NCERT Class 9 Solutions: Constructions (Chapter 11) Exercise 11.2

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Construction of triangle

Construction of Triangle

Construction of triangle

Q-1 Construct a triangle JAB in which

Solution:

Image Title: Triangle JAB

Image Description: Triangle JAB,

Give,

Step of construction:

1. A line segment AB of 7 cm is drawn.

2. At point A, an angle is constructed such that it is equal to .

3. A line segment cm is cut on AC (which is equal to ).

4. DB is joined and is made.

5. Let BE intersect AC at J.

Thus, required triangle is

Q-2 Construct a triangle DAB in which .

Solution:

Triangle DAB

Triangle DAB,AB=8cm,∠A=45°andAD-DB=3.5cm

Give

Steps of Construction:

1. A line segment cm is drawn and at point A, make an angle of i.e.

2. Cut the line segment (equal to ) on ray .

3. Join GB and draw the perpendicular bisector PQ of GB.

4. Let it intersect AC at point D. Join DB.

Thus, is the required triangle.

Q-3 Construct a triangle EAR in which

Solution:

Triangle EAR

Triangle EAR, AR=6cm,∠A=60°andPR–EA=2cm.

Give, Triangle EAR,

Steps of Construction:

1. A ray AX is drawn and cut off a line segment AR = 6 cm from it.

2. A ray AY is constructed making an angle of 60º with AR and YA is produced to form a line YAY'

3. Cut off a line segment AB = 2cm from AY'. RB is joined.

4. Draw perpendicular bisector of RB intersecting AY at a point E. ER is joined.

Thus, ΔEAR is the required triangle.

Q-4 Construct a triangle HFI in which

Solution:

Triangle HFI

Triangle HFI,∠F=30°,∠I=90°andHF+FI+IH=11cm.

Give, Triangle HFI

Steps of Construction:

1. A line segment cm is drawn. ( )

An angle, and an angle

2. And are bisected. The bisectors of these angles intersect each other at point H.

3. Perpendicular bisectors PQ of CH and RV of AH are constructed.

4. Let PQ intersect CA at F and RV intersect CA at I. FH and HI are joined.

Thus, is the required triangle.

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

Solution:

Triangle PAJ

Triangle PAJ,AJ=12cm,AQ=18cm

Give, right triangle base and othrside is

Steps of Construction:

1. A ray AX is drawn and a cut off a line segment is made on it.

2. is constructed.

3. Cut off a line segment cm is made on AY. JQ is joined.

4. Perpendicular bisector of JQ is constructed intersecting AQ at P. PJ is joined.

Thus, is the required triangle.