SOLUTION: Find the dimensions and area of the largest rectangle that can be inscribed in a right angle triangle, whose sides have lengths of 9cm, 12cm and a hypotenuse of 15cm.
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Question 1043677: Find the dimensions and area of the largest rectangle that can be inscribed in a right angle triangle, whose sides have lengths of 9cm, 12cm and a hypotenuse of 15cm.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find the dimensions and area of the largest rectangle that can be inscribed in a right angle triangle, whose sides have lengths of 9cm, 12cm and a hypotenuse of 15cm.
:
let x = the width of the rectangle
let y = the length
:
Draw out a right triangle with the rectangle inscribed therein, x by y
the height of the right triangle = 12 and the base = 9
:
We are interested in the small right triangles adjacent the inscribed rectangle
The tangent of the top angle = 9/12 which = of the small right triangle on top
Cross multiply
12x = 9(12-y)
12x = 108 - 9y
x = 9 - y
similarly, the tangent of the base angle = 12/9 =
Cross multiply
9y = 12(9-x)
9y = 108 - 12x
y = 12 - x
:
Area = xy
replace y with (12-x)
A = x(12-x)
A = 12x - x^2
A quadratic equation, max occurs at the axis of symmetry; x=-b/(2a)
x =
x =
x = +4.5 cm is the width of the rectangle with max area
:
Find y
y = 12 - *4.5
y = 12 - 6
y = 6 cm is length
:
Max area: 6 * 4.5 = 27 sq/cm
:
NOt sure if I explained this clearly, let me know. CK
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