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As shown in the figure, then angle CDB = ()
Choices:
A:54°
B:64°
C:27°
D:37° | C |
|
As shown in the figure, and point C is the midpoint of arc BD, passing point C to draw the perpendicular line EF of AD, then the length of CE is ()
Choices:
A:3
B:\frac{20}{3}
C:\frac{12}{5}
D:\frac{16}{5} | C |
|
As shown in the figure, then angle AOC = ()
Choices:
A:100°
B:110°
C:120°
D:130° | A |
|
As shown in the figure, the central angles of chords AB and CD are angle AOB, angle COD, and angle AOB is complementary to angle COD, then the length of chord AB is ()
Choices:
A:6
B:8
C:5√{2}
D:5√{3} | A |
|
As shown in the figure, Given that AB = 2 DE, then the degree of angle ABC is ()
Choices:
A:32°
B:24°
C:16°
D:48° | B |
|
then the degree of angle C is ()
Choices:
A:44°
B:22°
C:46°
D:36° | B |
|
As shown in the figure, then the degree of angle ACD is ()
Choices:
A:50°
B:30°
C:25°
D:20° | C |
|
so the width of the water surface AB is ()
Choices:
A:4m
B:5m
C:6m
D:8m | D |
|
As shown in the figure, then angle β = ()
Choices:
A:30°
B:40°
C:50°
D:65° | B |
|
the arc ACB is exactly a semicircle. the water surface width A′B′ in the bridge hole is ()
Choices:
A:√{15}m
B:2√{15}m
C:2√{17}m
D:no solution | B |
|
AD is the angle bisector of angle CAB, and crossing point C is the high line CE on the AD side in triangle ACD, then the degree of angle ECD is ()
Choices:
A:63°
B:45°
C:27°
D:18° | C |
|
As shown in the figure, AO is the height of the cone, then the length of AO is ()
Choices:
A:2.4
B:2.2
C:1.8
D:1.6 | A |
|
As shown in the figure, arc AB = arc BC = arc CD, then angle BDC = ()
Choices:
A:35°
B:45°
C:55°
D:70° | A |
|
As shown in the figure, and the area of the shaded part in the figure is ()
Choices:
A:πcm²
B:\frac{2}{3}πcm²
C:\frac{1}{2}cm²
D:\frac{2}{3}cm² | C |
|
then the bottom perimeter of the paper cap is ()
Choices:
A:2πcm
B:3πcm
C:4πcm
D:5πcm | C |
|
The area of paper required to make this paper cap is (the seams are ignored) ()
Choices:
A:60π
B:65π
C:78π
D:156π | B |
|
then the height of the paper is ()
Choices:
A:3cm
B:2√{2}cm
C:3√{2}cm
D:4√{2}cm | B |
|
As shown in the figure, then the radius of its bottom is ()
Choices:
A:3
B:4
C:5
D:6 | C |
|
As shown in the figure, then the area of this sector cardboard is ()
Choices:
A:240πcm^{2}
B:480πcm^{2}
C:1200πcm^{2}
D:2400πcm^{2} | A |
|
As shown in the figure, then the bottom area of the cone is ()
Choices:
A:3πcm^{2}
B:9πcm^{2}
C:16πcm^{2}
D:25πcm^{2} | B |
|
the handle AP always bisects the angle angle BAC formed by the two ribs in the same plane, and AE = AF, DE = DF, then the area of oil paper required to make such a paper umbrella is (don't remember the seam) ()
Choices:
A:48cm^{2}
B:70πcm^{2}
C:2400πcm^{2}
D:2500πcm^{2} | C |
|
As shown in the figure, then the height of the cone is ()
Choices:
A:6
B:8
C:3√{3}
D:4√{2} | D |
|
As shown in the figure, rotate triangle ABC around the line where AC is located to obtain a rotating body, then the lateral area of the rotating body is ()
Choices:
A:12π
B:15π
C:30π
D:60π | B |
|
As shown in the figure, then the radius of the sector is ()
Choices:
A:2
B:4
C:6
D:8 | B |
|
it is known that， then the radius of the bottom circle of the chimney cap is ()
Choices:
A:5cm
B:12cm
C:10cm
D:15cm | C |
|
As shown in the figure, then the lateral area of the cone is ()
Choices:
A:60π
B:80π
C:10π
D:96π | A |
|
then the radius of the sector is ()
Choices:
A:√{3}
B:√{6}
C:3
D:6 | C |
|
As shown in the figure, If OA and OB are overlapped to form a cone side, the diameter of the bottom of the cone is ()
Choices:
A:2cm
B:4cm
C:3cm
D:5cm | B |
|
As shown in the picture, and the lateral area of the cone-shaped tent roof (excluding the seams) is ()
Choices:
A:15πm^{2}
B:30πm^{2}
C:50πm^{2}
D:75πm^{2} | C |
|
As shown in the figure, then the diameter of the circle AD is ()
Choices:
A:5√{2}
B:10√{2}
C:15√{2}
D:20√{2} | B |
|
As shown in the figure, The circle with CD as the diameter intersects AD at point P. Then the length of AB is ()
Choices:
A:8
B:2√{10}
C:4√{3}
D:2√{13} | D |
|
As shown in the figure, then the height of the pavilion AB is approximately ()
Choices:
A:9m
B:9.6m
C:10m
D:10.2m | D |
|
As shown in the figure, and the straight line PS is perpendicular to the river. Choose an appropriate point T on the straight line a passing point S and perpendicular to PS. The intersection of PT and the straight line b passing point Q and perpendicular to PS is R. then the width of the river PQ is ()
Choices:
A:40m
B:120m
C:60m
D:180m | B |
|
As shown in the picture, It is known that AB perpendicular BD, CD perpendicular BD. Then the height of the ancient city wall CD is ()
Choices:
A:6m
B:8m
C:10m
D:12m | B |
|
As shown in the figure, point M is the midpoint of arc AB. then the degree of angle N is ()
Choices:
A:70°
B:40°
C:35°
D:20° | C |
|
A certain mathematics learning interest group measured the shadow length of Xiaoliang in the sun as 1.5. Knowing that Xiaoliang's height is 1.8. As shown in the figure, then the height of tree AB is ()
Choices:
A:10.8m
B:9m
C:7.5m
D:0.3m | A |
|
As shown in the picture, When foot D touches the ground, the head point E rises ()
Choices:
A:0.5m
B:0.6m
C:0.3m
D:0.9m | A |
|
As shown in the figure, AB parallel CD, then the distance between AB and CD is ().
Choices:
A:\frac{6}{5}
B:\frac{7}{6}
C:\frac{9}{5}
D:\frac{15}{2} | C |
|
As shown in the figure, the distance between the candle and the paper tube should be ()
Choices:
A:60cm
B:65cm
C:70cm
D:75cm | D |
|
As shown in the figure, Lin Dan, the athlete standing at M in the field, clicks the request from N to point B in the opponent. then when Lin Dan takes off, the distance from the hitting point to the ground NM = ()
Choices:
A:3.42m
B:1.9m
C:3m
D:3.24m | A |
|
then the height of the building is ()
Choices:
A:10m
B:12m
C:15m
D:22.5m | A |
|
As shown in the figure: then the length of the distance from E to the ground EF is ()
Choices:
A:2
B:2.2
C:2.4
D:2.5 | A |
|
As shown in the figure, then the edge length of the square DEFG is ()
Choices:
A:\frac{24}{5}cm
B:4cm
C:\frac{24}{7}cm
D:5cm | A |
|
The height h of the racket hit is ()
Choices:
A:\frac{8}{5}m
B:\frac{4}{3}m
C:1m
D:\frac{8}{15}m | D |
|
As shown in the figure, then the degree of angle BOC is ()
Choices:
A:60°
B:70°
C:120°
D:140° | B |
|
then the length of the ladder is ()
Choices:
A:5.60m
B:6.00m
C:6.10m
D:6.20m | B |
|
then the height of the flagpole AC is ()
Choices:
A:6m
B:7m
C:8.5m
D:9m | D |
|
As shown in the figure, then the height of the flag pole is ()
Choices:
A:12m
B:10m
C:8m
D:7m | A |
|
As shown in the figure, AC perpendicular CD, BD perpendicular CD, then the length of the line segment ED is ()
Choices:
A:\frac{20}{3}
B:\frac{10}{3}
C:7
D:\frac{14}{3} | A |
|
As shown in the figure, Xiaoming designed two right angles to measure the width of the river BC, then the width of the river BC is ()
Choices:
A:\frac{7}{2}m
B:\frac{17}{2}m
C:\frac{40}{7}m
D:11m | C |
|
As shown in the figure, and the height of the tree is ( ).
Choices:
A:3.4
B:5.1
C:6.8
D:8.5 | B |
|
As shown in the figure, then the length of the ladder is ()
Choices:
A:3.5m
B:4m
C:4.5m
D:5m | B |
|
As shown in the figure, then the height of the window AB is ()
Choices:
A:1.5m
B:1.6m
C:1.86m
D:2.16m | A |
|
As shown in the figure, the length of the ladder is ()
Choices:
A:3.85m
B:4.00m
C:4.40m
D:4.50m | C |
|
As shown in the figure, then the height of the flagpole is ()
Choices:
A:6.4m
B:7m
C:8m
D:9m | C |
|
As shown in the figure, the quadrilateral ABCD and A′B′C′D′ are similar. If OA′: A′A = 2.0:1.0, the area of the quadrilateral A′B′C′D′ is 12.0 ^ 2, then the area of the quadrilateral ABCD is ()
Choices:
A:24cm^{2}
B:27cm^{2}
C:36cm^{2}
D:54cm^{2} | B |
|
As shown in the figure, then cosB is equal to ()
Choices:
A:\frac{3}{5}
B:\frac{3}{4}
C:\frac{4}{5}
D:\frac{4}{3} | A |
|
As shown in the figure, then the value of sinB is equal to ()
Choices:
A:\frac{4}{3}
B:\frac{3}{4}
C:\frac{4}{5}
D:\frac{3}{5} | C |
|
As shown in the figure, then the value of cosA is ()
Choices:
A:\frac{3}{5}
B:\frac{5}{3}
C:\frac{4}{5}
D:\frac{3}{4} | A |
|
As shown in the figure, then the value of tanB is ()
Choices:
A:\frac{3}{5}
B:\frac{3}{4}
C:\frac{4}{5}
D:\frac{4}{3} | D |
|
As shown in the figure, , If the ratio of the distance from the bulb to the vertex of the triangle ruler to the distance from the bulb to the corresponding vertex of the triangular ruler projection is 2.0:5.0, Then the corresponding edge length of the projection triangle is ()
Choices:
A:8cm
B:20cm
C:3.2cm
D:10cm | B |
|
As shown in the figure, then the degree of the central angle angle BOC is ()
Choices:
A:40°
B:60°
C:80°
D:100° | C |
|
As shown in the figure, then the value of cosA is ()
Choices:
A:\frac{3}{5}
B:\frac{4}{5}
C:\frac{3}{4}
D:\frac{4}{3} | B |
|
As shown in the figure, then the value of sinA is ()
Choices:
A:\frac{3}{4}
B:\frac{4}{5}
C:\frac{3}{5}
D:\frac{5}{3} | C |
|
then the value of sinB is ()
Choices:
A:\frac{1}{2}
B:\frac{√{2}}{2}
C:\frac{√{3}}{2}
D:2 | C |
|
As shown in the figure, in ABC, then sinB is equal to ()
Choices:
A:\frac{3}{4}
B:\frac{3}{5}
C:\frac{4}{5}
D:\frac{5}{4} | B |
|
As shown in the figure: then the value of sinB is equal to ()
Choices:
A:\frac{3}{5}
B:\frac{4}{5}
C:\frac{3}{4}
D:\frac{4}{3} | B |
|
As shown in the figure, then the value of tanα is ()
Choices:
A:\frac{2}{3}
B:\frac{3}{2}
C:\frac{2√{13}}{13}
D:\frac{3√{13}}{13} | B |
|
As shown in the figure, then the length of BC is ()
Choices:
A:2
B:8
C:2√{5}
D:4√{5} | A |
|
As shown in the figure, then the value of sinB is ()
Choices:
A:\frac{2}{3}
B:\frac{3}{5}
C:\frac{3}{4}
D:\frac{4}{5} | D |
|
As shown in the figure, cosA = frac {3.0}{5.0}, then the length of BC is ()
Choices:
A:5cm
B:6cm
C:3cm
D:8cm | D |
|
As shown in the figure, E and F are the midpoints of AB and AD respectively. then tanC is equal to ()
Choices:
A:\frac{3}{4}
B:\frac{4}{3}
C:\frac{3}{5}
D:\frac{4}{5} | B |
|
As shown in the figure, then the degree of angle BAD is ()
Choices:
A:50°
B:40°
C:25°
D:35° | C |
|
As shown in the figure, OA = OB = OC, then angle DAO + angle DCO = ()
Choices:
A:90°
B:110°
C:120°
D:165° | D |
|
As shown in the figure, then the degree of angle AOB is ()
Choices:
A:40°
B:50°
C:80°
D:100° | C |
|
As shown in the figure, angle BAC + angle EAD = 180.0, then the radius of circle A is ()
Choices:
A:10
B:6
C:5
D:8 | C |
|
As shown in the figure
Choices:
A:3m
B:3.4m
C:4m
D:2.8m | A |
|
As shown in the figure, then the radius of circle O is ()
Choices:
A:3
B:2√{3}
C:6
D:4√{3} | C |
|
As shown in the figure, PB = AB, then the length of PA is ( )
Choices:
A:5
B:\frac{5√{3}}{2}
C:5√{2}
D:5√{3} | D |
|
As shown in the figure, then the value of AE^ 2 + CE^ 2 is ()
Choices:
A:1
B:2
C:3
D:4 | B |
|
As shown in the figure, then the length of BC is ()
Choices:
A:10
B:9
C:8
D:no solution | C |
|
As shown in the figure, then the maximum value of the distance from point O to vertex A is ()
Choices:
A:8
B:10
C:8√{2}
D:10√{2} | D |
|
As shown in the figure, then PC is ()
Choices:
A:4
B:4√{3}
C:6
D:2√{3} | D |
|
As shown in the figure, CD is tangent to the circle at point D, the length of the line segment CD is ()
Choices:
A:1
B:√{3}
C:2
D:2√{3} | D |
|
As shown in the figure, then the minimum area of the square PQRS is ()
Choices:
A:7
B:8
C:9
D:10 | B |
|
As shown in the figure, passing point C is the tangent of circle O, then the value of OD is ()
Choices:
A:1
B:√{3}
C:2
D:2√{3} | B |
|
As shown in the figure, the straight line AB is tangent to circle O at point A, then the length of OB is ()
Choices:
A:1
B:2
C:√{3}
D:2√{3} | B |
|
As shown in the figure, it is known that BA is the tangent of circle O, then the length of BC is ()
Choices:
A:2√{2}-1
B:√{2}
C:2√{2}-2
D:2-√{2} | C |
|
As shown in the figure, AB, AC, and BD are the tangents of circle O. then the length of BD is ()
Choices:
A:1.5
B:2
C:2.5
D:3 | B |
|
As shown in the figure, PA and PB are tangents of circle O, then the size of angle ACB is ()
Choices:
A:65°
B:60°
C:55°
D:50° | A |
|
As shown in the figure, and the tangent line of circle O passing through point A, then the length of AP is ()
Choices:
A:3
B:\frac{3}{2}
C:\frac{2}{3}√{3}
D:\frac{3}{2}√{3} | D |
|
As shown in the figure, the straight lines PA and PB are the two tangents of circle O, then the length of chord AB is ()
Choices:
A:5
B:10
C:10√{3}
D:5√{3} | B |
|
As shown in the figure, PA and PB are tangent to circle O, the tangent EF of circle O intersects PA and PB at points E and F respectively. | C |
|
Then the outer diameter of the round nut is ()
Choices:
A:12cm
B:24cm
C:6√{3}cm
D:12√{3}cm | D |
|
As shown in the figure, passing point P to draw the tangent PC of circle O, then the length of PC is ()
Choices:
A:1
B:√{3}
C:2
D:√{5} | B |
|
As shown in the figure, the chord AB of the great circle is tangent to the small circle, the area of the torus is ()
Choices:
A:16m^2
B:8πm^2
C:64m^2
D:16πm^2 | D |
|
As shown in the figure, circle O ia tangent to AB at point C, then S_triangle CDE is ()
Choices:
A:6√{5}
B:6√{3}
C:6√{2}
D:6 | B |
|
There is an isosceles triangle. Find the area of the figure.
Choices:
A:30
B:60
C:120
D:240 | B |
|
Find O X.
Choices:
A:5
B:12
C:13
D:26 | A |
|
There is an isosceles triangle. Find x. Round to the nearest tenth.
Choices:
A:18.8
B:23.2
C:25.9
D:44.0 | A |
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