LiAuto-DSR/GeoGramBench huggingface dataset:
The perpendicular height of the solid figure is 12m.

Hence, find the length of the diameter of the solid figure's base.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 h = 9 
 l = 11
 r = np.sqrt(l**2 - h**2)  
 
 
 plt.plot([0, -r], [0, -h], 'k', lw=3)
 plt.plot([0, r], [0, -h], 'k', lw=3)
 
 plt.plot([0, 0], [0, -h], 'k--', lw=2)
 
 
 theta = np.linspace(0, 2 * np.pi, 100)
 x = r * np.cos(theta)
 y = (r/2) * np.sin(theta) - h
 plt.plot(x, y, 'navy', lw=3)
 
 
 plt.plot([-1.5*r, -1.5*r], [0, -h], 'k', lw=2)
 plt.plot([-1.5*r-0.2, -1.5*r+0.2], [0, 0], 'k', lw=2)
 plt.plot([-1.5*r-0.2, -1.5*r+0.2], [-h, -h], 'k', lw=2)
 plt.text(-1.65*r, -h/2, '12 m', fontsize=14, va='center')
 
 
 plt.plot([0.5, 1.1*r], [0, -h], 'k', lw=3)
 plt.plot([0, 0.2*r-0.1], [0, 0.1*r], 'k', lw=3)
 plt.plot([1*r, 1.2*r], [-h-0.2, -h+0.2], 'k', lw=3)
 plt.text(r/2 + 0.6, -h/2, '13 m', fontsize=14)
 
 
 plt.plot([0, r], [-h, -h], 'k--', lw=2)
 plt.plot([0.1*r, 0.5, 0.1], [-h, -h+0.5, -h+0.5], 'teal', lw=4)
 plt.text(r/2, -h - 0.9, 'r', fontsize=14, ha='center')
 
 
 plt.plot([0], [-h], 'o', color='black')
 
 plt.axis('equal')
 plt.axis('off')
 plt.ylim(-15, 2)
 plt.xlim(-15, 15) 
[/python]

Answer: $10$
The diagram shows a solid figure with a slant height of 13 m. The radius of the base of the solid figure is denoted by $r$.

Find the value of $r$.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 h = 9
 s = 10
 
 r = (s**2 - h**2)**0.5
 
 theta = np.linspace(0, 2*np.pi, 200)
 x_circ = r * np.cos(theta)
 y_circ = r/2 * np.sin(theta)
 
 fig, ax = plt.subplots(figsize=(6,10))
 
 ax.plot(x_circ, y_circ, 'k')
 
 ax.plot([0, -r], [h, 0], 'k')
 ax.plot([0, r], [h, 0], 'k')
 
 ax.plot([0, 0], [0, h], 'k--')
 
 ax.plot([-r, r], [0, 0], 'b', linewidth=1)
 
 ax.plot([0.4, 0.4, 0], [0, 0.4, 0.4], 'b', linewidth=1)
 
 ax.text(-0.2, h/2, "$12m$", fontsize=14, color="k")
 ax.text(r-2.5, h/2, "$13m$", fontsize=14, color="k")
 ax.text(r/2-0.1, -0.7, "$r$", fontsize=14, color="b")
 ax.text(0-0.4, -0.5, "$O$", fontsize=14, color="b")
 
 ax.set_aspect('equal')
 ax.axis('off')
 plt.ylim(-r-1, h+1)
 plt.xlim(-r-2, r+2) 
[/python]

Answer: $5$
A solid figure has a side $GF$ of length 15cm as shown.

Now, we want to find $y$, the length of the diagonal $DF$.

Calculate $y$ to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig = plt.figure(figsize=(9,6))
 ax = fig.add_subplot(121, projection='3d')
 
 # Vertices as (x, y, z)
 A = [0, 0, 12]
 B = [0, 8, 12]
 C = [17, 8, 12]
 D = [17, 0, 12]
 E = [0, 0, 0]
 F = [0, 8, 0]
 G = [17, 8, 0]
 H = [17, 0, 0]
 
 edges = [
     (A,D), (D,C), (C,B), (B,A),
     (H,G), (E,H),
     (A,E), (D,H), (C,G)
 ]
 
 for e in edges:
     x, y, z = zip(*e)
     ax.plot(x, y, z, color='black')
 
 ax.plot([B[0],F[0]], [B[1],F[1]], [B[2],F[2]], ls='dashed', color='black')
 ax.plot([E[0],F[0]], [E[1],F[1]], [E[2],F[2]], ls='dashed', color='black')
 ax.plot([F[0],G[0]], [F[1],G[1]], [F[2],G[2]], ls='dashed', color='black')
 
 ax.plot([D[0],F[0]], [D[1],F[1]], [D[2],G[2]], ls='dotted', color='grey')
 ax.plot([H[0],F[0]], [H[1],F[1]], [H[2],G[2]], ls='dotted', color='grey')
 
 verts = {'A':A, 'B':B, 'C':C, 'D':D, 'E':E, 'F':F, 'G':G, 'H':H}
 for k, v in verts.items():
     ax.scatter(*v, color='navy')
     d = dict(A=(-0.5,-0.5,0.5), D=(0.5,-0.5,0.5), C=(0.5,0.5,0.5), B=(-0.5,0.5,0.5),
              H=(0.5,-0.5,-0.25), G=(0.7,0.7,-0.2), F=(-0.5,0.5,-0.2), E=(-0.8,-0.7,-0.2))
     o = d.get(k, (0,0,0))
     ax.text(v[0]+o[0], v[1]+o[1], v[2]+o[2], f'{k}', color='navy', fontsize=17, fontweight='bold')
 
 ax.text(1, 6, (A[2]+D[2])/2, r"$11\,\mathit{cm}$", fontsize=17, color='black', va='center', ha='right', rotation=90)
 ax.text(17, 6, 6, r"$11\,\mathit{cm}$", fontsize=17, color='black', va='center', ha='left', rotation=90)
 
 ax.text(7.5, 8.7, 0, r"$15\,\mathit{cm}$", fontsize=17, color='black', ha='center', va='bottom')
 
 ax.text((D[0]+F[0])/2, (D[1]+F[1])/2+0.5, (D[2]+F[2])/2+0.9, r"$y\,cm$", fontsize=17, color='black')
 ax.text((E[0]+G[0])/2-1, (E[1]+G[1])/2-1, (E[2]+G[2])/2, r"$z\,cm$", fontsize=17, color='black')
 
 ax.set_xlim(-3, 19)
 ax.set_ylim(-3, 12)
 ax.set_zlim(-3, 16)
 ax.set_xticks([])
 ax.set_yticks([])
 ax.set_zticks([])
 ax.set_box_aspect([2,1,1])
 ax.view_init(azim=-60, elev=22)
 plt.tight_layout()
 ax.axis('off') 
[/python]

Answer: $21.61$
The length of CD side is 12cm. 

Calculate the area of the triangular divider correct to two decimal places.

[python]
 import matplotlib.pyplot as plt
 from mpl_toolkits.mplot3d.art3d import Poly3DCollection
 import numpy as np
 
 A = np.array([0, 0, 0])
 B = np.array([20, 0, 0])
 C = np.array([20, 15, 0])
 D = np.array([20, 15, 15])
 E = np.array([0, 0, 15])
 F = np.array([20, 0, 15])
 G = np.array([0, 15, 0])
 H = np.array([0, 15, 15])
 
 fig = plt.figure(figsize=(8, 6))
 ax = fig.add_subplot(111, projection='3d')
 
 edges = [
     [A, B], [B, C], [C, G], [G, A],  # bottom
     [E, F], [F, D], [D, H], [H, E],  # top
     [A, E], [B, F], [C, D], [G, H]   # vertical
 ]
 for edge in edges:
     ax.plot(*zip(*edge), c='k')
 
 face = [A, C, D]
 ax.add_collection3d(Poly3DCollection([[A, C, D]], facecolors='cyan', linewidths=1, edgecolors='cyan', alpha=0.4))
 
 ax.plot([A[0], C[0]], [A[1], C[1]], [A[2], C[2]], color='r', linestyle='--', linewidth=2)
 
 def annotate_point(point, label, offset=(0,0.5,0)):
     ax.text(point[0]+offset[0], point[1]+offset[1], point[2]+offset[2], label, fontsize=15, color='navy')
 
 annotate_point(A, 'A')
 annotate_point(B, 'B')
 annotate_point(C, 'C')
 annotate_point(D, 'D')
 
 ax.text(8.5, -1.5, 0, r'$17cm$', fontsize=13)
 ax.text(8.5, 6, 0, r'$zcm$', fontsize=13)
 ax.text(20.8, 6, 0, r'$12cm$', fontsize=13)
 ax.text(20.8, 15, 6, r'$12cm$', fontsize=13)
 
 ax.set_xlim([0, 20])
 ax.set_ylim([0, 15])
 ax.set_zlim([0, 15])
 ax.set_box_aspect([20, 15, 15])
 
 ax.axis('off') 
[/python]

Answer: $124.85$
The figure shows a solid figure with radius of 6 centimeters.

Find the volume of the solid figure, rounding your answer to two decimal places.

[python]
 import numpy as np
 import matplotlib.pyplot as plt
 
 r = 10
 h = 12
 
 theta = np.linspace(0, 2*np.pi, 100)
 x = r * np.cos(theta)
 y = r/2 * np.sin(theta)
 
 fig, ax = plt.subplots(figsize=(6,6))
 ax.plot(x, y + h, 'k')
 ax.plot(x[50:], y[50:], 'k')
 
 ax.plot([r, r], [h, 0], 'k')
 ax.plot([-r, -r], [h, 0], 'k')
 
 ax.plot(x[(theta > np.pi)], y[(theta > np.pi)] + h, 'k--', alpha=0.5)
 
 ax.plot([0, r], [h, h], 'k')
 ax.text(r/2, h+0.5, '6 cm', fontsize=12, ha='center')
 
 ax.text(r+0.5, h/2, '8 cm', fontsize=12, va='center')
 
 ax.axis('equal')
 ax.axis('off') 
[/python]

Answer: $904.78$
Find the volume of the solid figure shown. The perpendicular height is 6 centimeters.

Round your answer to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(10, 10))
 
 center = [0, 0]
 radius = 1
 height = 5
 
 theta = np.linspace(0, 2 * np.pi, 200)
 x = radius * np.cos(theta)
 y = 0.5 * radius * np.sin(theta)
 
 ax.plot(x, y, 'k')
 
 ax.plot([0, -radius], [height, 0], 'k')
 ax.plot([0, radius], [height, 0], 'k')
 
 ax.plot([0, 0], [0, height], 'k')
 
 ax.plot([0, radius], [0, 0], 'k')
 
 ax.plot([0.2, 0.2, 0], [0, 0.2, 0.2], 'k')
 
 ax.text(0.1, height/2, r"$6cm$", fontsize=14, ha='left', va='center')
 ax.text(radius/2, -0.3, r"$2cm$", fontsize=14, ha='center')
 
 ax.set_xlim(-3, 3)
 ax.set_ylim(-1, 7)
 ax.set_aspect('equal')
 ax.axis('off') 
[/python]

Answer: $25.13$
Find the volume of the solid figure shown. The slant height is 8 centimeters.

Round your answer to two decimal places.

[python]
 import numpy as np
 import matplotlib.pyplot as plt
 
 radius = 3
 slant_height = 9
 height = np.sqrt(slant_height**2 - radius**2)
 
 theta = np.linspace(0, 2*np.pi, 100)
 x_circle = radius * np.cos(theta)
 y_circle = radius/2 * np.sin(theta)
 
 fig, ax = plt.subplots(figsize=(6, 6))
 
 ax.plot([0, -radius], [height, 0], 'k')
 ax.plot([0, radius], [height, 0], 'k')
 
 ax.plot(x_circle, y_circle, 'k')
 
 ax.plot([0, radius], [0, 0], 'k', lw=1)
 
 ax.text(radius * 0.5, height/2 + 0.1, '8 cm', fontsize=12, style='italic')
 ax.text(radius/4, -0.4, '2 cm', fontsize=12, style='italic')
 
 ax.plot(0, height, 'ko')
 
 ax.set_aspect('equal')
 ax.axis('off')
 plt.ylim(-2.5, height+1)
 plt.xlim(-3, 3) 
[/python]

Answer: $32.45$
The diameter is 3 centimeters.

Find the volume of the solid, correct to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 A = np.array([0, 0])
 B = np.array([4, 0]) 
 
 mid_BC = (A + B) / 2
 height = 9
 C = mid_BC + np.array([0, height])
 
 plt.plot([A[0], B[0]], [A[1], B[1]], 'k')
 plt.plot([A[0], C[0]], [A[1], C[1]], 'k')
 plt.plot([B[0], C[0]], [B[1], C[1]], 'k')
 
 plt.plot([mid_BC[0], C[0]], [mid_BC[1], C[1]], 'k--')
 
 theta = np.linspace(np.pi, 2 * np.pi, 100)
 r = 2
 x_semicircle = mid_BC[0] + r * np.cos(theta)
 y_semicircle = mid_BC[1] + r * np.sin(theta)
 plt.plot(x_semicircle, y_semicircle, 'k')
 
 plt.text(1, -0.5, r'$3cm$', fontsize=14)
 plt.text(3.75, 4, r'$9cm$', fontsize=14)
 plt.plot([2, 2.5], [0.5, 0.5], 'k')
 plt.plot([2.5, 2.5], [0.5, 0], 'k')
 
 plt.axis('equal')
 plt.axis('off') 
[/python]

Answer: $10.45$
Find the volume of the solid figure pictured here. The radius is 4.1 centimeters.

(Give your answer correct to 1 decimal place.)

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(5,6))
 
 ax.fill([0, 4.3, -4.3, 0], [0, -13.5, -13.5, 0], color='olive')
 
 theta = np.linspace(0, 2*np.pi, 100)
 x_ellipse = 4.3 * np.cos(theta)
 y_ellipse = -13.5 + 1 * np.sin(theta)
 ax.plot(x_ellipse[50:], y_ellipse[50:], color='black')
 ax.plot(x_ellipse[:50], y_ellipse[:50], color='black', linestyle="--")
 ax.fill(x_ellipse, y_ellipse, color="olive")
 
 ax.plot([0, 0], [0, -13.5], 'k--', lw=1)
 ax.plot([0, 4.3], [-13.5, -13.5], 'k--', lw=1)
 
 offset = 0.3
 ax.plot([0, offset], [-13.5, -13.5], color='black')
 ax.plot([offset, offset], [-13.5, -13.5+offset], color='black')
 ax.plot([0, offset], [-13.5+offset, -13.5+offset], color='black')
 
 ax.text(-1.18, -6.25, r"$12.5cm$", fontsize=14, va='center', color='black', style='italic')
 ax.text(2.05, -13.1, r"4.1cm", fontsize=14, ha='center', color='black', style='italic')
 
 ax.set_aspect('equal')
 plt.axis('off') 
[/python]

Answer: $220.0$
Find the volume of the solid figure pictured. The perpendicular height is 6.

(Give your answer correct to 2 decimal places.)

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(5,6))
 
 ax.fill([0, 4.1, -4.1, 0], [0, -12.5, -12.5, 0], color='olive')
 
 theta = np.linspace(0, 2*np.pi, 100)
 x_ellipse = 4.1 * np.cos(theta)
 y_ellipse = -12.5 + 1 * np.sin(theta)
 ax.plot(x_ellipse[50:], y_ellipse[50:], color='black')
 ax.plot(x_ellipse[:50], y_ellipse[:50], color='black', linestyle="--")
 ax.fill(x_ellipse, y_ellipse, color="olive")
 
 
 ax.plot([0, 0], [0, -12.5], 'k--', lw=1)
 ax.plot([-4.1, 0], [-12.5, -12.5], 'k--', lw=1)
 
 offset = 0.3
 #ax.plot([0, offset], [-12.5, -12.5], color='black')
 ax.plot([-offset, -offset], [-12.5, -12.5+offset], color='black')
 ax.plot([-offset, 0], [-12.5+offset, -12.5+offset], color='black')
 
 ax.text(0.2, -6.25, r"$6$", fontsize=14, va='center', color='black', style='italic')
 ax.text(-2.05, -13.1, r"$3$", fontsize=14, ha='center', color='black', style='italic')
 
 ax.set_aspect('equal')
 plt.axis('off') 
[/python]

Answer: $56.55$
Find the volume of the solid figure shown.

Round your answer to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(8,9))
 
 circle = plt.Circle((0, 0), 3.5, fill=False, linewidth=3, color='#2c3e50')
 ax.add_patch(circle)
 
 theta = np.linspace(0, 2*np.pi, 100)
 x = 3.5 * np.cos(theta)
 y = 1.75 * np.sin(theta)
 ax.plot(x, y, linestyle="--", color='#2c3e50')
 ax.plot(0, 0, 'o', color='#2c3e50', markersize=10)
 
 ax.plot([0, 3.5], [0, 0], color='#2c3e50', linestyle='-', linewidth=3)
 
 ax.text(1.75, 0.1, '3 cm', fontsize=16)
 
 ax.set_aspect('equal')
 ax.axis('off')
 ax.set_xlim(-3.5, 3.5)
 ax.set_ylim(-3.5, 3.5) 
[/python]

Answer: $113.10$
Find the volume of the solid figure shown.

Round your answer to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(5,5))
 
 circle = plt.Circle((0, 0), 3, fill=False, linewidth=3, color='#2c3e50')
 ax.add_patch(circle)
 
 theta = np.linspace(0, 2*np.pi, 100)
 y = 3 * np.cos(theta)
 x = 1.1 * np.sin(theta)
 ax.plot(x, y, linestyle="--", color='#2c3e50')
 ax.plot(0, 0, 'o', color='#2c3e50', markersize=10)
 
 ax.plot([0, 0], [3, -3], color='#2c3e50', linestyle='-', linewidth=3)
 
 ax.text(0.2, 0.1, '4 cm', fontsize=16)
 
 ax.set_aspect('equal')
 ax.axis('off')
 ax.set_xlim(-3.5, 3.5)
 ax.set_ylim(-3.5, 3.5) 
[/python]

Answer: $33.51$
The diameter of large solid figure is 12 centimeters. Find the volume of the solid.

Round your answer to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(8,8))
 
 R = 9
 r = 4.5
 
 theta = np.linspace(0, np.pi, 500)
 x_outer = R * np.cos(theta)
 y_outer = R * np.sin(theta)
 
 theta_inner = np.linspace(0, np.pi, 200)
 x_inner = r * np.cos(theta_inner)
 y_inner = r * np.sin(theta_inner) + (R-r)
 
 ax.plot(x_outer, -y_outer, color='navy', lw=3, zorder=1)
 
 ellipse_x = R * np.cos(np.linspace(0, 2*np.pi, 200))
 ellipse_y = 0.33 * R * np.sin(np.linspace(0, 2*np.pi, 200))
 ax.plot(ellipse_x, ellipse_y, color='navy', lw=3)
 
 dash_theta = np.linspace(0, 2*np.pi, 400)
 dash_x = r * np.cos(dash_theta)
 dash_y = 0.33 * r * np.sin(dash_theta)
 ax.plot(dash_x[:200], dash_y[:200], 'k--', lw=2)
 ax.plot(dash_x[200:], dash_y[200:], 'k', lw=2)
 
 ax.plot(x_inner, y_inner-R+r, color='navy', lw=3, zorder=2)
 
 ax.annotate('', xy=(-r, 5), xytext=(r, 5), arrowprops=dict(arrowstyle='<->', lw=1.8))
 ax.text(0, 5.25, "6 cm", ha='center', va='bottom', fontsize=14)
 
 ax.annotate('', xy=(-R, -9.2), xytext=(R, -9.2), arrowprops=dict(arrowstyle='<->', lw=1.8))
 
 ax.set_aspect('equal')
 ax.axis('off') 
[/python]

Answer: $508.94$
Find the volume of the solid shown, giving your answer correct to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(5,5))
 
 circle = plt.Circle((0, 0), 3, fill=False, linewidth=3, color='#2c3e50')
 ax.add_patch(circle)
 
 theta = np.linspace(0, 2*np.pi, 100)
 x = 3 * np.cos(theta)
 y = 1.5 * np.sin(theta)
 ax.plot(x, y, linestyle="--", color='#2c3e50')
 ax.plot(0, 0, 'o', color='#2c3e50', markersize=10)
 
 ax.plot([0, 3], [0, 0], color='#2c3e50', linestyle='-', linewidth=3)
 
 ax.text(1, 0.1, '8.8 cm', fontsize=16)
 
 ax.set_aspect('equal')
 ax.axis('off')
 ax.set_xlim(-3.5, 3.5)
 ax.set_ylim(-3.5, 3.5) 
[/python]

Answer: $2854.54$
Find the volume of the solid shown, giving your answer correct to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(5,5))
 
 circle = plt.Circle((0, 0), 3, fill=False, linewidth=3, color='#2c3e50')
 ax.add_patch(circle)
 
 theta = np.linspace(0, 2*np.pi, 100)
 x = 3 * np.cos(theta)
 y = 1.5 * np.sin(theta)
 ax.plot(x, y, linestyle="--", color='#2c3e50')
 ax.plot(0, 0, 'o', color='#2c3e50', markersize=10)
 
 ax.plot([0, 3], [0, 0], color='#2c3e50', linestyle='-', linewidth=3)
 
 ax.text(1, 0.1, '22.36m', fontsize=16)
 
 ax.set_aspect('equal')
 ax.axis('off')
 ax.set_xlim(-3.5, 3.5)
 ax.set_ylim(-3.5, 3.5) 
[/python]

Answer: $46827.83$
Find the volume of the following solid figure.

Round your answer to three decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(6,4))
 
 radius = 7
 
 theta = np.linspace(np.pi, 2*np.pi, 100)
 x_semi = radius * np.cos(theta)
 y_semi = radius * np.sin(theta)
 ax.fill(x_semi, -radius-y_semi, color='#C6E6EB', zorder=1)
 ax.plot(x_semi, -radius-y_semi, color='#22313F', linewidth=3)
 
 
 
 a = radius
 b = radius/3
 t = np.linspace(0, 2*np.pi, 100)
 x_ellipse = a * np.cos(t)
 y_ellipse = -radius + b * np.sin(t)
 ax.plot(x_ellipse, y_ellipse, color='#22313F', linewidth=2, linestyle='--')
 ax.fill_between(x_ellipse, y_ellipse, -radius, color='#8ED7DE', alpha=0.7, zorder=2)
 
 ax.plot([0, 0], [0, -radius], color='#22313F', linestyle='--', linewidth=2)
 
 ax.plot([0, a], [-radius, -radius], color='#22313F', linestyle='--', linewidth=2)
 
 ax.text(0.3, -radius/2, '6', fontsize=16)
 
 ax.set_aspect('equal')
 ax.axis('off') 
[/python]

Answer: $452.389$
Find the volume of the following solid figure.

Round your answer to three decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 fig, ax = plt.subplots(figsize=(6,4))
 
 radius = 6
 
 theta = np.linspace(np.pi, 2*np.pi, 100)
 x_semi = radius * np.cos(theta)
 y_semi = radius * np.sin(theta)
 ax.fill(x_semi, -radius-y_semi, color='#C6E6EB', zorder=1)
 ax.plot(x_semi, -radius-y_semi, color='#22313F', linewidth=3)
 
 
 
 a = radius
 b = radius/3
 t = np.linspace(0, 2*np.pi, 100)
 x_ellipse = a * np.cos(t)
 y_ellipse = -radius + b * np.sin(t)
 ax.plot(x_ellipse, y_ellipse, color='#22313F', linewidth=2, linestyle='--')
 ax.fill_between(x_ellipse, y_ellipse, -radius, color='#8ED7DE', alpha=0.7, zorder=2)
 
 ax.plot([0, 0], [0, -radius], color='#22313F', linestyle='--', linewidth=2)
 
 ax.plot([0, a], [-radius, -radius], color='#22313F', linestyle='--', linewidth=2)
 
 ax.text(-3.3, -radius/2, '$25.36cm$', fontsize=16)
 
 ax.set_aspect('equal')
 ax.axis('off') 
[/python]

Answer: $34159.095$
Consider the following solid figure with a height of 35 cm. Find the surface area of the solid figure.

Round your answer to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 radius = 20
 height = 70
 
 theta = np.linspace(0, 2*np.pi, 100)
 x = radius * np.cos(theta)
 y = radius/2 * np.sin(theta)
 
 fig, ax = plt.subplots(figsize=(5,8))
 
 ax.plot(x, y + height, 'k')
 
 ax.plot(x[:50], y[:50], 'k--')
 ax.plot(x[50:], y[50:], 'k')
 
 ax.plot([radius, radius], [height, 0], 'k')
 ax.plot([-radius, -radius], [height, 0], 'k')
 
 ax.annotate(
     '', xy=(radius + 4, height), xytext=(radius + 4, 0),
     arrowprops=dict(arrowstyle='-', lw=1.5)
 )
 ax.plot([radius+3, radius+5], [height, height], color="black")
 ax.plot([radius+3, radius+5], [0, 0], color="black")
 ax.text(radius + 6, height/2, f'{35}', va='center', fontsize=12)
 
 ax.annotate(
     '', xy=(0, height+0.5), xytext=(-radius-0.5, height+0.5),
     arrowprops=dict(arrowstyle='-', lw=1.5)
 )
 ax.plot([0, 0], [height-0.2, height+1.2], color="black")
 ax.plot([-radius, -radius], [height-0.2, height+1.2], color="black")
 ax.text(-radius/2, height + 1, f'{10}', ha='center', fontsize=12)
 
 ax.set_xlim(-radius*1.5, radius*1.5)
 ax.set_ylim(-radius, height + radius)
 ax.set_aspect('equal')
 ax.axis('off') 
[/python]

Answer: $2827.43$
The following solid figure has a base radius of 21 m.

Find the surface area.

Give your answer to the nearest two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 # Cylinder dimensions
 radius = 17
 length = 36
 
 fig, ax = plt.subplots(figsize=(8, 5))
 
 # Draw the side edges of the cylinder
 ax.plot([0, length], [radius, radius], 'k')
 ax.plot([0, length], [-radius, -radius], 'k')
 
 # Draw the front ellipse (solid)
 theta = np.linspace(0, 2 * np.pi, 100)
 x_front = length + radius/2 * np.cos(theta)
 y_front = radius * np.sin(theta)
 ax.plot(x_front, y_front, 'k')
 
 # Draw the back ellipse (dashed)
 x_back = 0 + radius/2 * np.cos(theta)
 y_back = radius * np.sin(theta)
 ax.plot(x_back, y_back, 'k--')
 
 # Draw the radius on the front face
 ax.plot([length, length], [0, radius], 'k')
 ax.plot(length, 0, 'bo', markersize=3) # Center point
 
 # Add dimension lines and labels
 # Length
 ax.annotate('', xy=(0, -radius-5), xytext=(length, -radius-5),
             arrowprops=dict(arrowstyle='<->', lw=1.5))
 ax.text(length/2, -radius-8, '49', ha='center', va='top', fontsize=12)
 
 # Radius
 ax.annotate('', xy=(length, 0), xytext=(length, radius),
             arrowprops=dict(arrowstyle='<->', lw=1.5))
 ax.text(length+3, radius/2, '21', va='center', fontsize=12)
 
 # Set aspect, limits and remove axes
 ax.set_aspect('equal')
 ax.axis('off')
 ax.set_xlim(-radius*0.7, length+radius*1.2)
 ax.set_ylim(-radius*1.3, radius*1.3)
 
 plt.show() 
[/python]

Answer: $9236.28$
Find the surface area of the given solid figure. The radius is 98. All measurements in the diagram are in mm.

Round your answer to two decimal places.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 # Cylinder parameters
 radius = 80
 height = 82
 
 # Plot setup
 fig, ax = plt.subplots(figsize=(7,5))
 ax.set_xlim(-120, 160)
 ax.set_ylim(-70, 90)
 
 # Angles for ellipse
 theta = np.linspace(0, 2*np.pi, 100)
 
 # Top ellipse (at y = height/2)
 x_top = radius * np.cos(theta)
 y_top = 0.25 * radius * np.sin(theta) + height/2
 
 # Bottom ellipse (at y = -height/2)
 x_bot = radius * np.cos(theta)
 y_bot = 0.25 * radius * np.sin(theta) - height/2
 
 # Draw top and bottom ellipses
 ax.plot(x_top, y_top, 'k')
 ax.plot(x_bot, y_bot, 'k', linestyle='dashed')
 
 # Draw sides
 ax.plot([-radius, -radius], [-height/2, height/2], 'k')
 ax.plot([radius, radius], [-height/2, height/2], 'k')
 
 # Draw radius annotation
 ax.plot([0, radius], [height/2, height/2], 'b', marker='o')
 ax.text(radius/2, height/2+7, '98', ha='center', va='bottom', fontsize=12)
 
 # Draw height annotation with double arrow
 ax.annotate(
     '', xy=(radius+15, height/2), xytext=(radius+15, -height/2),
     arrowprops=dict(arrowstyle='<->', linewidth=2))
 ax.text(radius+22, 0, '80', va='center', fontsize=12)
 
 # Remove axes
 ax.axis('off')
 
 plt.show() 
[/python]

Answer: $109603.88$
Find the height $h$ mm of this closed solid figure if its surface area (S) is 27288(mm)^2.

Round your answer to the nearest whole number.

[python]
 import matplotlib.pyplot as plt
 import numpy as np
 
 radius = 20
 height = 70
 
 theta = np.linspace(0, 2*np.pi, 100)
 x = radius * np.cos(theta)
 y = radius/2 * np.sin(theta)
 
 fig, ax = plt.subplots(figsize=(5,8))
 
 ax.plot(x, y + height, 'k')
 
 ax.plot(x[:50], y[:50], 'k--')
 ax.plot(x[50:], y[50:], 'k')
 
 ax.plot([radius, radius], [height, 0], 'k')
 ax.plot([-radius, -radius], [height, 0], 'k')
 
 ax.annotate(
     '', xy=(radius + 4, height), xytext=(radius + 4, 0),
     arrowprops=dict(arrowstyle='-', lw=1.5)
 )
 ax.plot([radius+3, radius+5], [height, height], color="black")
 ax.plot([radius+3, radius+5], [0, 0], color="black")
 ax.text(radius + 6, height/2, '$hmm$', va='center', fontsize=12)
 
 ax.annotate(
     '', xy=(0, height+0.5), xytext=(-radius-0.5, height+0.5),
     arrowprops=dict(arrowstyle='-', lw=1.5)
 )
 ax.plot([0, 0], [height-0.2, height+1.2], color="black")
 ax.plot([-radius, -radius], [height-0.2, height+1.2], color="black")
 ax.text(-radius/2, height + 1, '$43mm$', ha='center', fontsize=12)
 
 ax.set_xlim(-radius*1.5, radius*1.5)
 ax.set_ylim(-radius, height + radius)
 ax.set_aspect('equal')
 ax.axis('off') 
[/python]

Answer: $58$