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Activity 3.2 (Measurement of Solids)

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Name: JULY ROLAND P. CABRISOS
Year & Section: 1ST YEAR – BSCE_1N_C14
Date: MAY 10, 2021
Score:
Activity 3.2: Measurement of Solids
Direction: Answer the following problem and show your complete solution. For some problems,
leave your answers in terms of 𝝅.
1. Consider a regular pentagonal prism. Find the following:
360°
πœƒ=
𝑛
360°
πœƒ=
5
πœƒ = 72°
πœƒ
= 36°
2
4
tan 36° =
π‘Ž
𝒂 = πŸ“. πŸ“πŸ , 𝒃 = πŸ– , 𝒉 = 𝟐𝟎
i) Area of the base.
1
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ = 2 × × 5 × π‘Ž × π‘
2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ = 5 × 5.51 × π‘
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒔𝒆 = 𝟐𝟐𝟎. πŸ’ π’–π’π’Šπ’•πŸ
ii) Lateral surface area.
𝐿𝑆𝐴 = 5 × π‘ × β„Ž
𝐿𝑆𝐴 = 5 × 8 × 20
𝑳𝑺𝑨 = πŸ–πŸŽπŸŽ π’–π’π’Šπ’•πŸ
iii) Total surface area.
𝑇𝑆𝐴 = 5π‘Žπ‘ + 5π‘β„Ž
𝑇𝑆𝐴 = 5(5.51)(8) + 5(8)(20)
𝑻𝑺𝑨 = 𝟏𝟎𝟐𝟎. πŸ’ π’–π’π’Šπ’•πŸ
iv) Volume of the solid.
5
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = × π‘ × β„Ž
2
5
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = × 8 × 20
2
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 π’”π’π’π’Šπ’… = πŸ’πŸŽπŸŽ π’–π’π’Šπ’•πŸ‘
2. Consider a regular hexagonal pyramid. Find the following:
i) Area of the base.
3√3 2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ =
π‘Ž
2
3√3 2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ =
10
2
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒔𝒆 = πŸπŸ“πŸ—. πŸ–πŸ π’–π’π’Šπ’•πŸ
ii) Lateral surface area.
1
𝐿𝑆𝐴 = 6 × × π‘ × β„Ž
2
1
𝐿𝑆𝐴 = 6 × × 10 × 22
2
𝑳𝑺𝑨 = πŸ”πŸ”πŸŽ π’–π’π’Šπ’•πŸ
iii) Total surface area.
𝑇𝑆𝐴 = π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ + 𝐿𝑆𝐴
𝑇𝑆𝐴 = 259.81 + 660
𝑻𝑺𝑨 = πŸ—πŸπŸ—. πŸ–πŸ π’–π’π’Šπ’•πŸ
iv) Volume of the solid.
10⁄
2
π‘‘π‘Žπ‘›30° =
π‘Ž
5
π‘Ž=
π‘‘π‘Žπ‘›30°
𝒂 = πŸ“√πŸ‘
β„Ž = √𝑠 2 − π‘Ž2
β„Ž = √222 − (5√3)2
𝒉 = 𝟐𝟎. 𝟐𝟐
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = π‘Ž × π‘ × β„Ž
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 5√3 × 10 × 20.22
π‘½π’π’π’–π’Žπ’† = πŸπŸ•πŸ“πŸ. 𝟏𝟎 π’–π’π’Šπ’•πŸ‘
3. The solid to the right is a prism with a pyramid on its top. Find the following:
i) Area of the base.
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ = 𝑠 2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ = 42
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒔𝒆 = πŸπŸ” π’Šπ’.𝟐
ii) Lateral surface area of the prism.
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Ÿπ‘–π‘ π‘š = β„Ž × π‘π‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Ÿπ‘–π‘ π‘š = 5 × (4 + 4 + 4 + 4)
𝑳𝑺𝑨 𝒐𝒇 𝒕𝒉𝒆 π’‘π’“π’Šπ’”π’Ž = πŸ–πŸŽ π’Šπ’.𝟐
iii) Lateral surface area of the pyramid.
Finding the slant height (x)
π‘₯ = √62 + 42
𝒙 = πŸ•. 𝟐𝟏 π’Šπ’.
1
× π‘π‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿπ‘œπ‘“ π‘π‘Žπ‘ π‘’ × π‘ π‘™π‘Žπ‘›π‘‘ β„Žπ‘’π‘–π‘”β„Žπ‘‘ (π‘₯)
2
1
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘ = × (4 + 4 + 4 + 4) × 7.21
2
𝑳𝑺𝑨 𝒐𝒇 𝒕𝒉𝒆 π’‘π’šπ’“π’‚π’Žπ’Šπ’… = πŸ“πŸ•. πŸ”πŸ– π’Šπ’.𝟐
iv) Total surface area of the entire solid.
𝑇𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = 𝐿𝑆𝐴 π‘œπ‘“ π‘π‘Ÿπ‘–π‘ π‘š + 𝐿𝑆𝐴 π‘œπ‘“ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘
𝑇𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = 80 + 57.68
𝑻𝑺𝑨 𝒐𝒇 𝒕𝒉𝒆 π’†π’π’•π’Šπ’“π’† π’”π’π’π’Šπ’… = πŸπŸ‘πŸ•. πŸ”πŸ– π’Šπ’.𝟐
v) Volume of the prism.
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Ÿπ‘–π‘ π‘š = β„Ž × π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘Žπ‘ π‘’
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Ÿπ‘–π‘ π‘š = 5 × 16
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 π’‘π’“π’Šπ’”π’Ž = πŸ–πŸŽ π’Šπ’.πŸ‘
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘ =
vi) Volume of the pyramid.
π‘™π‘€β„Ž
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘ =
3
4×4×6
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘ =
3
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 π’‘π’šπ’“π’‚π’Žπ’Šπ’… = πŸ‘πŸ π’Šπ’.πŸ‘
vii) Volume of the entire solid.
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘Ÿπ‘–π‘ π‘š + π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = 80 + 32
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 π’†π’π’•π’Šπ’“π’† π’”π’π’π’Šπ’… = 𝟏𝟏𝟐 π’Šπ’.πŸ‘
4. The solid to the right is a cube with a cone cut out. Find the following:
i) Area of the base of the cube.
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ = π‘Ž2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘Žπ‘ π‘’ = 62
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒔𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒖𝒃𝒆 = πŸ‘πŸ” π’„π’ŽπŸ
ii) Lateral surface area of the cube.
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ 𝑐𝑒𝑏𝑒 = 4π‘Ž2
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ 𝑐𝑒𝑏𝑒 = 4 × 62
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ 𝑐𝑒𝑏𝑒 = πŸπŸ’πŸ’ π’„π’ŽπŸ
iii) Area of the circle of the cone.
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘–π‘Ÿπ‘π‘™π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹π‘Ÿ 2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘–π‘Ÿπ‘π‘™π‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹ × 22
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 π’„π’Šπ’“π’„π’π’† 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒏𝒆 = πŸ’π… π’„π’ŽπŸ
iv) Lateral surface area of the cone.
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹π‘Ÿ√π‘Ÿ 2 + β„Ž2
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹ × 2√22 + 62
𝑳𝑺𝑨 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒏𝒆 = πŸ’π…√𝟏𝟎 π’„π’ŽπŸ
v) Total surface area of the entire solid.
𝑇𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = 6π‘Ž2
𝑇𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = 6 × 62
𝑻𝑺𝑨 𝒐𝒇 𝒕𝒉𝒆 π’†π’π’•π’Šπ’“π’† π’”π’π’π’Šπ’… = πŸπŸπŸ” π’„π’ŽπŸ
vi) Volume of the cube.
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ 𝑐𝑒𝑏𝑒 = π‘Ž3
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ 𝑐𝑒𝑏𝑒 = 63
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒖𝒃𝒆 = πŸπŸπŸ” π’„π’ŽπŸ‘
vii) Volume of the cone.
1
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹π‘Ÿ 2 β„Ž
3
1
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹ × 22 × 6
3
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒏𝒆 = πŸ–π… π’„π’ŽπŸ‘
viii) Volume of the entire solid.
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ 𝑐𝑒𝑏𝑒 + π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘œπ‘›π‘’
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = 216 + 8πœ‹
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 π’†π’π’•π’Šπ’“π’† π’”π’π’π’Šπ’… = πŸπŸ’πŸ. πŸπŸ‘ π’„π’ŽπŸ‘
5. The solid to the right is a hemisphere with a cone on its top. Find the following:
i) Area of the great circle (or the base of the cone).
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘”π‘Ÿπ‘’π‘Žπ‘‘ π‘π‘–π‘Ÿπ‘π‘™π‘’ = πœ‹π‘Ÿ 2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘”π‘Ÿπ‘’π‘Žπ‘‘ π‘π‘–π‘Ÿπ‘π‘™π‘’ = πœ‹ × 392
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 π’ˆπ’“π’†π’‚π’• π’„π’Šπ’“π’„π’π’† = πŸπŸ“πŸπŸπ… 𝒇𝒕.𝟐
ii) Lateral surface area of the hemisphere.
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ β„Žπ‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ = 2πœ‹π‘Ÿ 2
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ β„Žπ‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ = 2πœ‹ × 392
𝑳𝑺𝑨 𝒐𝒇 𝒕𝒉𝒆 π’‰π’†π’Žπ’Šπ’”π’‘π’‰π’†π’“π’† = πŸ‘πŸŽπŸ’πŸπ… 𝒇𝒕.𝟐
iii) Lateral surface area of the cone.
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹π‘Ÿ√β„Ž2 + π‘Ÿ 2
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹ × 39√(81 − 39)2 + 392
𝑳𝑺𝑨 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒏𝒆 = πŸπŸπŸ•π… √πŸ‘πŸ”πŸ“ 𝒇𝒕.𝟐
iv) Total surface area of the entire solid.
𝑇𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = 𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ β„Žπ‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ + 𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’
𝑇𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = 3042πœ‹ 𝑓𝑑.2 + 117πœ‹ √365 𝑓𝑑.2
𝑻𝑺𝑨 𝒐𝒇 𝒕𝒉𝒆 π’†π’π’•π’Šπ’“π’† π’”π’π’π’Šπ’… = πŸπŸ”πŸ“πŸ•πŸ—. 𝟏𝟎 𝒇𝒕.𝟐
vi) Volume of the hemisphere.
2
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ β„Žπ‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ = πœ‹ × π‘Ÿ 3
3
2
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ β„Žπ‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ = πœ‹ × (39)3
3
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 π’‰π’†π’Žπ’Šπ’”π’‘π’‰π’†π’“π’† = πŸ‘πŸ—πŸ“πŸ’πŸ”π… 𝒇𝒕.πŸ‘
vii) Volume of the cone.
β„Ž
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹π‘Ÿ 2
3
(81 − 39)
2
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘’ = πœ‹(39)
3
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒐𝒏𝒆 = πŸπŸπŸπŸ—πŸ’π… 𝒇𝒕.πŸ‘
viii) Volume of the entire solid.
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ β„Žπ‘’π‘šπ‘–π‘ π‘β„Žπ‘’π‘Ÿπ‘’ + π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘π‘œπ‘›π‘’
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘’π‘›π‘‘π‘–π‘Ÿπ‘’ π‘ π‘œπ‘™π‘–π‘‘ = 39546πœ‹ 𝑓𝑑.3 + 21294πœ‹ 𝑓𝑑.3
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 π’†π’π’•π’Šπ’“π’† π’”π’π’π’Šπ’… = πŸ”πŸŽπŸ–πŸ’πŸŽπ… 𝒇𝒕.πŸ‘
7. Below are two similar square pyramids with a volume ratio of 8:27. The base lengths are equal
to the heights. Use this to answer the following questions.
i)
What is the scale factor?
Given that the ratio of their volume is 8: 27 which means that the ratio of their
𝟏
𝟏
base lengths will be (πŸ–)πŸ‘ ∢ (πŸπŸ•)πŸ‘ or 2 : 3 which is the scale factor.
ii)
What is the ratio of the total surface areas?
The ratio of the total surface area will be 𝟐𝟐 ∢ πŸ‘πŸ 𝒐𝒓 πŸ’ ∢ πŸ— .
iii)
Find 𝒉, 𝒙, and π’š.
As it is given that the base length is equal to height, it means that for smaller
pyramid, y = 8
from the scale factor, we can say that
8 ∢ π‘₯=2∢3
8×3
π‘₯=
2
π‘₯ = 12
So, x = 12 and consequently, h = 12.
iv)
Find π’˜ and 𝒛.
w is the hypotenuse of a right triangle whose perpendicular is 8 and base is half
of 8 which is 4
So, 𝑀 = √82 + 42
π’˜ = πŸ’√πŸ“
Similarly, z is the hypotenuse of a right triangle whose perpendicular is 12 and
base is half of 12 which is 6
So, 𝑧 = √122 + 62
𝒛 = πŸ”√πŸ“
v)
What is volume of smaller pyramid?
1
× π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘Žπ‘ π‘’ × β„Ž
3
1
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘ π‘šπ‘Žπ‘™π‘™π‘’π‘Ÿ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘ = × (8 × 8) × 8
3
πŸ“πŸπŸ
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 π’”π’Žπ’‚π’π’π’†π’“ π’‘π’šπ’“π’‚π’Žπ’Šπ’… =
π’–π’π’Šπ’•πŸ‘
πŸ‘
π‘‰π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘‘β„Žπ‘’ π‘ π‘šπ‘Žπ‘™π‘™π‘’π‘Ÿ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘ =
vi)
What is volume of larger pyramid?
Using the scale factor,
512
βˆΆπ‘‰
3
27 × 512
𝑉=
3×8
π‘½π’π’π’–π’Žπ’† 𝒐𝒇 𝒕𝒉𝒆 π’π’‚π’“π’ˆπ’†π’“ π’‘π’šπ’“π’‚π’Žπ’Šπ’… = πŸ“πŸ•πŸ” π’–π’π’Šπ’•πŸ‘
8 ∢ 27 =
v) Find the measure of one lateral surface area of smaller pyramid.
1
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘ π‘šπ‘Žπ‘™π‘™π‘’π‘Ÿ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘ = × π‘π‘Žπ‘ π‘’ × β„Žπ‘’π‘–π‘”β„Žπ‘‘
2
1
𝐿𝑆𝐴 π‘œπ‘“ π‘‘β„Žπ‘’ π‘ π‘šπ‘Žπ‘™π‘™π‘’π‘Ÿ π‘π‘¦π‘Ÿπ‘Žπ‘šπ‘–π‘‘ = × 8 × 8
2
𝑳𝑺𝑨 𝒐𝒇 𝒕𝒉𝒆 π’”π’Žπ’‚π’π’π’†π’“ π’‘π’šπ’“π’‚π’Žπ’Šπ’… = πŸ‘πŸ π’–π’π’Šπ’•πŸ
v) Find the measure of one lateral surface area of larger pyramid.
Using the scale factor,
4 ∢ 9 = 32 ∢ 𝐴
9 × 32
𝐴=
4
𝑳𝑺𝑨 𝒐𝒇 π’π’‚π’“π’ˆπ’†π’“ π’‘π’šπ’“π’‚π’Žπ’Šπ’… = πŸ•πŸ π’–π’π’Šπ’•πŸ
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