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11. 如图,在$4×4$的正方形网格中,$\triangle MNP$绕某点旋转一定的角度,得到$\triangle M_{1}N_{1}P_{1}$,则其旋转中心是(
A. 点$E$
B. 点$F$
C. 点$G$
D. 点$H$
C
)A. 点$E$
B. 点$F$
C. 点$G$
D. 点$H$
答案
C
12. 如图,$ABCD和DCGH$是两块全等的正方形铁皮,绕某点旋转后使它们重合,则存在的旋转中心有(
A. 1 个
B. 2 个
C. 3 个
D. 4 个
C
)A. 1 个
B. 2 个
C. 3 个
D. 4 个
答案
C
13. 如图所示,将图形①以点$O$为旋转中心,每次顺时针旋转$90^{\circ}$,则第 2026 次旋转后的图形是
③
(在下列各图中选填正确图形的序号即可)答案
③
14. 如图,将$\triangle ABC绕点A按逆时针方向旋转后到达\triangle ADE$的位置,设$DE与AC$,$BC分别交于点O$,$F$.
(1)若$\triangle ABC$的周长为 24,$AD = 6$,$AE = 8$,求$BC$的长;
(2)若$∠BAC = 72^{\circ}$,$∠DAC = 32^{\circ}$,求$∠EFC$的大小.
(1)
(2)
(1)若$\triangle ABC$的周长为 24,$AD = 6$,$AE = 8$,求$BC$的长;
(2)若$∠BAC = 72^{\circ}$,$∠DAC = 32^{\circ}$,求$∠EFC$的大小.
(1)
10
(2)
40°
答案
解 (1)由旋转的性质,得 $AB = AD = 6$, $AC = AE = 8$, $\therefore AB + AC = 6 + 8 = 14$.
$\because \triangle ABC$ 的周长为 24,
$\therefore AB + AC + BC = 24$,
$\therefore BC = 24 - (AB + AC) = 10$.
(2)$\because \angle BAC = 72^{\circ}$, $\angle DAC = 32^{\circ}$,
$\therefore \angle BAD = \angle BAC - \angle DAC = 72^{\circ} - 32^{\circ} = 40^{\circ}$. 由旋转的性质可知, $\angle C = \angle E$, $\angle CAE = \angle BAD = 40^{\circ}$.
$\because \angle COF = 180^{\circ} - \angle EFC - \angle C$, $\angle AOE = 180^{\circ} - \angle CAE - \angle E$, $\angle COF = \angle AOE$, $\therefore \angle EFC = \angle CAE = 40^{\circ}$.
$\because \triangle ABC$ 的周长为 24,
$\therefore AB + AC + BC = 24$,
$\therefore BC = 24 - (AB + AC) = 10$.
(2)$\because \angle BAC = 72^{\circ}$, $\angle DAC = 32^{\circ}$,
$\therefore \angle BAD = \angle BAC - \angle DAC = 72^{\circ} - 32^{\circ} = 40^{\circ}$. 由旋转的性质可知, $\angle C = \angle E$, $\angle CAE = \angle BAD = 40^{\circ}$.
$\because \angle COF = 180^{\circ} - \angle EFC - \angle C$, $\angle AOE = 180^{\circ} - \angle CAE - \angle E$, $\angle COF = \angle AOE$, $\therefore \angle EFC = \angle CAE = 40^{\circ}$.
