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chapter 1 58

# chapter 1 58 - El PRINCIPLES OF PROBLEM SOLVING H we see...

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Unformatted text preview: El PRINCIPLES OF PROBLEM SOLVING H we see that g (4) (y) % (h) h b a . P2 — 100 . . Refer to Example 1. where we obtained It 2 T. The 100 came from 4 times the area of the mangle. P2 — 4 6h In this case. the area of the triangle is %(h)(12) : 6h. Thus. h = % 2 2Ph : P2 — 24h :> 2 P2 _ 2 _ : 2Ph+24hiP :> h(2P+24)—P => h 2P+24' 2m~1ifx2§ m+5 ifmZ—S .lZm—ll: and im—l—Sl: . 1~2zc ifm<§ —m—51far<~5 By using the area formula for a triangle. % (base) (height). in two ways. E h . Since 42 + y2 = hz. (a). soa : 4v};2 — 16 y : t/h2 — 16.anda : ——. h Therefore. we consider the three cases a: < —5, —5 g as < %. and ac 2 % Ifa: < 75. we must have 1 2m ( as 5) — 3 S a: — 3. which is false. since we are considering :c < ~5. If—5§m<%.wemusthavel~2\$~(a§+5):3 (i) x:—%. IfmZ %.wemusthave2a:—1~(:E+5):3 4:} \$29. So the two solutions ofthe equation areas : —§ andm 2 9. 26—1 ifol 55—3 ifx23 .fw—ll: ‘ and|m~3|= l—m if\$<1 3~sc ifac<3 Therefore. we consider the three cases .1‘ < 1. 1 S a: < 3. and a: 2 3. Ifcc<1.wemusthave1—a:~(3~a:)25 (a) 027.whichisfalse. lfl 3x < 3. wemusthavem~ 1 — (3—2:) 2 5 <:> :c 2 §.whichisfalsebecausea: < 3. Ifzc23.wemusthavea:—1~(mﬁ3)25 4:) 225.whichisfalse. All three cases lead to falsehoods. so the inequality has no solution. 59 ...
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