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(a) The areas of two similar triangles are 36 and 25 If a median of the smaller triangle has length 10, find the length of the corresponding median of the larger (912) (b) Corresponding altitudes of two similar triangles have lengths 3 and 4 If the area of the larger triangle is 112, find the area of the smaller (c) Two similar polygons have perimeters of 32 and 24 If the area of the smaller is 27, find the area of the larger (d) The areas of two similar pentagons are 88 and 22 If a diagonal of the larger has length 5, find the length of the corresponding diagonal of the smaller (e) In two similar polygons, the ratio of the lengths of two corresponding sides is 23:1 If the area of the smaller is 15, find the area of the larger (911) (911).

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THE MEAN-VALUE THEOREM AND THE SIGN OF THE DERIVATIVE Consider the root c = 1 + 2 3. Since 3 < 3, 3 1<1+ Thus, c is in (1, 3) and f (c) = 0. 2 2 3 < 1 + (3) = 1 + 2 = 3 3 3

(c) H3(ejw) =

1. Angle bisector of a triangle: An angle bisector of a triangle is a segment or ray that bisects an angle and extends to the opposite side.

(w) = tan-'

Thus BD, the angle bisector of /B in Fig. 1-31, bisects /B, making /1 > /2.

Therefore,

rl(w) = -

2. Median of a triangle: A median of a triangle is a segment from a vertex to the midpoint of the opposite side.

17.2 Verify the mean-value theorem for f (x) = x 3 6x 2 4x + 30 on the interval [4, 6].

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(1 - a c o s w ) a c o ~ w - (asinw)'

Fig. 1-31

+ (-)2

Fig. 1-32

FOURIER ANALYSIS which, after simplification, becomes q(w) = ( I - a c o s ~ ) a c o ~ w - ( a s i n w )- a 2 - a c o s w ~ I +az-2acosw (1 - a cos w ) ~ ( a sin w)Z

f is differentiable and, therefore, continuous for all x. f (6) = (6)3 6(6)2 4(6) + 30 = 216 216 24 + 30 = 6 f (4) = (4)3 6(4)2 4(4) + 30 = 64 96 16 + 30 = 18 whence, f (6) f (4) 6 ( 18) 24 = = = 12 6 4 6 4 2

3. Perpendicular bisector of a side: A perpendicular bisector of a side of a triangle is a line that bisects and is perpendicular to a side.

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I suggest you use Google as there is a lot of information on the topic: http://en. lmgtfy.com/?q=ucc+ ean - 128 +barcode+generator[^]. —SA.

Another way to solve this problem is to use the expression for the group delay derived in Rob. 2.18. With Hl(eJ") = 1 - a c o s w + j a s i n w we see that HR(eJw) 1 - a cos w = Because the unit sample response is h(n) = S(n) - aS(n then Hl(eJw)= a sin w

4. Altitude to a side of a triangle: An altitude of a triangle is a segment from a vertex perpendicular to the opposite side.

- 1)

g(n) = nh(n) = -a6(n - I )

Thus BD, the altitude to AC in Fig. 1-33, is perpendicular to AC and forms right angles 1 and 2. Each angle bisector, median, and altitude of a triangle extends from a vertex to the opposite side.

which is the same as before. (b) Having found the group delay for Hl(ejW)= 1 - ae-J", we may easily derive the group delay for H2(eJW), which is the inverse of Hl(eJw):

We must therefore nd some c in (4, 6) such that f (c) = 12. Now, f (x) = 3x 2 12x 4, so that c will be a solution of 3x 2 12x 4 = 12 By the quadratic formula, 144 4(3)( 16) 12 144 + 192 12 336 = = 2(3) 6 6 12 4 21 12 16 21 = = 2 2 21 = 3 6 6 Choose c = 2 + 2 21. Since 4 < 21 < 5, 3 x= 12 4<2+ Thus, c is in (4, 6) and f (c) = 12. 8 2 2 2 = 2 + (4) < 2 + 21 < 2 + (5) < 2 + 4 = 6 3 3 3 3 or 3x 2 12x 16 = 0

5. Altitudes of obtuse triangle: In an obtuse triangle, the altitude drawn to either side of the obtuse angle falls outside the triangle.

Specifically, because 1 Hz(eJ") = Hz(eJ") &(w) = -@I(w) and, therefore,

Thus in obtuse triangle ABC (shaded) in Fig. 1-34, altitudes BD and CE fall outside the triangle. In each case, a side of the obtuse angle must be extended.

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