A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $7$ $7$ and $8$ $8$ and the pyramid's height is $5$ $5$ . If one of the base's corners has an angle of $\frac{π}{4}$ $pi/4$ , what is the pyramid's surface area?

Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $7$ $7$ and $8$ $8$ and the pyramid's height is $5$ $5$ . If one of the base's corners has an angle of $\frac{π}{4}$ $pi/4$ , what is the pyramid's surface area?

1 Answer

Impact of this question

1722 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-21T05:08:21

total surface area pyramid $= 28 \sqrt{2} + 2 \sqrt{498} + 7 \sqrt{33}$ $=28sqrt2+2sqrt(498)+7sqrt(33)$
total surface area pyramid $= 124.4417455$ $=124.4417455" "$ square units

Explanation:

The center of the parallelogram is at the intersection of the diagonals.

Compute half-length of the longer diagonal $d_{L}$ $d_L$ by cosine law for sides

$d_{L} = \sqrt{8^{2} + 7^{2} - 2 \cdot (7) \cdot (8) \cdot cos (135^{\circ})}$ $d_L=sqrt(8^2+7^2-2*(7)*(8)*cos (135^@))$
$\frac{d_{L}}{2} = \frac{1}{2} \cdot \sqrt{113 + 56 \sqrt{2}}$ $d_L/2=1/2*sqrt(113+56sqrt(2))$

Let x be one of the lengths of edges of the triangular face which is directly on top of the longer diagonal as seen on top of the pyramid

$x = \sqrt{{(\frac{d_{L}}{2})}^{2} + h^{2}}$ $x=sqrt((d_L/2)^2+h^2)$
$x = \sqrt{{(\frac{1}{2} \cdot \sqrt{113 + 56 \sqrt{2}})}^{2} + 5^{2}}$ $x=sqrt((1/2*sqrt(113+56sqrt(2)))^2+5^2)$
$x = 8.54687018$ $x=8.54687018$

Compute half-length of the shorter diagonal $d_{S}$ $d_S$ by cosine law for sides

$d_{S} = \sqrt{8^{2} + 7^{2} - 2 \cdot (7) \cdot (8) \cdot cos (45^{\circ})}$ $d_S=sqrt(8^2+7^2-2*(7)*(8)*cos (45^@))$
$\frac{d_{S}}{2} = \frac{1}{2} \cdot \sqrt{113 - 56 \sqrt{2}}$ $d_S/2=1/2*sqrt(113-56sqrt(2))$

Let y be one of the lengths of edges of the triangular face which is directly on top of the shorter diagonal as seen on top of the pyramid

$y = \sqrt{{(\frac{d_{S}}{2})}^{2} + h^{2}}$ $y=sqrt((d_S/2)^2+h^2)$
$y = \sqrt{{(\frac{1}{2} \cdot \sqrt{113 - 56 \sqrt{2}})}^{2} + 5^{2}}$ $y=sqrt((1/2*sqrt(113-56sqrt(2)))^2+5^2)$
$y = 5.783684823$ $y=5.783684823$

Compute area $A_{8}$ $A_8$ of triangular face with sides 8, x, and y, by Heron's Formula

where $s_{8} = \frac{1}{2} \cdot (8 + x + y) = 11.1652775$ $s_8=1/2*(8+x+y)=11.1652775$

$A_{8} = \sqrt{s_{8} \cdot (s_{8} - 8) \cdot (s_{8} - x) \cdot (s_{8} - y)}$ $A_8=sqrt(s_8*(s_8-8)*(s_8-x)*(s_8-y))$
$A_{8} = \sqrt{11.1652775 \cdot (11.1652775 - 8) \cdot (11.1652775 - 8.54687018) (11.1652775 - 5.783684823)}$ $A_8=sqrt(11.1652775*(11.1652775-8)*(11.1652775-8.54687018)(11.1652775-5.783684823))$
$A_{8} = \sqrt{498}$ $A_8=sqrt(498)$

Compute area $A_{7}$ $A_7$ of triangular face with sides 7, x, and y, by Heron's Formula

where $s_{7} = \frac{1}{2} \cdot (7 + x + y) = 10.6652775$ $s_7=1/2*(7+x+y)=10.6652775$

$A_{7} = \sqrt{s_{7} \cdot (s_{7} - 7) \cdot (s_{7} - x) \cdot (s_{7} - y)}$ $A_7=sqrt(s_7*(s_7-7)*(s_7-x)*(s_7-y))$
$A_{7} = \sqrt{10.6652775 \cdot (10.6652775 - 7) \cdot (10.6652775 - 8.54687018) (10.6652775 - 5.783684823)}$ $A_7=sqrt(10.6652775*(10.6652775-7)*(10.6652775-8.54687018)(10.6652775-5.783684823))$
$A_{7} = \frac{7 \cdot \sqrt{33}}{2}$ $A_7=(7*sqrt(33))/2$

Compute Total Area $T_{A} :$ $T_A:$

$T_{A} = a r e a b a s e + 2 \cdot A_{8} + 2 \cdot A_{7}$ $T_A=area base+2*A_8+2*A_7$

$T_{A} = (7 \cdot {sin 45}^{\circ}) \cdot (8) + 2 \cdot \sqrt{498} + 2 \cdot \frac{7 \cdot \sqrt{33}}{2}$ $T_A=(7*sin 45^@)*(8)+2*sqrt(498)+2*(7*sqrt(33))/2$
$T_{A} = 28 \sqrt{2} + 2 \sqrt{498} + 7 \sqrt{33}$ $T_A=28sqrt2+2sqrt(498)+7sqrt(33)$
$T_{A} = 124.4417455$ $T_A=124.4417455" "$ square units

God bless....I hope the explanation is useful.

Science

Math

Humanities

... and beyond

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $7$ $7$ and $8$ $8$ and the pyramid's height is $5$ $5$ . If one of the base's corners has an angle of $\frac{π}{4}$ $pi/4$ , what is the pyramid's surface area?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 77 and 88 and the pyramid's height is 55 . If one of the base's corners has an angle of π4pi/4, what is the pyramid's surface area?

1 Answer

Explanation:

Related questions

Impact of this question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $7$ $7$ and $8$ $8$ and the pyramid's height is $5$ $5$ . If one of the base's corners has an angle of $\frac{π}{4}$ $pi/4$ , what is the pyramid's surface area?