Trắc nghiệm Ôn tập Toán 10 Chương 2 Hình học có đáp án (Thông hiểu)

A. ${15}^{\circ }$

B. ${60}^{\circ }$

C. ${120}^{\circ }$

D. ${135}^{\circ }$

A. $3\stackrel{\to }{GG\text{'}}=\stackrel{\to }{AA\text{'}}+\stackrel{\to }{BB\text{'}}+\stackrel{\to }{CC\text{'}}$

B. $3\stackrel{\to }{GG\text{'}}=\stackrel{\to }{AC\text{'}}+\stackrel{\to }{BA\text{'}}+\stackrel{\to }{CB\text{'}}$

C. $3\stackrel{\to }{GG\text{'}}=\stackrel{\to }{AB\text{'}}+\stackrel{\to }{BC\text{'}}+\stackrel{\to }{CA\text{'}}$

D. $3\stackrel{\to }{GG\text{'}}=\stackrel{\to }{A\text{'}A}+\stackrel{\to }{B\text{'}B}+\stackrel{\to }{C\text{'}C}$

A. Có ba góc nhọn

B. Có một góc vuông

C. Có một góc tù

D. Đều

A. $\stackrel{\to }{CI}=\stackrel{\to }{CA}-3\stackrel{\to }{CB}$

B. $\stackrel{\to }{CI}=\frac{1}{2}\left(3\stackrel{\to }{CB}-\stackrel{\to }{CA}\right)$

C. $\stackrel{\to }{CI}=\frac{1}{2}\left(\stackrel{\to }{CA}-3\stackrel{\to }{CB}\right)$

D. $\stackrel{\to }{CI}=3\stackrel{\to }{CB}-\stackrel{\to }{CA}$

A. $-\frac{3}{8}$

B. $-\frac{1}{6}$

C. $\frac{3}{2}$

D. $-\frac{1}{2}$

A. $\frac{9a^2}{2}$

B. $-\frac{9a^2}{2}$

C. 0

D. $9a^2$

A. $\stackrel{\to }{AB}=\left(4;2\right),\stackrel{\to }{BC}=\left(2;-4\right)$

B. $\stackrel{\to }{AB}\perp \stackrel{\to }{BC}$

C. Tam giác ABC vuông cân tại A

D. Tam giác ABC vuông cân tại B

A. 7

B. 5

C. -7

D. -5

A. $\sqrt{21}$

B. $\sqrt{61}$

C. 21

D. 61

A. $A=\frac{1}{2}\left(c^2-a^2-b^2\right)$

B. $A=\frac{1}{2}\left(2c^2-a^2-b^2\right)$

C. $A=\frac{1}{2}\left(3c^2-a^2-b^2\right)$

D. $A=\frac{1}{2}\left(3c^2+a^2-b^2\right)$

A. P = -1

B. $P=3a^2$

C. $P=-3a^2$

D. $P=2a^2$

A. $\stackrel{\to }{AB}=\left(-3;-2\right),\stackrel{\to }{AC}=\left(3;-\frac{9}{2}\right)$

B. $\stackrel{\to }{AB}.\stackrel{\to }{AC}=0$

C. $\left|\stackrel{\to }{AB}\right|=\sqrt{13}$

D. $\left|\stackrel{\to }{BC}\right|=\frac{\sqrt{13}}{2}$

A. Hai góc $\widehat{BAD},\widehat{BCD}$ phụ nhau

B. Góc $\widehat{BCD}$ là góc nhọn

C. $\cos \left(\stackrel{\to }{AB},\stackrel{\to }{AD}\right)=\cos \left(\stackrel{\to }{CB},\stackrel{\to }{CD}\right)$

D. Hai góc $\widehat{BAD},\widehat{BCD}$ bù nhau

A. $\stackrel{\to }{CI}=\stackrel{\to }{CA}-3\stackrel{\to }{CB}$

B. $\stackrel{\to }{CI}=\frac{1}{2}\left(3\stackrel{\to }{CB}-\stackrel{\to }{CA}\right)$

C. $\stackrel{\to }{CI}=\frac{1}{2}\left(\stackrel{\to }{CA}-3\stackrel{\to }{CB}\right)$

D. $\stackrel{\to }{CI}=3\stackrel{\to }{CB}-\stackrel{\to }{CA}$

A. $-\frac{3}{8}$

B. $-\frac{1}{6}$

C. $\frac{3}{2}$

D. $-\frac{1}{2}$

A. $\frac{9a^2}{2}$

B. $-\frac{9a^2}{2}$

C. 0

D. $9a^2$

A. 7

B. 5

C. -7

D. -5

A. $\sqrt{21}$

B. $\sqrt{61}$

C. 21

D. 61

A. $A=\frac{1}{2}\left(c^2-a^2-b^2\right)$

B. $A=\frac{1}{2}\left(2c^2-a^2-b^2\right)$

C. $A=\frac{1}{2}\left(3c^2-a^2-b^2\right)$

D. $A=\frac{1}{2}\left(3c^2+a^2-b^2\right)$

A. P = -1

B. $P=3a^2$

C. $P=-3a^2$

D. $P=2a^2$