几何 - 例题解析



答案 - D

解释

A line has no points.

答案 - B

解释

A line segment has two end points.

答案 - A

解释

A ray has one end point.

题4 - 大于180°但小于360°的角称为

A - 锐角

B - 钝角

C - 平角

D - 优角

答案 - B

解释

An angle which is greater than 180°  but less than 360° is called a reflex angle.

题5 - 62°的余角是。

A - 118°

B - 28°

C - 38°

D - 48°

答案 - B

解释

Complement of 62°= (90° – 62°) = 28°.

题6 - 60°的补角是

A - 30°

B - 40°

C - 120°

D - 300°

答案 - B

解释

Supplement of 60° = (180°-60°) =120°.

题7 - 72°40'的余角是

A - 107°20'

B - 27°20'

C - 17°20'

D - 12°40'

答案 - C

解释

Complement of 72° 40' = (90°-72° 40') =17° 20'.

题8 - 一个角是其补角的五分之一。该角的度数是

A - 15°

B - 30°

C - 75°

D - 150°

答案 - B

解释

x = 1/5 (180 – x )⇒ 5x = 180 – x ⇒ 6x = 180 ⇒ x = 30°.

题9 - 如果一个角是它自身的余角,则它的度数是

A - 30°

B - 45°

C - 60°

D - 90°

答案 - B

解释

x=(90-x) ⇒ 2x = 90 ⇒ x = 45° .

题10 - 图中所示射线构成多少个角?

q 10

A - 5

B - 6

C - 8

D - 10

答案 - D

解释

The angle are ∠AOB , ∠BOC,∠COD,∠DOE,∠AOC,∠AOD, ∠AOE,∠BOD,∠BOD,∠COE. 
Thus , 10 angle are  formed.

题11 - 一个角比它的余角大24°。该角的度数是

A - 57°

B - 47°

C - 53°

D - 66°

答案 - A

解释

x – (90-x ) = 24 ⇒ 2x = 114 ⇒ x = 57
∴ Required angle is 57°.

题12 - 一个角比它的补角小32°。该角的度数是

A - 37°

B - 74°

C - 48°

D - 66°

答案 - A

解释

(180 –X) – X = 32 ⇒  2x = 180 – 32 = 148 ⇒ x = 74. 
Required angle is 74°.

题13 - 两个互补角的比是3:2。较小的角的度数是

A - 108°

B - 81°

C - 72°

D - 66°

答案 - C

解释

Let the measures of the angle be (3x)° and (2x)°. Then, 
3x+2x=180 ⇒ 5x = 180 ⇒ x = 36.
Smaller angle = (2x)° = (2*36)° = 72°. 

题14 - 在给定图形中,AOB是一条直线,∠AOC = 68°,∠BOC = x°。x的值是

q 14

A - 120°

B - 22°

C - 112°

D - 132°

答案 - A

解释

Since ∠AOB is a straight angle , we have
X+ 68 = 180 ⇒ x= (180-68)°  = 120°

题15 - 在给定图形中,AOB是一条直线,∠AOC = (3x+20)°,∠BOC =(4x-36)°。x的值是

q 15

A - 32°

B - 22°

C - 26°

D - 24°

答案 - B

解释

Since ∠AOB is a straight angle , we have 
∠AOC + ∠ BOC  =180° 
⇒ 3x + 20 +4x – 36 = 180
⇒ 7x = 164 ⇒ x = 22.

题16 - 在给定图形中,AOB是一条直线,∠AOC = (3x-8)°,∠COD =50°,∠BOD =(x+10)°。x的值是

q 16

A - 32°

B - 42°

C - 36°

D - 52°

答案 - A

解释

Since ∠AOB is a straight angle , we have
∠AOC  + ∠ COB  + ∠ BOD  = 180° 
⇒ (3X – 8)° + 50° + (X+ 10)° = 180° 
⇒ 4X = 128 ⇒ X = 32. 
aptitude_geometry.htm
广告