Constructions – Exercise 11.2 – (MATHEMATICS) – 9th Class

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Exercise 11.2

1. Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.

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Sol. Steps of construction:
(i) Draw a line segment BC of length 7 cm.
(ii) Draw an angle, say XBC, of measure 75° at the end B of BC.

(iii) Cut a line segment BD = 13 cm from the ray BX.
(iv) Join DC and draw the perpendicular bisector of it to intersect BD at A.
(v) Join AC.
Then, ΔABC is the required triangle.


2. Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB — AC = 3.5 cm.

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Sol. Steps of construction:
(i) Draw a line segment BC of length 8 cm.
(ii) Make an angle, say XBC, of measure 45° at the end B of BC.
(iii) Cut a line segment BD of length 3.5 cm from the ray BX.

(iv) Join CD and draw the perpendicular bisector of it to intersect BX at A.
(v) Join AC.
Then, ΔABC is the required triangle.


3. Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR — PQ = 2 cm.

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Sol. Steps of construction:
(i) Draw a line segment QR of length 6 cm.
(ii) Make an angle, say XQR, of measure 60° at the end Q of QR

(iii) Cut a line segment QS of length 2 cm from PQ produced.
(iv) Join SR and draw the perpendicular bisector of it to intersect the QX at P.
(v) Join PR.
Then, ΔPQR is the required triangle.


4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

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Sol. Steps of construction:
(i) Draw a line segment PQ of length 11 cm.
(ii) Draw angles of 30° at P and 90° at Q on the same side of PQ.
(iii) Bisect these angles. Let the bisectors intersect each other at a point X.

(iv) Draw perpendicular bisectors of PX and QX. Let these bisectors intersect the line segment PQ at Y and Z respectively.
(v) Join XY and XZ.
Then, ΔXYZ is the required triangle.


5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

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Sol. Steps of construction:
(i) Draw a line segment BC of length 12 cm.
(ii) Make an angle BCD of 90′ at C.
(iii) Cut a line segment CE of length 18 cm.
(iv) Join BE and draw the perpendicular bisector of it. Let the bisector intersect CE at A.
(v) Join AB.
Then, ΔABC is the required triangle.


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