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In the figure given below, rectangle $CDEF$ with perimeter $32$ has the maximum area. Find the area of the triangle $ABF$

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So, I tried the following

$P = 2W+2H$ where $P$ is given $32$. I am not able figure out which will be next step & how can I find Area of Triangle $ABF$

Jeel Shah
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SSK
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  • $w+h=16\ \frac { d }{ dw } \left( w(16-w) \right) =0\Rightarrow 16-2w=0\ w=8\ $ – newzad May 19 '13 at 03:12
  • can please explain bit more. – SSK May 19 '13 at 03:17
  • If you want to draw a rectangle with maximum area for a given perimeter you must draw square. Then width and height of rectangle are equal, which is $8$. – newzad May 19 '13 at 03:23
  • Yes , then how we can find the area of the triangle – SSK May 19 '13 at 03:25
  • $CF = BF = 8, AB = 8\sqrt(3)$, now I suppose that you can calculate area of triangle. – newzad May 19 '13 at 03:29
  • I am seriously so confused right now ! How do you found 8 and why you using Sqrt of 3 . Please can explain me in details .. It will be great help in maths – SSK May 19 '13 at 03:31
  • Have you had differential calculus? If so, a common exercise is to show that for a rectangle of specified area to have the greatest possible area, it must be a square. If you haven't had calculus, you can note that $ \ w = 16 - h \ $ and therefore that the area of the rectangle is $ \ h \cdot (16-h) \ = \ 16h - h^2 \ $ and find the largest value that function can have (or just use the geometrical fact I've mentioned). Then note that triangle CFB is a right isosceles triangle; knowing the length of BF, you will then need trig to find the length of AB, triangle ABF's base. – colormegone May 19 '13 at 03:57
  • Okay , i will read some more calculus . I tried to figure out ans is 32 Sqrt 2 – SSK May 19 '13 at 04:01

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Here the rectangle with perimieter $32$ has maximum area.Using calculus (or any other method) that it must be a rectangle.

As perimeter is $32$ the each side length is $8$ . so now each side is $8$ units. So Cf= $8$ units . As triangle BFC IS Isosceles with right angle, so Side BF = $8$ now triangle ABF is 30-60-90 triangle so side BA = BF $\times \sqrt {3}$

Now as we know area of this triangle = $\frac {1}{2} \times BA \times BF$ = $\frac {1}{2} \times 8 \times \sqrt {3} \times 8$ = $32 \sqrt {3}$ units

Rusty
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