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5. 如图，在$△ ABC$中，$AB = AC = 10$，$BD⊥ AC$于$D$，$CD = 2$，则$BD$的长为(
A.$4$
B.$5$
C.$6$
D.$8$
C
)A.$4$
B.$5$
C.$6$
D.$8$
答案
5. C
6. 如图，每个小方格都是边长为$1$的正方形，试计算五边形$ABCDE$的周长和面积。
答案
6. 解:周长为$ 5 + 2\sqrt{2} + \sqrt{29} + \sqrt{13} + \sqrt{17} $;面积为24.
7. 如图，在$Rt△ ABC$中，$∠ A = 90^{\circ}$，$BD$平分$∠ ABC$，交$AC$与点$D$，且$AB = 4$，$BD = 5$，求点$D$到$BC$的距离。
答案
7. 解:过点$ D $作$ DE ⊥ BC $,垂足为$ E $.
在$ \mathrm{Rt}△ ABC $中,$ ∠ A = 90^{\circ} $,
$ \therefore AD ⊥ AB $,$ DE ⊥ BC $.
$ \because BD $平分$ ∠ ABC $,
$ \therefore AD = DE $.
在$ \mathrm{Rt}△ ABD $中,$ ∠ A = 90^{\circ} $,
$ AD = \sqrt{BD^{2} - AB^{2}} = \sqrt{5^{2} - 4^{2}} = \sqrt{9} = 3 $,
$ \therefore DE = AD = 3 $.
在$ \mathrm{Rt}△ ABC $中,$ ∠ A = 90^{\circ} $,
$ \therefore AD ⊥ AB $,$ DE ⊥ BC $.
$ \because BD $平分$ ∠ ABC $,
$ \therefore AD = DE $.
在$ \mathrm{Rt}△ ABD $中,$ ∠ A = 90^{\circ} $,
$ AD = \sqrt{BD^{2} - AB^{2}} = \sqrt{5^{2} - 4^{2}} = \sqrt{9} = 3 $,
$ \therefore DE = AD = 3 $.
1. 已知直角$△ ABC$的两直角边分别为$6$，$8$，分别以其三边为直径作半圆，求图中阴影部分的面积。
答案
1. 解:$ \because $直角$ △ ABC $的两直角边分别为6,8,
$ \therefore AB = \sqrt{6^{2} + 8^{2}} = 10 $.
$ \therefore $以$ BC $为直径的半圆的面积为$ \dfrac{1}{2}π · (\dfrac{8}{2})^{2} = 8π $.
以$ AC $为直径的圆的半圆的面积为$ \dfrac{1}{2}π · (\dfrac{6}{2})^{2} = \dfrac{9}{2}π $,
以$ AB $为直径的半圆的面积为$ \dfrac{1}{2}π · (\dfrac{10}{2})^{2} = \dfrac{25}{2}π $,
$ △ ABC $的面积为$ \dfrac{1}{2}AC · BC = 24 $
$ \therefore $阴影部分的面积为
$ S_{\mathrm{阴影}} = 8π + \dfrac{9}{2}π + 24 - \dfrac{25}{2}π = 24 $
$ \therefore AB = \sqrt{6^{2} + 8^{2}} = 10 $.
$ \therefore $以$ BC $为直径的半圆的面积为$ \dfrac{1}{2}π · (\dfrac{8}{2})^{2} = 8π $.
以$ AC $为直径的圆的半圆的面积为$ \dfrac{1}{2}π · (\dfrac{6}{2})^{2} = \dfrac{9}{2}π $,
以$ AB $为直径的半圆的面积为$ \dfrac{1}{2}π · (\dfrac{10}{2})^{2} = \dfrac{25}{2}π $,
$ △ ABC $的面积为$ \dfrac{1}{2}AC · BC = 24 $
$ \therefore $阴影部分的面积为
$ S_{\mathrm{阴影}} = 8π + \dfrac{9}{2}π + 24 - \dfrac{25}{2}π = 24 $
2. 如图，$Rt△ ABC$中，$∠ C = 90^{\circ}$，$∠ A = 30^{\circ}$，$BD$是$∠ ABC$的平分线，$AD = 20$，求$BC$的长。
答案
2. 解:在$ \mathrm{Rt}△ ABC $中,$ ∠ C = 90^{\circ} $,$ ∠ A = 30^{\circ} $,
$ \therefore ∠ B = 180^{\circ} - ∠ C - ∠ A = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ} $,
$ \because BD $平分$ ∠ ABC $,
$ \therefore ∠ ABD = ∠ CBD = \dfrac{1}{2}∠ ABC = \dfrac{1}{2} × 60^{\circ} = 30^{\circ} $,
$ \therefore ∠ A = ∠ ABD $,
$ \therefore BD = AD = 20 $.
在$ \mathrm{Rt}△ BCD $中,$ ∠ C = 90^{\circ} $,$ ∠ DBC = 30^{\circ} $,
$ \therefore CD = \dfrac{1}{2}BD = \dfrac{1}{2} × 20 = 10 $.
由勾股定理可得:
$ BC = \sqrt{BD^{2} - CD^{2}} = \sqrt{20^{2} - 10^{2}} = \sqrt{300} = 10\sqrt{3} $.
$ \therefore ∠ B = 180^{\circ} - ∠ C - ∠ A = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ} $,
$ \because BD $平分$ ∠ ABC $,
$ \therefore ∠ ABD = ∠ CBD = \dfrac{1}{2}∠ ABC = \dfrac{1}{2} × 60^{\circ} = 30^{\circ} $,
$ \therefore ∠ A = ∠ ABD $,
$ \therefore BD = AD = 20 $.
在$ \mathrm{Rt}△ BCD $中,$ ∠ C = 90^{\circ} $,$ ∠ DBC = 30^{\circ} $,
$ \therefore CD = \dfrac{1}{2}BD = \dfrac{1}{2} × 20 = 10 $.
由勾股定理可得:
$ BC = \sqrt{BD^{2} - CD^{2}} = \sqrt{20^{2} - 10^{2}} = \sqrt{300} = 10\sqrt{3} $.
3. 如图，在$△ ABC$中，$∠ C = 60^{\circ}$，$AB = 4\sqrt{3}$，$AC = 4$，$AD$是$BC$边上的高，求$BC$的长。
答案
3. 解:$ \because AD $是$ BC $边上的高,
$ \therefore ∠ ADC = ∠ ADB = 90^{\circ} $.
在$ \mathrm{Rt}△ ADC $中,
$ \because ∠ D = 90^{\circ} $,$ ∠ C = 60^{\circ} $,
$ \therefore ∠ CAD = 30^{\circ} $.
$ \because AC = 4 $,
$ \therefore CD = \dfrac{1}{2}AC = \dfrac{1}{2} × 4 = 2 $.
由勾股定理可得:
$ AD = \sqrt{AC^{2} - CD^{2}} = \sqrt{4^{2} - 2^{2}} = \sqrt{12} = 2\sqrt{3} $.
在$ \mathrm{Rt}△ ABD $中,$ ∠ ADB = 90^{\circ} $,$ AB = 4\sqrt{3} $,
由勾股定理可得:
$ BD = \sqrt{AB^{2} - AD^{2}} = \sqrt{48 - 12} = \sqrt{36} = 6 $,
$ \therefore BC = CD + BD = 2 + 6 = 8 $.
$ \therefore ∠ ADC = ∠ ADB = 90^{\circ} $.
在$ \mathrm{Rt}△ ADC $中,
$ \because ∠ D = 90^{\circ} $,$ ∠ C = 60^{\circ} $,
$ \therefore ∠ CAD = 30^{\circ} $.
$ \because AC = 4 $,
$ \therefore CD = \dfrac{1}{2}AC = \dfrac{1}{2} × 4 = 2 $.
由勾股定理可得:
$ AD = \sqrt{AC^{2} - CD^{2}} = \sqrt{4^{2} - 2^{2}} = \sqrt{12} = 2\sqrt{3} $.
在$ \mathrm{Rt}△ ABD $中,$ ∠ ADB = 90^{\circ} $,$ AB = 4\sqrt{3} $,
由勾股定理可得:
$ BD = \sqrt{AB^{2} - AD^{2}} = \sqrt{48 - 12} = \sqrt{36} = 6 $,
$ \therefore BC = CD + BD = 2 + 6 = 8 $.
