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USING UNDEFINED TERMS AND DEFINITIONS A point has no dimension. It is usually represented by a small dot. Point A A • USING UNDEFINED TERMS AND DEFINITIONS A line extends in one dimension. It is usually represented by a straight line with two arrowheads to indicate that the line extends without end in two directions. Line or AB Collinear points are points that lie on the same line. USING UNDEFINED TERMS AND DEFINITIONS A plane extends in two dimensions. It is usually represented by a shape that looks like a table or a wall, however you must imagine that the plane extends without end. USING UNDEFINED TERMS AND DEFINITIONS Coplanar points are points that lie on the same plane. A B C M Plane M or plane ABC Naming Collinear and Coplanar Points • Name three points that are collinear. H • Name four points that are coplanar. G • Name three points that are not collinear. E F D SOLUTION • Points D, E, F lie on the same line, so they are collinear. • Points D, E, F, and G lie on the same plane, so they are coplanar. Also, D, E, F, and H are coplanar. • There are many correct answers. For instance, points H, E, and G do not lie on the same line. USING UNDEFINED TERMS AND DEFINITIONS Another undefined concept in geometry is the idea that a point on a line is between two other points on the line. You can use this idea to define other important terms in geometry. USING UNDEFINED TERMS AND DEFINITIONS Consider the line AB (symbolized by AB). The line segment or segment AB (symbolized by AB) consists of the endpoints A and B, and all points on AB that are between A and B. USING UNDEFINED TERMS AND DEFINITIONS The ray AB (symbolized by AB) consists of the initial point A and all points on AB that lie on the same side of A as point B. Note that AB is the same as BA, and AB is the same as BA. However, AB and BA are not the same. They have different initial points and extend in different directions. USING UNDEFINED TERMS AND DEFINITIONS If C is between A and B, then CA and CB are opposite rays. Like points, segments and rays are collinear if they lie on the same plane. So, any two opposite rays are collinear. Segments, rays, and lines are coplanar if they lie on the same plane. Drawing Lines, Segments, and Rays Draw three noncollinear points J, K, L. Then draw JK, KL and LJ. SOLUTION 1 Draw J, K, and L 2 Draw JK. 3 Draw KL. 4 Draw LJ. K J L Drawing Opposite Rays Draw two lines. Label points on the lines and name two pairs of opposite rays. SOLUTION Points M, N, and X are collinear and X is between M and N. So, XM and XN are opposite rays. Points P, Q, and X are collinear and X is between P and Q. So, XP and XQ are opposite rays. SKETCHING INTERSECTIONS OF LINES AND PLANES Two or more geometric figures intersect if they have one or more points in common. The intersection of the figures is the set of points the figures have in common. Sketching Intersections Sketch a line that intersects a plane at one point. SOLUTION Draw a plane and a line. Emphasize the point where they meet. Dashes indicate where the line is hidden by the plane. Sketching Intersections Sketch two planes that intersect in a line. SOLUTION Draw two planes. Emphasize the line where they meet. Dashes indicate where one plane is hidden by the other plane CONGRUENCE OF ANGLES THEOREM THEOREM 2.2 Properties of Angle Congruence Angle congruence is r ef lex ive, sy mme tric, and transitive. Here are some examples. REFLEX IVE For any angle A, SYMMETRIC If A B, then TRANSITIVE If A B and B C, then A A B A A C Transitive Property of Angle Congruence Prove the Transitive Property of Congruence for angles. SOLUTION To prove the Transitive Property of Congruence for angles, begin by drawing three congruent angles. Label the vertices as A, B, and C. A B GIVEN B, C PROVE A C B A C Transitive Property of Angle Congruence A B GIVEN B, C Statements A B 1 PROVE A C Reasons B, C Given 2 m A=m B Definition of congruent angles 3 m B=m C Definition of congruent angles 4 m A=m C Transitive property of equality 5 A C Definition of congruent angles Using the Transitive Property This two-column proof uses the Transitive Property. GIVEN m 3 = 40°, PROVE m 1 = 40° 1 2, Statements 1 2 m 1 3 3 m 1=m 4 m 1 = 40° 3 Reasons 3 = 40°, 2 3 1 2 2, Given Transitive property of Congruence 3 Definition of congruent angles Substitution property of equality Proving Theorem 2.3 THEOREM THEOREM 2.3 Right Angle Congruence Theorem All right angles are congruent. You can prove Theorem 2.3 as shown. GIVEN 1 and PROVE 1 2 are right angles 2 Proving Theorem 2.3 GIVEN 1 and PROVE 1 2 are right angles 2 Statements 1 and 1 Reasons 2 are right angles 2 m 1 = 90°, m 3 m 1=m 4 1 2 2 2 = 90° Given Definition of right angles Transitive property of equality Definition of congruent angles PROPERTIES OF SPECIAL PAIRS OF ANGLES THEOREMS THEOREM 2.4 Congruent Supplements Theorem If two angles are supplementary to the same angle (or to congruent angles) then they are congruent. 1 2 3 PROPERTIES OF SPECIAL PAIRS OF ANGLES THEOREMS THEOREM 2.4 Congruent Supplements Theorem If two angles are supplementary to the same angle (or to congruent angles) then they are congruent. 1 2 3 If m 1 + m 2 = 180° and m 2 + m 3 = 180° then 1 3 1 PROPERTIES OF SPECIAL PAIRS OF ANGLES THEOREMS THEOREM 2.5 Congruent Complements Theorem If two angles are complementary to the same angle (or to congruent angles) then the two angles are congruent. 5 4 6 PROPERTIES OF SPECIAL PAIRS OF ANGLES THEOREMS THEOREM 2.5 Congruent Complements Theorem If two angles are complementary to the same angle (or to congruent angles) then the two angles are congruent. 4 5 4 If m 4 + m 5 = 90° and m 5 + m 6 = 90° then 4 6 6 Proving Theorem 2.4 GIVEN 1 and 2 are supplements 3 and 4 are supplements 1 4 PROVE 2 3 Statements 1 and 2 are supplements 3 and 4 are supplements 1 4 1 2 Reasons m m 1+m 3+m 2 = 180° 4 = 180° Given Definition of supplementary angles Proving Theorem 2.4 GIVEN 1 and 2 are supplements 3 and 4 are supplements 1 4 PROVE 2 Statements 3 Reasons 3 m m 1+m 3+m 2= 4 Transitive property of equality 4 m 1=m 4 Definition of congruent angles 5 m m 1+m 3+m 2= 1 Substitution property of equality Proving Theorem 2.4 GIVEN 1 and 2 are supplements 3 and 4 are supplements 1 4 PROVE 2 Statements 6 7 m 2=m 2 3 3 Reasons 3 Subtraction property of equality Definition of congruent angles PROPERTIES OF SPECIAL PAIRS OF ANGLES POSTULATE POSTULATE 12 Linear Pair Postulate If two angles form a linear pair, then they are supplementary. m 1+m 2 = 180° Proving Theorem 2.6 THEOREM THEOREM 2.6 Vertical Angles Theorem Vertical angles are congruent 1 3, 2 4 Proving Theorem 2.6 GIVEN PROVE 5 and 6 and 5 6 are a linear pair, 7 are a linear pair 7 Statements Reasons 1 5 and 6 and 6 are a linear pair, 7 are a linear pair Given 2 5 and 6 and 6 are supplementary, 7 are supplementary Linear Pair Postulate 3 5 7 Congruent Supplements Theorem PROPERTIES OF PARALLEL LINES POSTULATE POSTULATE 15 Corresponding Angles Postulate If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent. 1 2 1 2 PROPERTIES OF PARALLEL LINES THEOREMS ABOUT PARALLEL LINES THEOREM 3.4 Alternate Interior Angles If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent. 3 4 3 4 PROPERTIES OF PARALLEL LINES THEOREMS ABOUT PARALLEL LINES THEOREM 3.5 Consecutive Interior Angles If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles are supplementary. 5 6 m 5+m 6 = 180° PROPERTIES OF PARALLEL LINES THEOREMS ABOUT PARALLEL LINES THEOREM 3.6 Alternate Exterior Angles If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent. 7 8 7 8 PROPERTIES OF PARALLEL LINES THEOREMS ABOUT PARALLEL LINES THEOREM 3.7 Perpendicular Transversal If a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other. j k Proving the Alternate Interior Angles Theorem Prove the Alternate Interior Angles Theorem. SOLUTION GIVEN p || q PROVE 1 Statements 2 Reasons 1 p || q Given 2 1 3 3 2 Vertical Angles Theorem 4 1 Transitive property of Congruence 3 2 Corresponding Angles Postulate Using Properties of Parallel Lines Given that m 5 = 65°, find each measure. Tell which postulate or theorem you use. SOLUTION m 6 = m 5 = 65° m 7 = 180° – m m 8 = m 5 = 65° Corresponding Angles Postulate m 9 = m 7 = 115° Alternate Exterior Angles Theorem Vertical Angles Theorem 5 = 115° Linear Pair Postulate PROPERTIES OF SPECIAL PAIRS OF ANGLES Using Properties of Parallel Lines Use properties of parallel lines to find the value of x. SOLUTION m m 4 = 125° 4 + (x + 15)° = 180° 125° + (x + 15)° = 180° x = 40° Corresponding Angles Postulate Linear Pair Postulate Substitute. Subtract. Estimating Earth’s Circumference: History Connection Over 2000 years ago Eratosthenes estimated Earth’s circumference by using the fact that the Sun’s rays are parallel. When the Sun shone exactly down a vertical well in Syene, he measured the angle the Sun’s rays made with a vertical stick in Alexandria. He discovered that m 2 1 50 of a circle Estimating Earth’s Circumference: History Connection m 2 1 50 of a circle Using properties of parallel lines, he knew that m 1= m 2 He reasoned that m 1 1 50 of a circle Estimating Earth’s Circumference: History Connection m 1 1 50 of a circle The distance from Syene to Alexandria was believed to be 575 miles 1 50 of a circle Earth’s circumference 575 miles Earth’s circumference 50(575 miles) Use cross product property 29,000 miles How did Eratosthenes know that m 1=m 2? Estimating Earth’s Circumference: History Connection How did Eratosthenes know that m 1=m SOLUTION Because the Sun’s rays are parallel, Angles 1 and 2 are alternate interior angles, so 1 2 By the definition of congruent angles, m 1=m 2 2? SSS AND SAS CONGRUENCE POSTULATES If all six pairs of corresponding parts (sides and angles) are congruent, then the triangles are congruent. If Sides are congruent and Angles are congruent 1. AB DE 4. A D 2. BC EF 5. B E 3. AC DF 6. C F then Triangles are congruent ABC DEF SSS AND SAS CONGRUENCE POSTULATES POSTULATE POSTULATE 19 Side - Side - Side (SSS) Congruence Postulate If three sides of one triangle are congruent to three sides of a second triangle, then the two triangles are congruent. If Side S MN QR Side S NP RS Side S PM SQ then MNP QRS SSS AND SAS CONGRUENCE POSTULATES The SSS Congruence Postulate is a shortcut for proving two triangles are congruent without using all six pairs of corresponding parts. Using the SSS Congruence Postulate Prove that PQW TSW. SOLUTION Paragraph Proof The marks on the diagram show that PQ TS, PW TW, and QW SW. So by the SSS Congruence Postulate, you know that PQW TSW. SSS AND SAS CONGRUENCE POSTULATES POSTULATE POSTULATE 20 Side-Angle-Side (SAS) Congruence Postulate If two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the two triangles are congruent. If Side S Angle A Side S PQ WX Q X QS XY then PQS WXY Using the SAS Congruence Postulate Prove that AEB DEC. 1 2 3 1 Statements Reasons AE DE, BE CE Given 1 2 AEB DEC 2 Vertical Angles Theorem SAS Congruence Postulate MODELING A REAL-LIFE SITUATION Proving Triangles Congruent You are designing the window shown in the drawing. You want to make DRA congruent to DRG. You design the window so that DR AG and RA RG. ARCHITECTURE Can you conclude that DRA DRG ? D SOLUTION GIVEN PROVE DR AG RA RG DRA A DRG R G Proving Triangles Congruent GIVEN DR AG RA RG DRA PROVE D DRG A Statements R G Reasons Given 1 DR AG 2 DRA and DRG are right angles. If 2 lines are , then they form 4 right angles. 3 DRA 4 RA RG Given 5 DR DR Reflexive Property of Congruence 6 DRA DRG SAS Congruence Postulate DRG Right Angle Congruence Theorem Congruent Triangles in a Coordinate Plane Use the SSS Congruence Postulate to show that ABC FGH. SOLUTION AC = 3 and FH = 3 AC FH AB = 5 and FG = 5 AB FG Congruent Triangles in a Coordinate Plane Use the distance formula to find lengths BC and GH. d= BC = (x 2 – x1 ) 2 + ( y2 – y1 ) 2 (– 4 – (– 7)) 2 + (5 – 0 ) 2 d= GH = (x 2 – x1 ) 2 + ( y2 – y1 ) 2 (6 – 1) 2 + (5 – 2 ) 2 = 32 + 52 = 52 + 32 = 34 = 34 Congruent Triangles in a Coordinate Plane BC = 34 and GH = 34 BC GH All three pairs of corresponding sides are congruent, ABC FGH by the SSS Congruence Postulate. USING MEDIANS OF A TRIANGLE A median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. D USING MEDIANS OF A TRIANGLE The three medians of a triangle are concurrent. The point of concurrency is called the centroid of the triangle. The centroid is always inside the triangle. centroid acute triangle centroid centroid right triangle obtuse triangle The medians of a triangle have a special concurrency property. USING MEDIANS OF A TRIANGLE THEOREM THEOREM 5.7 Concurrency of Medians of a Triangle The medians of a triangle intersect at a point that is two-thirds of the distance from each vertex to the midpoint of the opposite side. If P is the centroid of ABC, then 2 AD 3 BP = 2 BF 3 CP = 2 CE 3 AP = P USING MEDIANS OF A TRIANGLE The centroid of a triangle can be used as its balancing point. A triangular model of uniform thickness and density will balance at the centroid of the triangle. Using the Centroid of a Triangle P is the centroid of QRS shown below and PT = 5. Find RT and RP. SOLUTION Because P is the centroid, RP = 2 RT. 3 1 Then PT = RT – RP = RT 3 1 Substituting 5 for PT, 5 = RT, so RT = 15. 3 Then RP = 2 RT = 2 (15) = 10. 3 3 So, RP = 10 and RT = 15. Finding the Centroid of a Triangle Find the coordinates of the centroid of JKL. (5, 8) SOLUTION The centroid is two thirds of the distance from each vertex to the midpoint of the opposite side. Choose the median KN. Find the coordinates of N, the midpoint of JL. The coordinates of N are 3 + 7 , 6 + 10 2 2 L = (5, 8) (7, 10) N P (5, 2) K = 10 , 16 2 2 (3, 6) J Finding the Centroid of a Triangle Find the coordinates of the centroid of JKL. (5, 8) SOLUTION Find the distance from vertex K to midpoint N. The distance from K (5,2) to N(5,8) is 8 – 2, or 6 units. Determine the coordinates of 2 the centroid, which is • 6, or 3 4 units up from vertex K along (3, 6) (7, 10) N (5, 6) P L J M (5, 2) K the median KN. The coordinates of the centroid P are (5, 2 + 4), or (5, 6) [Yellow coordinates appear.] USING ALTITUDES OF A TRIANGLE An altitude of a triangle is the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side. An altitude can lie inside, on, or outside the triangle. Every triangle has three altitudes. The lines containing the altitudes are concurrent and intersect at a point called the orthocenter of the triangle. Drawing Altitudes and Orthocenters Where is the orthocenter of an acute triangle? SOLUTION Draw an example. The three altitudes intersect at G, a point inside the triangle. Drawing Altitudes and Orthocenters Where is the orthocenter of a right triangle? SOLUTION The two legs, LM and KM, are also altitudes. They intersect at the triangle’s right angle. This implies that the orthocenter is on the triangle at M, the vertex of the right angle of the triangle. Drawing Altitudes and Orthocenters Where is the orthocenter of an obtuse triangle? SOLUTION The three lines that contain the altitudes intersect at W, a point outside the triangle. USING ALTITUDES OF A TRIANGLE THEOREM THEOREM 5.8 Concurrency of Altitudes of a Triangle The lines containing the altitudes of a triangle are concurrent. If AE, BF, and CD are the altitudes of ABC, then the lines AE, BF, and CD intersect at some point H. PROPERTIES OF SPECIAL PARALLELOGRAMS A rhombus is a parallelogram with four congruent sides. A rectangle is a parallelogram with four right angles. | | | | | | | | A square is a parallelogram with four congruent sides and four right angles. PROPERTIES OF SPECIAL PARALLELOGRAMS The Venn diagram shows the relationships among parallelograms, rhombuses, rectangles, and squares. Each shape has the properties of every group that it belongs to. For instance, a square is a rectangle, a rhombus, and a parallelogram, so it has all of the properties of each of those shapes. PROPERTIES OF SPECIAL PARALLELOGRAMS parallelograms rhombuses rectangles squares Describing a Special Parallelogram Decide whether the statement is always, sometimes, or never true. A rhombus is a rectangle. SOLUTION The statement is sometimes true. In the Venn Diagram, the regions for rhombuses and rectangles overlap. If the rhombus is a square, it is a rectangle. Help Describing a Special Parallelogram Decide whether the statement is always, sometimes, or never true. A parallelogram is a rectangle. SOLUTION The statement is sometimes true. Some parallelograms are rectangles. In the Venn diagram, you can see that some of the shapes in the parallelogram box are in the region for rectangles, but many aren’t. Help Using Properties of Special Parallelograms ABCD is a rectangle. What else do you know about ABCD? A B D C SOLUTION Because ABCD is a rectangle, it has four right angles by the definition. The definition also states that rectangles are parallelograms, so ABCD has all the properties of a parallelogram: 1 Opposite sides are parallel and congruent. 2 Opposite angles are congruent and consecutive angles are supplementary. 3 Diagonals bisect each other. Using Properties of Special Parallelograms ABCD is a rectangle. What else do you know about ABCD? A B D C A rectangle is defined as a parallelogram with four right angles. But any quadrilateral with four right angles is a rectangle because any quadrilateral with four right angles is a parallelogram. Using Properties of Special Parallelograms COROLLARIES ABOUT SPECIAL QUADRILATERALS RHOMBUS COROLLARY A quadrilateral is a rhombus if and only if it has four congruent sides. RECTANGLE COROLLARY A quadrilateral is a rectangle if and only if it has four right angles. SQUARE COROLLARY A quadrilateral is a square if and only if it is a rhombus and a rectangle. You can use these corollaries to prove that a quadrilateral is a rhombus, rectangle, or square without proving first that the quadrilateral is a parallelogram. Using Properties of a Rhombus In the diagram, PQRS is a rhombus. What is the value of y? P Q 2y + 3 S SOLUTION 5y – 6 All four sides of a rhombus are congruent, so RS = PS. 5y – 6 = 2y + 3 Equate lengths of congruent sides. 5y = 2y + 9 Add 6 to each side. 3y = 9 Subtract 2y from each side. y=3 Divide each side by 3. R USING DIAGONALS OF SPECIAL PARALLELOGRAMS THEOREMS THEOREM 6.11 A parallelogram is a rhombus if and only if its diagonals are perpendicular. ABCD is a rhombus if and only if AC BD USING DIAGONALS OF SPECIAL PARALLELOGRAMS THEOREMS THEOREM 6.12 B C A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles. A ABCD is a rhombus if and only if AC bisects BD bisects D DAB and ADC and BCD and CBA USING DIAGONALS OF SPECIAL PARALLELOGRAMS THEOREM S THEOREM 6.13 A B D C A parallelogram is a rectangle if and only if its diagonals are congruent. ABCD is a rectangle if and only if AC BD USING DIAGONALS OF SPECIAL PARALLELOGRAMS You can rewrite Theorem 6.11 as a conditional statement and its converse. Conditional statement: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. Converse: If a parallelogram is a rhombus, then its diagonals are perpendicular. To prove the theorem, you must prove both statements. USING SIMILARITY THEOREMS THEOREM S THEOREM 8.2 Side-Side-Side (SSS) Similarity Theorem If the corresponding sides of two triangles are proportional, then the triangles are similar. P A Q R If AB = BC = CA RP PQ QR then ABC ~ PQR. B C USING SIMILARITY THEOREMS THEOREM S THEOREM 8.3 Side-Angle-Side (SAS) Similarity Theorem If an angle of one triangle is congruent to an angle of a second triangle and the lengths of the sides including these angles are proportional, then the triangles are similar. If X XY M and ZX = MN PM then XYZ ~ MNP. M X P Z Y N Proof of Theorem 8.2 GIVEN RS = ST = TR LM MN NL PROVE RST ~ LMN SOLUTION Paragraph Proof M L S P N Q R T Locate P on RS so that PS = LM. Draw PQ so that PQ RT. Then RST ~ PSQ, by the AA Similarity Postulate, and RS = ST = TR . PS SQ QP Because PS = LM, you can substitute in the given proportion and find that SQ = MN and QP = NL. By the SSS Congruence Theorem, it follows that PSQ LMN. Use the definition of congruent triangles and the AA Similarity Postulate to conclude that RST ~ LMN. Using the SSS Similarity Theorem Which of the following three triangles are similar? 12 A E C 6 6 9 F 14 G J 4 8 D 6 B 10 H SOLUTION To decide which of the triangles are similar, consider the ratios of the lengths of corresponding sides. Ratios of Side Lengths of ABC and DEF AB 6 3 = = , DE 4 2 Shortest sides CA 12 3 = = , FD 8 2 Longest sides BC 9 3 = = EF 6 2 Remaining sides Because all of the ratios are equal, ABC ~ DEF Using the SSS Similarity Theorem Which of the following three triangles are similar? 12 A E C 6 6 9 F 14 G J 4 8 D 6 B 10 H SOLUTION To decide which of the triangles are similar, consider the ratios of the lengths of corresponding sides. Ratios Lengths of ABC and and GHJ ABC is not SinceofSide ABC is similar to DEF AB 6 to GHJ, DEF CA is12 6 similar not= similar = , = = 1 , GH 6 JG 14 7 Shortest sides Longest sides BC . to GHJ = 9 HJ 10 Remaining sides Because all of the ratios are not equal, ABC and DEF are not similar. Using the SAS Similarity Theorem Use the given lengths to prove that RST ~ PSQ. SOLUTION GIVEN SP = 4, PR = 12, SQ = 5, QT = 15 PROVE RST ~ PSQ S Use the SAS Similarity Theorem. Find the ratios of the lengths of the corresponding sides. Paragraph Proof 4 P SR SP + PR 4 + 12 16 = = = = 4 SP SP 4 4 ST SQ + QT 5 + 15 20 = = = = 4 SQ SQ 5 5 12 R The side lengths SR and ST are proportional to the corresponding side lengths of PSQ. Because S is the included angle in both triangles, use the SAS Similarity Theorem to conclude that RST ~ PSQ. 5 Q 15 T USING SIMILAR TRIANGLES IN REAL LIFE Using a Pantograph SCALE DRAWING As you move the tracing pin of a pantograph along a figure, the pencil attached to the far end draws an enlargement. P R Q T S USING SIMILAR TRIANGLES IN REAL LIFE Using a Pantograph As the pantograph expands and contracts, the three brads and the tracing pin always form the vertices of a parallelogram. P R Q T S USING SIMILAR TRIANGLES IN REAL LIFE Using a Pantograph The ratio of PR to PT is always equal to the ratio of PQ to PS. Also, the suction cup, the tracing pin, and the pencil remain collinear. P R Q T S Using a Pantograph P R How can you show that PRQ ~ PTS? T Q SOLUTION S PR PQ You know that PT = PS . Because P P, you can apply the SAS Similarity Theorem to conclude that PRQ ~ PTS. Using a Pantograph P In the diagram, PR is 10 inches and RT is 10 inches. The length of the cat, RQ, original print 2.4 inches. Find in thethe length TS in theisenlargement. 10" R 2.4" 10" T Q SOLUTION Because the triangles are similar, you can set up a proportion to find the length of the cat in the enlarged drawing. PR RQ = Write proportion. PT TS 10 = 2.4 Substitute. 20 TS Solve for TS. TS = 4.8 So, the length of the cat in the enlarged drawing is 4.8 inches. S Finding Distance Indirectly ROCK CLIMBING are at an indoor climbing wall. To estimate of SimilarYou triangles can be used to find distances thatthe areheight difficult the wall, youto place a mirror on the floor 85 feet from the base of the wall. Then measure directly. you walk backward until you can see the top of the wall centered in the mirror. You are 6.5 feet from the mirror and your eyes are 5 feet above the ground. Use similar triangles to estimate the height of the wall. D B 5 ft Not drawn to scale A 6.5 ft C 85 ft E Finding Distance Indirectly Use similar triangles to estimate the height of the wall. SOLUTION Due to the reflective property of mirrors, you can reason that ACB ECD. D Using the fact that ABC and EDC are right triangles, you can apply the AA Similarity Postulate to conclude that these two triangles are similar. B 5 ft A 6.5 ft C 85 ft E Finding Distance Indirectly Use similar triangles to estimate the height of the wall. SOLUTION DE EC Ratios of lengths of = corresponding sides are equal. BA AC So, the height of the wall is about 65 feet. DE = 85 Substitute. 5 6.5 D Multiply each side by 5 and simplify. 65.38 DE B 5 ft A 6.5 ft C 85 ft E