Surface areas of solids

Geometry Level 4

The above are the top, front and side views of a solid. Find the surface area of this solid. If your answer is of the form a + b c a+b\sqrt{c} , where a a , b b and c c are positive integers with c c being square-free, give your answer as a + b + c a+b+c .

Note:

  1. All blue lines are dimension lines.
  2. All dimensions are unitless.


The answer is 291.

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1 solution

Relevant wiki: Surface Area - Problem Solving

The solid is composed of a right prism and regular pyramid. The surface area of the solid is equal to the surface area of the prism plus the lateral area of the pyramid minus the area of the base of the prism.

The surface area of the prism is 2 [ ( 9 ) ( 9 ) + ( 3 ) ( 9 ) + ( 3 ) ( 9 ) ] = 270 2[(9)(9)+(3)(9)+(3)(9)]=270 . The area of the base of the pyramid is ( 4 ) ( 4 ) = 16 (4)(4)=16 . The lateral area of the regular pyramid is 1 2 \dfrac{1}{2} multiplied by the perimeter of the base multiplied by the slant height. The slant height can be computed by Pythagorean Theorem and it is 29 \sqrt{29} . Therefore, the lateral area of the regular pyramid is 1 2 ( 4 ) ( 4 ) ( 29 = 8 29 \dfrac{1}{2}(4)(4)(\sqrt{29}=8\sqrt{29} .

Thus, the surface area of the solid is 270 + 8 29 − 16 = 254 + 8 29 270+8\sqrt{29}-16=254+8\sqrt{29} .

Finally,

a + b + c = 254 + 8 + 29 = a+b+c=254+8+29= 291 \boxed{\large\color{#D61F06}291}

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