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電子課本網(wǎng) 第13頁

第13頁

信息發(fā)布者:
?$6$?
?$30$?
?$21$?或?$11$?
解:∵?$∠A C B=90°, ∠A=30°, $?∴?$B C=\frac {1}{2}\ \mathrm {A} B=2 $?
∴?$A C=\sqrt {A B^2-B C^2}=2 \sqrt {3} $?
∴?$S_{\triangle A B C}=\frac {1}{2}\ \mathrm {A} C· B C=\frac {1}{2} ×2 ×2 \sqrt {3}=2 \sqrt {3} $?
?$\text { 又 } S_{\triangle A B C}=\frac{1}{2} A B· C D, $?∴?$C D=\frac {2 \sqrt {3}}{2}=\sqrt {3} .$?

解:∵?$A B \perp A C, ∠B=45° \text {, } $?
∴?$∠A C B=45° \text {, } $?
∵?$A D //B C \text {, } $?
∴?$∠D A C=∠A C B=45° \text {, } $?
∵?$A D \perp D C \text {, } $?
∴?$∠A C D=90°-45°=45° \text {, } $?
∴?$∠A C D=∠C A D \text {, } $?
∴?$A D=C D=2 \mathrm{cm} \text {, } $?
?$\text { 在Rt } \triangle A D C \text { 中, } A C=\sqrt{2^{2}+2^{2}}=2 \sqrt{2}(\mathrm{cm}), $?
∵?$∠A C B=∠B \text {, } $?
∴?$A C=A B=2 \sqrt{2} \mathrm{cm} \text {, } $?
?$\text { 在Rt } \triangle A B C \text { 中, } B C=\sqrt{(2 \sqrt{2})^{2}+(2 \sqrt{2})^{2}}=4(\mathrm{cm}) . $?


解:∵?$A D \perp B C $?
∴?$∠A D B=∠A D C=90° $?
∵?$A D=6, S_{\triangle A B C}=\frac {1}{2}\ \mathrm {B} C ·A D=42 $?
∴?$B C=\frac {42 ×2}{A D}=\frac {42 ×2}{6}=14 $?
∵?$∠B=45° $?
∴?$\triangle A B D \text { 是等腰直角三角形 } $?
∴?$B D=A D=6 $?
∴?$C D=B C-B D=14-6=8 $?
∴?$A C=\sqrt {A D^2+C D^2}=\sqrt {6^2+8^2} =10$?