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Zorluk: ZorSurface Area and Volume of 3D Solids

A solid metal trophy is formed by mounting a right circular cone of slant height 5 cm5\text{ cm} on top of a right circular cylinder of height 8 cm8\text{ cm}. Both the cone and the cylinder share a common base radius of 3 cm3\text{ cm}. If the trophy is completely melted down and recast to form a solid pyramid with a square base of side length 6 cm6\text{ cm}, what is the vertical height of the pyramid?

  1. A
    9π cm9\pi\text{ cm}
  2. 7π cm7\pi\text{ cm}Cevap
  3. C
    \(\frac{7\pi}{3}\text{ cm}\)
  4. D
    3π cm3\pi\text{ cm}

Cevap

7π cm7\pi\text{ cm}
First, find the height of the cone using the Pythagorean theorem: hcone=52−32=4 cmh_{\text{cone}} = \sqrt{5^2 - 3^2} = 4\text{ cm}. The volume of the conical section is 13π(32)(4)=12π cm3\frac{1}{3}\pi (3^2)(4) = 12\pi\text{ cm}^3, and the volume of the cylindrical section is π(32)(8)=72π cm3\pi (3^2)(8) = 72\pi\text{ cm}^3. The total volume melted is 12π+72π=84π cm312\pi + 72\pi = 84\pi\text{ cm}^3. For the recast pyramid with square base area 62=36 cm26^2 = 36\text{ cm}^2, its volume is 13(36)h=12h\frac{1}{3}(36)h = 12h. Equating the volumes (12h=84π12h = 84\pi) yields a vertical height of 7π cm7\pi\text{ cm}.

Adım Adım Çözüm

1
Calculate the vertical height of the conical top
hcone=52−32=4 cmh_{\text{cone}} = \sqrt{5^2 - 3^2} = 4\text{ cm}
The slant height ll, radius rr, and vertical height hconeh_{\text{cone}} form a right-angled triangle where hcone=l2−r2h_{\text{cone}} = \sqrt{l^2 - r^2}.
2
Compute the total volume of the composite metal solid
Vtotal=12π+72π=84π cm3V_{\text{total}} = 12\pi + 72\pi = 84\pi\text{ cm}^3
The total volume is the sum of the cone volume Vcone=13π(32)(4)=12π cm3V_{\text{cone}} = \frac{1}{3}\pi (3^2)(4) = 12\pi\text{ cm}^3 and the cylinder volume Vcylinder=π(32)(8)=72π cm3V_{\text{cylinder}} = \pi (3^2)(8) = 72\pi\text{ cm}^3.
3
Equate the total volume to the volume of the square pyramid and solve for its height
h=7π cmh = 7\pi\text{ cm}
The base area of the square pyramid is A=62=36 cm2A = 6^2 = 36\text{ cm}^2. Its volume is Vpyramid=13(36)h=12hV_{\text{pyramid}} = \frac{1}{3}(36)h = 12h. Setting 12h=84π12h = 84\pi gives h=84π12=7π cmh = \frac{84\pi}{12} = 7\pi\text{ cm}.

Anahtar Kavram

Conservation of volume during recasting of 3D solids and combined solid geometry
Tahmini Süre:2m 0s
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Konular

Surface Area and Volume of 3D Solids

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