The Pythagorean Theorem says that if a right triangle has legs of length and , and a hypotenuse of length , then . Note that the Pythagorean Theorem is one of the formulas given to you in the beginning of each math section.
The converse of the Pythagorean Theorem is also true: If a triangle has sides with length , , and satisfying , then the triangle is a right triangle.
More specifically, we have the following.
if and only if the angle opposite the side of length is greater than 90 degrees.
if and only if the angle opposite the side of length is less than 90 degrees.
Try to answer the following question using the converse of the Pythagorean Theorem together with the Triangle Rule. Do not check the solution until you have attempted this question yourself.
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L EVEL 4: G EOMETRY
1. In the figure above, and . If , what is one possible length of ̅̅̅̅?
Solution
We have 12 – 9 = 3 and 12 + 9 = 21. So by the Triangle Rule, 3 < PQ < 21.
Using the converse of the Pythagorean Theorem, (PQ)2 > (PR)2 + (RQ)2. So we have (PQ)2 > 122 + 92 = 144 + 81 = 225, and therefore PQ > 15.
Putting the two rules together we have 15 < PQ < 21.
For example, we can grid in 16.
Now try to solve each of the following problems. The answers to these problems, followed by full solutions are at the end of this lesson. Do not look at the answers until you have attempted these problems yourself.
Please remember to mark off any problems you get wrong.
L EVEL 3: G EOMETRY
2. In the figure above, what is the area of square ?
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3. In right triangle above, what is the length of side ̅̅̅̅ ?
L EVEL 4: G EOMETRY
4. In the triangle above, and . What is the area of the triangle?
L EVEL 5: G EOMETRY
5. If is an integer less than 6, how many different triangles are there with sides of length 5, 9 and ?
(A) One (B) Two (C) Three (D) Four (E) Five
6. The lengths of the sides of a triangle are , 16 and 31, where is the shortest side. If the triangle is not isosceles, what is a possible value of ?
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7. In the figure above, and are squares, and are the midpoints of ̅̅̅̅ and ̅̅̅̅, respectively, and . If , what is the length of ̅̅̅̅?
(A) √ (B) √ (C) √ (D) √ (E)
8. Points and lie in a plane. If the distance between and is 18 and the distance between and is 11, which of the following could be the distance between and ?
I. 7 II. 28 III. 29 (A) I only (B) II only (C) III only (D) I and III only (E) I, II, and III
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9. The lengths of the sides of an isosceles triangle are 22, , and . If is an integer, what is the smallest possible perimeter of the triangle?
(A) 30 (B) 31 (C) 32 (D) 34 (E) 46
10. In , the length of side ̅̅̅̅ is 16 and the length of side ̅̅̅̅ is 17. What is the least possible integer length of side ̅̅̅̅?
11. In the triangle above, and . Point lies on between and so that . Which of the following cannot be the length of ?
(A) 1 (B) 2 (C) 3 (D) 5 (E) 7
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12. In the -plane above . What is the value of ?
Answers
1. 15 < PQ < 21 5. A 9. E 2. 146 6. 15 < x < 16 10. 2
3. 8 7. A 11. E
4. 240 8. E 12. 21
Full Solutions
2.
* Solution using the Pythagorean Theorem: Let x be the length of a side of the square. So AD = x. By the Pythagorean Theorem
x2 = 112 + 52 = 121 + 25 =146.
But x2 is precisely the area of the square. Therefore the answer is 146.
3.
* Solution using the Pythagorean Theorem: c2 = a2 + b2 = 23 + 41 = 64.
Therefore AC = c = 8.
4.
* We choose ST as the base, and draw altitude RP from vertex R to base ST.
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In an isosceles triangle the altitude is equal to the median. It follows that TP = ST = 10. Note that 10 = 2·5 and 26 = 2·13. Using the Pythagorean triple 5, 12, 13, we have that RP = 2·12 = 24.
Area = ( )bh = ( )(20)(24) = 240.
Remarks:
(1) An altitude of a triangle is perpendicular to the base. A median of a triangle splits the base into two equal parts. In an isosceles triangle, the altitude and median are equal (when you choose the base that is not one of the equal sides).
(2) We chose ST to be the base because it is the side that is not one of the equal sides.
(3) 3, 4, 5 and 5, 12, 13 are the two most common Pythagorean triples.
These sets of numbers satisfy the Pythagorean Theorem.
(4) If you do not remember the Pythagorean triples it is no big deal. Just use the Pythagorean Theorem. In this case,
102 + b2 = 262 100 + b2 = 676 b2 = 676 – 100 = 576
b = 24.
5.
* Solution using the triangle rule: By the triangle rule, 9 – 5 < x < 9 + 5.
That is, 4 < x < 14. Since x is an integer less than 6, x can only be 5. So there is one possibility, choice (A).
6.
* Solution using the triangle rule: By the triangle rule, x lies between 31 – 16 = 15 and 31 + 16 = 47. That is, we have 15 < x < 47.
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But we are also given that x is the length of the shortest side of the triangle. So x < 16.Therefore we can grid in any number between 15 and 16. For example, we can grid in 15.1.
7.
* Let’s label the given figure with what we know.
By the Pythagorean Theorem, c2 = 12 + ( )2 = 1 + = . So c = √ = √ . Then UV = QT = c – = √ = √ , choice (A).
8.
* Solution using the triangle rule: In this case, if Q, R and S form a triangle, then the length of QS is between 18 – 11 = 7 and 18 + 11 = 29.
The extreme cases 7 and 29 form straight lines. In this problem that is fine, so the distance between Q and S is between 7 and 29, inclusive.
Thus, the answer is choice (E).
9.
Solution using the triangle rule: Using the triangle rule we have that m – m < 22 < m + m. That is, 0 < 22 < 2m. So m > = 11. Therefore the smallest integer that m can be is m = 12, and it follows that the smallest possible perimeter of the triangle is 22 + 12 + 12 = 46, choice (E).
* A slightly quicker solution: For this particular question we actually only need that the third side of the triangle is less than the sum of the other two sides. So we have that 22 < m + m = 2m, and so m > = 11.
Once again, it follows that we should let m = 12, and thus the perimeter is 22 + 12 + 12 = 46, choice (E).
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10.
* Solution using the triangle rule: By the triangle rule, we have that 17 – 16 < AB < 17 + 16. That is, 1 < AB < 33. Therefore the least possible integer length of side AB is 2.
11.
*Let’s draw SU.
Now just note that RS is the hypotenuse of triangle RSU. Thus SU must be less than 6. So SU cannot be 7, and the answer is choice (E).
12.
* Solution using strategy 28: To get from O to A we go up 3 units, right 18 units. So the slope of OA is = . Since AB is perpendicular to OA, we have that the slope of AB is -6 = - . Thus, for every unit we move right along AB, we must move down 6 units. Equivalently, for every unit we move left along AB, we must move up 6 units. To get from 18 to 15 we must move left 3 units. Therefore we must move up 3·6 = 18 units. Since we are starting at 3, k = 3 + 18 = 21.
An algebraic solution using slopes: We can do all of this algebraically using the slope formula as follows.
The slope of OA is = (see the remark at the end of the first solution to problem 7 from Lesson 3). So the slope of AB is -6 because OA and AB are perpendicular. We can also compute the slope of AB using the slope formula as follows.
mAB = – – = –
Now set these equal to each other and solve for k (or guess and check).
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– = -6 k – 3 = 18
k = 21
A solution using two applications of the Pythagorean Theorem: We form two right triangles and use the given points to write down three lengths as shown in the picture below.
We can now find OA using the Pythagorean Theorem.
OA2 = 182 + 32 = 333.
So OA = √ , and therefore AB = √ also since OA = AB is given.
Finally, we can use the Pythagorean Theorem one more time to find b.
32 + b2 = AB2 9 + b2 = 333 b2 = 324
b = 18 So k = 3 + 18 = 21.
85 L ESSON 8
C OUNTING
Reminder: Before beginning this lesson remember to redo the problems from Lesson 4 that you have marked off. Do not
“unmark” a question unless you get it correct.
Strategy 21 – Writing a list
Sometimes the easiest way to count the number of possibilities is to simply list them all. When doing this it is important that we have a systematic way of forming our list. This will reduce the likelihood of missing something, or listing something twice.
Try to answer the following question by writing a list. Do not check the solution until you have attempted this question yourself.