# Activity 3.2 (Measurement of Solids)

```Name: JULY ROLAND P. CABRISOS
Year &amp; 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,
1. Consider a regular pentagonal prism. Find the following:
360&deg;
π=
π
360&deg;
π=
5
π = 72&deg;
π
= 36&deg;
2
4
tan 36&deg; =
π
π = π. ππ , π = π , π = ππ
i) Area of the base.
1
π΄πππ ππ π‘βπ πππ π = 2 &times; &times; 5 &times; π &times; π
2
π΄πππ ππ π‘βπ πππ π = 5 &times; 5.51 &times; π
π¨πππ ππ πππ ππππ = πππ. π πππππ
ii) Lateral surface area.
πΏππ΄ = 5 &times; π &times; β
πΏππ΄ = 5 &times; 8 &times; 20
π³πΊπ¨ = πππ πππππ
iii) Total surface area.
πππ΄ = 5ππ + 5πβ
πππ΄ = 5(5.51)(8) + 5(8)(20)
π»πΊπ¨ = ππππ. π πππππ
iv) Volume of the solid.
5
ππππ’ππ = &times; π &times; β
2
5
ππππ’ππ = &times; 8 &times; 20
2
π½πππππ ππ πππ πππππ = πππ πππππ
2. Consider a regular hexagonal pyramid. Find the following:
i) Area of the base.
3√3 2
π΄πππ ππ π‘βπ πππ π =
π
2
3√3 2
π΄πππ ππ π‘βπ πππ π =
10
2
π¨πππ ππ πππ ππππ = πππ. ππ πππππ
ii) Lateral surface area.
1
πΏππ΄ = 6 &times; &times; π &times; β
2
1
πΏππ΄ = 6 &times; &times; 10 &times; 22
2
π³πΊπ¨ = πππ πππππ
iii) Total surface area.
πππ΄ = π΄πππ ππ π‘βπ πππ π + πΏππ΄
πππ΄ = 259.81 + 660
π»πΊπ¨ = πππ. ππ πππππ
iv) Volume of the solid.
10⁄
2
π‘ππ30&deg; =
π
5
π=
π‘ππ30&deg;
π = π√π
β = √π  2 − π2
β = √222 − (5√3)2
π = ππ. ππ
ππππ’ππ = π &times; π &times; β
ππππ’ππ = 5√3 &times; 10 &times; 20.22
π½πππππ = ππππ. ππ πππππ
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ππ ππ π‘βπ πππ π = π  2
π΄πππ ππ π‘βπ πππ π = 42
π¨πππ ππ πππ ππππ = ππ ππ.π
ii) Lateral surface area of the prism.
πΏππ΄ ππ π‘βπ ππππ π = β &times; πππππππ‘ππ
πΏππ΄ ππ π‘βπ ππππ π = 5 &times; (4 + 4 + 4 + 4)
π³πΊπ¨ ππ πππ πππππ = ππ ππ.π
iii) Lateral surface area of the pyramid.
Finding the slant height (x)
π₯ = √62 + 42
π = π. ππ ππ.
1
&times; πππππππ‘ππππ πππ π &times; π ππππ‘ βπππβπ‘ (π₯)
2
1
πΏππ΄ ππ π‘βπ ππ¦πππππ = &times; (4 + 4 + 4 + 4) &times; 7.21
2
π³πΊπ¨ ππ πππ πππππππ = ππ. ππ ππ.π
iv) Total surface area of the entire solid.
πππ΄ ππ π‘βπ πππ‘πππ π ππππ = πΏππ΄ ππ ππππ π + πΏππ΄ ππ ππ¦πππππ
πππ΄ ππ π‘βπ πππ‘πππ π ππππ = 80 + 57.68
π»πΊπ¨ ππ πππ ππππππ πππππ = πππ. ππ ππ.π
v) Volume of the prism.
ππππ’ππ ππ π‘βπ ππππ π = β &times; ππππ ππ πππ π
ππππ’ππ ππ π‘βπ ππππ π = 5 &times; 16
π½πππππ ππ πππ πππππ = ππ ππ.π
πΏππ΄ ππ π‘βπ ππ¦πππππ =
vi) Volume of the pyramid.
ππ€β
ππππ’ππ ππ π‘βπ ππ¦πππππ =
3
4&times;4&times;6
ππππ’ππ ππ π‘βπ ππ¦πππππ =
3
π½πππππ ππ πππ πππππππ = ππ ππ.π
vii) Volume of the entire solid.
ππππ’ππ ππ π‘βπ πππ‘πππ π ππππ = π£πππ’ππ ππ ππππ π + π£πππ’ππ ππ ππ¦πππππ
ππππ’ππ ππ π‘βπ πππ‘πππ π ππππ = 80 + 32
π½πππππ ππ πππ ππππππ πππππ = πππ ππ.π
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ππ ππ π‘βπ πππ π = π2
π΄πππ ππ π‘βπ πππ π = 62
π¨πππ ππ πππ ππππ ππ πππ ππππ = ππ πππ
ii) Lateral surface area of the cube.
πΏππ΄ ππ π‘βπ ππ’ππ = 4π2
πΏππ΄ ππ π‘βπ ππ’ππ = 4 &times; 62
πΏππ΄ ππ π‘βπ ππ’ππ = πππ πππ
iii) Area of the circle of the cone.
π΄πππ ππ π‘βπ ππππππ ππ π‘βπ ππππ = ππ 2
π΄πππ ππ π‘βπ ππππππ ππ π‘βπ ππππ = π &times; 22
π¨πππ ππ πππ ππππππ ππ πππ ππππ = ππ πππ
iv) Lateral surface area of the cone.
πΏππ΄ ππ π‘βπ ππππ = ππ√π 2 + β2
πΏππ΄ ππ π‘βπ ππππ = π &times; 2√22 + 62
π³πΊπ¨ ππ πππ ππππ = ππ√ππ πππ
v) Total surface area of the entire solid.
πππ΄ ππ π‘βπ πππ‘πππ π ππππ = 6π2
πππ΄ ππ π‘βπ πππ‘πππ π ππππ = 6 &times; 62
π»πΊπ¨ ππ πππ ππππππ πππππ = πππ πππ
vi) Volume of the cube.
ππππ’ππ ππ π‘βπ ππ’ππ = π3
ππππ’ππ ππ π‘βπ ππ’ππ = 63
π½πππππ ππ πππ ππππ = πππ πππ
vii) Volume of the cone.
1
ππππ’ππ ππ π‘βπ ππππ = ππ 2 β
3
1
ππππ’ππ ππ π‘βπ ππππ = π &times; 22 &times; 6
3
π½πππππ ππ πππ ππππ = ππ πππ
viii) Volume of the entire solid.
ππππ’ππ ππ π‘βπ πππ‘πππ π ππππ = ππππ’ππ ππ ππ’ππ + ππππ’ππ ππ ππππ
ππππ’ππ ππ π‘βπ πππ‘πππ π ππππ = 216 + 8π
π½πππππ ππ πππ ππππππ πππππ = πππ. ππ πππ
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ππ ππ π‘βπ πππππ‘ ππππππ = ππ 2
π΄πππ ππ π‘βπ πππππ‘ ππππππ = π &times; 392
π¨πππ ππ πππ πππππ ππππππ = πππππ ππ.π
ii) Lateral surface area of the hemisphere.
πΏππ΄ ππ π‘βπ βππππ πβπππ = 2ππ 2
πΏππ΄ ππ π‘βπ βππππ πβπππ = 2π &times; 392
π³πΊπ¨ ππ πππ ππππππππππ = πππππ ππ.π
iii) Lateral surface area of the cone.
πΏππ΄ ππ π‘βπ ππππ = ππ√β2 + π 2
πΏππ΄ ππ π‘βπ ππππ = π &times; 39√(81 − 39)2 + 392
π³πΊπ¨ ππ πππ ππππ = ππππ √πππ ππ.π
iv) Total surface area of the entire solid.
πππ΄ ππ π‘βπ πππ‘πππ π ππππ = πΏππ΄ ππ π‘βπ βππππ πβπππ + πΏππ΄ ππ π‘βπ ππππ
πππ΄ ππ π‘βπ πππ‘πππ π ππππ = 3042π ππ‘.2 + 117π √365 ππ‘.2
π»πΊπ¨ ππ πππ ππππππ πππππ = πππππ. ππ ππ.π
vi) Volume of the hemisphere.
2
ππππ’ππ ππ π‘βπ βππππ πβπππ = π &times; π 3
3
2
ππππ’ππ ππ π‘βπ βππππ πβπππ = π &times; (39)3
3
π½πππππ ππ πππ ππππππππππ = ππππππ ππ.π
vii) Volume of the cone.
β
ππππ’ππ ππ π‘βπ ππππ = ππ 2
3
(81 − 39)
2
ππππ’ππ ππ π‘βπ ππππ = π(39)
3
π½πππππ ππ πππ ππππ = ππππππ ππ.π
viii) Volume of the entire solid.
ππππ’ππ ππ π‘βπ πππ‘πππ π ππππ = ππππ’ππ ππ βππππ πβπππ + ππππ’ππ ππ ππππ
ππππ’ππ ππ π‘βπ πππ‘πππ π ππππ = 39546π ππ‘.3 + 21294π ππ‘.3
π½πππππ ππ πππ ππππππ πππππ = ππππππ ππ.π
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
π
π
base lengths will be (π)π βΆ (ππ)π 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 ππ βΆ ππ ππ π βΆ π .
iii)
Find π, π, and π.
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 βΆ π₯=2βΆ3
8&times;3
π₯=
2
π₯ = 12
So, x = 12 and consequently, h = 12.
iv)
Find π and π.
w is the hypotenuse of a right triangle whose perpendicular is 8 and base is half
of 8 which is 4
So, π€ = √82 + 42
π = π√π
Similarly, z is the hypotenuse of a right triangle whose perpendicular is 12 and
base is half of 12 which is 6
So, π§ = √122 + 62
π = π√π
v)
What is volume of smaller pyramid?
1
&times; ππππ ππ πππ π &times; β
3
1
ππππ’ππ ππ π‘βπ π ππππππ ππ¦πππππ = &times; (8 &times; 8) &times; 8
3
πππ
π½πππππ ππ πππ πππππππ πππππππ =
πππππ
π
ππππ’ππ ππ π‘βπ π ππππππ ππ¦πππππ =
vi)
What is volume of larger pyramid?
Using the scale factor,
512
βΆπ
3
27 &times; 512
π=
3&times;8
π½πππππ ππ πππ ππππππ πππππππ = πππ πππππ
8 βΆ 27 =
v) Find the measure of one lateral surface area of smaller pyramid.
1
πΏππ΄ ππ π‘βπ π ππππππ ππ¦πππππ = &times; πππ π &times; βπππβπ‘
2
1
πΏππ΄ ππ π‘βπ π ππππππ ππ¦πππππ = &times; 8 &times; 8
2
π³πΊπ¨ ππ πππ πππππππ πππππππ = ππ πππππ
v) Find the measure of one lateral surface area of larger pyramid.
Using the scale factor,
4 βΆ 9 = 32 βΆ π΄
9 &times; 32
π΄=
4
π³πΊπ¨ ππ ππππππ πππππππ = ππ πππππ
```