sat math: shortest distance between two points

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SAT Math: Shortest Distance Between Two Points April 26th, 2010 by Marty Rafson Share Euclid is credited as the founder of most of our modern geometry. Though he lived approximately 300 years B.C., his foundation of geometry is still the basis of the geometry we teach in high school and, therefore, the basis for SAT questions on geometry. One of the most common geometry questions on recent SAT’s goes right back to one of Euclid’s most basic ideas. What is the shortest distance between two points? Euclid said the shortest distance between two points is a straight line. This was one of his fundamental axioms, or postulates statements that are accepted without proof because they are so obviously true. Consider this situation: You need to travel from point A to point B. According to Euclid, the shortest distance would be along the straight line AB. However, now let’s suppose that you had to divert first to a point C that is not along this straight path. You would first travel from A to C (AC) and then from C to B (CB). Clearly, this second trip which takes us first to point C is the longer trip. If you draw a triangle to represent this situation, we can rename the result: AC plus CB must be longer than directly from A to B, namely AB. Thus, based on Euclid’s axiom, we have proven a theorem: The sum of two sides of a triangle must be greater than the third side. Someone at the Educational Testing Service has decided that this is a critical concept and a question on it has appeared on most recent SAT’s. Problem: “Two sides of a triangle are 5 and 9. If the length of the third side is an integer, find one possible length of the third side.” Let’s try 3. This doesn’t work since 5+3=8 which is not greater than 9. Perhaps we should try 4. Now, 5+4=9 but is not greater than 9 so this is not good. If we try 5, the sum of any two sides is greater than the third side. Note that 6,7,8,9,10,11,12,13 also work. When we reach 14, 5+9=14 but is not greater so that doesn’t work. Problem: “Two sides of an isosceles triangle are 2 and 5. Find every possible perimeter of the triangle.” Isosceles triangles have two equal sides. So, the triangle could have sides of 5, 5 and 2 giving us a perimeter of 12. However, many students would then suggest that there’s another isosceles triangle with sides of 2, 2 and 5. Sadly, they would be wrong. Since the sum of any two sides must be greater than the third side and 2+2 is not greater than 5, this second triangle doesn’t even exist. The only possible perimeter is 12. The sum of any two sides of a triangle must be greater than the third side. Make sure you know this when you prep for the SAT . Tags: SAT Math , SAT Prep , SAT Preparation , Test Prep

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Page 1: SAT Math: Shortest Distance Between Two Points

SAT Math: Shortest Distance Between Two Points

April 26th, 2010 by Marty Rafson Share

Euclid is credited as the founder of most of our modern geometry. Though he lived

approximately 300 years B.C., his foundation of geometry is still the basis of the geometry we

teach in high school and, therefore, the basis for SAT questions on geometry. One of the most

common geometry questions on recent SAT’s goes right back to one of Euclid’s most basic

ideas.

What is the shortest distance between two points? Euclid said the shortest distance between two

points is a straight line. This was one of his fundamental axioms, or postulates – statements that

are accepted without proof because they are so obviously true.

Consider this situation: You need to travel from point A to point B. According to Euclid, the

shortest distance would be along the straight line AB. However, now let’s suppose that you had

to divert first to a point C that is not along this straight path. You would first travel from A to C

(AC) and then from C to B (CB). Clearly, this second trip which takes us first to point C is the

longer trip. If you draw a triangle to represent this situation, we can rename the result: AC plus

CB must be longer than directly from A to B, namely AB. Thus, based on Euclid’s axiom, we

have proven a theorem: The sum of two sides of a triangle must be greater than the third side.

Someone at the Educational Testing Service has decided that this is a critical concept and a

question on it has appeared on most recent SAT’s.

Problem: “Two sides of a triangle are 5 and 9. If the length of the third side is an integer, find

one possible length of the third side.”

Let’s try 3. This doesn’t work since 5+3=8 which is not greater than 9. Perhaps we should try 4.

Now, 5+4=9 but is not greater than 9 so this is not good. If we try 5, the sum of any two sides is

greater than the third side. Note that 6,7,8,9,10,11,12,13 also work. When we reach 14, 5+9=14

but is not greater so that doesn’t work.

Problem: “Two sides of an isosceles triangle are 2 and 5. Find every possible perimeter of the

triangle.” Isosceles triangles have two equal sides. So, the triangle could have sides of 5, 5 and 2

giving us a perimeter of 12. However, many students would then suggest that there’s another

isosceles triangle with sides of 2, 2 and 5. Sadly, they would be wrong. Since the sum of any two

sides must be greater than the third side and 2+2 is not greater than 5, this second triangle

doesn’t even exist. The only possible perimeter is 12.

The sum of any two sides of a triangle must be greater than the third side. Make sure you know

this when you prep for the SAT.

Tags: SAT Math, SAT Prep, SAT Preparation, Test Prep

Page 2: SAT Math: Shortest Distance Between Two Points

Marty Rafson wrote the ESC math curriculum and has been an SAT math teacher, tutor, and

curriculum developer for 30 years. He has been a high school math teacher for 36 years and a

math department chairman for 25 years. He was also an adjunct professor at City College of New

York School of Education.

This entry was posted by Marty Rafson on Monday, April 26th, 2010 at 7:43 am and is filed under SAT Math Prep.

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