《2025年阳光同学分层设计九年级数学全一册人教版福建专版》

5. 如图,在$\triangle ABC$中,$\angle BAC = 90^{\circ}$,$AB = AC$,点D,E均在边BC上,且$\angle BAD + \angle EAC = 45^{\circ}$,若$BD = 3$,$EC = 6$,求DE的长.

$3\sqrt{5}$

答案： $3\sqrt{5}$

6. 如图,在等腰直角三角形ABC中,$\angle ACB = 90^{\circ}$,P为$\triangle ABC$内一点,且$PA = 5$,$PC = 2\sqrt{2}$,$\angle BPC = 135^{\circ}$,求PB的长.

3

答案： 3

7. 如图,P为正方形ABCD内一点,$AP = 2\sqrt{2}$,$PB = 1$,$PC = \sqrt{10}$,求四边形ABCP的面积.$\frac{5}{2}$

答案： $\frac{5}{2}$

8. 如图,在四边形ABCD中,$\angle BCD = 45^{\circ}$,连接对角线AC,BD,$\angle ADB = 90^{\circ}$,$AD = BD$,若$CB = 2$,$CD = 4$,求CA的长.

答案：
解:如图,将$\triangle DAC$绕点D逆时针旋转$90^{\circ}$得到$\triangle DBF$,连接CF.
　　　　　
根据旋转的性质可知，$\triangle DAC\cong\triangle DBF$.
$\therefore AC=BF$, $DC=DF=4$, $\angle ADC=\angle BDF$.
$\therefore\angle ADB+\angle BDC=\angle BDC +\angle CDF$.
$\therefore\angle CDF=\angle ADB=90^{\circ}$.
$\therefore CF=\sqrt{CD^{2}+DF^{2}}=4\sqrt{2}$, $\angle DCF=\angle DFC=\frac{1}{2}\times90^{\circ}=45^{\circ}$.
$\because\angle BCD=45^{\circ}$,
$\therefore\angle BCF=\angle BCD+\angle DCF=90^{\circ}$.
$\therefore CA=BF=\sqrt{BC^{2}+CF^{2}}=\sqrt{2^{2}+(4\sqrt{2})^{2}}=6$.