Given: Regular pyramid DO = 6, m∠ADO = 42º, AO ⊥ (FBCD) Find: Volume and Surface Area of ABCDF

Question

Given: Regular pyramid DO = 6, m∠ADO = 42º, AO ⊥ (FBCD) Find: Volume and Surface Area of ABCDF

Answer 1

Follow the link on GeoGebra for a diagram

https://www.geogebra.org/3d/yb3z4n9e

Given: Regular pyramid ABCDF =>

Base BCDF is a square and the four (4) lateral faces are congruent isosceles triangles

1. Surface Area of ABCDF: S_ABCDF = S_BCDF + 4 • S_ADC

S_BCDF = 2 • S_FDC = 2 • ((|FC| • |DO|) / 2) = |FC| • |DO| = 2 • |DO| • |DO| = 72

Given: DO = 6, m∠ADO = 42º, AO ⊥ (BCDF) =>

|AD| = |DO| / cos∠ADO = 6 / cos42º ≈ 8.07

|AO| = |DO| • tan∠ADO = 6tan42º ≈ 5.40

Let point H be the midpoint of segment [DC] => [AH] ⊥ [DC] =>

|HO| = (√2 / 2) • |DO| = (√2 / 2) • 6 = 3√2 ≈ 4.24

|AH|² = |AO|² + |HO|² ≈ 5.40² + 4.24² = 47.14

|AH| ≈ 6.87

|DH|² = |AD|² - |AH|² ≈ 8.07² - 47.84 = 17.28

|DH| ≈ 4.16

S_ADC = (|DC| • |AH|) / 2 = (2 • |DH| • |AH|) / 2 ≈ 4.16 • 6.87 = 28.56

S_ABCDF = S_BCDF + 4 • S_ADC ≈ 72 + 4 • 28.56 = 186.24

2. Volume of ABCDF: V_ABCDF = (1/3) • S_BCDF • |AO|

V_ABCDF = (1/3) • S_BCDF • |AO| ≈ (72 • 5.40) / 3 = 129.60

Answer: Volume ≈ 129.60 units³ and Surface Area ≈ 186.24 units²

Answer 2

You need to find the height which is AO from the triangle ADO having one side = 6 and one angle =42 degrees. Using the tangent of the angle and cross multiply, you will get:

Tan ADO = AO/OD

Height of the pyramid: AO = OD Tan ADO = 6 Tan 42 = 5.4

Now you need the area of the base which is a square and is two times the area of the isosceles right triangle BDC.

Area of the triangle BDC = Base x height / 2 = 12 x 6 / 2 = 36

Area of the square FBCD is twice of it which is 2 x 36 = 72

Volume of AFBCD = Area of the base x height / 3 = 72 x 5.4 / 3 = 129.6

To get the Surface Area you need to find the side of the base and the Slant Height of the pyramid. Perimeter of the base x Slant Height / 2.

The side of the base is calculated from the right triangle ODC which the hypothenuse is 6 x 2 = 12. So to find its leg X:

X^2 + X^2 = 12^2

2X^2 = 144

X^2 = 144/2 and X = 12 Radical 2 / 2 = 6 Radical 2 which is CD.

To find the Slant Height we need to draw the height of the triangle ADC which is AK and join O to K. So

OK = 6 radical 2 /2 = 3 radical 2

AK^2 = AO^2 + OK^2 = 5.4^2 + (3 Radical 2)^2

AK=6.8673

Surface Area = Perimeter of the base x Slant height / 2 = 4 x 6 Radical 2 x 6.8673 / 2 = 116.54