Lesson Using proportions to solve word problems in Geometry
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<H2>Using proportions to solve word problems in Geometry</H2> <H3>Problem 1</H3>Sid wants to measure the height of a tree. He sights the top of the tree, using a mirror that is lying flat on the ground. The mirror is 7.2 meters from the tree, and Sid is standing 1.2 meters from the mirror, as shown in the figure. His eyes are 1.5 m above the ground. How tall is the tree? <B>Solution</B> <pre> In the Figure below, AB represents the tree; CD represents Sid and ==== represents the mirror. The rays of light are shown by points. According to the optics law (the reflection law), angle AMB is equal to angle CMD, so the right-angled triangles AMB and CMD are similar. From the triangles similarity, we have this proportion {{{abs(AB)/abs(AM)}}} = {{{abs(CD)/abs(MC)}}}. Substituting the given values there, we have {{{x/7.2}}} = {{{1.5/1.2}}}, where x is the height of the tree. It gives the solution for the tree's height x = {{{(7.2*1.5)/1.2}}} = 6*1.5 = 9 meters. <U>ANSWER</U>. The height of the tree is 9 meters. </pre> Solved. <pre> B + /|\ . / | \ . / /|\ \ . D / | \ . o / /|\ \ . . _|_ / | \ . . / | \ | . . | +---------------------=====------/ \-- A M C F i g u r e </pre> <H3>Problem 2</H3>A figurine of the Statue of Liberty is 10 inches tall. The height of the Statue of Liberty is 150 feet. What is the scale factor of the dilation if the figurine is the image of the statue? <B>Solution</B> <pre> The scale factor is the ratio of two similar/corresponding linear measurements, expressed in consistent units. Use {{{10/12}}} feet for the 10 inches of the figurine and 150 feet for the original statue. So the scale factor of the dilation (from large to small) is {{{150/((10/12))}}} = {{{(150*12)/10}}} = 15*12 = 180. <U>ANSWER</U> </pre> <H3>Problem 3</H3>A spotlight is placed 2 feet from a 1-foot tall vase. A shadow 5 feet tall is cast on a wall. Find the distance of the vase from the wall. <B>Solution</B> <pre> It is about proportions . . . You have two similar (right-angled) triangles. The legs of the smaller triangle are 2 ft and 1 ft, where 1 ft is the height of the vase. The legs of the greater triangle are (2+x) ft and 5 ft. The unknown x is the distance of the vase from the wall. The proportion (from the triangles similarity) is {{{1/2}}} = {{{5/(2+x)}}}. From the proportion 2+x = 2*5 = 10 =============> x = 10-2 = 8. <U>ANSWER</U>. The distance from the vase to the wall is 8 ft. </pre> <H3>Problem 4</H3>On the number line x = 1/4 and y = 11/12. The point z divides the segment from x to y into two parts such that the distance from x to z is 3/8 of the distance from z to y. Find the distance from z to y. <B>Solution</B> <pre> The distance between the given points x = 1/4 and y = 11/12 is {{{11/12 - 1/4}}} = {{{11/12 - 3/12}}} = {{{8/12}}} = {{{2/3}}}. From the problem's description, point z is located BETWEEN points x and y. +---------------------------------------------------------+ | Let d be the distance from z to y: it is precisely | | the unknown quantity under the problem's question. | +---------------------------------------------------------+ Then the distance from x to z is {{{2/3}}} - {{{d}}}. You are given that the distance from x to z is 3/8 of the distance from z to y. In mathematical terms, it means that {{{2/3}}} - {{{d}}} = {{{(3/8)*d}}}. +-------------------------------------------+ | Thus you just have an equation for d | | to solve it and to find d. | +-------------------------------------------+ Multiply both sides by 24 to rid of the denominators. You will get then 2*8 - 24d = 3*3*d 16 = 9d + 24d 16 = 33d d = 16/33. Thus the distance d from z to y is {{{16/33}}}. <U>ANSWER</U> </pre> My other lessons on <B>proportions</B> in this site are - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/-Proprtions.lesson>Proportions</A> - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-solve-word-problems.lesson>Using proportions to solve word problems</A> - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-solve-word-problems-in-Physics.lesson>Using proportions to solve word problems in Physics</A> - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-solve-Chemistry-problems.lesson>Using proportions to solve Chemistry problems</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Typical-problems-on-proportions.lesson>Typical problems on proportions</A> - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-estimate-the-number-of-fish-in-a-lake.lesson>Using proportions to estimate the number of fish in a lake</A> - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/HOW-TO-algebreze-and-solve-this-problem-using-proportions.lesson>HOW TO algebraize and solve these problems using proportions</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-solve-some-nice-simple-Travel-and-Distance-problems.lesson>Using proportions to solve some nice simple Travel and Distance problems</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Miscellaneous-problems-on--proportions.lesson>Advanced problems on proportions</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Problems-on-proportions-for-mental-solution.lesson>Problems on proportions for mental solution</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/Selected-problems-on-proportions-from-the-archive.lesson>Selected problems on proportions from the archive</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Entertainment-problems-on-proportions.lesson>Entertainment problems on proportions</A> - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/OVERVIEW-of-lessons-on-proportions.lesson>OVERVIEW of lessons on proportions</A>
