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Ex 7.4, 8 ABCD is a rectangle formed by the points A(–1, –1), B(–1, 4), C(5, 4) and D(5, –1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer Point P P is the mid point of AB. Applying mid point Formula, Coordinates of P = ((𝑥_1 +〖 𝑥〗_2)/2, (𝑦_1 + 𝑦_2)/2) Point Q Q is the mid point of BC. Applying mid point Formula, Coordinates of Q = ((𝑥_1 +〖 𝑥〗_2)/2, (𝑦_1 + 𝑦_2)/2) Here 𝑥_1=−1 𝑦_1=−1 𝑥_2=−1 𝑦_2=4 Coordinates of P = ((−1 − 1)/2, (−1 + 4)/2) = ((−2)/2,3/2) = (−1, 3/2) Here 𝑥_1=−1 𝑦_1=4 𝑥_2=5 𝑦_2=4 Coordinates of Q = ((−1 + 5)/2, (4 + 4)/2) = (4/2,8/2) = (2, 4) Point R R is the mid point of CD. Applying mid point Formula, Coordinates of R = ((𝑥_1 +〖 𝑥〗_2)/2, (𝑦_1 + 𝑦_2)/2) Here 𝑥_1=5 𝑦_1=−1 𝑥_2=5 𝑦_2=4 Coordinates of R = ((5 + 5)/2, (−1 + 4)/2) = (10/2,3/2) = (5, 3/2) Point S S is the mid point of AD. Applying mid point Formula, Coordinates of S = ((𝑥_1 +〖 𝑥〗_2)/2, (𝑦_1 + 𝑦_2)/2) Here 𝑥_1=−1 𝑦_1=−1 𝑥_2=5 𝑦_2=−1 Coordinates of S = ((−1 + 5)/2, (−1 −1)/2) = (4/2,(−2)/2) = (2, −1) Thus, Vertices of quadrilateral PQRS are P (−1, 3/2), Q (2, 4), R (5, 3/2) and 5 (2, 1) Now, finding all sides of quadrilateral PQRS Applying distance formula, √((𝑥_2 − 𝑥_1 )^2+(𝑦_2−𝑦_1 )^2 ) For PQ Here 𝑥_1=−1 𝑦_1=3/2 𝑥_2=2 𝑦_2=4 PQ = √((2−(−1))^2+(4 − 3/2)^2 ) = √((2+1)^2+((8 − 3)/2)^2 ) = √((3)^2+(5/2)^2 )= √(9+25/4) = √((36 + 25)/4) = √(61/4) units For QR Here 𝑥_1=2 𝑦_1=4 𝑥_2=5 𝑦_2=3/2 QR =√((5−2)^2+(3/2 −4)^2 ) = √((3)^2+((3 − 8)/2)^2 ) = √(9+((−5)/2)^2 )= √(9+25/4) = √((36 + 25)/4)= √(61/4) units For RS Here 𝑥_1=5 𝑦_1=3/2 𝑥_2=2 𝑦_2=−1 PQ = √((2−5)^2+(−1 − 3/2)^2 ) = √((−3)^2+((−2 − 3)/2)^2 ) = √((−3)^2+((−5)/2)^2 ) = √(9+25/4) = √((36 + 25)/4)= √(61/4) units For SP Here 𝑥_1=2 𝑦_1=−1 𝑥_2=−1 𝑦_2=3/2 PQ = √((−1−2)^2+(3/2+1)^2 ) = √((−3)^2+((3 + 2)/2)^2 ) = √((−3)^2+(5/2)^2 ) = √(9+25/4) = √((36 + 25)/4) = √(61/4) units Thus, PQ = QR = RS = SP = √(61/4) units ∴, PQRS is a square or a rhombus. Now to check whether PQRS is a square or a rectangle So, we find length of Diagonols RS and SQ For RS 𝑥_1=−1 𝑦_1=3/2 𝑥_2=5 𝑦_2=3/2 For SQ 𝑥_1=2 𝑦_1=−1 𝑥_2=2 𝑦_2=4 PR = √((5+1)^2+( 3/2−3/2)^2 ) = √((6)^2+(0)^2 ) = √((6)^2 ) = 6 units PR = √((2−2)^2+(4+1)^2 ) = √((0)^2+(5)^2 ) = √((5)^2 ) = 5 units Since PQ ≠ SQ ∴ Diagonal of the quadrilateral are not equal. So, PQRS is not a square ∴ PQRS is a Rhombus.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.