Question

7.) In the given figure , ABCD is a square and ∠PQR = 90° . If PB = QC = DR , prove that
i) QB = RC
ii) PQ = QR
iii) ∠QPR = 45°​

Answer 1

Answer:

We know that the line segment QB can be written as QB = BC – QC Since ABCD is a square we know that BC = DC and QC = DR So we get QB = CD – DR From the figure we get QB = RC (ii) Consider △ PBQ and △ QCR It is given that PB = QC Since ABCD is a square we get ∠ PBQ = ∠ QCR = 90o By SAS congruence criterion △ PBQ ≅ △ QCR PQ = QR (c. p. c. t) (iii) We know that PQ = QR Consider △ PQR From the figure we know that ∠ QPR and ∠ QRP are base angles of isosceles triangle ∠ QPR = ∠ QRP We know that the sum of all the angles in a triangle is 180o ∠ QPR + ∠ QRP + ∠ PRQ = 180o By substituting the values in the above equation ∠ QPR + ∠ QRP + 90o = 180o On further calculation ∠ QPR + ∠ QRP = 180o – 90o By subtraction ∠ QPR + ∠ QRP = 90o We know that ∠ QPR = ∠ QRP So we get ∠ QPR + ∠ QPR = 90o By addition 2 ∠ QPR = 90o By division ∠ QPR = 45oRead more on Sarthaks.com - https://www.sarthaks.com/725692/in-the-given-figure-abcd-is-a-square-and-pqr-90-if-pb-qc-dr-prove-that

Answer 2

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