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Lesson 8 – Similar and Equivalent Figures and Solids

Mathematics 506 - Technical & Scientific Option
Lesson 8 – Similar and Equivalent Figures and Solids Review
Area and Volume of Solids
Area
Prism:
𝐴𝐿 = 𝑃𝐡 × π‘•
Volume
𝑉 = 𝐴𝐡 × π‘•
𝐴𝑇 = 2𝐴𝐡 + 𝐴𝐿
𝑃𝐡 × π‘ 
2
𝐴𝑇 = 𝐴𝐡 + 𝐴𝐿
𝐴𝐡 × π‘•
3
Pyramid:
𝐴𝐿 =
𝑉=
Cylinder:
𝐴𝐿 = 𝑃𝐡 × π‘• = 2πœ‹π‘Ÿπ‘•
𝑉 = 𝐴𝐡 × π‘•
𝐴𝑇 = 2𝐴𝐡 + 𝐴𝐿
Cone:
𝐴𝐿 = πœ‹π‘Ÿπ‘ 
𝐴𝐿 = 𝐴𝐡 + 𝐴𝐿
Sphere:
𝐴𝑇 = 4πœ‹π‘Ÿ 2
𝑉=
𝐴𝐡 × π‘•
3
4πœ‹π‘Ÿ 3
𝑉=
3
Mathematics 506 - Technical & Scientific Option
Similar Solids
Using k as the scale factor of two similar solids, we get the following ratios:
π‘˜1 = π‘Ÿπ‘Žπ‘‘π‘–π‘œ π‘œπ‘“ 𝑙𝑒𝑛𝑔𝑑𝑕𝑠 (𝑒π‘₯.
π‘˜2 = π‘Ÿπ‘Žπ‘‘π‘–π‘œ π‘œπ‘“ π‘Žπ‘Ÿπ‘’π‘Žπ‘  (𝑒π‘₯.
π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  1
π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  2
𝐴𝐡 1
𝐴𝐡 2
π‘˜3 = π‘Ÿπ‘Žπ‘‘π‘–π‘œ π‘œπ‘“ π‘£π‘œπ‘™π‘’π‘šπ‘’π‘  (𝑒π‘₯.
)
)
𝑉1
𝑉2
)
Equivalent Figures
Two plane figures (ie 2 dimensional) are equivalent if they have the same area.
Ex.
4 π‘π‘š
9 π‘π‘š
6 π‘π‘š
π΄π‘†π‘žπ‘’π‘Žπ‘Ÿπ‘’ = 6 × 6 = 36π‘π‘š2
π΄π‘…π‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ = 9 × 4 = 36π‘π‘š2
Equivalent Solids
Two solids are considered equivalent if they have the same volume.
3 π‘π‘š
Ex.
4 π‘π‘š
3 π‘π‘š
6 π‘π‘š
2
𝑉 = πœ‹π‘Ÿ 𝑕 = 36πœ‹ = 113.1 π‘π‘š
3
πœ‹π‘Ÿ 2 𝑕
𝑉=
= 36πœ‹ = 113.1 π‘π‘š3
3

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