Question

2.)Construct APQR such that ZP = 80°, ZR = 60°,
QR = 9.3 cm and construct its circumcircle.​

Answer 1

Answer:

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