1 / 8

Chapter 6 Section 3

Chapter 6 Section 3. Rectangles. P. Q. Warm-Up. Given: T riangle PRS is congruent to triangle RPQ. Prove: PQRS is a parallelogram. S. R. Triangle PRS is congruent to triangle RPQ. Given. <SRP is congruent to < QPR. CPCTC.

kellie-travis
Télécharger la présentation

Chapter 6 Section 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6Section 3 Rectangles

  2. P Q Warm-Up Given: Triangle PRS is congruent to triangle RPQ. Prove: PQRS is a parallelogram. S R Triangle PRS is congruent to triangle RPQ Given <SRP is congruent to < QPR CPCTC If alternate interior angles are congruent then the lines are parallel PQ is parallel to RS PQ is congruent to RS CPCTC One pair of opposite sides is congruent and parallel. PQRS is a parallelogram

  3. Vocabulary Rectangle- A quadrilateral with four right angles. Theorem 6-9- If a parallelogram is a rectangle, then its diagonals are congruent. AC is congruent to BD Theorem 6-10- If the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. B A D C

  4. Example 1: If quadrilateral JKLM is a rectangle, LP = 3x + 7, and MK = 26, find the value of x. M L P Since it is a rectangle, the diagonals are congruent. LJ = MK Since a rectangle is also a parallelogram, the diagonals bisect each other. MK = LJ MK = LP + PJ MK = LP + LP MK = 2(LP) 26 = 2(3x + 7) 26 = 6x + 14 12 = 6x 2 = x J K

  5. Example 2: Quadrilateral MNOP is a rectangle. Find the value of x. A) MO = 2x – 8, NP = 23 Diagonals are congruent. NP = MO 23 = 2x – 8 31 = 2x 15.5 = x B) MO = 4x – 13, PC = x + 7 The diagonals are congruent. MO = PN Since a rectangle is also a parallelogram, the diagonals bisect each other. MO = PN MO = PC + CN MO = PC + PC MO = 2(PC) 4x - 13 = 2(x + 7) 4x - 13 = 2x + 14 2x - 13 = 14 2x = 27 x = 13.5 N M C O P

  6. Example 3: Use rectangle KLMN and the given information to solve each problem. A) m<1 = 70. Find m<2, m<5, and m<6. Find m<2 m<1 + m<2 = 90 70 + m<2 = 90 m<2 = 20 Find m<5 Alternate interior angles are congruent. m<5 = m<1 m<5 = 70 Find m<6 Alternate interior angles are congruent. m<6 = m<2 m<6 = 20 L K 8 1 7 2 C 9 10 6 3 5 4 M N

  7. Example 3: Use rectangle KLMN and the given information to solve each problem. B) m<9 =128. Find m<6, m<7, and m<8. Find m<6 KC is congruent to CN because the the diagonals are congruent and they bisect each other. So triangle KCN is an isosceles triangle. <6 is congruent to <7. All the angles in a triangle add up to 180. 180 = m<9 + m<6 + m<7 180 = 128 + 2(m<6) 52 = 2(m<6) 26 = m<6 Find m<7 <6 is congruent to <7 m<6 = m<7 26 = m<7 L K 8 1 7 2 C 9 10 6 3 5 4 M N Find m<8 90 = m<7 + m<8 90 = 26 + m<8 64 = m<8

  8. Example 4: Determine whether parallelogram ABCD is a rectangle, given A(-6,9), B(5,10), C(6,-1), and D(-5,-2) Method 1 Using diagonals: Find the length of AC and BD. The distance formula is d=√((x2 – x1)2 + (y2 – y1)2) AC d=√((x2 – x1)2 + (y2 – y1)2) d=√((-6 – 6)2+ (9 – -1)2) d=√((-6 – 6)2+ (9 + 1)2) d=√((-12)2+ (10)2) d=√(144 + 100) d=√(244) BD d=√((x2 – x1)2 + (y2 – y1)2) d=√((5 – -5)2+ (10 – -2)2) d=√((5 + 5)2+ (10 + 2)2) d=√((10)2+ (12)2) d=√(100 + 144) d=√(244) The diagonals are congruent. So it is a rectangle. CD m = (y2 – y1)/(x2 – x1) m = (-1 – -2)/(6– -5) m = (-1 + 2)/(6 + 5) m = 1/11 Method 2 Using slopes: Find the slope of AB, BC,CD, and AD. The slope formula is m = (y2– y1)/(x2 – x1) AB m = (y2 – y1)/(x2 – x1) m = (9– 10)/(-6– 5) m = (-1)/(-11) m = 1/11 BC m = (y2 – y1)/(x2 – x1) m = (10 – -1)/(5 – 6) m = (10 + 1)/(5 - 6) m = 11/(-1) m = -11 AD m = (y2 – y1)/(x2 – x1) m = (9 – -2)/(-6 – -5) m = (9 + 2)/(-6 + 5) m = 11/(-1) m = -11 AB is perpendicular to BC, BC is perpendicular to CD, CD is perpendicular to AD, and AD is perpendicular to AB.

More Related