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<image>As shown in the figure, and angle AOC = 126.0, then angle CDB = ()
Choices:
A:54°
B:64°
C:27°
D:37°
|
C
|
mathverse_vi_503
|
|
<image>As shown in the figure, if the radius of circle O is 2.5, the length of AC is 4.0, then the length of CE is ()
Choices:
A:3
B:\frac{20}{3}
C:\frac{12}{5}
D:\frac{16}{5}
|
C
|
mathverse_vi_508
|
|
<image>As shown in the figure, and it is known that angle ABC = 130.0, then angle AOC = ()
Choices:
A:100°
B:110°
C:120°
D:130°
|
A
|
mathverse_vi_513
|
|
<image>As shown in the figure, it is known that the radius of circle O is 5.0, chord CD = 8.0, then the length of chord AB is ()
Choices:
A:6
B:8
C:5√{2}
D:5√{3}
|
A
|
mathverse_vi_518
|
|
<image>As shown in the figure, Given that AB = 2 DE, angle E = 16.0, then the degree of angle ABC is ()
Choices:
A:32°
B:24°
C:16°
D:48°
|
B
|
mathverse_vi_523
|
|
<image>angle AOD = 136.0, then the degree of angle C is ()
Choices:
A:44°
B:22°
C:46°
D:36°
|
B
|
mathverse_vi_528
|
|
<image>As shown in the figure, If angle BOD = 130.0, then the degree of angle ACD is ()
Choices:
A:50°
B:30°
C:25°
D:20°
|
C
|
mathverse_vi_533
|
|
<image>the distance CD from the top of the round arch bridge to the water surface is 8.0, and the arch radius OC is 5.0, so the width of the water surface AB is ()
Choices:
A:4m
B:5m
C:6m
D:8m
|
D
|
mathverse_vi_538
|
|
<image>As shown in the figure, it is known that angle α = 130.0, then angle β = ()
Choices:
A:30°
B:40°
C:50°
D:65°
|
B
|
mathverse_vi_543
|
|
<image>As shown in the figure, when the width of the water surface AB in the circular bridge hole is 8.0, When the water surface rises 1.0, the water surface width A′B′ in the bridge hole is ()
Choices:
A:√{15}米
B:2√{15}米
C:2√{17}米
D:不能计算
|
B
|
mathverse_vi_548
|
|
<image>angle CAB = 90.0, angle ABC = 72.0, then the degree of angle ECD is ()
Choices:
A:63°
B:45°
C:27°
D:18°
|
C
|
mathverse_vi_553
|
|
<image>As shown in the figure, the bottom radius of the cone OB = 0.7, the length of AB is 2.5, then the length of AO is ()
Choices:
A:2.4
B:2.2
C:1.8
D:1.6
|
A
|
mathverse_vi_558
|
|
<image>As shown in the figure, if angle BEC = 110.0, then angle BDC = ()
Choices:
A:35°
B:45°
C:55°
D:70°
|
A
|
mathverse_vi_563
|
|
<image>As shown in the figure, in the sector OAB with a radius of 1.0, 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
|
mathverse_vi_568
|
|
<image>Use a sector paper sheet with a central angle of 120.0 and a radius of 6.0, then the bottom perimeter of the paper cap is ()
Choices:
A:2πcm
B:3πcm
C:4πcm
D:5πcm
|
C
|
mathverse_vi_573
|
|
<image>The length of its generatrix l is 13.0 and its height h is 12.0. 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
|
mathverse_vi_578
|
|
<image>Use a sector piece of paper with a central angle of 120.0 and a radius of 3.0, then the height of the paper is ()
Choices:
A:3cm
B:2√{2}cm
C:3√{2}cm
D:4√{2}cm
|
B
|
mathverse_vi_583
|
|
<image>As shown in the figure, the expanded figure of the lateral surface of a cone is a semicircle with a radius of 10.0, then the radius of its bottom is ()
Choices:
A:3
B:4
C:5
D:6
|
C
|
mathverse_vi_588
|
|
<image>As shown in the figure, use a sector cardboard with a radius of 24.0 to make a conical hat. If the radius of the bottom surface of the conical hat is 10.0, then the area of this sector cardboard is ()
Choices:
A:240πcm^{2}
B:480πcm^{2}
C:1200πcm^{2}
D:2400πcm^{2}
|
A
|
mathverse_vi_593
|
|
<image>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
|
mathverse_vi_598
|
|
<image>If AB = 50.0, AD = 14.0, 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
|
mathverse_vi_603
|
|
<image>As shown in the figure, a sector with a central angle of 120.0 and a radius of 6.0, then the height of the cone is ()
Choices:
A:6
B:8
C:3√{3}
D:4√{2}
|
D
|
mathverse_vi_608
|
|
<image>As shown in the figure, AC = 4.0, BC = 3.0, 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
|
mathverse_vi_613
|
|
<image>As shown in the figure, If the radius of the circle is 1.0 and the central angle of the sector is equal to 90.0, then the radius of the sector is ()
Choices:
A:2
B:4
C:6
D:8
127
|
B
|
mathverse_vi_618
|
|
<image>it is known that OA = 30.0, angle AOB = 120.0, then the radius of the bottom circle of the chimney cap is ()
Choices:
A:5cm
B:12cm
C:10cm
D:15cm
|
C
|
mathverse_vi_623
|
|
<image>As shown in the figure, it is known that the radius of the bottom surface of the cone is 6.0, and the length of the generatrix is 10.0, then the lateral area of the cone is ()
Choices:
A:60π
B:80π
C:10π
D:96π
|
A
|
mathverse_vi_628
|
|
<image>If the radius of the circle is 1.0. The central angle of the sector is equal to 120.0, then the radius of the sector is ()
Choices:
A:√{3}
B:√{6}
C:3
D:6
|
C
|
mathverse_vi_633
|
|
<image>As shown in the figure, a radius of 6.0. the diameter of the bottom of the cone is ()
Choices:
A:2cm
B:4cm
C:3cm
D:5cm
|
B
|
mathverse_vi_638
|
|
<image>As shown in the picture, the length of the generatrix of the cone-shaped tent roof is AB = 10.0, the bottom radius is BO = 5.0, 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
|
mathverse_vi_643
|
|
<image>As shown in the figure, the length of chord AB is 10.0, and the angle of circumference angle ACB = 45.0, then the diameter of the circle AD is ()
Choices:
A:5√{2}
B:10√{2}
C:15√{2}
D:20√{2}
|
B
|
mathverse_vi_648
|
|
<image>As shown in the figure, angle C = 90.0. If AC = 6.0. The minimum length of the line segment BP is 2.0. Then the length of AB is ()
Choices:
A:8
B:2√{10}
C:4√{3}
D:2√{13}
|
D
|
mathverse_vi_653
|
|
<image>As shown in the figure, DE = BC = 0.6, and measure CG = 12.0, and measure GE = 2.0, Xiaoming's height EF = 1.6, then the height of the pavilion AB is approximately ()
Choices:
A:9米
B:9.6米
C:10米
D:10.2米
|
D
|
mathverse_vi_658
|
|
<image>As shown in the figure, If QS = 60.0, ST = 120.0, QR = 80.0, then the width of the river PQ is ()
Choices:
A:40m
B:120m
C:60m
D:180m
|
B
|
mathverse_vi_663
|
|
<image>As shown in the picture, And it is measured that AB = 1.4, BP = 2.1, PD = 12.0. Then the height of the ancient city wall CD is ()
Choices:
A:6米
B:8米
C:10米
D:12米
|
B
|
mathverse_vi_668
|
|
<image>As shown in the figure, if angle AOB = 140.0, then the degree of angle N is ()
Choices:
A:70°
B:40°
C:35°
D:20°
|
C
|
mathverse_vi_673
|
|
<image>As shown in the figure, a certain mathematics learning interest group measured the length of the tree's shadow BC in the sun as 9.0. At the same moment, they also measured the shadow length of Xiaoliang in the sun as 1.5. Knowing that Xiaoliang's height is 1.8, then the height of tree AB is ()
Choices:
A:10.8m
B:9m
C:7.5m
D:0.3m
|
A
|
mathverse_vi_678
|
|
<image>As shown in the picture, It is known that the height of the support column AB is 0.3, the length of the pedal DE is 1.0, and the distance from the support point A to the foot D is 0.6. When foot D touches the ground, the head point E rises ()
Choices:
A:0.5米
B:0.6米
C:0.3米
D:0.9米
|
A
|
mathverse_vi_683
|
|
<image>As shown in the figure, AB = 2.0, CD = 5.0, the distance between point P and CD is 3.0, 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
|
mathverse_vi_688
|
|
<image>As shown in the figure, Xiaoqiang made a small hole imaging device in which the length of the paper tube is 15.0. He prepared a candle with a length of 20.0. To get an image with a height of 4.0, the distance between the candle and the paper tube should be ()
Choices:
A:60cm
B:65cm
C:70cm
D:75cm
|
D
|
mathverse_vi_693
|
|
<image>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. It is known that the net height OA = 1.52, OB = 4.0, OM = 5.0, 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
|
mathverse_vi_698
|
|
<image>Xiao Ming first measured the shadow length BA of the building on the ground as 15.0 (as shown in the figure), and then set up a benchmark with a height of 2.0 at A, and measured the shadow length AC of the benchmark as 3.0, then the height of the building is ()
Choices:
A:10米
B:12米
C:15米
D:22.5米
|
A
|
mathverse_vi_703
|
|
<image>As shown in the figure: AB is 6.0, the length of CD is 3.0, 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
|
mathverse_vi_708
|
|
<image>As shown in the figure, if BC = 12.0, AH = 8.0, 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
|
mathverse_vi_713
|
|
<image>It is known that the net height is 0.8, and the horizontal distance from the hitting point to the net is 4.0. When the ball is played, the ball can hit the net and the landing point is exactly 6.0 away from the net. Then the height h of the racket hit is ()
Choices:
A:\frac{8}{5}米
B:\frac{4}{3}米
C:1米
D:\frac{8}{15}米
|
D
|
mathverse_vi_718
|
|
<image>As shown in the figure, angle ABO = 40.0, angle ACO = 30.0, then the degree of angle BOC is ()
Choices:
A:60°
B:70°
C:120°
D:140°
|
B
|
mathverse_vi_723
|
|
<image>As shown in the figure, the foot of the ladder is away from the wall 2.0, the point D on the ladder is away from the wall 1.8, the length of BD is 0.6, then the length of the ladder is ()
Choices:
A:5.60米
B:6.00米
C:6.10米
D:6.20米
|
B
|
mathverse_vi_728
|
|
<image>a school math interest group erected a benchmark DF with a length of 1.5 at point F. the length of the shadow EF of DF is measured as 1.0, and then measure the length of the shadow BC of the flagpole AC to be 6.0, then the height of the flagpole AC is ()
Choices:
A:6米
B:7米
C:8.5米
D:9米
|
D
|
mathverse_vi_733
|
|
<image>As shown in the figure, Xiaodong uses a bamboo pole with a length of 3.2 as a measuring tool to measure the height of the school flagpole, and moves the bamboo pole so that the shadow on the top of the pole and the flag pole falls on the same point on the ground. At this time, the distance between the bamboo pole and this point is 8.0, and the distance from the flag pole is 22.0, then the height of the flag pole is ()
Choices:
A:12m
B:10m
C:8m
D:7m
|
A
|
mathverse_vi_738
|
|
<image>As shown in the figure, and AC = 3.0, BD = 6.0, CD = 10.0, 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
|
mathverse_vi_743
|
|
<image>As shown in the figure, he measured AB = 2.0, BD = frac {7.0}{3.0}, CE = 9.0, then the width of the river BC is ()
Choices:
A:\frac{7}{2}米
B:\frac{17}{2}米
C:\frac{40}{7}米
D:11米
|
C
|
mathverse_vi_748
|
|
<image>As shown in the figure, a student saw a tree by the lake. He visually observed that the distance between himself and the tree is 20.0, and the reflection of the top of the tree in the water is 5.0 far away from him. The student's height is 1.7, and the height of the tree is ( ).
Choices:
A:3.4
B:5.1
C:6.8
D:8.5
|
B
|
mathverse_vi_753
|
|
<image>As shown in the figure, the distance from the foot of the ladder B to the foot of the wall C is 1.6, the distance from the point D on the ladder to the wall is 1.4, and the length of the ladder is 0.5, then the length of the ladder is ()
Choices:
A:3.5m
B:4m
C:4.5m
D:5m
|
B
|
mathverse_vi_758
|
|
<image>As shown in the figure, the length of the shadow of the window frame AB on the ground DE = 1.8, the distance from the lower eaves of the window to the ground BC = 1.0, EC = 1.2, then the height of the window AB is ()
Choices:
A:1.5m
B:1.6m
C:1.86m
D:2.16m
|
A
|
mathverse_vi_763
|
|
<image>As shown in the figure, the foot of the ladder B is away from the wall 1.6, the point D on the ladder is away from the wall 1.4, the length of BD is 0.55, then the length of the ladder is ()
Choices:
A:3.85米
B:4.00米
C:4.40米
D:4.50米
|
C
|
mathverse_vi_768
|
|
<image>As shown in the figure, the student Xiao Li whose height is 1.6 wants to measure the height of the school's flagpole. When he stands at C, the shadow of the top of his head coincides with the shadow of the top of the flagpole, and AC = 2.0, BC = 8.0, then the height of the flagpole is ()
Choices:
A:6.4米
B:7米
C:8米
D:9米
|
C
|
mathverse_vi_773
|
|
<image> 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
|
mathverse_vi_778
|
|
<image>As shown in the figure, if AC = 4.0, BC = 3.0, then cosB is equal to ()
Choices:
A:\frac{3}{5}
B:\frac{3}{4}
C:\frac{4}{5}
D:\frac{4}{3}
|
A
|
mathverse_vi_783
|
|
<image>As shown in the figure, AC = 4.0, BC = 3.0, 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
|
mathverse_vi_788
|
|
<image>As shown in the figure, AC = 3.0, BC = 4.0, then the value of cosA is ()
Choices:
A:\frac{3}{5}
B:\frac{5}{3}
C:\frac{4}{5}
D:\frac{3}{4}
|
A
|
mathverse_vi_793
|
|
<image>As shown in the figure, AB = 10.0, AC = 8.0, then the value of tanB is ()
Choices:
A:\frac{3}{5}
B:\frac{3}{4}
C:\frac{4}{5}
D:\frac{4}{3}
|
D
|
mathverse_vi_798
|
|
<image>As shown in the figure, and the length of one edge of the triangle ruler is 8.0, Then the corresponding edge length of the projection triangle is ()
Choices:
A:8cm
B:20cm
C:3.2cm
D:10cm
|
B
|
mathverse_vi_803
|
|
<image>As shown in the figure, given the angle of circumference angle BAC = 40.0, then the degree of the central angle angle BOC is ()
Choices:
A:40°
B:60°
C:80°
D:100°
|
C
|
mathverse_vi_808
|
|
<image>As shown in the figure, AC = 4.0, AB = 5.0, then the value of cosA is ()
Choices:
A:\frac{3}{5}
B:\frac{4}{5}
C:\frac{3}{4}
D:\frac{4}{3}
|
B
|
mathverse_vi_813
|
|
<image>As shown in the figure, AB = 5.0, AC = 4.0, then the value of sinA is ()
Choices:
A:\frac{3}{4}
B:\frac{4}{5}
C:\frac{3}{5}
D:\frac{5}{3}
|
C
|
mathverse_vi_818
|
|
<image>AB = 2.0, BC = 1.0, then the value of sinB is ()
Choices:
A:\frac{1}{2}
B:\frac{√{2}}{2}
C:\frac{√{3}}{2}
D:2
|
C
|
mathverse_vi_823
|
|
<image>As shown in the figure, AC = 3.0, AB = 4.0, then sinB is equal to ()
Choices:
A:\frac{3}{4}
B:\frac{3}{5}
C:\frac{4}{5}
D:\frac{5}{4}
|
B
|
mathverse_vi_828
|
|
<image>As shown in the figure: AC = 8.0, AB = 10.0, 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
|
mathverse_vi_833
|
|
<image>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
|
mathverse_vi_838
|
|
<image>As shown in the figure, AC = 4.0, tanA = frac {1.0}{2.0}, then the length of BC is ()
Choices:
A:2
B:8
C:2√{5}
D:4√{5}
|
A
|
mathverse_vi_843
|
|
<image>As shown in the figure, AC = 4.0, AB = 5.0, then the value of sinB is ()
Choices:
A:\frac{2}{3}
B:\frac{3}{5}
C:\frac{3}{4}
D:\frac{4}{5}
|
D
|
mathverse_vi_848
|
|
<image>As shown in the figure, the hypotenuse of triangle ABC AB = 10.0, cosA = frac {3.0}{5.0}, then the length of BC is ()
Choices:
A:5cm
B:6cm
C:3cm
D:8cm
|
D
|
mathverse_vi_853
|
|
<image>As shown in the figure, If EF = 2.0, BC = 5.0, CD = 3.0, then tanC is equal to ()
Choices:
A:\frac{3}{4}
B:\frac{4}{3}
C:\frac{3}{5}
D:\frac{4}{5}
|
B
|
mathverse_vi_858
|
|
<image>As shown in the figure, angle BOD = 50.0, then the degree of angle BAD is ()
Choices:
A:50°
B:40°
C:25°
D:35°
|
C
|
mathverse_vi_863
|
|
<image>As shown in the figure, angle ABC = angle ADC = 65.0, then angle DAO + angle DCO = ()
Choices:
A:90°
B:110°
C:120°
D:165°
|
D
|
mathverse_vi_868
|
|
<image>As shown in the figure, if angle OBC = 40.0, then the degree of angle AOB is ()
Choices:
A:40°
B:50°
C:80°
D:100°
|
C
|
mathverse_vi_873
|
|
<image>As shown in the figure, the known chord BC = 8.0, DE = 6.0, angle BAC + angle EAD = 180.0, then the radius of circle A is ()
Choices:
A:10
B:6
C:5
D:8
|
C
|
mathverse_vi_878
|
|
<image>As shown in the figure, the cross section of a tunnel is a semicircle with a radius of 3.4, and a truck with a width of 3.2 can pass through the tunnel.
Choices:
A:3m
B:3.4m
C:4m
D:2.8m
|
A
|
mathverse_vi_883
|
|
<image>As shown in the figure, if angle ACB = 30.0, AB = 6.0, then the radius of circle O is ()
Choices:
A:3
B:2√{3}
C:6
D:4√{3}
|
C
|
mathverse_vi_888
|
|
<image>As shown in the figure, angle C = 30.0, the radius of circle O is 5.0, then the length of PA is ( )
Choices:
A:5
B:\frac{5√{3}}{2}
C:5√{2}
D:5√{3}
|
D
|
mathverse_vi_893
|
|
<image>As shown in the figure, knowing that the radius of circle O is 1.0, then the value of AE^ 2 + CE^ 2 is ()
Choices:
A:1
B:2
C:3
D:4
|
B
|
mathverse_vi_898
|
|
<image>As shown in the figure, if OC = 5.0, AC = 6.0, then the length of BC is ()
Choices:
A:10
B:9
C:8
D:无法确定
|
C
|
mathverse_vi_903
|
|
<image>As shown in the figure, angle XOY = 45.0, where AB = 10.0, 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
|
mathverse_vi_908
|
|
<image>As shown in the figure, the radius of circle O is 2.0, then PC is ()
Choices:
A:4
B:4√{3}
C:6
D:2√{3}
|
D
|
mathverse_vi_913
|
|
<image>As shown in the figure, in the circle O with a radius of 2.0, given that angle DAC = 30.0, the length of the line segment CD is ()
Choices:
A:1
B:√{3}
C:2
D:2√{3}
|
D
|
mathverse_vi_918
|
|
<image>As shown in the figure, Circle O is a circle with a radius of 1.0, the distance from point O to line L is 3.0, then the minimum area of the square PQRS is ()
Choices:
A:7
B:8
C:9
D:10
|
B
|
mathverse_vi_923
|
|
<image>As shown in the figure, if angle E = 30.0, CE = 6.0, then the value of OD is ()
Choices:
A:1
B:√{3}
C:2
D:2√{3}
|
B
|
mathverse_vi_928
|
|
<image>As shown in the figure, the radius of circle O is 1.0, if angle OBA = 30.0, then the length of OB is ()
Choices:
A:1
B:2
C:√{3}
D:2√{3}
|
B
|
mathverse_vi_933
|
|
<image>As shown in the figure, If angle B = 45.0 and the length of AB is 2.0, then the length of BC is ()
Choices:
A:2√{2}-1
B:√{2}
C:2√{2}-2
D:2-√{2}
|
C
|
mathverse_vi_938
|
|
<image>As shown in the figure, AB, AC, and BD are the tangents of circle O, and the tangent points are P, C, and D respectively. If AB = 5.0, AC = 3.0, then the length of BD is ()
Choices:
A:1.5
B:2
C:2.5
D:3
|
B
|
mathverse_vi_943
|
|
<image>As shown in the figure, given that angle P = 50.0, then the size of angle ACB is ()
Choices:
A:65°
B:60°
C:55°
D:50°
|
A
|
mathverse_vi_948
|
|
<image>As shown in the figure, angle B = 30.0, OP = 3.0, then the length of AP is ()
Choices:
A:3
B:\frac{3}{2}
C:\frac{2}{3}√{3}
D:\frac{3}{2}√{3}
|
D
|
mathverse_vi_953
|
|
<image>As shown in the figure, If angle APB = 120.0, the radius of circle O is 10.0, then the length of chord AB is ()
Choices:
A:5
B:10
C:10√{3}
D:5√{3}
|
B
|
mathverse_vi_958
|
|
<image>As shown in the figure, if the length of PA is 2.0, then the perimeter of triangle PEF is ()
Choices:
A:8
B:6
C:4
D:2
|
C
|
mathverse_vi_963
|
|
<image>Angle CAB = 60.0, if AD = 6.0, then the outer diameter of the round nut is ()
Choices:
A:12cm
B:24cm
C:6√{3}cm
D:12√{3}cm
|
D
|
mathverse_vi_968
|
|
<image>As shown in the figure, If AO = OB = PB = 1.0, then the length of PC is ()
Choices:
A:1
B:√{3}
C:2
D:√{5}
|
B
|
mathverse_vi_973
|
|
<image>As shown in the figure, if the measured length of AB is 8.0, the area of the torus is ()
Choices:
A:16平方米
B:8π平方米
C:64平方米
D:16π平方米
|
D
|
mathverse_vi_978
|
|
<image>As shown in the figure, angle BCE = 60.0, DC = 6.0, DE = 4.0, then S_triangle CDE is ()
Choices:
A:6√{5}
B:6√{3}
C:6√{2}
D:6
|
B
|
mathverse_vi_983
|
|
<image>Find the area of the figure.
Choices:
A:30
B:60
C:120
D:240
|
B
|
mathverse_vi_988
|
|
<image>Circle O has a radius of 13 inches. Chord C D is 24 inches long. Find O X.
Choices:
A:5
B:12
C:13
D:26
|
A
|
mathverse_vi_993
|
|
<image>There is a triangle with side length 32. The base angle is 54 degrees. Find x. Round to the nearest tenth.
Choices:
A:18.8
B:23.2
C:25.9
D:44.0
|
A
|
mathverse_vi_998
|
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