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24. (7分)利用折纸可以作出角平分线,如图1折叠,则OC为∠AOB的平分线,如图2、图3,折叠长方形纸片,OC,OD均是折痕,折叠后,点A落在点A',点B落在点B'.
(1)如图2,若点B'恰好落在OA'上,且∠AOC=32°,则∠BOD=
(2)如图3,当点B'在∠COA'的内部时,若∠AOC=44°,∠BOD=61°,求∠A'OB'的度数.
(1)如图2,若点B'恰好落在OA'上,且∠AOC=32°,则∠BOD=
$58°$
;(2)如图3,当点B'在∠COA'的内部时,若∠AOC=44°,∠BOD=61°,求∠A'OB'的度数.
答案
【点拨】本题考查折叠的性质,平角的定义,角的和差计算,掌握从图形中找出角之间的关系是解题的关键.
【解析】(1)由题意知$∠AOC = ∠A'OC$,$∠BOD = ∠B'OD$,因为$∠AOC + ∠A'OC + ∠BOD + ∠B'OD = 180°$,$∠AOC = 32°$,所以$∠BOD = \frac{1}{2} × (180° - 2 × 32°) = 58°$. 故答案为$58°$.
(2)由题意知$∠AOC = ∠A'OC$,$∠BOD = ∠B'OD$,因为$∠AOC + ∠A'OC + ∠A'OD + ∠BOD = 180°$,$∠AOC = 44°$,$∠BOD = 61°$,所以$∠A'OD = 180° - 2 × 44° - 61° = 31°$,所以$∠A'OB' = ∠B'OD - ∠A'OD = 61° - 31° = 30°$.
【解析】(1)由题意知$∠AOC = ∠A'OC$,$∠BOD = ∠B'OD$,因为$∠AOC + ∠A'OC + ∠BOD + ∠B'OD = 180°$,$∠AOC = 32°$,所以$∠BOD = \frac{1}{2} × (180° - 2 × 32°) = 58°$. 故答案为$58°$.
(2)由题意知$∠AOC = ∠A'OC$,$∠BOD = ∠B'OD$,因为$∠AOC + ∠A'OC + ∠A'OD + ∠BOD = 180°$,$∠AOC = 44°$,$∠BOD = 61°$,所以$∠A'OD = 180° - 2 × 44° - 61° = 31°$,所以$∠A'OB' = ∠B'OD - ∠A'OD = 61° - 31° = 30°$.
25. (11分)【阅读理解】
若$x$满足$(32 - x)(x - 12) = 100$,求$(32 - x)^2 + (x - 12)^2$的值.
解:设$32 - x = a, x - 12 = b$,则$(32 - x)(x - 12) = ab = 100, a + b = (32 - x) + (x - 12) = 20, (32 - x)^2 + (x - 12)^2 = a^2 + b^2 = (a + b)^2 - 2ab = 20^2 - 2 × 100 = 200$.
我们把这种方法叫换元法. 利用换元法达到简化方程的目的,体现了转化的数学思想.
【解决问题】
(1)若$x$满足$(100 - x)(x - 95) = 5$,则$(100 - x)^2 + (x - 95)^2 = \_\_\_\_\_\_$;
(2)若$x$满足$(2\,023 - x)^2 + (x - 2\,000)^2 = 229$,求$(2\,023 - x)(x - 2\,000)$的值;
(3)如图,在长方形$ABCD$中,$AB = 24\ \mathrm{cm}$,点$E,F$是边$BC,CD$上的点,$EC = 12\ \mathrm{cm}$,且$BE = DF = x\ \mathrm{cm}$,分别以$FC,CB$为边在长方形$ABCD$外侧作正方形$CFGH$和$CBMN$,若长方形$CBQF$的面积为$320\ \mathrm{cm}^2$,求图中阴影部分的面积和.
若$x$满足$(32 - x)(x - 12) = 100$,求$(32 - x)^2 + (x - 12)^2$的值.
解:设$32 - x = a, x - 12 = b$,则$(32 - x)(x - 12) = ab = 100, a + b = (32 - x) + (x - 12) = 20, (32 - x)^2 + (x - 12)^2 = a^2 + b^2 = (a + b)^2 - 2ab = 20^2 - 2 × 100 = 200$.
我们把这种方法叫换元法. 利用换元法达到简化方程的目的,体现了转化的数学思想.
【解决问题】
(1)若$x$满足$(100 - x)(x - 95) = 5$,则$(100 - x)^2 + (x - 95)^2 = \_\_\_\_\_\_$;
(2)若$x$满足$(2\,023 - x)^2 + (x - 2\,000)^2 = 229$,求$(2\,023 - x)(x - 2\,000)$的值;
(3)如图,在长方形$ABCD$中,$AB = 24\ \mathrm{cm}$,点$E,F$是边$BC,CD$上的点,$EC = 12\ \mathrm{cm}$,且$BE = DF = x\ \mathrm{cm}$,分别以$FC,CB$为边在长方形$ABCD$外侧作正方形$CFGH$和$CBMN$,若长方形$CBQF$的面积为$320\ \mathrm{cm}^2$,求图中阴影部分的面积和.
答案
【点拨】本题考查完全平方公式的应用.
【解析】(1)设$100 - x = a$,$x - 95 = b$,则$(100 - x)(x - 95) = ab = 5$,$a + b = (100 - x) + (x - 95) = 5$,所以$(100 - x)^2 + (x - 95)^2 = a^2 + b^2 = (a + b)^2 - 2ab = 5^2 - 2 × 5 = 15$. 故答案为15.
(2)设$2\ 023 - x = a$,$x - 2\ 000 = b$,则$(2\ 023 - x)^2 + (x - 2\ 000)^2 = a^2 + b^2 = 229$,$a + b = (2\ 023 - x) + (x - 2\ 000) = 23$,所以$(2\ 023 - x)(x - 2\ 000) = ab = \frac{1}{2}[(a + b)^2 - (a^2 + b^2)] = \frac{1}{2} × (23^2 - 229) = 150$.
(3)由题意,得$FC = (24 - x)$cm,$BC = (12 + x)$cm,因为长方形 CBQF 的面积为$320\ \mathrm{cm}^2$,所以$(24 - x)(12 + x) = 320$,设$24 - x = a$,$12 + x = b$,则$(24 - x)(x + 12) = ab = 320$,$a + b = (24 - x) + (x + 12) = 36$,所以阴影部分的面积和$=(24 - x)^2 + (x + 12)^2$
$=a^2 + b^2$
$=(a + b)^2 - 2ab$
$=36^2 - 2 × 320$
$=656(\mathrm{cm}^2)$.
【解析】(1)设$100 - x = a$,$x - 95 = b$,则$(100 - x)(x - 95) = ab = 5$,$a + b = (100 - x) + (x - 95) = 5$,所以$(100 - x)^2 + (x - 95)^2 = a^2 + b^2 = (a + b)^2 - 2ab = 5^2 - 2 × 5 = 15$. 故答案为15.
(2)设$2\ 023 - x = a$,$x - 2\ 000 = b$,则$(2\ 023 - x)^2 + (x - 2\ 000)^2 = a^2 + b^2 = 229$,$a + b = (2\ 023 - x) + (x - 2\ 000) = 23$,所以$(2\ 023 - x)(x - 2\ 000) = ab = \frac{1}{2}[(a + b)^2 - (a^2 + b^2)] = \frac{1}{2} × (23^2 - 229) = 150$.
(3)由题意,得$FC = (24 - x)$cm,$BC = (12 + x)$cm,因为长方形 CBQF 的面积为$320\ \mathrm{cm}^2$,所以$(24 - x)(12 + x) = 320$,设$24 - x = a$,$12 + x = b$,则$(24 - x)(x + 12) = ab = 320$,$a + b = (24 - x) + (x + 12) = 36$,所以阴影部分的面积和$=(24 - x)^2 + (x + 12)^2$
$=a^2 + b^2$
$=(a + b)^2 - 2ab$
$=36^2 - 2 × 320$
$=656(\mathrm{cm}^2)$.
