Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3 cm and 4 cm if two sides of the rectangle lie along the legs.

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Melinda McCombs · Accepted Answer

Step 1Step 2Let x and y be the side lengths of the rectangle, then the area isA=xyand by similar triangles (ABC and FBE, note: BF is 3−y)x4=3−y33x4=3−y3x4−3=−yy=3−34xStep 3Substitute for y in the area equation, find A′A=x(3−34x)=3x−34x2A′=3−32xFind where it is 00=3−32x32x=3x=2Step 4Find te corresponding y (with the similar triangle equation)y=3−34(2)=3−32=6−32=32So the maximum rectangle area isA=2⋅32=3cm2

Becky Harrison · Accepted Answer

Step 1  The largest possible rectangle must have a vertex that touches the hypotenuse of the triangle at one point.  Let us put the right angle of the triangle at the point (0,0), another vertex at (8,0) and the final vertex at (0,3)  We can represent points on the hypotenuse parametrically as:  (8t, 3(1−t))  where t∈[0,1]  Then the area of the rectangle with vertices:  (0,0),(8t,0),(8t,3(1−t)),(0,3(1−t)) is:  f(t)=8t⋅3(1−t)=24t(1−t)=24(t−t2)=24(14−(t−12)2)  This takes its maximum value when t=12 and f(t)=24⋅14=6

Find the area of the largest rectangle that can be inscribed in a right triangle

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