# The base of pyramid PABCD is the diamond ABCD with the lengths of the diagonals equal to 6 cm, 8 cm.pyramid height PD has a length equal to 2 cm.Determine the total surface area of the pyramid

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embizze | Certified Educator

The base of the pyramid is a rhombus (diamond), so the sides of the base are congruent.

The surface area of a pyramid is the area of the base plus the area of the sides. Each of the sides is a triangle.

(1) The area of the base: the area of a rhombus can be found by `A=1/2d_1d_2` where `d_1,d_2` are the lengths of the diagonals. Thus the area of the base is `A=1/2(6)(8)=24"cm"^2`

**Alternatively, the diagonal dissect the rhombus into 4 congruent right triangles. Since the diagonals bisect each other, the legs of the right triangles are 3 and 4. So the area of each triangle is 1/2(3)(4)=6; since there are four triangles the total area is 4(6)=24.**

(2) The sides are 4 congruent triangles. The base of the triangle is 5. (The base of the sides of the triangle is the hypotenuse of the right triangle formed by the intersection of the diagonals. 3-4-5 is a primitive Pythagorean triple, or you can use the Pythagorean theorem to find the hypotenuse: `3^2+4^2=25=c^2==>c=5` .)

We can find the lateral area by `1/2pl` where p is the perimeter and l is the slant height, or we can find the area of 1 triangle and multiply by 4. (It is the same calculation.) So we need the slant height.

If we drop an altitude from the intersection of the diagonals to a side of the rhombus,h, we find that `h=12/5` .

**Drop an altitude from the right angle of a 3-4-5 right triangle to the hypotenuse. The altitude cuts the hypotenuse into 2 pieces. The legs of the triangle are the means between the segments on the hypotenuse adjacent to the leg and the hypotenuse. Thus `3^2=5x==>x=9/5` or `4^2=5x==>x=16/5` , so the hypotenuse is cut into lengths of `9/5,16/5` . The altitude,h, is the mean between these two segments or `h^2=(9/5)(16/5)==>h=12/5`

Or using trigonometry label the angle between the leg of length 3 and the hypotenuse `alpha` . Then `alpha=tan^(-1)4/3~~53.13^@` . And the altitude is found by `sinalpha=h/3 ==>h=3sin(tan^(-1)4/3)=2.4=12/5` **

Now the height of the pyramid is 2, and the altitude to a side is `12/5` , so using the Pythagorean Theorem we get:

`2^2+(12/5)^2=l^2==>l^2=244/25==>l=(2sqrt(61))/5`

The lateral area is `L=1/2(20)((2sqrt(61))/5)=((20sqrt(61))/5)`

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The surface area is `24+4sqrt(61)`

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