alex.gellatly wrote:Is an inscribed triangle always less than half of the area like this?
Also, if there were more answer choices less than 14 is there a way to solve this with out the use of a formula?
Thanks
The area of a rectangle = bh.
The area of a triangle = (1/2)bh.
A triangle WITHIN a rectangle cannot have a greater bh than the rectangle.
Thus, the greatest possible area of a triangle inscribed inside a rectangle = (1/2)(base of the rectangle)(height of the rectangle) = 1/2 the area of the rectangle.
Here is such a triangle:
Since the triangle and the rectangle have the same base and height, the triangle = 1/2 the rectangle.
The problem at hand does not offer such a triangle:
Since ∆PQR does not have the same base and height as rectangle STRU, the area of ∆PQR must be LESS than 1/2 the area of the rectangle STRU.
Here's an easy way to calculate the area of ∆PQR:
Area of PQR = rectangle STRU - (∆SQP + ∆QTR + ∆RUP).
Rectangle STRU = 7*4 = 28.
∆SQP = (1/2)(4*3) = 6.
∆QTR = (1/2)(1*7) = 3.5
∆RUP = (1/2)(3*4) = 6.
Thus, ∆PQR = 28 - (6 + 3.5 + 6) = 28 - 15.5 = 12.5.
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