This question here is a pretty tough one and is actually not relevant for CAT. However it is an interesting one, can help you apply a nice maximising approach and can also introduce you to a new formula for area of triangle.

Please dont hesitate to have a look at the hint, it only gives a rarely known formula for area of triangle and yet leaves a lot of work for you to do.

Q.

One side of a triangle has length 75. Of the other two sides, the length of one is double the length of the other. What is the maximum possible area for this triangle?

A. 1125

B. \(\frac{1875\sqrt{3}}{2}\)

C. 1875

D. 1350

Area of a triangle, A, having sides a, b, c is given by …

$$A^2=\frac{\left(a+b+c\right)\left(a+b–c\right)\left(a–b+c\right)\left(–a+b+c\right)}{16}$$

C. 1875

The sides of the triangle can be taken as 75, x, 2x.

Using the formula given in the hint,

$$A^2=\frac{\left(75+3x\right)\left(75-x\right)\left(75+x\right)\left(3x-75\right)}{16}$$

$$A^2=\frac{\left(9x^2-{75}^2\right)\left({75}^2-x^2\right)}{16}$$

To maximise the area, we will need to maximise the above expression. We are going to consider the numerator as product of two terms and we will manipulate the second term such that the sum of the two terms is a constant.

$$A^2=\frac{\left(9x^2-{75}^2\right)\left(9 \times {75}^2-9x^2\right)}{16 \times 9}$$

Now the sum of the two terms in the numerator is a constant, 8 × 752.

Thus, the product of the two terms will be maximum when each is equal, and each will then be equal to 4 × 752.

Thus,

$$\text{max of }A^2=\frac{{\left(4 \times {75}^2\right)}^2}{16 \times 9}$$

$$\text{max of }A=\frac{\left(4 \times {75}^2\right)}{4 \times 3}$$

i.e. maximum area = 25 × 75 = 1875.

And this will occur when, 9x2 – 752 = 4 × 752

i.e. 9x2 = 5 × 752

i.e. x = 25√5