1. 1 PROBLEM 1A PROBLEM 2A PROBLEM 3A PROBLEM 4A PROBLEM 1B PROBLEM 4B PROBLEM 2B PROBLEM 3B SQUARE PROPERTIES PROBLEM 5 Standard 7 RHOMBUS PROPERTIES END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. 2 STANDARD 7: Students prove and use theorems involving the properties of parallel lines cut by a transversal, the properties of quadrilaterals, and properties of circles. ESTÁNDAR 7: Los estudiantes prueban y usan teoremas involucrando las propiedades de líneas paralelas cortadas por una transversal, las propiedades de cuadriláteros, y las propiedades de círculos. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 3 1. Two pairs of parallel sides. 2. All sides are congruent. 5. Diagonals bisect each other 7. Opposite angles are congruent. 8. Consecutive angles are supplementary. mAmB+=180° mBmC+= mCmD+= mDmA+= A D B C 4. Diagonals are congruent 3. All angles are right. 6. Diagonals form a right angle SQUARE Standard 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 5X+5 7X – 19 Since all sides are congruent: 5X + 5 = 7X – 19 -5X -5X 5 = 2X – 19 +19 24 = 2X 2 2 X=12 Standard 7 D G F E DEFG is a square DG = 5X + 5 EF = 7X – 19 Find the value for X and the lenght of the side. Now since all sides are congruent, we need to find the length of just one side: DG = 5X + 5 = 5( ) + 5 12 = 60 + 5 = 65 The length of the side is 65. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 9X – 3 6X + 24 Since all sides are congruent: 9X – 3 = 6X + 24 -6X -6X 3X – 3 = 24 +3 3x = 27 3 3 X=9 Standard 7 K H I J HIJK is a square KH =9X – 3 IJ = 6X + 24 Find the value for X and the lenght of the side. Now since all sides are congruent, we need to find the length of just one side: KH = 9X – 3 = 9( ) – 3 9 = 81 – 3 = 78 The length of the side is 78. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. 6 1. Two pairs of parallel sides. 2. All sides are congruent. 4. Diagonals bisect each other 6. Opposite angles are congruent and bisected by diagonals. 7. Consecutive angles are supplementary. mAmB+=180° mBmC+= mCmD+= mDmA+= A D B C 3. Diagonals are NOT congruent 5. Diagonals form a right angle RHOMBUS Standard 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 7 (25X+15)° It is a rhombus, because it has 4 congruent sides, so diagonals are perpendicular: 25X + 15 = 90° -15 25X = 75 25 X=3 Standard 7 Find the value for X in the figure below. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 8 (10X+20)° It is a rhombus, because it has 4 congruent sides, so diagonals are perpendicular: 10X + 20 = 90° -20 10X = 70 10 X=7 Standard 7 Find the value for X in the figure below. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. 9 F G H I 5X+2 15X – 38 i. Observing the figure we see that: GFH = m FHI m 5X + 2 = 15X – 38 -5X -5X 2 = 10X – 38 +38 40 = 10X 10 10 X=4 GFH= m ii. Finding angles by substituting X value: 5X+2 =5( )+2 4 = 20 + 2 = 22° 22° 44° and consecutive angles are supplementary: G= m 180-44° + G m = 180°44° G= m 136° I= m 136° So: 136° iii. Finding measure of vertices: 22° F= m 2( ) = 44° so, opposite angles are H= m 44° -44 -44 Standard 7 FGHI is a rhombus in which and What is the value for X and the measure for each vertex. GFH= m 5X + 2 FHI= m 15X – 38 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. 10 G D E F 3X+45 11X + 5 i. Observing the figure we see that: DGE = m GEF m 3X + 45 = 11X + 5 -3X -3X 45 = 8X + 5 -5 40 = 8X 8 8 X=5 DGE= m ii. Finding angles by substituting X value: 3X+45 =3( )+45 5 = 15 + 45 = 60° 60° 120° and consecutive angles are supplementary: D= m 180-120° + D m = 180°120° D= m 60° F= m 60° So: 60° iii. Finding measure of vertices: 60° G= m 2( ) = 120° so, opposite angles are E= m 120° -120 -120 Standard 7 DEFG is a rhombus in which and What is the value for X and the measure for each vertex. DGE= m 3X+45 GEF= m 11X+5 Note: Observe that acute and obtuse angles in figure are swicthed. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11. 11 R S T U X - 1 2 5X+5 From the figure: URS= m RTU m 2 5X+5 X – 1 2 =2( ) 2 X – 1 =10X + 10 -10X -10 X -10X – 11= 0 2 -10 -11 (1)(-11) 1+(-11)= -10 (X+1)(X – 11)= 0 X + 1= 0 X – 11 =0 X= -1 +11 X=11 What about X = 11? URS= m X - 1 2 11 = 121 - 1 = 120° 120° opposite angles are T= m URS m = 120° 120° and consecutive angles are supplementary: + S m = 180°120° -120 -120° S= m 60° opposite angles are U= m S m m 60° URS= m ( ) - 1 2 Standard 7 RSTU is a rhombus with and What is the value for X, and the measure of each vertex. URS= m X – 1 2 RTU= m 5X+5 URS= m X - 1 2 = 1 – 1 = 0° URS= m ( ) –1 2 We can’t have a rhombus with a 0° angle! Is there a vertex with X= -1? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12. 12 P Q R S X - 6 2 5X+9 From the figure: SPQ= m PRS m 2 5X+9 X - 6 2 =2( ) 2 X - 6 =10X + 18 -10X -18 X - 10X -24 = 0 2 -10 -24 (1)(-24) 1+(-24)= -23 (2)(-12) 2+(-12)= -10 (X+2)(X-12)= 0 X+2= 0 X-12=0 -2 X= -2 +12 X=12 What about X=12? SPQ= m X - 6 2 12 = 144 - 6 = 138° 138° opposite angles are R= m SPQ m = 138° 138° and consecutive angles are supplementary: + Q m = 180°138° -138 -138° Q= m 42° opposite angles are S= m Q m m 42° SPQ= m ( ) - 6 2 Standard 7 PQRS is a rhombus in which and What is the value for X and find the measure of each vertex. SPQ= m X – 6 2 PRS= m 5X + 9 SPQ= m X - 6 2 -2 = 4 – 6 = -2° URS= m ( ) –6 2 We can’t have a rhombus with a NEGATIVE ANGLE! Can we have a vertex with X= -2? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13. 13 x y 4 2 6-2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 -10 A B C D a) Let’s prove it has 4 congruent sides: Side AB: A(3,2), B(7,5) AB= ( - ) + ( - ) 22 AB= ( -4 ) + ( -3 ) 22 = 16 + 9 = 25 AB=5 Side BC:B(7,5), C(10,1) BC= ( - ) +( - ) 22 BC= ( -3 ) + ( 4 ) 22 = 9 + 16 = 25 BC=5 Distance Formula: d = (x –x ) + (y –y ) 2 2 1 1 2 2 y 1 y 2 x 1 x 2 y 1 y 2 x 1 x 2 3 7 2 5 710 5 1 5 5 5 5 Take notes Since it is a parallelogram: opposite sides are congruent CD=5DA=5 and Standard 7 ABCD is a parallelogram whose vertices are A(3,2), B(7,5), C(10,1), D(6,-2). Is it that ABCD is a square? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14. 14 x y 4 2 6-2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 -10 A B C D b) Let’s prove it has 4 right angles: 5 5 5 5 m= y 1 y 2 x 1 x 2 - - Slope Formula: Segment AB:A(3,2), B(7,5) y 1 y 2 x 1 x 2 3 7 2 5 = -3 -4 = 3 4 Segment BC:B(7,5), C(10,1) y 1 y 2 x 1 x 2 7 10 5 1 = 4 -3 3 4 = -1 Are AB and BC Perpendicular? Since the product of the slopes is -1 they are perpendicular: Opposite angles in a parallelogram are congruent: Consecutive angles are supplementary, so: C+ m D= m 180° C+ m 90° = 180° -90 C= m 90° Finally Opposite angles are congruent: IT IS A SQUARE! B= m 90° D= m 90° A= m 90° m = - - AB m = - - BC Standard 7 = 3 4 - = 4 3 - 12 - ABCD is a parallelogram whose vertices are A(3,2), B(7,5), C(10,1), D(6,-2). Is it that ABCD is a square? PRESENTATION CREATED BY SIMON PEREZ. All rights reserved