Question

Ted bought a new TV set. He decided to keep it on a shelf in the corner of the room at some height so that the view gets better. If AB = AC = 48 inches, is the space enough for a TV with an area of the base (as isosceles right triangle shape) equal to 1200 square inches so that no part of the TV lies outside the shelf? If not, what should be the minimum length of AB?

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• A. Yes
• B. No, 48.5
• C. No, 49
• D. No, 49.5

Answer 1

The correct option is C

No, 49

The area of the TV is $1200i{n}^2$

Now, area of a triangle is half the area of the corresponding parallelogram

Since it is given that the triangle is right angled and isoceles, the corresponding parallelogram has adjacent sides equal and all angles are ${90}^{\circ }$

Thus, the figure is a square

For the TV to fit in the triangular base, the square must have at least twice the area of the TV

Thus, area of the square must be at least $2400i{n}^2$

Hence, the side of the square must be equal to $\sqrt{2400}=48.89in=49in$

Thus, AB must have a minimum length of 49 inches.