《2025年大联考单元期末测试卷九年级数学全一册沪科版》

15. 若 $\frac{a + 2}{3} = \frac{b}{4} = \frac{c + 5}{6}$,且 $2a - b + 3c = 21$,试求 $a:b:c$.

答案： 15.解:设$\frac{a + 2}{3} = \frac{b}{4} = \frac{c + 5}{6} = k$，
则$a = 3k - 2$，$b = 4k$，$c = 6k - 5$，(3分)
$\therefore 2(3k - 2) - 4k + 3(6k - 5) = 21$，
解得$k = 2$，(6分)
$\therefore a = 6 - 2 = 4$，$b = 8$，$c = 7$，
$\therefore a:b:c = 4:8:7$.(8分)

16. 如图,已知△ABC 的边 BC 在直线 MN 上,将△ABC 绕点 A 按逆时针方向旋转,使点 C 落在直线 MN 上的点 C'处,得到△AB'C'.
(1) 请用尺规作图作出△AB'C',并标明字母(保留作图痕迹,不写作法).
(2) 若∠ACB = 118°,则∠BC'B' =56°.

答案：
16.解:
(1)如图所示，
△AB'C'即为所求.
(3分)

(2)$\because \angle ACB = 118^{\circ}$，
$\therefore \angle ACC' = 62^{\circ}$，
(4分)
由旋转可得$\triangle ABC \cong \triangle AB'C'$，
$\therefore \angle AC'B' = \angle ACB = 118^{\circ}$，$AC = AC'$，(6分)
$\therefore \angle AC'C = \angle ACC' = 62^{\circ}$，
$\therefore \angle BC'B′ = \angle AC'B' - \angle AC'C = 118^{\circ} - 62^{\circ} = 56^{\circ}$.(8分)

17. 如图,四边形 ABCD 是⊙O 的内接矩形,点 E,F 分别在射线 AB,AD 上,OE = OF,且点 C,E,F 在一条直线上,EF 与⊙O 相切于点 C.
(1) 求证:矩形 ABCD 是正方形.
(2) 若 OF = 10,求正方形 ABCD 的面积.

答案：
17.解:
(1)证明:如图，连接AC.
$\because$四边形ABCD是$\odot O$的内接矩形，
$\therefore AC$是$\odot O$的直径.

$\because EF$与$\odot O$相切于点$C$，$\therefore AC \perp EF$.
$\because OE = OF$，$\therefore CF = CE$.
$\because$四边形ABCD是矩形，$\therefore \angle FAE = 90^{\circ}$，
$\therefore AC = \frac{1}{2}EF = CF = CE$，$\therefore \angle CAE = 45^{\circ}$.
$\because \angle ABC = 90^{\circ}$，$\therefore \angle ACB = 45^{\circ}$，$\therefore AB = CB$，
$\therefore$矩形ABCD是正方形.(4分)
(2)$\because OC = \frac{1}{2}AC$，$AC = CF$，$\therefore CF = 2OC$.
$\because OF = 10$，$OF^{2} = OC^{2} + CF^{2}$，$\therefore 10^{2} = OC^{2} + 4OC^{2}$，$\therefore OC = 2\sqrt{5}$，
$\therefore AB = \sqrt{2}OC = 2\sqrt{10}$，$\therefore AB^{2} = 40$，
$\therefore$正方形ABCD的面积是40.(8分)