Explanation Given information : That ABCD is a quadrilateral in a plane in which P , Q , R and S are the mid points of A B ¯ , B C ¯ , C D ¯ , and D A ¯ , so that PQRS is also a quadrilateral as shown in the given figure, Here, it is joined point A to C . Formula used : Here it is used the concept of mid-point theorem, that states that, “ In a triangle, a line joining the mid points of any two points of this triangle, is always half and parallel to its third remaining side. Proof: By mid-point theorem, in Δ A B C , P Q | | A C , P Q = A C 2 ...... ( i ) Also, in Δ A C D , S R | | A C , S R = A C 2

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 1, Problem 74SP

To determine

To prove that PQRS is a parallelogram.

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Chapter 1, Problem 73SP

Chapter 1, Problem 75SP

Chapter 1 Solutions

Calculus

Ch. 1.1 - Prob. 1PS Ch. 1.1 - Prob. 2PS Ch. 1.1 - Prob. 3PS Ch. 1.1 - Prob. 4PS Ch. 1.1 - Prob. 5PS Ch. 1.1 - Prob. 6PS Ch. 1.1 - Prob. 7PS Ch. 1.1 - Prob. 8PS Ch. 1.1 - Prob. 9PS Ch. 1.1 - Prob. 10PS

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Prob. 16PS Ch. 1.5 - Prob. 17PS Ch. 1.5 - Prob. 18PS Ch. 1.5 - Prob. 19PS Ch. 1.5 - Prob. 20PS Ch. 1.5 - Prob. 21PS Ch. 1.5 - Prob. 22PS Ch. 1.5 - Prob. 23PS Ch. 1.5 - Prob. 24PS Ch. 1.5 - Prob. 25PS Ch. 1.5 - Prob. 26PS Ch. 1.5 - Prob. 27PS Ch. 1.5 - Prob. 28PS Ch. 1.5 - Prob. 29PS Ch. 1.5 - Prob. 30PS Ch. 1.5 - Prob. 31PS Ch. 1.5 - Prob. 32PS Ch. 1.5 - Prob. 33PS Ch. 1.5 - Prob. 34PS Ch. 1.5 - Prob. 35PS Ch. 1.5 - Prob. 36PS Ch. 1.5 - Prob. 37PS Ch. 1.5 - Prob. 38PS Ch. 1.5 - Prob. 39PS Ch. 1.5 - Prob. 40PS Ch. 1.5 - Prob. 41PS Ch. 1.5 - Prob. 42PS Ch. 1.5 - Prob. 43PS Ch. 1.5 - Prob. 44PS Ch. 1.5 - Prob. 45PS Ch. 1.5 - Prob. 46PS Ch. 1.5 - Prob. 47PS Ch. 1.5 - Prob. 48PS Ch. 1.5 - Prob. 49PS Ch. 1.5 - Prob. 50PS Ch. 1.5 - Prob. 51PS Ch. 1.5 - Prob. 52PS Ch. 1.5 - Prob. 53PS Ch. 1.5 - Prob. 54PS Ch. 1.5 - Prob. 55PS Ch. 1.5 - Prob. 56PS Ch. 1.5 - Prob. 57PS Ch. 1.5 - Prob. 58PS Ch. 1.5 - Prob. 59PS Ch. 1.5 - Prob. 60PS Ch. 1 - Prob. 1PE Ch. 1 - Prob. 2PE Ch. 1 - Prob. 3PE Ch. 1 - Prob. 4PE Ch. 1 - Prob. 5PE Ch. 1 - Prob. 6PE Ch. 1 - Prob. 7PE Ch. 1 - Prob. 8PE Ch. 1 - Prob. 9PE Ch. 1 - Prob. 10PE Ch. 1 - Prob. 11PE Ch. 1 - Prob. 12PE Ch. 1 - Prob. 13PE Ch. 1 - Prob. 14PE Ch. 1 - Prob. 15PE Ch. 1 - Prob. 16PE Ch. 1 - Prob. 17PE Ch. 1 - Prob. 18PE Ch. 1 - Prob. 19PE Ch. 1 - Prob. 20PE Ch. 1 - Prob. 21PE Ch. 1 - Prob. 22PE Ch. 1 - Prob. 23PE Ch. 1 - Prob. 24PE Ch. 1 - Prob. 25PE Ch. 1 - Prob. 26PE Ch. 1 - Prob. 27PE Ch. 1 - Prob. 28PE Ch. 1 - Prob. 29PE Ch. 1 - Prob. 30PE Ch. 1 - Prob. 1SP Ch. 1 - Prob. 2SP Ch. 1 - Prob. 3SP Ch. 1 - Prob. 4SP Ch. 1 - Prob. 5SP Ch. 1 - Prob. 6SP Ch. 1 - Prob. 7SP Ch. 1 - Prob. 8SP Ch. 1 - Prob. 9SP Ch. 1 - Prob. 10SP Ch. 1 - Prob. 11SP Ch. 1 - Prob. 12SP Ch. 1 - Prob. 13SP Ch. 1 - Prob. 14SP Ch. 1 - Prob. 15SP Ch. 1 - Prob. 16SP Ch. 1 - Prob. 17SP Ch. 1 - Prob. 18SP Ch. 1 - Prob. 19SP Ch. 1 - Prob. 20SP Ch. 1 - Prob. 21SP Ch. 1 - Prob. 22SP Ch. 1 - Prob. 23SP Ch. 1 - Prob. 24SP Ch. 1 - Prob. 25SP Ch. 1 - Prob. 26SP Ch. 1 - Prob. 27SP Ch. 1 - Prob. 28SP Ch. 1 - Prob. 29SP Ch. 1 - Prob. 30SP Ch. 1 - Prob. 31SP Ch. 1 - Prob. 32SP Ch. 1 - Prob. 33SP Ch. 1 - Prob. 34SP Ch. 1 - Prob. 35SP Ch. 1 - Prob. 36SP Ch. 1 - Prob. 37SP Ch. 1 - Prob. 38SP Ch. 1 - Prob. 39SP Ch. 1 - Prob. 40SP Ch. 1 - Prob. 41SP Ch. 1 - Prob. 42SP Ch. 1 - Prob. 43SP Ch. 1 - Prob. 44SP Ch. 1 - Prob. 45SP Ch. 1 - Prob. 46SP Ch. 1 - Prob. 47SP Ch. 1 - Prob. 48SP Ch. 1 - Prob. 49SP Ch. 1 - Prob. 50SP Ch. 1 - Prob. 51SP Ch. 1 - Prob. 52SP Ch. 1 - Prob. 53SP Ch. 1 - Prob. 54SP Ch. 1 - Prob. 55SP Ch. 1 - Prob. 56SP Ch. 1 - Prob. 57SP Ch. 1 - Prob. 58SP Ch. 1 - Prob. 59SP Ch. 1 - Prob. 60SP Ch. 1 - Prob. 61SP Ch. 1 - Prob. 62SP Ch. 1 - Prob. 63SP Ch. 1 - Prob. 64SP Ch. 1 - Prob. 65SP Ch. 1 - Prob. 66SP Ch. 1 - Prob. 67SP Ch. 1 - Prob. 68SP Ch. 1 - Prob. 69SP Ch. 1 - Prob. 70SP Ch. 1 - Prob. 71SP Ch. 1 - Prob. 72SP Ch. 1 - Prob. 73SP Ch. 1 - Prob. 74SP Ch. 1 - Prob. 75SP Ch. 1 - Prob. 76SP Ch. 1 - Prob. 77SP Ch. 1 - Prob. 78SP Ch. 1 - Prob. 79SP Ch. 1 - Prob. 80SP Ch. 1 - Prob. 81SP Ch. 1 - 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To prove that PQRS is a parallelogram.

Solution Summary: The author explains the concept of mid-point theorem, which states that a line joining the mid points of any two points is always half and parallel to its third remaining side.

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To prove that PQRS is a parallelogram.

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Chapter 1 Solutions