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Published bySamson Whitmer Modified over 3 years ago

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Essential Questions How do I use length and midpoint of a segment?

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Vocabulary coordinate midpoint distance bisect length segment bisector construction between congruent segments

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The distance between any two points is the absolute value of the difference of the coordinates. If the coordinates of points A and B are a and b, then the distance between A and B is |a – b| or |b – a|. The distance between A and B is also called the length of AB, or AB. AB = |a – b| or |b - a| A a B b

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**Congruent segments are segments that have the same length**

Congruent segments are segments that have the same length. In the diagram, PQ = RS, so you can write PQ RS. This is read as “segment PQ is congruent to segment RS.” Tick marks are used in a figure to show congruent segments.

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In order for you to say that a point B is between two points A and C, all three points must lie on the same line, and AB + BC = AC.

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**Example 1: Using the Segment Addition Postulate**

M is between N and O. Find NO. NM + MO = NO Seg. Add. Postulate 17 + (3x – 5) = 5x + 2 Substitute the given values 3x + 12 = 5x + 2 Simplify. – – 2 Subtract 2 from both sides. 3x + 10 = 5x Simplify. –3x –3x Subtract 3x from both sides. 10 = 2x Divide both sides by 2. 2 5 = x

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Example 1 Continued M is between N and O. Find NO. NO = 5x + 2 = 5(5) + 2 Substitute 5 for x. = 27 Simplify.

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Check It Out! Example 2 E is between D and F. Find DF. DE + EF = DF Seg. Add. Postulate (3x – 1) + 13 = 6x Substitute the given values 3x + 12 = 6x – 3x – 3x Subtract 3x from both sides. 12 = 3x Simplify. x 3 = Divide both sides by 3. 4 = x

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**Check It Out! Example 2 Continued**

E is between D and F. Find DF. DF = 6x = 6(4) Substitute 4 for x. = 24 Simplify.

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The midpoint M of AB is the point that bisects, or divides, the segment into two congruent segments. If M is the midpoint of AB, then AM = MB. So if AB = 6, then AM = 3 and MB = 3.

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**Example 3: Recreation Application**

The map shows the route for a race. You are at X, 6000 ft from the first checkpoint C. The second checkpoint D is located at the midpoint between C and the end of the race Y. The total race is 3 miles. How far apart are the 2 checkpoints? XY = 3(5280 ft) Convert race distance to feet. = 15,840 ft

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Example 3 Continued XC + CY = XY Seg. Add. Post. Substitute 6000 for XC and 15,840 for XY. CY = 15,840 – – 6000 Subtract 6000 from both sides. CY = 9840 Simplify. D is the mdpt. of CY, so CD = CY. = 4920 ft The checkpoints are 4920 ft apart.

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**Example 4: Using Midpoints to Find Lengths**

D is the midpoint of EF, ED = 4x + 6, and DF = 7x – 9. Find ED, DF, and EF. E D 4x + 6 7x – 9 F Step 1 Solve for x. ED = DF D is the mdpt. of EF. 4x + 6 = 7x – 9 Substitute 4x + 6 for ED and 7x – 9 for DF. –4x –4x Subtract 4x from both sides. 6 = 3x – 9 Simplify. Add 9 to both sides. 15 = 3x Simplify.

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Example 4 Continued D is the midpoint of EF, ED = 4x + 6, and DF = 7x – 9. Find ED, DF, and EF. E D 4x + 6 7x – 9 F Step 2 Find ED, DF, and EF. ED = 4x + 6 DF = 7x – 9 EF = ED + DF = 4(5) + 6 = 7(5) – 9 = = 26 = 26 = 52

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Check It Out! Example 5 S is the midpoint of RT, RS = –2x, and ST = –3x – 2. Find RS, ST, and RT. R S T –2x –3x – 2 Step 1 Solve for x. RS = ST S is the mdpt. of RT. –2x = –3x – 2 Substitute –2x for RS and –3x – 2 for ST. +3x +3x Add 3x to both sides. x = –2 Simplify.

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**Check It Out! Example 5 Continued**

S is the midpoint of RT, RS = –2x, and ST = –3x – 2. Find RS, ST, and RT. R S T –2x –3x – 2 Step 2 Find RS, ST, and RT. RS = –2x ST = –3x – 2 RT = RS + ST = –2(–2) = –3(–2) – 2 = 4 + 4 = 4 = 4 = 8

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The set of all points between the sides of the angle is the interior of an angle. The exterior of an angle is the set of all points outside the angle. Angle Name R, SRT, TRS, or 1 You cannot name an angle just by its vertex if the point is the vertex of more than one angle. In this case, you must use all three points to name the angle, and the middle point is always the vertex.

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Check It Out! Example 6 Write the different ways you can name the angles in the diagram. RTQ, T, STR, 1, 2

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**The measure of an angle is usually given in degrees**

The measure of an angle is usually given in degrees. Since there are 360° in a circle, one degree is of a circle. When you use a protractor to measure angles, you are applying the following postulate.

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You can use the Protractor Postulate to help you classify angles by their measure. The measure of an angle is the absolute value of the difference of the real numbers that the rays correspond with on a protractor. If OC corresponds with c and OD corresponds with d, mDOC = |d – c| or |c – d|.

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Check It Out! Example 7 Use the diagram to find the measure of each angle. Then classify each as acute, right, or obtuse. a. BOA b. DOB c. EOC mBOA = 40° BOA is acute. mDOB = 125° DOB is obtuse. mEOC = 105° EOC is obtuse.

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**Congruent angles are angles that have the same measure**

Congruent angles are angles that have the same measure. In the diagram, mABC = mDEF, so you can write ABC DEF. This is read as “angle ABC is congruent to angle DEF.” Arc marks are used to show that the two angles are congruent. The Angle Addition Postulate is very similar to the Segment Addition Postulate that you learned in the previous lesson.

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Check It Out! Example 8 mXWZ = 121° and mXWY = 59°. Find mYWZ. mYWZ = mXWZ – mXWY Add. Post. mYWZ = 121 – 59 Substitute the given values. mYWZ = 62 Subtract.

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**An angle bisector is a ray that divides an angle into two congruent angles.**

JK bisects LJM; thus LJK KJM.

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**Example 9: Finding the Measure of an Angle**

KM bisects JKL, mJKM = (4x + 6)°, and mMKL = (7x – 12)°. Find mJKM.

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Example 9 Continued Step 1 Find x. mJKM = mMKL Def. of bisector (4x + 6)° = (7x – 12)° Substitute the given values. Add 12 to both sides. 4x = 7x Simplify. –4x –4x Subtract 4x from both sides. 18 = 3x Divide both sides by 3. 6 = x Simplify.

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Example 9 Continued Step 2 Find mJKM. mJKM = 4x + 6 = 4(6) + 6 Substitute 6 for x. = 30 Simplify.

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Check It Out! Example 10 Find the measure of each angle. QS bisects PQR, mPQS = (5y – 1)°, and mPQR = (8y + 12)°. Find mPQS. Step 1 Find y. Def. of bisector Substitute the given values. 5y – 1 = 4y + 6 Simplify. y – 1 = 6 Subtract 4y from both sides. y = 7 Add 1 to both sides.

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**Check It Out! Example 10 Continued**

Step 2 Find mPQS. mPQS = 5y – 1 = 5(7) – 1 Substitute 7 for y. = 34 Simplify.

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Check It Out! Example 11 Find the measure of each angle. JK bisects LJM, mLJK = (-10x + 3)°, and mKJM = (–x + 21)°. Find mLJM. Step 1 Find x. LJK = KJM Def. of bisector (–10x + 3)° = (–x + 21)° Substitute the given values. +x x Add x to both sides. Simplify. –9x + 3 = 21 –3 –3 Subtract 3 from both sides. –9x = 18 Divide both sides by –9. x = –2 Simplify.

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**Check It Out! Example 11 Continued**

Step 2 Find mLJM. mLJM = mLJK + mKJM = (–10x + 3)° + (–x + 21)° = –10(–2) + 3 – (–2) + 21 Substitute –2 for x. = Simplify. = 46°

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Assignment Segments page 11-13 #12, 16-23, 28-29, 31-32, 36-40 Angles page 18-20 #16-18, 27, 29-38, 41-44

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