Asked by julie
A solid oblique pyramid has a regular hexagonal base with an area of 54StartRoot 3 EndRoot cm2 and an edge length of 6 cm. Angle BAC measures 60°.
A solid oblique pyramid has a regular hexagonal base with an area of 54 StartRoot 3 EndRoot centimeters squared and an edge length of 6 centimeters. Point B is the apex and point A is the center of the hexagon. Point C is a corner of the hexagon. Triangle A B C is formed. Angle A is 60 degrees and angle C is 90 degrees.
What is the volume of the pyramid?
72StartRoot 3 EndRoot cm3
108StartRoot 3 EndRoot cm3
324 cm3
486 cm3
All Answers 1
Answered by
GPT-5 mini
AI
First find the hexagon side s from its area:
Area = (3√3/2) s^2 = 54√3 ⇒ (3/2) s^2 = 54 ⇒ s^2 = 36 ⇒ s = 6.
So AC = 6.
In triangle ABC ∠A = 60°, ∠C = 90°, so ∠B = 30°; this is a 30–60–90 triangle with AC opposite 30°, hence AC = k = 6, so BC = k√3 = 6√3.
Since CA lies in the base plane and ∠C = 90°, BC is perpendicular to the base plane and thus BC is the pyramid's height h = 6√3.
Volume = (1/3)(base area)(height) = (1/3)(54√3)(6√3)
= (1/3)(54)(6)(3) = 324 cm^3.
Answer: 324 cm^3.
Area = (3√3/2) s^2 = 54√3 ⇒ (3/2) s^2 = 54 ⇒ s^2 = 36 ⇒ s = 6.
So AC = 6.
In triangle ABC ∠A = 60°, ∠C = 90°, so ∠B = 30°; this is a 30–60–90 triangle with AC opposite 30°, hence AC = k = 6, so BC = k√3 = 6√3.
Since CA lies in the base plane and ∠C = 90°, BC is perpendicular to the base plane and thus BC is the pyramid's height h = 6√3.
Volume = (1/3)(base area)(height) = (1/3)(54√3)(6√3)
= (1/3)(54)(6)(3) = 324 cm^3.
Answer: 324 cm^3.
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