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Answers
Step-by-step explanation:
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.
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