangle, so we can express AD in terms of 'a' using the trigonometric ratios.
We can write Sin ∠ACD = AD/a which gives us AD = a * Sin ∠ACD.
Hence, area of triangle ABC = \(\frac{1}{2}\) * b * a * Sin ∠ACD. This expression would be maximum when Sin ∠ACD is maximum which is when angle ∠ACD = 90 and Sin ∠ACD = 1.
Thus, it tells us that sides AC and BC are perpendicular to each other and hence form the two legs of a right angled triangle.
For two sides given the maximum area would be if these two sides form the legs of a right angled triangleComing back to the question, the sides given to us are 6 and 3 and we need to find the perimeter of the this right angled triangle with 6 & 3 being the legs of the right angled triangle. Using the phythagoras theorem we can calculate the hypotenuse which is the third side as 3√5.
Thus the perimeter will be 6+3+3√5 = 9 + 3√5
Answer is option D
Hope it helps
Regards
Harsh
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