Difference between revisions of "2007 iTest Problems/Problem 37"
(Created page with "== Problem == Rob is helping to build the set for a school play. For one scene, he needs to build a multi-colored tetrahedron out of cloth and bamboo. He begins by fitting three...") |
Rockmanex3 (talk | contribs) (Solution to Problem 37 (first solution with 3D drawings!)) |
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== Solution == | == Solution == | ||
| + | <asy> | ||
| + | |||
| + | import three; | ||
| + | unitsize(1cm); | ||
| + | size(200); | ||
| + | draw((0,0,0)--(0,0,10)); | ||
| + | draw((0,0,0)--(0,30,0)); | ||
| + | draw((0,0,0)--(40,0,0)); | ||
| + | draw((40,0,0)--(0,30,0)--(0,0,10)--(40,0,0),dotted); | ||
| + | label("a",(0,10,0),S); | ||
| + | label("b",(15,0,0),S); | ||
| + | label("c",(0,0,5),E); | ||
| + | |||
| + | currentprojection=orthographic(1,1/2,1/2); | ||
| + | |||
| + | </asy> | ||
| + | |||
| + | Let the three lengths of the rods be <math>a</math>, <math>b</math>, and <math>c</math>. Since the hypotenuse of the three right triangles are longer than the legs of the right triangle, the blue piece uses the lengths of the hypotenuses. For the red, yellow, and green pieces, right triangles are used, so write three equations based on the area. | ||
| + | <cmath>ab = 120</cmath> | ||
| + | <cmath>bc = 40</cmath> | ||
| + | <cmath>ac = 30</cmath> | ||
| + | Solving this equation yields <math>a = 3\sqrt{10}</math>, <math>b = 4\sqrt{10}</math>, and <math>c = \sqrt{10}</math>. By the [[Pythagorean Theorem]], the three side lengths of the blue piece are <math>5\sqrt{10}</math>, <math>10</math>, and <math>\sqrt{170}</math>. | ||
| + | |||
| + | <asy> | ||
| + | |||
| + | import three; | ||
| + | unitsize(1cm); | ||
| + | size(200); | ||
| + | draw((0,0,0)--(0,0,10)); | ||
| + | draw((0,0,0)--(0,30,0)); | ||
| + | draw((0,0,0)--(40,0,0)); | ||
| + | draw((40,0,0)--(0,30,0)--(0,0,10)--(40,0,0),dotted); | ||
| + | draw((0,0,0)--(96/5,72/5,0)--(0,0,10),dotted); | ||
| + | label("$5\sqrt{10}$",(20,15,0),SE); | ||
| + | label("$\sqrt{170}$",(15,0,5),NW); | ||
| + | label("10",(0,20,5),NE); | ||
| + | |||
| + | currentprojection=orthographic(1,1/2,1/2); | ||
| + | |||
| + | </asy> | ||
| + | |||
| + | Draw the altitude of the triangle with lengths <math>3\sqrt{10}</math> and <math>4\sqrt{10}</math>. Because the area of the red piece (one with lengths <math>3\sqrt{10}</math> and <math>4\sqrt{10}</math>) is <math>60</math>, the length of the altitude is <math>\frac{12\sqrt{10}}{5}</math>. The hypotenuse of the triangle with length <math>\sqrt{10}</math> and <math>\frac{12\sqrt{10}}{5}</math> is <math>\frac{13\sqrt{10}}{5}</math> and is perpendicular to the side with length <math>5\sqrt{10}</math> (since it is on the same plane as the altitude), so the area of the blue piece is <math>\frac{\frac{13\sqrt{10}}{5} \cdot 5\sqrt{10}}{2} = \boxed{65}</math>. | ||
| + | |||
| + | |||
| + | ==See Also== | ||
| + | {{iTest box|year=2007|num-b=36|num-a=38}} | ||
| + | |||
| + | [[Category:Intermediate Geometry Problems]] | ||
Latest revision as of 18:55, 15 June 2018
Problem
Rob is helping to build the set for a school play. For one scene, he needs to build a multi-colored tetrahedron out of cloth and bamboo. He begins by fitting three lengths of bamboo together, such that they meet at the same point, and each pair of bamboo rods meet at a right angle. Three more lengths of bamboo are then cut to connect the other ends of the first three rods. Rob then cuts out four triangular pieces of fabric: a blue piece, a red piece, a green piece, and a yellow piece. These triangular pieces of fabric just fill in the triangular spaces between the bamboo, making up the four faces of the tetrahedron. The areas in square feet of the red, yellow, and green pieces are 
, and 
respectively. If the blue piece is the largest of the four sides, find the number of square feet in its area.
Solution
![[asy] import three; unitsize(1cm); size(200); draw((0,0,0)--(0,0,10)); draw((0,0,0)--(0,30,0)); draw((0,0,0)--(40,0,0)); draw((40,0,0)--(0,30,0)--(0,0,10)--(40,0,0),dotted); label("a",(0,10,0),S); label("b",(15,0,0),S); label("c",(0,0,5),E); currentprojection=orthographic(1,1/2,1/2); [/asy]](http://latex.artofproblemsolving.com/f/c/5/fc522a6043ee11eb362c256925a4139fc16f1cc3.png)
Let the three lengths of the rods be 
, 
, and 
. Since the hypotenuse of the three right triangles are longer than the legs of the right triangle, the blue piece uses the lengths of the hypotenuses. For the red, yellow, and green pieces, right triangles are used, so write three equations based on the area.
![\[ab = 120\]](http://latex.artofproblemsolving.com/3/2/7/32705fccdb289b329e4cfff968c3505338d7c960.png)
![\[bc = 40\]](http://latex.artofproblemsolving.com/1/d/c/1dc072d929b9af6a87295fe4f6f30b3089e44055.png)
![\[ac = 30\]](http://latex.artofproblemsolving.com/1/2/9/129a240f1ad94eff1b04117555885b9262216be1.png)
Solving this equation yields 
, 
, and 
. By the Pythagorean Theorem, the three side lengths of the blue piece are 
, 
, and 
.
![[asy] import three; unitsize(1cm); size(200); draw((0,0,0)--(0,0,10)); draw((0,0,0)--(0,30,0)); draw((0,0,0)--(40,0,0)); draw((40,0,0)--(0,30,0)--(0,0,10)--(40,0,0),dotted); draw((0,0,0)--(96/5,72/5,0)--(0,0,10),dotted); label("$5\sqrt{10}$",(20,15,0),SE); label("$\sqrt{170}$",(15,0,5),NW); label("10",(0,20,5),NE); currentprojection=orthographic(1,1/2,1/2); [/asy]](http://latex.artofproblemsolving.com/2/e/3/2e3d5cf7e29a189c48d71d5cbf6c7f040c0be585.png)
Draw the altitude of the triangle with lengths 
and 
. Because the area of the red piece (one with lengths and ) is 
, the length of the altitude is 
. The hypotenuse of the triangle with length 
and is 
and is perpendicular to the side with length (since it is on the same plane as the altitude), so the area of the blue piece is 
.
See Also
| 2007 iTest (Problems, Answer Key) | ||
| Preceded by: Problem 36 |
Followed by: Problem 38 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4 | ||
