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Wewaii [24]
3 years ago
12

A regular square pyramid has base edges of length 16 and its lateral faces are inclined 30­° to the base of the pyramid. What is

the (1) height of the pyramid (2) volume of the pyramid?

Mathematics
1 answer:
Sedaia [141]3 years ago
8 0

Answer:

(1) 13.86 units.

(2) 1182.72 cubic units.

Step-by-step explanation:

Please find the attachment.

We have been given that a regular square pyramid has base edges of length 16 and its lateral faces are inclined 30­° to the base of the pyramid.

(1) We can find height of pyramid using tan.

\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}

The length of opposite side will half the length of square base.

\frac{16}{2}=8

\text{tan}(30^{\circ})=\frac{8}{h}

h=\frac{8}{\text{tan}(30^{\circ})}

h=13.8564064605420367

h\approx 13.86

Therefore, the height of the pyramid will be 13.86 units.

(2). We know that volume of pyramid is 1/3 the product of base area and height.

V=\frac{1}{3}*bh

V=\frac{1}{3}*16*16*13.86

V=\frac{1}{3}*3548.16

V=1182.72

Therefore, the volume of the pyramid would be 1182.72 cubic units.

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castortr0y [4]

Answer:

no, it cannot

Step-by-step explanation:

a difference of square is: a² - b² = (a - b)(a + b)

looking at the expression (5z+3)(-5z-3), we see that it does not fit the criteria of the breakdown of a perfect square, as (-5z-3) has a negative <em>a</em> term (-5z)

if we FOILed (5z+3)(-5z-3) out, we would get:

-25z² - 30z - 9, which is not a difference of squares

8 0
3 years ago
What is the distance between B to A<br> Please add the method.<br> The angle is 61 (a bit blurry)
deff fn [24]

Answer:

24.03 units (nearest hundredth)

Step-by-step explanation:

The distance between B and A is:  AB = AH + HB

We have been given AH, so we just need to find the measure of HB.

First, find the angle AOH using tan trig ratio:

\sf \tan(\theta)=\dfrac{O}{A}

where:

  • \theta is the angle
  • O is the side opposite the angle
  • A is the side adjacent the angle

Given:

  • \theta = ∠AOH
  • O = AH = 8
  • A = OH = 19.80

\implies \sf \tan(\angle AOH)=\dfrac{8}{19.8}

\implies \sf \angle AOH = 22.00069835^{\circ}

   ∠BOA = ∠BOH + ∠AOH

⇒ ∠BOH = ∠BOA - ∠AOH

⇒ ∠BOH = 61° - 22.00069835°

               = 38.99930165°

Now we can find HB by again using the tan trig ratio:

Given:

  • \theta = ∠BOH = 38.99930165°
  • O = HB
  • A = OH = 19.80

Substituting given values:

\implies \sf \tan(38.99930165^{\circ})=\dfrac{HB}{19.80}

\implies \sf HB=19.80 \tan(38.99930165^{\circ})

\implies \sf HB=16.03332427

Therefore:

   AB = AH + BH

⇒ AB = 8 + 16.03332427

         = 24.03 units (nearest hundredth)

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