1. 1 PROBLEM 1A PROBLEM 2A PROBLEM 3A PROBLEM 4A PROBLEM 1B PROBLEM 4B PROBLEM 2B PROBLEM 3B TRAPEZOID PROPERTIES PROBLEM 5APROBLEM 5B Standard 7 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. Exactly one pair of parallel sides. 2. One pair of congruent sides. 4. Diagonals DO NOT bisect each other 5. Base angles are congruent. 6. Opposite angles are supplementary. mAmC+=180° mBmD+= A D B C 3. Diagonals are congruent ISOSCELES TRAPEZOID 7. Line connecting midpoints of congruent sides is called MEDIAN. M b 1 b 2 b 1 b 2 and are bases of trapezoid Mis the median M b 1 b 2 2 = + M = b 1 b 2 + 1 2 Standard 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 M b 1 b 2 2 = + M = b 1 b 2 + 1 2 1. Exactly one pair of parallel sides. TRAPEZOID 2. Line connecting midpoints of non-congruent sides is called MEDIAN. b 1 b 2 and are bases of trapezoid Mis the median M b 1 b 2 Standard 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 PARALLELOGRAM 1.TWO PAIRS OF CONGRUENT SIDES 2.TWO PAIRS OF PARALLEL SIDES 3.OPPOSITE ANGLES CONGRUENT 4.CONSECUTIVE ANGLES SUPPLEMENTARY 5.DIAGONALS INTERSECT AT MIDPOINT ISOSCELES TRAPEZOID 1.EXACTLY ONE PAIR OF PARALLEL SIDES 2.TWO CONGRUENT SIDES CALLED LEGS 3.TWO CONGRUENT DIAGONALS 4.TWO CONGRUENT BASE ANGLES 5.OPPOSITE ANGLES ARE SUPPLEMENTARY TRAPEZOID 1.EXACTLY ONE PAIR OF PARALLEL SIDES Do they share properties? Standard 7 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. 6 TU = (PQ + SR) 1 2 TU = (31 + 43) 1 2 = (74) 1 2 TU= 37 31 43 37 Standard 7 Q P T U R S PQRS is an isosceles trapezoid with median TU. The trapezoid has bases 31 cm and 43 cm. Find the measure of its median. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 7 FC = (GB + ED) 1 2 FC = (34 + 52) 1 2 = (86) 1 2 FC= 43 34 52 43 Standard 7 B G F C D E BDEG is an isosceles trapezoid with median FC. The trapezoid has bases 34 cm and 52 cm. Find the measure of its median. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 8 4X-2 16X+2 40 RS = (NO + QP) 1 2 = ( + ) 1 2 4X-2 16X+2 40 40 = (20X) 1 2 40 = 10X 10 X=4 NO=4X-2 =4( )-2 4 =16-2 NO= 14 QP=16X+2 =16( )+2 4 = 64+2 QP=66 Standard 7 Given: NO = 4X – 2, RS= 40, QP= 16X + 2 Find: The measure of x, the measure of NO, and the measure of QP O N R S P Q NOPQ is an isosceles trapezoid with median RS. Verifying: RS = (NO + QP) 1 2 = ( + ) 1 2 14 66 40 40 = (80) 1 2 40 = 40 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. 9 5X+5 7X+3 64 HI = (EF + DG) 1 2 = ( + ) 1 2 5X+5 7X+3 64 64 = (12X+8) 1 2 60 = 6X 6 X=10 EF=5X+5 =5( )+5 10 =50+5 EF= 55 DG=7X+3 =7( )+3 10 = 70+3 DG=73 Standard 7 Given: EF = 5X + 5, HI= 64, DG= 7X + 3 Find: Find the value of x, measure of EF, and the measure of DG F E H I G D DEFG is an isosceles trapezoid with median HI. Verifying: HI = (EF + DG) 1 2 = ( + ) 1 2 55 73 64 64 = (128) 1 2 64 = 64 64 = 12X+8 2 64 = 6X + 4 -4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. 10 Given: NO=40.2, RS=63.8 Find: The measure of QP 40.2 87.4 RS= (NO + QP) 1 2 = ( + QP ) 1 2 40.2 QP 87.4 Standard 7 O N R S P Q NOPQ is an isosceles trapezoid with median RS. 63.8 2 2 127.6 = 40.2 + QP -40.2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11. 11 Given: EF=35.6, HI=62.8 Find: The measure of DG 35.6 90 HI= (EF + DG) 1 2 = ( + DG ) 1 2 35.6 DG = 90 Standard 7 62.8 2 2 125.6 = 35.6 + DG -35.6 F E H I G D DEFG is an isosceles trapezoid with median HI. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12. 12 3X+5 115° This is an ISOSCELES TRAPEZOID, so opposite angles are supplementary: m CGF + m CDF = 180° 3X + 5 + 115 = 180° 3X + 120 = 180° -120° -120° 3X = 60° 3 X=20 m CGF = 3X+5 = 3( )+5 20 = 60+5 m CGF = 65° 65° Base angles are congruent. 65° Corresponding angles are congruent:. 65° m DEH = 65° Standard 7 D C H E F G CDFG is an isosceles trapezoid with median HE. Given:, Find: the value of x and CGF= m 3X+5 CDF= m 115 DEH m m GFE = 65° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13. 13 5X+7 113° This is an ISOSCELES TRAPEZOID, so opposite angles are supplementary: m PON + m PLN = 180° 5X + 7 + 113 = 180° 5X + 120 = 180° -120° -120° 5X = 60° 5 X=12 m PON = 5X+7 = 5( )+7 12 = 60+7 m PON = 67° 67° Base angles are congruent. 67° Corresponding angles are congruent:. 67° m QRL = 67° Standard 7 L P Q R N O LNOP is an isosceles trapezoid with median QR Given:, Find: the value of x and PON= m 5X+7 PLN= m 113 QRL m m ONL = 67° PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14. 14 4. Given: m TNL= 9X+3, m SRL=14X-27 Find: the value of X and m NLR 9X+3 14X-27 Corresponding angles are congruent. 14X-27 Base angles are congruent m TNL = m NLU 9X+3 = 14X-27 -9X 3 = 5X - 27 +27 30 = 5X 5 X = 6 m SRL = 14X-27 = 14( )-27 6 = 84 - 27 m SRL = 57° 57° Consecutive angles to the congruent segment are supplementary: m NLR + m SRL = 180° m NLR +57 = 180° -57 m NLR= 123° 123° Standard 7 U T N L R S RSTU is an isosceles trapezoid with median NL. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15. 15 4. Given: m TNL= 6X+4, m SRL= 4X+30 Find: the value of X and m NLR 6X+4 4X+30 Corresponding angles are congruent. 4X+30 Base angles are congruent m TNL = m NLU 6X+4 = 4X + 30 -6X 4 = -2X +30 -30 -26 = -2X -2 X = 13 m SRL = 4X+30 = 4( )+30 13 = 52 + 30 m SRL = 82° 82° Consecutive angles to the congruent segment are supplementary: m NLR + m SRL = 180° m NLR +82 = 180° -82 m NLR= 98° 98° Standard 7 U T N L R S RSTU is an isosceles trapezoid with median NL. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved