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二、填空题(本大题共10小题,每小题2分,共20分)
7. 已知$3^{m}=2,9^{n}=5$,则$3^{2m-2n}=$
7. 已知$3^{m}=2,9^{n}=5$,则$3^{2m-2n}=$
$\frac{4}{5}$
.答案
本题考查同底数幂的除法,幂的乘方.
解析
∵3ᵐ = 2,9ⁿ = 5,
∴3^{2m-2n} = (3ᵐ)² ÷ (3²)ⁿ = 2² ÷ 5 = $\frac{4}{5}$.
故答案为$\frac{4}{5}$.
解析
∵3ᵐ = 2,9ⁿ = 5,
∴3^{2m-2n} = (3ᵐ)² ÷ (3²)ⁿ = 2² ÷ 5 = $\frac{4}{5}$.
故答案为$\frac{4}{5}$.
8. 若一个多边形的每个外角都为$36°$,则这个多边形的内角和是
1440
°.答案
本题考查多边形内角和与外角和定理.
解析
∵多边形的每个外角都为36°,360°÷36° = 10,
∴多边形的边数为10,则这个多边形的内角和为(10 - 2)×180° = 1440°.
故答案为1 440.
解析
∵多边形的每个外角都为36°,360°÷36° = 10,
∴多边形的边数为10,则这个多边形的内角和为(10 - 2)×180° = 1440°.
故答案为1 440.
9. 一个多边形除了一个内角之外,其余各内角的度数和为$1510°$,则这个多边形的边数为
11
.答案
本题考查多边形内角和定理.
解析设这个多边形的边数为n,除去的内角为α(0°＜α＜180°),则(n - 2)·180° = 1 510° + α.
∵1 510°÷180° = 8……70°,
∴n - 2 = 8 + 1,解得n = 11,即这个多边形的边数为11.故答案为11.
解析设这个多边形的边数为n,除去的内角为α(0°＜α＜180°),则(n - 2)·180° = 1 510° + α.
∵1 510°÷180° = 8……70°,
∴n - 2 = 8 + 1,解得n = 11,即这个多边形的边数为11.故答案为11.
10. 如图, 在 $△ ABC$ 中, $AD$ 是 $BC$ 边上的高, $AE$ 是 $∠ BAC$ 的平分线, $∠ B = 50°$, $∠ C = 70°$, 则 $∠ DAE =$
10
$°$.答案
本题考查三角形内角和定理,角平分线的性质.
解析
∵AE是∠BAC的平分线,且∠B = 50°,∠C = 70°,
∴∠BAE = ∠EAC = $\frac{1}{2}$(180° - ∠B - ∠C) = $\frac{1}{2}×(180° - 50° - 70°) = 30°$.
∵∠ADC = 90°,∠C = 70°,
∴∠DAC = 90° - 70° = 20°,
∴∠DAE = ∠EAC - ∠DAC = 30° - 20° = 10°. 故答案为10.
解析
∵AE是∠BAC的平分线,且∠B = 50°,∠C = 70°,
∴∠BAE = ∠EAC = $\frac{1}{2}$(180° - ∠B - ∠C) = $\frac{1}{2}×(180° - 50° - 70°) = 30°$.
∵∠ADC = 90°,∠C = 70°,
∴∠DAC = 90° - 70° = 20°,
∴∠DAE = ∠EAC - ∠DAC = 30° - 20° = 10°. 故答案为10.
11. 如图,五边形ABCDE是正五边形,$l_{1}// l_{2}$.若$∠1=20°$,则$∠2=$
$3-2x≥-1$
56°
.$3-2x≥-1$
答案
本题考查多边形内角和定理,平行线的性质,正确作出辅助线是解题的关键.
解析如题图,连接AC.
∵五边形ABCDE是正五边形,
∴∠B = ∠BAE = 108°,
∴∠ACB = ∠CAB = 36°,
∴∠CAE = 108° - 36° = 72°.
∵l₁//l₂,
∴∠2 + ∠ACB = ∠1 + ∠CAE,即∠2 + 36° = 20° + 72°,解得∠2 = 56°. 故答案为56°.
解析如题图,连接AC.
∵五边形ABCDE是正五边形,
∴∠B = ∠BAE = 108°,
∴∠ACB = ∠CAB = 36°,
∴∠CAE = 108° - 36° = 72°.
∵l₁//l₂,
∴∠2 + ∠ACB = ∠1 + ∠CAE,即∠2 + 36° = 20° + 72°,解得∠2 = 56°. 故答案为56°.
12. 已知关于$ x $的一元一次不等式组$\begin{cases} \\ 3x - 1 ＞ 2m \end{cases}$的整数解有且仅有4个，则$ m $的取值范围是________。
答案
本题考查解一元一次不等式(组).
解析解3 - 2x ≥ -1,得x ≤ 2,解3x - 1 ＞ 2m,得x ＞ $\frac{2m + 1}{3}$.
∵不等式组的整数解有且仅有4个,
∴这4个整数解为2,1,0,-1,
∴-2 ≤ $\frac{2m + 1}{3}$ ＜ -1,解得$-\frac{7}{2} ≤ m ＜ -2$. 故答案为$-\frac{7}{2} ≤ m ＜ -2$.
解析解3 - 2x ≥ -1,得x ≤ 2,解3x - 1 ＞ 2m,得x ＞ $\frac{2m + 1}{3}$.
∵不等式组的整数解有且仅有4个,
∴这4个整数解为2,1,0,-1,
∴-2 ≤ $\frac{2m + 1}{3}$ ＜ -1,解得$-\frac{7}{2} ≤ m ＜ -2$. 故答案为$-\frac{7}{2} ≤ m ＜ -2$.
13. 如图, $∠A+∠B+∠C+∠D+∠E+∠F+∠G=$
180°
.答案
本题考查三角形的内角和定理与外角性质.
解析如图,∠1 = ∠EOF + ∠GOF = ∠B + ∠F + ∠C + ∠G,∠2 = ∠A + ∠D.
∵∠1 + ∠2 + ∠E = 180°,
∴∠A + ∠B + ∠C + ∠D + ∠E + ∠F + ∠G = 180°. 故答案为180°.
14. 如图，BD，CE为$△ ABC$的两条角平分线，则$∠ 1$，$∠ 2$，$∠ A$之间的数量关系为________。
答案
本题考查三角形的内角和定理与外角性质.
解析
∵BD,CE为△ABC的两条角平分线,
∴∠ABD = $\frac{1}{2}$∠ABC,∠ACE = $\frac{1}{2}$∠ACB.
∵∠1 = ∠ACE + ∠A,∠2 = ∠ABD + ∠A,
∴∠1 + ∠2 = ∠ACE + ∠A + ∠ABD + ∠A = $\frac{1}{2}$∠ACB + ∠A + $\frac{1}{2}$∠ABC + ∠A = $\frac{1}{2}$∠ACB + $\frac{1}{2}$∠ABC + $\frac{1}{2}$∠A + $\frac{3}{2}$∠A = $\frac{1}{2}$(∠ACB + ∠ABC + ∠A) + $\frac{3}{2}$∠A = 90° + $\frac{3}{2}$∠A,即∠1 + ∠2 - $\frac{3}{2}$∠A = 90°. 故答案为∠1 + ∠2 - $\frac{3}{2}$∠A = 90°.
解析
∵BD,CE为△ABC的两条角平分线,
∴∠ABD = $\frac{1}{2}$∠ABC,∠ACE = $\frac{1}{2}$∠ACB.
∵∠1 = ∠ACE + ∠A,∠2 = ∠ABD + ∠A,
∴∠1 + ∠2 = ∠ACE + ∠A + ∠ABD + ∠A = $\frac{1}{2}$∠ACB + ∠A + $\frac{1}{2}$∠ABC + ∠A = $\frac{1}{2}$∠ACB + $\frac{1}{2}$∠ABC + $\frac{1}{2}$∠A + $\frac{3}{2}$∠A = $\frac{1}{2}$(∠ACB + ∠ABC + ∠A) + $\frac{3}{2}$∠A = 90° + $\frac{3}{2}$∠A,即∠1 + ∠2 - $\frac{3}{2}$∠A = 90°. 故答案为∠1 + ∠2 - $\frac{3}{2}$∠A = 90°.
15. 如图,在$△ ABC$中,点$D,E$分别在$AB,BC$上,$BD=3AD$,$EC=2BE$,$AE,CD$相交于点$O$.若$△ ABC$的面积为$S$,则四边形$BEOD$的面积是$\underline{\hspace{5em}}$.(用含$S$的代数式表示)
答案
本题考查等高倍底模型,三角形的面积.
解析如题图,连接OB.
∵BD = 3AD,
∴S△BOD = 3S△AOD,S△BCD = 3S△ACD.
∵EC = 2BE,
∴S△OCE = 2S△OBE,S△ACE = 2S△ABE. 设S△AOD = a,S△OBE = b,S△AOC = c,则S△BOD = 3a,S△OCE = 2b.
∵S△BCD = S△BOD + S△OBE + S△OCE,S△ACD = S△AOD + S△AOC,
∴S△BOD + S△OBE + S△OCE = 3(S△AOD + S△AOC),即3a + b + 2b = 3(a + c),
∴b = c.
∵S△ACE = S△AOC + S△OCE,S△ABE = S△AOD + S△BOD + S△OBE,
∴S△AOC + S△OCE = 2(S△AOD + S△BOD + S△OBE),即c + 2b = 2(a + 3a + b),
∴c = 8a.
∵S△ABC = S△AOD + S△BOD + S△OBE + S△OCE + S△AOC,
∴a + 3a + b + 2b + c = S,
∴4a + 3b + c = 4a + 3·8a + 8a = 36a = S,
∴a = $\frac{1}{36}$S,
∴S四边形BEOD = S△BOD + S△OBE = 3a + b = 3a + 8a = 11a = $\frac{11}{36}$S. 故答案为$\frac{11}{36}$S.
解析如题图,连接OB.
∵BD = 3AD,
∴S△BOD = 3S△AOD,S△BCD = 3S△ACD.
∵EC = 2BE,
∴S△OCE = 2S△OBE,S△ACE = 2S△ABE. 设S△AOD = a,S△OBE = b,S△AOC = c,则S△BOD = 3a,S△OCE = 2b.
∵S△BCD = S△BOD + S△OBE + S△OCE,S△ACD = S△AOD + S△AOC,
∴S△BOD + S△OBE + S△OCE = 3(S△AOD + S△AOC),即3a + b + 2b = 3(a + c),
∴b = c.
∵S△ACE = S△AOC + S△OCE,S△ABE = S△AOD + S△BOD + S△OBE,
∴S△AOC + S△OCE = 2(S△AOD + S△BOD + S△OBE),即c + 2b = 2(a + 3a + b),
∴c = 8a.
∵S△ABC = S△AOD + S△BOD + S△OBE + S△OCE + S△AOC,
∴a + 3a + b + 2b + c = S,
∴4a + 3b + c = 4a + 3·8a + 8a = 36a = S,
∴a = $\frac{1}{36}$S,
∴S四边形BEOD = S△BOD + S△OBE = 3a + b = 3a + 8a = 11a = $\frac{11}{36}$S. 故答案为$\frac{11}{36}$S.
16. 如图,已知线段AB与直线BC的夹角∠ABC=75°,D是直线BC上的一个动点,平移线段AB,使点B移到点D的位置,得到线段DE,连接BE,再将△BDE沿BE折叠,点D落在点F处:若BF平分∠ABE,则∠BED=
50或70
°.答案
本题考查图形的平行与折叠,角平分线的性质,平行线的性质,分类讨论是解题的关键.
解析如图1,点D在点B的右侧.
∵BF平分∠ABE,
∴∠ABF = ∠EBF.
∵将△BDE沿BE折叠,点D落在点F处,
∴∠EBF = ∠EBD.
∵∠ABC = 75°,
∴∠ABF = ∠EBF = ∠EBD = $\frac{1}{3}$∠ABC = $\frac{1}{3}×75° = 25°$.
∵平移线段AB得到线段DE,
∴DE//AB,
∴∠BED = ∠ABE = 25° + 25° = 50°;
如图2,点D在点B的左侧.
∵BF平分∠ABE,
∴∠ABF = ∠EBF.
∵将△BDE沿BE折叠,点D落在点F处,
∴∠EBF = ∠EBD.
∵∠ABC = 75°,
∴∠ABD = 180° - ∠ABC = 180° - 75° = 105°,
∴∠ABF = ∠EBF = ∠EBD = $\frac{1}{3}$∠ABD = $\frac{1}{3}×105° = 35°$.
∵DE//AB,
∴∠BED = ∠ABE = 35° + 35° = 70°. 故答案为50或70.
三、解答题(本大题共7小题,共68分.解答应写出过程)
17. (12分)计算.
(1)$(-\dfrac{1}{4})^{-1} + (-2)^2 × 5^0 - (\dfrac{1}{2})^{-2}$;
(2)$(-2x^2)^3 - (x^3)^2 + x^8 ÷ x^2$;
(3)$(a + 2)^2 - (a + 1)(a - 3)$;
(4)$(x + y + 2z)(x + y - 2z)$.
17. (12分)计算.
(1)$(-\dfrac{1}{4})^{-1} + (-2)^2 × 5^0 - (\dfrac{1}{2})^{-2}$;
(2)$(-2x^2)^3 - (x^3)^2 + x^8 ÷ x^2$;
(3)$(a + 2)^2 - (a + 1)(a - 3)$;
(4)$(x + y + 2z)(x + y - 2z)$.
答案
本题考查幂的运算,整式乘法.
解析(1) $(-\frac{1}{4})^{-1} + (-2)^2 × 5^0 - (\frac{1}{2})^{-2}$
$= -4 + 4 × 1 - 4$
$= -4.$
(2) $(-2x^2)^3 - (x^3)^2 + x^8 ÷ x^2$
$= -8x^6 - x^6 + x^6$
$= -8x^6.$
(3) $(a + 2)^2 - (a + 1)(a - 3)$
$= a^2 + 4a + 4 - (a^2 - 2a - 3)$
$= a^2 + 4a + 4 - a^2 + 2a + 3$
$=6a + 7.$
(4) $(x + y + 2z)(x + y - 2z)$
$= (x + y)^2 - (2z)^2$
$= x^2 + 2xy + y^2 - 4z^2.$
解析(1) $(-\frac{1}{4})^{-1} + (-2)^2 × 5^0 - (\frac{1}{2})^{-2}$
$= -4 + 4 × 1 - 4$
$= -4.$
(2) $(-2x^2)^3 - (x^3)^2 + x^8 ÷ x^2$
$= -8x^6 - x^6 + x^6$
$= -8x^6.$
(3) $(a + 2)^2 - (a + 1)(a - 3)$
$= a^2 + 4a + 4 - (a^2 - 2a - 3)$
$= a^2 + 4a + 4 - a^2 + 2a + 3$
$=6a + 7.$
(4) $(x + y + 2z)(x + y - 2z)$
$= (x + y)^2 - (2z)^2$
$= x^2 + 2xy + y^2 - 4z^2.$
