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9.6 Area of an equilateral triangle In Fig. 9-11, find the area of (a) an equilateral triangle whose perimeter is 24; (b) a rhombus in which the shorter diagonal has length 12 and an angle measures 60 ; (c) a regular hexagon with a side of length 6.

Note that this problem may also be worked in the reverse direction. For example, given a frequency response function such as

Fig. 9-11

a difference equation may be easily found that will implement this system. First, dividing numerator and denominator by 2 and rewriting the frequency response as follows,

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Because the DTFT maps convolution in the time domain into multiplication in the frequency domain, the DTFT provides an alternative to performing convolutions in the time domain. The following example illustrates the procedure.

1 2 4 s 23 1 4 (64) 23

Let x = AC; then BC = 9 x. By the Pythagorean theorem, SC 2 = 6 The time spent rowing is, therefore, T1 = and the time spent walking will be T2 = (9 x)/5 36 + x 2 4 [hours]

let us find the response of the system to the input x ( n ) = Bnu(n),where (a1< 1, the system is the convolution of x ( n ) with h(n),

16 23.

y(n) = h(n)* x(n)

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(b) Since m/A 60 , ^ADB is equilateral and s d 12. The area of the rhombus is twice the area of 2 A 1 B (144) 23 72 23. ^ABD. Hence A 2 A 1s2 23 B 4 4 (c) A side s of the inscribed hexagon subtends a central angle of measure 1(360 ) 6 60 . Then, since OA OB radius R of the circumscribed circle, m/OAB m/OBA 60 . Thus ^ AOB is equilateral. 6 A 1 B (36 23) 54 23. Area of hexagon 6(area of ^AOB) 6 A 1s2 23 B 4 4

Y ( e J W= H ( e J w W ( e J w= ) )

The area of a trapezoid equals one-half the product of the length of its altitude and the sum of the lengths of its bases. (A proof of this theorem is given in 16.) Thus if h 20, b 27, and b 23 in Fig. 9-12, then A 1(20)(27 23) 500. 2

Therefore, all that is required is to find the inverse DTFT of Y (el"). This may be done easily by expanding Y (ej") as follows:

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We have to minimize T (x) on the interval [0, 9], since x can vary from 0 (at A) to 9 (at B), T (x) = Dx = = (36 + x 2 )1/2 4 + Dx x 9 5 5 [by the power chain rule]

The area of a trapezoid equals the product of the lengths of its altitude and median. Since 1 br) and m 1 (b b ), A hm. 2 h(b 2

where A and B are constants that are to be determined. Expressing the right-hand side of this expansion over a common denominator,

In Chap. 1 we looked at methods for solving difference equations in the "time domain." The DTFT may be used to solve difference equations in the "frequency domain" provided that the initial conditions are zero. The procedure is simply to transform the difference equation into the frequency domain by taking the DTFT of each term in the equation, solving for the desired term, and finding the inverse DTFT.

Fig. 9-12

EXAMPLE 2.7.3 Let us solve the following LCCDE for y ( n ) assuming zero initial conditions,

9.7 Area of a trapezoid (a) Find the area of a trapezoid if the bases have lengths 7.3 and 2.7, and the altitude has length 3.8. (b) Find the area of an isosceles trapezoid if the bases have lengths 22 and 10, and the legs have length 10. (c) Find the bases of an isosceles trapezoid if the area is 52 23, the altitude has length 4 23, and each leg has length 8.

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