A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is $7$ $7$ , its base has sides of length $3$ $3$ , and its base has a corner with an angle of $\frac{3 π}{4}$ $(3 pi)/4$ . What is the pyramid's surface area?

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A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is $7$ $7$ , its base has sides of length $3$ $3$ , and its base has a corner with an angle of $\frac{3 π}{4}$ $(3 pi)/4$ . What is the pyramid's surface area?

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Answer 1 · 2018-03-27T04:53:37

The total surface area is (approximately) $45.075 {units}^{2}$ $45.075 " units"^2$ .

Explanation:

The pyramid is composed of 5 pieces: 1 base (a rhombus) and 4 sides (congruent triangles). (They're congruent because each triangle has a "base" of length 3, and one side is from the tip to a wide corner, while the other side is from the tip to a narrow corner.)

The surface area of the whole pyramid is

$A_{pyramid} = A_{rhombus} + 4 A_{triangle}$ $A_"pyramid"= A_"rhombus" + 4A_"triangle"$

Step 1: The Rhombus

The area of a rhombus is $A = b h$ $A = bh$ . This rhombus's base is 3, while its height is $3 sin (\frac{π}{4})$ $3sin (pi/4)$ . (Draw the rhombus and you'll see $sin (\frac{π}{4}) = \frac{opp}{adj} = \frac{h}{3} .$ $sin(pi/4) = "opp"/"adj" = h/3.$ ) So the area is

$A_{rhombus} = 3 sin (\frac{π}{4}) = \frac{3}{\sqrt{2}}$ $A_"rhombus" = 3sin(pi/4) = 3/sqrt2$

Step 2: The (4) Triangles

The area of a triangle is $A = \frac{1}{2} b h .$ $A = 1/2bh.$ Each triangle has a base of length 3, but we don't know their heights. These can be found by picturing a right triangle whose height is the height of the pyramid (7) and whose base is the distance from the centre of the rhombus to the centre of one of the bottom sides (half of 3, which is 1.5).

Then use Pythagorean theorem to find the "slant" height of the triangle side:

$a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$

${(\frac{3}{2})}^{2} + 7^{2} = c^{2}$ $(3/2)^2+7^2 = c^2$

$\frac{9}{4} + 49 = c^{2}$ $9/4+49 = c^2$

$\frac{205}{4} = c^{2}$ $205/4 = c^2$

$\frac{\sqrt{205}}{2} = c$ $sqrt205/2 = c$

Then:

$A_{triangle} = \frac{1}{2} (3) (\frac{\sqrt{205}}{2}) = \frac{3 \sqrt{205}}{4}$ $A_"triangle" = 1/2(3)(sqrt205/2)=(3sqrt205)/4$

Step 3: Add them together!

$A_{pyramid} = A_{rhombus} + 4 A_{triangle}$ $A_"pyramid"= A_"rhombus" + 4A_"triangle"$

$A_{pyramid} = \frac{3 \sqrt{2}}{2} + 4 (\frac{3 \sqrt{205}}{4})$ $color(white)(A_"pyramid")= (3sqrt2)/2+4((3sqrt205)/4)$

$A_{pyramid} = \frac{3 \sqrt{2}}{2} + 3 \sqrt{205}$ $color(white)(A_"pyramid")= (3sqrt2)/2+3sqrt205$

$A_{pyramid} \approx 45.075$ $color(white)(A_"pyramid")~~ 45.075$

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A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is $7$ $7$ , its base has sides of length $3$ $3$ , and its base has a corner with an angle of $\frac{3 π}{4}$ $(3 pi)/4$ . What is the pyramid's surface area?

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Explanation:

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A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is $7$ $7$ , its base has sides of length $3$ $3$ , and its base has a corner with an angle of $\frac{3 π}{4}$ $(3 pi)/4$ . What is the pyramid's surface area?