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Constructions : Exercise 11.2 (Mathematics NCERT Class 9th)


Q.1    Construct a triangle ABC in which  BC = 7 cm, \angle B = {75^o} and  AB + AC = 13 cm
Sol.       Steps of Construction :

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1.Draw a ray BX and cut off a line segment BC = 7 cm
2. Construct \angle XBY ={ 75^o}

3. From BY, cut off BD = 13 cm
4. Join CD.
5. Draw the perpendicular  bisect of CD, intersecting BA at A.
6. Join AC.
The  triangle ABC thus obtained is the required triangle.


Q.2      Construct a triangle ABC in which  BC = 8 cm, \angle B = {45^o} and AB – AC = 3.5 cm
Sol.

Steps of Construction :
1. Draw  a ray BX and cut off a line  segment BC = 8 cm from it.
2. Construct \angle YBC = {45^o}
3. Cut off a line  segment BD = 3.5 cm from BY.
4. Join CD.                     
5. Draw perpendicular bisector of CD intersecting  BY at a point A.
6. Join AC

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Then  ABC is the  required triangle.


Q.3       Construct a triangle PQR in which  QR = 6 cm \angle Q = {60^o} and PR – PQ = 2cm
Sol.

Steps of Construction :
1. Draw a ray QX and cut off a line segment QR = 6 cm  from it.
2. Construct  a ray QY making  an angle  of 60º with QR and produce YQ to form a line YQY'
3. Cut off  a line segment QS = 2cm  from QY'.
4. Join RS.
5. Draw perpendicular bisector  of RS intersecting QY at a point  P.
6. Join  PR.

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Then  PQR is the required triangle.


Q.4       Construct a triangle XYZ in which \angle Y = {30^o}\angle Z = {90^o} and XY + YZ + ZX = 11 cm.
Sol.

Steps  of Construction :
1. Draw  a line segment PQ = 11 cm

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2. At P,  draw a ray PL  such  that \angle LPQ = {1 \over 2} \times {30^o} = {15^o}
3. At Q, draw ray QM such that \angle MQP = {1 \over 2} \times 90^o = {45^o} intersecting  PL at X.
4. Draw perpendicular bisectors of XP and XQ intersecting  PQ in Y and Z respectively.
Then  \Delta XYZ is the required  triangle.
Note : -  For  clarity in figure, method of drawing angles of 15º and 45º have not been shown. Students should draw these angles with the help of ruler and  compass only by the method as shown earlier.


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

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1. Draw a ray BX and cut off a line segment  BC = 12 cm

2. Construct \angle XBY = {90^o}
3. From by  cut off a line segment BD = 18 cm .
4. Join CD.
5. Draw the perpendicular  bisector of CD intersecting BD at A.
6. Join AC
Then  ABC is the  required triangle.  



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