LiAuto-DSR/GeoGramBench huggingface dataset:
Find the surface area of the solid figure shown. The height is 21 cm.

[python]
import matplotlib.pyplot as plt

fig, ax = plt.subplots(figsize=(4, 3))


A = [0, 4.8]
B = [-2.5, 0]
C = [2.5, 0]
E = [10, 10]
F = [12.5, 5.2]
ax.plot([A[0], B[0]], [A[1], B[1]], color='black')
ax.plot([B[0], C[0]], [B[1], C[1]], color='black')
ax.plot([C[0], A[0]], [C[1], A[1]], color='black')

ax.plot([E[0], F[0]], [E[1], F[1]], color='black')


ax.plot([A[0], E[0]], [A[1], E[1]], color='black')
ax.plot([C[0], F[0]], [C[1], F[1]], color='black')

ax.text(-1, -1, "$\\bf{12\\ cm}$", fontsize=11)
ax.text(1.7, 2.4, "$\\bf{10\\ cm}$", fontsize=11)
ax.plot([B[0] - 0.5, B[0] - 0.5], [A[1], 0], color='black')
ax.plot([B[0] - 0.7, B[0] - 0.3], [A[1], A[1]], color='black')
ax.plot([B[0] - 0.7, B[0] - 0.3], [0, 0], color='black')

ax.plot([0, 0.5, 0.5], [0.5, 0.5, 0], color='black')


ax.plot([0, 0], [A[1], 0], color='black', linestyle='--')

ax.plot([C[0], F[0]], [C[1], F[1]], color='black')

ax.plot([C[0], F[0]], [C[1], F[1]], color='black')
ax.plot([B[0]/2 - 0.4, B[0]/2 + 0.4], [2.6, 2.3], color='black')
ax.plot([C[0]/2 - 0.4, C[0]/2 + 0.4], [2.3, 2.6], color='black')

ax.text(B[0] - 3, 2.4, "$\\bf{8\\ cm}$", fontsize=11)


ax.axis('off')
plt.tight_layout()
plt.show()[/python]

Answer: $768$
Find the surface area of the solid figure shown. The height is 20 cm.

[python]
import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots(figsize=(5,7))

A = np.array([0, 0])
B = np.array([10, 0])
C = np.array([3.6, -4.8])

A0 = C * 0.9
B0 = A0 + 0.1* (B - C)
C0 = C + 0.1* (B - C)


dy = 12
A1 = A + np.array([0, dy])
B1 = B + np.array([0, dy])
C1 = C + np.array([0, dy])

A10 = A0 + np.array([0, dy])
B10 = B0 + np.array([0, dy])
C10 = C0 + np.array([0, dy])

for p1, p2 in [(B,C), (C,A),
               (A1,B1), (B1,C1), (C1,A1),
               (A,A1), (B,B1), (C,C1),
               (A0, B0), (B0, C0), (A10, B10), (B10, C10)]:
    ax.plot([p1[0], p2[0]],[p1[1], p2[1]],'k', lw=1.5)

ax.plot([A[0], B[0]], [A[1], B[1]], 'k--')
ax.text(1, -2.5, r'$6\,cm$', fontsize=14, ha='center', va='top', style='italic')
ax.text(4, dy+0.45, r'$10\,cm$', fontsize=14, ha='left', va='center', style='italic')
ax.text(9, -2.65, r'$8\,cm$', fontsize=14, ha='right', va='center', style='italic')

ax.axis('equal')
ax.axis('off')
plt.show()[/python]

Answer: $528$
Find the surface area of the solid prism. The bottom face has edges of 18cm and 8cm.

[python]
import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots(figsize=(8,4))

A = np.array([0, 0])
B = np.array([14, 0])
C = np.array([12, 3])
D = np.array([2, 3])

w = 5
A1 = A + np.array([w, w])
B1 = B + np.array([w, w])
C1 = C + np.array([w, w])
D1 = D + np.array([w, w])

front = np.array([A,B,C,D,A])
ax.plot(front[:,0], front[:,1], 'k', lw=3)
back = np.array([B1,C1,D1])
ax.plot(back[:,0], back[:,1], 'k', lw=3)
ax.plot([B[0],B1[0]], [B[1],B1[1]], 'k', lw=3)
ax.plot([C[0],C1[0]], [C[1],C1[1]], 'k', lw=3)
ax.plot([D[0],D1[0]], [D[1],D1[1]], 'k', lw=3)

ax.plot([7,7],[0,3],'k--', lw=2)
ax.text(7.3,1.1,'3 cm',fontsize=13, va='center')

ax.text(7, 3.5, '10 cm', fontsize=13, ha='center', va='bottom')
ax.text(0, 2.2, '5 cm', fontsize=13, ha='right', va='center')
ax.text(18, 7, '5 cm', fontsize=13, ha='left', va='center')

ax.set_aspect('equal')
ax.axis('off')
ax.set_xlim(-2,20)
ax.set_ylim(-1,10)
plt.tight_layout()
plt.show()[/python]

Answer: $338$
Find the surface area of this solid. The thickness of the solid is 3cm.

[python]
import matplotlib.pyplot as plt
import numpy as np


dx, dy = -3, 3

A = np.array([0, 0])
B = np.array([w1, 0])
C = np.array([w1-w2, h1])
D = np.array([0, h1])

E = np.array([w1, h2])
F = np.array([w1-w2, h2])

A_ = A + np.array([dx, dy])
B_ = B + np.array([dx, dy])
C_ = C + np.array([dx, dy])
D_ = D + np.array([dx, dy])
F_ = F + np.array([dx, dy])
E_ = E + np.array([dx, dy])

ax = plt.gca()
for p1, p2 in [(A,B),(B,E),(E,F),(F,C), (C, D), (D, A)]:
    ax.plot([p1[0], p2[0]], [p1[1], p2[1]], color='k')

for p1, p2 in [(A,A_),(C,C_),(D,D_), (E,E_), (F,F_), (A_, D_), (D_, C_), (F_, C_), (E_, F_)]:
    ax.plot([p1[0],p2[0]], [p1[1],p2[1]], color='k')


ax.plot([E[0], F[0]], [E[1], F[1]], color='k')

ax.text(-4, h1/2+dy, '8cm', va='center', ha='right', fontsize=13, fontstyle='italic')
ax.text(w1/2, -1.4, '17cm', va='top', ha='center', fontsize=13, fontstyle='italic')
ax.text(w1 + 1, h2/2, '16cm', va='center', ha='left', fontsize=13, fontstyle='italic')
ax.text(w1-w2/2+dx, h2+dy+0.6, '4cm', va='bottom', ha='center', fontsize=13, fontstyle='italic')

ax.set_aspect('equal')
ax.axis('off')
plt.xlim(-6, 28)
plt.ylim(-4, 22)
plt.tight_layout()
plt.show()[/python]

Answer: $534$
Find the surface area of the following solid prism. The right rectangle surface has one edge of 10cm.

Give your answer to the nearest one decimal place.

[python]
import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots(figsize=(7,7))

height = 10
depth = 3
width = 13

dx, dy = 5, 3.5

A = np.array([0,0])
B = np.array([width,0])
C = np.array([width, depth])
D = np.array([0,height])

A1 = A + np.array([dx,dy])
B1 = B + np.array([dx,dy])
C1 = C + np.array([dx,dy])
D1 = D + np.array([dx,dy])


ax.plot([A[0],B[0]],[A[1],B[1]],'k')
ax.plot([B[0],C[0]],[B[1],C[1]],'k')
ax.plot([C[0],D[0]],[C[1],D[1]],'k')
ax.plot([D[0],A[0]],[D[1],A[1]],'k')
ax.plot([A1[0],B1[0]],[A1[1],B1[1]],'k--')
ax.plot([B1[0],C1[0]],[B1[1],C1[1]],'k')
ax.plot([B1[0],B[0]],[B1[1],B[1]],'k')
ax.plot([C1[0],D1[0]],[C1[1],D1[1]],'k')
ax.plot([D1[0],A1[0]],[D1[1],A1[1]],'k--')
ax.plot([D[0],D1[0]],[D[1],D1[1]],'k')
ax.plot([C[0],C1[0]], [C[1],C1[1]], 'k')

ax.plot([A[0], A1[0]], [A[1], A1[1]], 'k--')

ax.text(width/2-0.8, -1.1, r'$13\,cm$', fontsize=15, style='italic')
ax.text(-1.0, height/2, r'$10\,cm$', fontsize=15, style='italic', va='center', ha='right')
ax.text(width+dx+0.5, dy+depth/2, r'$3\,cm$', fontsize=15, style='italic', va='center', ha='left')
ax.text((A1[0]+B1[0])/2+0.5, (C1[1]+D1[1])/2+1, r'$y\,cm$', fontsize=15, style='italic', ha='center')

ax.set_aspect('equal')
ax.axis('off')
plt.xlim(-4, width+dx+depth+2)
plt.ylim(-3, max(height,dy+height)+depth+2)
plt.tight_layout()
plt.show()[/python]

Answer: $577.0$
Find the total surface area of the solid figure shown.

[python]
import matplotlib.pyplot as plt

fig, ax = plt.subplots(figsize=(4, 3))


A = [0, 4.8]
B = [-2.5, 0]
C = [2.5, 0]
E = [10, 10]
F = [12.5, 5.2]
ax.plot([A[0], B[0]], [A[1], B[1]], color='black')
ax.plot([B[0], C[0]], [B[1], C[1]], color='black')
ax.plot([C[0], A[0]], [C[1], A[1]], color='black')

ax.plot([E[0], F[0]], [E[1], F[1]], color='black')


ax.plot([A[0], E[0]], [A[1], E[1]], color='black')
ax.plot([C[0], F[0]], [C[1], F[1]], color='black')

ax.text(-1, -1, "$\\bf{12\\ cm}$", fontsize=11)
ax.text(1.7, 2.4, "$\\bf{10\\ cm}$", fontsize=11)
ax.plot([B[0] - 0.5, B[0] - 0.5], [A[1], 0], color='black')
ax.plot([B[0] - 0.7, B[0] - 0.3], [A[1], A[1]], color='black')
ax.plot([B[0] - 0.7, B[0] - 0.3], [0, 0], color='black')

ax.plot([0, 0.5, 0.5], [0.5, 0.5, 0], color='black')


ax.plot([0, 0], [A[1], 0], color='black', linestyle='--')

ax.plot([C[0], F[0]], [C[1], F[1]], color='black')

ax.plot([C[0], F[0]], [C[1], F[1]], color='black')
ax.plot([B[0]/2 - 0.4, B[0]/2 + 0.4], [2.6, 2.3], color='black')
ax.plot([C[0]/2 - 0.4, C[0]/2 + 0.4], [2.3, 2.6], color='black')

ax.text(B[0] - 3, 2.4, "$\\bf{x\\ cm}$", fontsize=11)


ax.axis('off')
plt.tight_layout()
plt.show()[/python]

Answer: $608$
Find the surface area of the figure shown. The height is 3 cm.

Give your answer to the nearest two decimal places.

[python]
import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots(figsize=(7,7))

height = 10
width = 15
width1 = 8

dx, dy = 5, 3.5

A = np.array([0,0])
B = np.array([width,0])
C = np.array([width1, height])
D = np.array([0, height])

A1 = A + np.array([dx,dy])
B1 = B + np.array([dx,dy])
C1 = C + np.array([dx,dy])
D1 = D + np.array([dx,dy])


ax.plot([A[0],B[0]],[A[1],B[1]],'k')
ax.plot([B[0],C[0]],[B[1],C[1]],'k')
ax.plot([C[0],D[0]],[C[1],D[1]],'k')
ax.plot([D[0],A[0]],[D[1],A[1]],'k')
ax.plot([B1[0],C1[0]],[B1[1],C1[1]],'k')
ax.plot([B1[0],B[0]],[B1[1],B[1]],'k')
ax.plot([C1[0],D1[0]],[C1[1],D1[1]],'k')
ax.plot([D[0],D1[0]],[D[1],D1[1]],'k')
ax.plot([C[0],C1[0]], [C[1],C1[1]], 'k')


ax.text(width/2-0.8, -1.1, r'$12\,cm$', fontsize=15, style='italic')
ax.text(-1.0, height/2, r'$5\,cm$', fontsize=15, style='italic', va='center', ha='right')
ax.text((A[0]+C[0])/2+0.5, C[1]+0.5, r'$9\,cm$', fontsize=15, style='italic', ha='center')

ax.set_aspect('equal')
ax.axis('off')
plt.xlim(-4, width+dx+depth+2)
plt.ylim(-3, max(height,dy+height)+depth+2)
plt.tight_layout()
plt.show()[/python]

Answer: $200.49$
Find the surface area of the figure shown.

[python]
import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots(figsize=(8,6))

bar_width = 16
bar_depth = 3
bar_height = 3
vert_width = 4
vert_height = 7
vert_depth = 1

dx, dy = 2, 1.2

A = np.array([0, 0])
B = np.array([bar_width, 0])
C = np.array([bar_width, bar_height])
D = np.array([0, bar_height])

A1 = A + np.array([dx, dy])
B1 = B + np.array([dx, dy])
C1 = C + np.array([dx, dy])
D1 = D + np.array([dx, dy])

left_margin = (bar_width - vert_width) / 2
right_margin = left_margin + vert_width

E = np.array([left_margin, 0])
F = E + np.array([vert_width, 0])
G = F + np.array([0, -vert_height])
H = E + np.array([0, -vert_height])

E1 = E + np.array([dx, dy])
F1 = F + np.array([dx, dy])
G1 = G + np.array([dx, dy])
H1 = H + np.array([dx, dy])

for X, Y in [(A,B), (B,C), (C,D), (D,A), (B,B1), (C,C1), (D,D1), (B1,C1), (C1,D1)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')

for X, Y in [(F,G),(G,H),(H,E),(G,G1)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')

ax.plot([F1[0],G1[0]], [F1[1]-1.3,G1[1]], 'k') 


ax.text(bar_width/2, 2.5, r"$12\,cm$", ha='center', va='top', fontsize=14, style='italic')
ax.text(-0.7, bar_height/2, r"$1\,cm$", ha='right', va='center', fontsize=14, style='italic')
ax.text(bar_width+dx-1, bar_height/2-dy-0.5, r"$1\,cm$", ha='left', va='center', fontsize=14, style='italic')
ax.text((left_margin+vert_width/2 + 4), -vert_height-0.5, r"$1\,cm$",  ha='center', va='bottom', fontsize=14, style='italic')

ax.text((left_margin+vert_width/2), -vert_height-1, r"$4\,cm$",  ha='center', va='bottom', fontsize=14, style='italic')
ax.text(left_margin-0.2, -vert_height/2, r"$7\,cm$", ha='right', va='center', fontsize=14, style='italic')

ax.plot([bar_width/6, bar_width/6], [0.35, -0.35], 'k')
ax.plot([bar_width*5/6, bar_width*5/6], [0.35, -0.35], 'k')

ax.set_aspect('equal')
ax.axis('off')
plt.xlim(-2, 20)
plt.ylim(-15, 10)
plt.tight_layout()
plt.show()[/python]

Answer: $120$
Find the surface area of the solid figure. The height is 6 cm.

Round your answer to one decimal place.

[python]
import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots(figsize=(5,4))

a = 10
h = a * np.sqrt(3) / 2

O = np.array([0, 0])
E = np.array([0, 10])
A = np.array([-a/2 - h/2, -h/2])
B = np.array([a - h/2, -h/2])

dx, dy = h/2, h
A1 = A + np.array([dx, dy])
B1 = B + np.array([dx, dy])


for X, Y in [(A, A1), (B,B1), (A,B), (A,E), (A1,E), (B,E), (B1,E)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')


ax.plot([E[0],0],[E[1],0], 'k--')
ax.plot([A1[0],B1[0]],[A1[1],B1[1]], 'k--')


ax.text((B[0]+B1[0])/2+0.5, 0, r"$10\,cm$", fontsize=12, style='italic')
ax.plot([(A[0]+B[0])/2, (A[0]+B[0])/2], [A[1]-0.3, A[1]+0.3], 'k')
ax.plot([(A1[0]+B1[0])/2, (A1[0]+B1[0])/2], [A1[1]-0.3, A1[1]+0.3], 'k')
ax.plot([(A[0]+A1[0])/2-0.25, (A[0]+A1[0])/2+0.25], [0.20, -0.20], 'k', lw=2)
ax.plot([(B[0]+B1[0])/2-0.25, (B[0]+B1[0])/2+0.25], [0.20, -0.20], 'k', lw=2)

ax.plot([0, 1, 1], [1, 1, 0], color='black')

ax.set_aspect('equal')
ax.axis('off')
plt.xlim(-a, a)
plt.ylim(-a, h+2)
plt.tight_layout()
plt.show()[/python]

Answer: $256.2$
Find the surface area of the solid shown. The radius is 3cm.

Round your answer to two decimal places.

[python]
import matplotlib.pyplot as plt
import matplotlib.patches as patches

fig, ax = plt.subplots(figsize=(4,3))


ellipse = patches.Ellipse((0, 0), 16, 6, fill=False, color='black', linewidth=2)
ax.add_patch(ellipse)
ax.plot([-8, 0, 8], [0, 14, 0], color='black')
ax.plot([0, 8], [0, 0], color='black')

ax.text(4.5, 7.5, r"$10\,cm$", fontsize=12, style='italic')

ax.set_aspect('equal')
ax.set_xlim(-12, 12)
ax.set_ylim(-4, 18)
ax.axis('off')

plt.show()[/python]

Answer: $122.52$
The solid figure has a diameter of 4cm. Find the surface area of the solid formed, correct to two decimal places.

[python]
import matplotlib.pyplot as plt
import matplotlib.patches as patches

fig, ax = plt.subplots(figsize=(5,5))


width = 20
height = 4
ellipse1 = patches.Ellipse((0, 0), width, height, fill=False, color='black', linewidth=2)
ellipse2 = patches.Ellipse((0, 10), width/2, height/2, fill=False, color='black', linewidth=2)
ax.add_patch(ellipse1)
ax.add_patch(ellipse2)
ax.plot([-width/2, -width/4, width/4, width/2, -width/2], [0, 10, 10, 0, 0], color='black')
ax.plot([0, 8], [0, 0], color='black')
ax.plot([-width/4, 0, width/4], [10, 20, 10], 'k--', color='black')
ax.plot([1, width/2 + 1.5], [21, 0.5], color='black')
ax.plot([0.5, 1.5], [20.75, 21.5], color='black')
ax.plot([width/4 + 0.75,  width/4 + 1.75], [10.5, 11.25], color='black')
ax.plot([width/2 + 1, width/2 + 2], [0, 0.5], color='black')

ax.plot([0, 0], [10, 10], 'ko')
ax.plot([0, 0], [0, 0], 'ko')

ax.text(width/3 + 2.5, 5, r"$8\,cm$", fontsize=12, style='italic')
ax.text(width/4 - 0.5, 15, r"$8\,cm$", fontsize=12, style='italic')
ax.text(-1, 7.5, r"$2\,cm$", fontsize=11, style='italic')

ax.set_aspect('equal')
ax.set_xlim(-width, width)
ax.set_ylim(-4, 30)
ax.axis('off')

plt.show()[/python]

Answer: $91.11$
Find the surface area of the solid figure. The perpendicular height is 8mm.

Round your answer to two decimal places.

[python]
import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots(figsize=(5,4))

a = 11
h = a * np.sqrt(3) / 2

O = np.array([0, 0])
E = np.array([0, 20])
A = np.array([-a/2 - h/2, -h/2])
B = np.array([a - h/2, -h/2])

dx, dy = h/2, h
A1 = A + np.array([dx, dy])
B1 = B + np.array([dx, dy])


for X, Y in [(B,B1), (A,B), (A,E), (B,E), (B1,E)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')


ax.plot([E[0],0],[E[1],0], 'k--')
ax.plot([A1[0],B1[0]],[A1[1],B1[1]], 'k--')
ax.plot([A[0], A1[0]], [A[1], A1[1]], 'k--')
ax.plot([A1[0], E[0]], [A1[1], E[1]], 'k--')


ax.text((B[0]+B1[0])/2+1, 0, r"$10\,mm$", fontsize=11, style='italic')
ax.text(-dx + 1, A[1]-1.5, r"$10\,mm$", fontsize=11, style='italic')
ax.text(0-4, E[1]/2-2, r"$8\,mm$", fontsize=11, style='italic')

ax.plot([0, 1, 1], [1, 1, 0], color='black')

ax.set_aspect('equal')
ax.axis('off')
plt.xlim(-a, a)
plt.ylim(-a, 20)
plt.tight_layout()
plt.show()[/python]

Answer: $288.6$
Find the surface area of the solid figure given. The slant height is 13.7cm.

Give your answer correct to 2 decimal places.

[python]
import matplotlib.pyplot as plt
import matplotlib.patches as patches

fig, ax = plt.subplots(figsize=(5,4))


ellipse = patches.Ellipse((0, 0), 7, 20, fill=False, color='black', linewidth=2)
ax.add_patch(ellipse)
E = np.array([-25, 0])
ax.plot([0, E[0], 0], [10, 0, -10], color='black', linewidth=2)
ax.plot([0, 0], [0, -10], 'k--', linewidth=2)

ax.text(-2, 1, r"$9.1\,cm$", fontsize=10, style='italic')
ax.text(E[0]/2-3, 6.5, r"$13.7\,cm$", fontsize=10, style='italic')
ax.plot([0, 0], [0, 0], 'ko')

ax.set_aspect('equal')
ax.set_xlim(-28, 5)
ax.set_ylim(-16, 18)
ax.axis('off')

plt.show()[/python]

Answer: $651.82$
Find the surface area of the solid figure shown.

Round your answer to two decimal places.

[python]
import matplotlib.pyplot as plt

fig, ax = plt.subplots(figsize=(4, 3))

h = 8

A = np.array([0, 4.8])
B = np.array([-2.5, 0])
C = np.array([2.5, 0])
E = np.array([10, 10])
F = np.array([12.5, 5.2])
B1 = B + np.array([0, -h])
C1 = C + np.array([0, -h])
F1 = F + np.array([0, -h])


for X, Y in [(B,B1), (C,C1), (B1,C1), (C1,F1), (F,F1)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k', color='black')


ax.plot([A[0], B[0]], [A[1], B[1]], color='black')
ax.plot([B[0], C[0]], [B[1], C[1]], color='black')
ax.plot([C[0], A[0]], [C[1], A[1]], color='black')

ax.plot([E[0], F[0]], [E[1], F[1]], color='black')


ax.plot([A[0], E[0]], [A[1], E[1]], color='black')
ax.plot([C[0], F[0]], [C[1], F[1]], color='black')

ax.plot([B[0] - 0.5, B[0] - 0.5], [A[1], 0], color='black')
ax.plot([B[0] - 0.7, B[0] - 0.3], [A[1], A[1]], color='black')
ax.plot([B[0] - 0.7, B[0] - 0.3], [0, 0], color='black')

ax.plot([0, 0.5, 0.5], [0.5, 0.5, 0], color='black')


ax.plot([0, 0], [A[1], 0], color='black', linestyle='--')

ax.plot([C[0], F[0]], [C[1], F[1]], color='black')

ax.plot([C[0], F[0]], [C[1], F[1]], color='black')
ax.plot([B[0]/2 - 0.4, B[0]/2 + 0.4], [2.6, 2.3], color='black')
ax.plot([C[0]/2 - 0.4, C[0]/2 + 0.4], [2.3, 2.6], color='black')

ax.text(B[0] - 3, 2.4, "$\\bf{3\\ m}$", fontsize=11)
ax.text(B[0] - 3, -h/2, "$\\bf{6\\ m}$", fontsize=11)
ax.text(-1, -h-1.5, "$\\bf{5\\ m}$", fontsize=11)
ax.text((C1[0] + F1[0])/2, (C1[1] + F1[1])/2-1, "$\\bf{10\\ m}$", fontsize=11)


ax.axis('off')
plt.tight_layout()
plt.show()[/python]

Answer: $323.10$
Find the surface area of the figure shown, rounded to two decimal places. The height is 12 cm.

[python]
import matplotlib.pyplot as plt
import matplotlib.patches as patches

fig, ax = plt.subplots(figsize=(6,6))

r = 9
main = patches.Arc(O, r, r, theta1=90, theta2=360, color='black', linewidth=2)
ax.add_patch(main)

dx, dy = 12, 6
dxy = np.array([dx, dy])
O = np.array([0, 0])
O1 = O + np.array([dx, dy])
backend1 = patches.Arc(O1, r, r, theta1=90, theta2=120, color='black', linewidth=2)
backend2 = patches.Arc(O1, r, r, theta1=120, theta2=300, color='black', linewidth=2, linestyle='--')
backend3 = patches.Arc(O1, r, r, theta1=300, theta2=360, color='black', linewidth=2)
ax.add_patch(backend1)
ax.add_patch(backend2)
ax.add_patch(backend3)

A = [0, r/2]
B = [r/2, 0]
C = [-r/4, r*1.7/4]
D = [r/4, -r*1.7/4]

A1 = A + dxy
B1 = B + dxy
C1 = C + dxy
D1 = D + dxy

for X, Y in [(O,O1), (A, O), (B, O), (A1, O1), (B1, O1), (A, A1), (B, B1), (C, C1), (D, D1)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')


ax.text(1, -1, "$\\bf{5\\ cm}$", fontsize=10, style='italic')

ax.set_aspect('equal')
ax.set_xlim(-12, 20)
ax.set_ylim(-6, 12)
ax.axis('off')

plt.show()[/python]

Answer: $520.55$
The middle solid has radius of 6 and height of 11.
Find the surface area of the solid.

Round your answer to two decimal places.

[python]
import matplotlib.pyplot as plt
import numpy as np
import matplotlib.patches as patches

fig, ax = plt.subplots(figsize=(8,4))

width, height = 15, 20

A = np.array([0, 0])
B = np.array([width, 0])
D = A + np.array([0, height])
C = B + np.array([0, height])

dx, dy = -5, 6
dxy = np.array([dx, dy])
A1 = A + dxy
B1 = B + dxy
C1 = C + dxy
D1 = D + dxy

x0 = np.array([25, 0])
A2 = A + x0
B2 = B + x0
C2 = C + x0
D2 = D + x0

A3, B3, C3, D3 = A1 + x0,  B1 + x0, C1 + x0, D1 + x0

for X, Y in [(A, B), (B, C), (C, D), (D, A)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')

for X, Y in [(C1, D1), (D1, A1), (D, D1), (C, C1), (A, A1), (A2, A3)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')

for X, Y in [(A2, B2), (B2, C2), (C2, D2), (D2, A2)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')

for X, Y in [(C3, D3), (D3, A3), (D2, D3), (C2, C3)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')

for X, Y in [(A1, B1), (B1, C1), (A3, B3), (B3, C3), (B, B1), (B2, B3)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k--')

main1 = patches.Arc(((B1[0] + B[0])/2, ((B1[1] + C1[1])/2 + (B[1] + C[1])/2)/2), 3, 13, theta1=0, theta2=360, color='black', linewidth=2, linestyle='--')
ax.add_patch(main1)

main2 = patches.Arc(((A2[0] + A3[0])/2, ((A1[1] + D1[1])/2 + (A[1] + D[1])/2)/2), 3, 13, theta1=270, theta2=90, color='black', linewidth=2)
main3 = patches.Arc(((A2[0] + A3[0])/2, ((A1[1] + D1[1])/2 + (A[1] + D[1])/2)/2), 3, 13, theta1=90, theta2=270, color='black', linewidth=2, linestyle='--')
ax.add_patch(main2)
ax.add_patch(main3)

ax.plot([(B1[0] + B[0])/2, (A2[0] + A3[0])/2], [((B1[1] + C1[1])/2 + (B[1] + C[1])/2)/2 + 6.5, ((B1[1] + C1[1])/2 + (B[1] + C[1])/2)/2 + 6.5], 'k')
ax.plot([(B1[0] + B[0])/2, (A2[0] + A3[0])/2], [((B1[1] + C1[1])/2 + (B[1] + C[1])/2)/2 - 6.5, ((B1[1] + C1[1])/2 + (B[1] + C[1])/2)/2 - 6.5], 'k')

ax.plot([B[0], A2[0]], [-2, -2], 'k')
ax.plot([B[0], B[0]], [-1, -3], 'k')
ax.plot([A2[0], A2[0]], [-1, -3], 'k')
ax.plot([(A2[0] + A3[0])/2, (A2[0] + A3[0])/2], [((A1[1] + D1[1])/2 + (A[1] + D[1])/2)/2, ((A1[1] + D1[1])/2 + (A[1] + D[1])/2)/2+6.5], 'k')

ax.text(B2[0] + 1, C[1]/2, r"$20\,cm$", fontsize=12, style='italic')
ax.text((B2[0] + A2[0])/2, -2, r"$25\,cm$", fontsize=12, style='italic')
ax.text((B2[0] + B3[0])/2, (C3[1] + C2[1])/2, r"$33\,cm$", fontsize=12, style='italic')

ax.set_aspect('equal')
ax.axis('off')
plt.xlim(-8, 60)
plt.ylim(-5, 30)
plt.tight_layout()
plt.show()[/python]

Answer: $8128.50$
Given the following solid. The length of the base is 10cm.
Find the volume of the solid.

[python]
import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots(figsize=(5,4))

a = 10
h = a * np.sqrt(3) / 2

O = np.array([0, 0])
E = np.array([0, 10])
A = np.array([-a/2 - h/2, -h/2])
B = np.array([a - h/2, -h/2])

dx, dy = h/2, h
A1 = A + np.array([dx, dy])
B1 = B + np.array([dx, dy])


for X, Y in [(A, A1), (B,B1), (A,B), (A,E), (A1,E), (B,E), (B1,E)]:
    ax.plot([X[0],Y[0]], [X[1],Y[1]], 'k')


ax.plot([E[0],0],[E[1],0], 'k--')
ax.plot([A1[0],B1[0]],[A1[1],B1[1]], 'k--')


ax.text((B[0]+B1[0])/2+0.5, 0, r"$10\,cm$", fontsize=12, style='italic')
ax.text(-3, 2, r"$6\,cm$", fontsize=12, style='italic')
ax.plot([(A[0]+B[0])/2, (A[0]+B[0])/2], [A[1]-0.3, A[1]+0.3], 'k')
ax.plot([(A1[0]+B1[0])/2, (A1[0]+B1[0])/2], [A1[1]-0.3, A1[1]+0.3], 'k')
ax.plot([(A[0]+A1[0])/2-0.25, (A[0]+A1[0])/2+0.25], [0.20, -0.20], 'k', lw=2)
ax.plot([(B[0]+B1[0])/2-0.25, (B[0]+B1[0])/2+0.25], [0.20, -0.20], 'k', lw=2)

ax.plot([0, 1, 1], [1, 1, 0], color='black')

ax.set_aspect('equal')
ax.axis('off')
plt.xlim(-a, a)
plt.ylim(-a, h+2)
plt.tight_layout()
plt.show()[/python]

Answer: $200$
Consider the two similar solids shown. The larger solid has a radius of 12 cm.


Find the volume of solid A, in simplest exact form.

[python]
import matplotlib.pyplot as plt
import matplotlib.patches as patches

fig, ax = plt.subplots(figsize=(6,6))

r = 6
R = 10
OA = [0, 0]
OB = [10, 0]
mainA1 = patches.Arc(OA, r, r, theta1=00, theta2=360, color='black', linewidth=2)
mainA2 = patches.Arc(OA, r, 2, theta1=00, theta2=360, color='black', linewidth=2)
ax.add_patch(mainA1)
ax.add_patch(mainA2)
ax.plot([0, r/2], [0, 0], color='black')

mainB1 = patches.Arc(OB, R, R, theta1=0, theta2=360, color='black', linewidth=2)
mainB2 = patches.Arc(OB, r, R, theta1=0, theta2=360, color='black', linewidth=2)
ax.add_patch(mainB1)
ax.add_patch(mainB2)
ax.plot([OB[0], OB[0]], [0, R/2], color='black')


ax.text(-1, -0.1, r"$3$", fontsize=10, style='italic')
ax.text(0, r/2+1, r"$A$", fontsize=12, style='italic')
ax.text(OB[0], R/2+1, r"$B$", fontsize=12, style='italic')


ax.set_aspect('equal')
ax.set_xlim(-5, 20)
ax.set_ylim(-6, 6)
ax.axis('off')

plt.show()[/python]

Answer: $36$
Consider the two solids shown. The larger solid has a radius of 12 cm.
What is the ratio of the volume of solid B to solid A?

[python]
import matplotlib.pyplot as plt
import matplotlib.patches as patches

fig, ax = plt.subplots(figsize=(6,6))

r = 6
R = 10
OA = [0, 0]
OB = [10, 0]
mainA1 = patches.Arc(OA, r, r, theta1=00, theta2=360, color='black', linewidth=2)
mainA2 = patches.Arc(OA, r, 2, theta1=00, theta2=360, color='black', linewidth=2)
ax.add_patch(mainA1)
ax.add_patch(mainA2)
ax.plot([0, r/2], [0, 0], color='black')

mainB1 = patches.Arc(OB, R, R, theta1=0, theta2=360, color='black', linewidth=2)
mainB2 = patches.Arc(OB, r, R, theta1=0, theta2=360, color='black', linewidth=2)
ax.add_patch(mainB1)
ax.add_patch(mainB2)
ax.plot([OB[0], OB[0]], [0, R/2], color='black')


ax.text(-1, -0.1, r"$3$", fontsize=10, style='italic')
ax.text(0, r/2+1, r"$A$", fontsize=12, style='italic')
ax.text(OB[0], R/2+1, r"$B$", fontsize=12, style='italic')


ax.set_aspect('equal')
ax.set_xlim(-5, 20)
ax.set_ylim(-6, 6)
ax.axis('off')

plt.show()[/python]

Answer: $64$