8. 已知在$△ABC$中,$AB= 17,AC= 10$,BC边上的高$AD= 8$,则边BC的长为()
A.9
B.21
C.6或15
D.9或21

D

9. 在平行四边形ABCD中,若$∠A= 30^{\circ }$,则$∠B= $.

$150^{\circ}$

10. 把直线$y= -2x-1$沿x轴向右平移3个单位长度,所得直线的函数解析式为.

$y = - 2x + 5$

11. 若已知a,b为实数,且$\sqrt {a-5}+\sqrt {10-2a}= b+4$,则$a+b=$.

1

12. 某航空公司规定,乘客所携带行李的重量x(kg)与运费y(元)满足如图所示的函数图象,那么每位乘客最多可免费携带kg的行李.

20

13. 如图,在四边形ABCD中,$AD// BC,AD= 12cm,BC= 8cm$,P,Q分别从A,C同时出发,P以1 cm/s的速度由A向D运动,Q以2 cm/s的速度由C出发向B运动,s后四边形ABQP是平行四边形.

$\frac{8}{3}$

14. 计算:$4\sqrt {2}(\sqrt {\frac {1}{8}}-\sqrt {6})-\sqrt {48}÷\sqrt {3}+(\sqrt {3}+1)^{2}.$

解：原式$ = 4 \sqrt { 2 × \frac { 1 } { 8 } } - 4 \sqrt { 2 × 6 } - \sqrt { 48 ÷ 3 } + 3 + 2 \sqrt { 3 } + 1 = 2 - 8 \sqrt { 3 } - 4 + 4 + 2 \sqrt { 3 } = 2 - 6 \sqrt { 3 }$.

15. 如图,在平行四边形ABCD中,$AE⊥BC$于点E,$AF⊥CD$于点F,$∠EAF= 60^{\circ },BE= 2,DF= 3$,求AB=,BC=及平行四边形ABCD的面积=.

解：$\because AE \perp BC$于点$E$，$AF \perp CD$于点$F$，
$\therefore \angle AEC = \angle AFC = 90^{\circ}$.
$\because \angle EAF = 60^{\circ}$，
$\therefore \angle C = 360^{\circ} - \angle AEC - \angle AFC - \angle EAF = 120^{\circ}$，
$\therefore \angle B = 60^{\circ}$，$\therefore \angle BAE = 30^{\circ}$.
$\therefore AB = 2BE = 4$.
$\because \angle D = \angle B = 60^{\circ}$，$\therefore \angle DAF = 30^{\circ}$.
$\therefore AD = 2DF = 6$.
$\therefore BC = AD = 6$.
在$\mathrm{Rt}\triangle ADF$中，$AF = \sqrt { 36 - 9 } = 3 \sqrt { 3 }$.
$\therefore$平行四边形$ABCD$的面积$ = CD\cdot AF = 4 × 3 \sqrt { 3 } = 12 \sqrt { 3 }$.