Trắc nghiệm Toán 9 Bài Ôn tập chương (có đáp án): Hệ thức lượng trong tam giác vuông (phần 2)

A . $A{H}^2=BH.CH$

B . $A{B}^2=BH.BC$

C . $\frac{1}{A{H}^2}=\frac{1}{A{B}^2}+\frac{1}{A{C}^2}$

D . AH. AB = BC. AC

A. sin B = $\frac{AH}{AB}$

B . cos C = $\frac{AC}{BC}$

C . tan B = $\frac{AC}{AB}$

D . tan C = $\frac{AH}{AC}$

A . ${\sin }^2\alpha +co{s}^2\alpha =1$

B . $\tan \alpha .cot\alpha =1$

C . $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$

D . Cả A, B, C đều đúng.

A . sin $\alpha$ > sin $\beta$ .

B . cos $\alpha$ < cos $\beta$ .

C . tan $\alpha$ < tan $\beta$ .

D . cot $\alpha$ < cot $\beta$ .

A . $2\sqrt{6}$

B . $\sqrt{6}$

C . $3\sqrt{6}$

D . 27

A . $\frac{1}{3}$

B . 3

C . $\sqrt{3}$

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

A . BH = 2cm, CH = 3,2cm, AC = 4cm, AH = 2,4cm

B . BH = 1,8cm; CH = 3,2cm; AC = 4cm; AH = 2,4cm

C . BH = 1,8cm; CH = 3,2cm; AC = 3cm; AH = 2,4cm

D . BH = 1,8cm; CH = 3,2cm; AC = 4cm; AH = 4,2cm

A . AC = 37,2cm; AB = 33,4cm; $\hat{C}={32}^0$

B . AC = 37,2cm; AB = 33,5cm; $\hat{C}={42}^0$

C . AB = 37,2cm; AC = 33,5cm; $\hat{C}={42}^0$

D . AC = 37,2cm; AB = 33,5cm; $\hat{C}={42}^0$

A . 21,3cm

B . 24cm

C . 22,3cm

D . 23,2cm

A . $\frac{\sqrt{93}}{14}$

B . $14\sqrt{93}$

C . $\frac{14\sqrt{93}}{93}$

D . $\frac{14}{17}$

A . 1

B . 2

C . 4

D . −1

A . 25cm

B . 25,7cm

C . 26cm

D . 12,9cm

A . x = 0,75

B . x = 4,5

C . x = $4\sqrt{3}$

D . x = 4

A . M = 5

B . M = $-\frac{5}{4}$

C . M = −5

D . M = $\frac{1}{5}$

A . x = 30; y = 28

B . x = $2\sqrt{481}$ ; y = $\frac{225}{8}$

C . x = 18; y = 40

D . x = 40; y = 18

A . ${45}^0$

B . ${30}^0$

C . ${60}^0$

D . ${90}^0$

A . $\frac{15\sqrt{74}}{7}$

B . $3\sqrt{74}$ cm

C . 22cm

D . 21cm

A . $\frac{48}{35}$ cm

B . 7,2cm

C . $\frac{60}{7}$ cm

D . $\frac{48}{25}$ cm

A . 25

B . 16

C. 9

D . 25 + 48sin $\alpha$ .cos $\alpha$

A . $\frac{89}{891}$

B . $\frac{89}{159}$

C . $\frac{89}{459}$

D . $-\frac{89}{459}$

A . $\tan {28}^0$ < $\tan {33}^0$ < $cot{40}^0$ < $cot{55}^0$ < $cot{70}^0$

B . $\tan {28}^0$ < $cot{70}^0$ < $\tan {33}^0$ < $cot{55}^0$ < $cot{40}^0$

C . $cot{70}^0$ < $\tan {28}^0$ < $\tan {33}^0$ < $cot{55}^0$ < $cot{40}^0$

D . $cot{70}^0$ > $\tan {28}^0$ > $\tan {33}^0$ > $cot{55}^0$ > $cot{40}^0$

A . 0

B . 1

C . −1

D . 2

A . $\frac{a^2}{2\sin \alpha .\cos \alpha }$

B . $\frac{a^2}{\sin \alpha .\cos \alpha }$

C . $\frac{a}{2\sin \alpha .\cos \alpha }$

D . $\frac{2a^2}{\sin \alpha .\cos \alpha }$

A . $12\sqrt{2}$

B . $12\sqrt{3}$

C . 12

D . $12\sqrt{6}$

A . $BC=3\sqrt{3}cm$ , $\hat{A}={120}^0$ , $\hat{D}={60}^0$

B . $BC=4\sqrt{3}cm$ , $\hat{A}={120}^0$ , $\hat{D}={60}^0$

C . $BC=3\sqrt{3}cm$ , $\hat{A}={135}^0$ , $\hat{D}={45}^0$

D . $BC=4\sqrt{3}cm$ , $\hat{A}={135}^0$ , $\hat{D}={45}^0$

A . BE = $10\sqrt{2}$ cm

B . BE = 10cm

C . BE = $10\sqrt{3}$ cm

D . BE = 20cm

A . 6,52m

B . 6,06m

C . 5,86m

D . 5,38m

A . 80 $c{m}^2$

B . 90 $c{m}^2$

C . 100 $c{m}^2$

D . 120 $c{m}^2$

A . BC = 25; $\hat{B}={36}^{0}52\text{'}$ ; $\hat{C}={53}^{0}8\text{'}$

B. BC = 25; $\hat{B}={53}^{0}8\text{'}$ ; $\hat{C}={53}^{0}8\text{'}$

C . BC = 25; $\hat{B}={36}^{0}52\text{'}$ ; $\hat{C}={53}^{0}8\text{'}$

D . BC = 25; $\hat{B}={36}^{0}52\text{'}$ ; $\hat{C}={53}^{0}8\text{'}$

A . $\hat{B}={53}^{0}8\text{'}$ ; $\hat{C}={36}^{0}52\text{'}$ ; AH = 3,6cm

B . $\hat{B}={36}^{0}52\text{'}$ ; $\hat{C}={53}^{0}8\text{'}$ ; AH = 3,6cm

C. $\hat{B}={48}^{0}35\text{'}$ ; $\hat{C}={41}^{0}25\text{'}$ ; AH = 3,6cm

D . $\hat{B}={41}^{0}25\text{'}$ ; $\hat{C}={48}^{0}35\text{'}$ ; AH = 3,6cm

A . AC = 8,57cm; BC = 10,46cm; $\hat{C}={55}^0$ .

B . AC = 4,9cm; BC = 7,75cm; $\hat{C}={55}^0$ .

C . AC = 4,2cm; BC = 7,32cm; $\hat{C}={55}^0$ .

D . AC = 3,44cm; BC = 6,92cm; $\hat{C}={55}^0$ .

A . AB = 5cm, AC = $\frac{15}{4}$ cm; AM = $\frac{25}{8}$ cm; $S_{∆ABC}=\frac{75}{8}c{m}^2$ .

B . AB = 5cm, AC = 3cm; AM = 4cm; $S_{∆ABC}=\frac{39}{4}c{m}^2$ .

C . AB = $\frac{14}{3}$ cm, AC = $\frac{14}{4}$ cm; AM = 3cm; $S_{∆ABC}=\frac{75}{8}c{m}^2$ .

D . AB = $\frac{14}{3}$ cm, AC = 3 cm; AM = $\frac{27}{8}$ cm; $S_{∆ABC}=9c{m}^2$

A . $\hat{B}={45}^0$ ; $\hat{C}={45}^0$ ; AH = $\sqrt{3}$

B . $\hat{B}={50}^0$ ; $\hat{C}={40}^0$ ; AH = 2

C . $\hat{B}={30}^0$ ; $\hat{C}={60}^0$ ; AH = 4

D . $\hat{B}={60}^0$ ; $\hat{C}={30}^0$ ; AH = 2

A . MN = 12cm; MP = 19,5cm, NP = $\frac{3\sqrt{13}}{2}$ cm

B . MN = 13cm; MP = 19,5cm, NP = $\frac{3\sqrt{13}}{2}$ cm

C . MN = 13cm; MP = 17,5cm, NP = $\frac{3\sqrt{13}}{2}$ cm

D . MN = 13cm; MP = 19,5cm, NP = $\frac{5\sqrt{13}}{2}$ cm