Construct A PQR with perimeter 9.5 cm and each of base angle is 80°. Write the type of APQR. [APQR is isosceles (i.e. PQ = PR)]
Answers
Step-by-step explanation:
i. As shown in the figure, take point T and S on line QR, such that QT = PQ and RS = PR ….(i) QT + QR + RS = TS [T-Q-R, Q-R-S] ∴ PQ + QR + PR = TS …..(ii) [From (i)] Also, PQ + QR + PR = 9.5 cm ….(iii) [Given] ∴ TS = 9.5 cm ii. In ∆PQT PQ = QT [From (i)] ∴ ∠QPT = ∠QTP = x° ….(iv) [Isosceles triangle theorem] In ∆PQT, ∠PQR is the exterior angle. ∴ ∠QPT + ∠QTP = ∠PQR [Remote interior angles theorem] ∴ x + x = 70° [From (iv)] ∴ 2x = 70° x = 35° ∴ ∠PTQ = 35° ∴ ∠T = 35° Similarly, ∠S = 40° iii. Now, in ∆PTS ∠T = 35°, ∠S = 40° and TS = 9.5 cm Hence, ∆PTS can be drawn. iv. Since, PQ = TQ, ∴ Point Q lies on perpendicular bisector of seg PT. Also, RP = RS ∴ Point R lies on perpendicular bisector of seg PS. Points Q and R can be located by drawing the perpendicular bisector of PT and PS respectively. ∴ ∆PQR can be drawn. Steps of construction: i. Draw seg TS of length 9.5 cm. ii. From point T draw ray making angle of 35°. iii. From point S draw ray making angle of 40°. iv. Name the point of intersection of two rays as P. v. Draw the perpendicular bisector of seg PT and seg PS intersecting seg TS in Q and R respectively. vi. Join PQ and PR. Hence, ∆PQR is the required triangle.