Question

Construct ΔPQR, in which ∠Q = 70°, ∠R = 80° and PQ + QR + PR = 9.5 cm.

Answer 1

Rough figure:

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:

1. Draw seg TS of length 9.5 cm.
2. From point T draw ray making angle of 35°.
3. From point S draw ray making angle of 40°.
4. Name the point of intersection of two rays as P.
5. Draw the perpendicular bisector of seg PT and seg PS intersecting seg TS in Q and R respectively.
6. Join PQ and PR.

Therefore, △PQR is the required triangle.