# what is the slant height for the given pyramid to the nearest whole unit? with a pyramid base of 12cm and a height of 8cm. MY ANSWER: 14

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## To find the slant height of a pyramid, you can use the Pythagorean theorem. The slant height is the hypotenuse of a right triangle formed by the height of the pyramid and half the length of the base.

In this case, the height (h) of the pyramid is given as 8 cm, and the base (b) is given as 12 cm. The half-length of the base is 12/2 = 6 cm.

Now you can use the Pythagorean theorem:

c^2 = a^2 + b^2

where c is the slant height, a is the height, and b is half the length of the base.

Plugging in the values, we have:

c^2 = 8^2 + 6^2
c^2 = 64 + 36
c^2 = 100

To find the value of c, we take the square root of both sides:

c = sqrt(100)
c = 10

Therefore, the slant height of the given pyramid is 10 cm, not 14 cm.

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## To find the slant height of a pyramid, we can use the Pythagorean theorem. The slant height is the hypotenuse of a right triangle, with the base length as one leg and the height as the other leg.

In this case, the base length is 12 cm and the height is 8 cm. We can substitute these values into the Pythagorean theorem:

slant height = √(base length^2 + height^2)

slant height = √(12^2 + 8^2)

slant height = √(144 + 64)

slant height = √208

To the nearest whole unit, the slant height is approximately 14 cm.

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## Interpretation #1 : Slant height is the edge of the pyramid

Consider the right-angled (internal) triangle formed by the height , half the diagonal of the base and hypotenuse being the slant height.

let the slant height be x

let the diagonal of the base be d
d^2 = 1^2 + 12^2 = 288
d^2 = 288
d = 12√2
d/2 = 6√2

then x^2 = (d/2)^2 + 8^2 = 72 + 64 = 136
x+ √136 = 2√34 or appr 11.66

Interpretation #2 : slant height is the altitude of one of the triangles:

Consider one of the triangles whose base runs from the centre of the pyramid to the midpoint of a base side, with height 8 and hypotenuse as your "slant height"
base = 6
slant height = hypotenuse = y
height =8

where y^2 = 8^2 + 6^2 = 100
y = √100 = 10

Check: notice I can obtain the edge length from this as well

x^2 = y^2 + 6^2
x^2 = 100+36 = 136
x = √136 (same as above)