Chapter 11. CONSTRUCTIONS

Class 10 Maths Chapter 11. CONSTRUCTIONS Exercise 11.1 Solutions

In each of the following , give the justification of the construction also :

1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8 . Measure the two parts .

Solution: Step of construction :

(i) We draw any ray AX making an acute angle with AB .

(ii) We draw a ray BY parallel to AX by making  .

(iii) locate the points and  on AX and and  on BY such that  .

(iv) join  and it intersect AB at a point C .

Then, 

Justification :  Since  [by construction ]

Then, 

2. Construct a triangle of sides 4 cm , 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle .

Solution: Given, a triangle of sides 4 cm , 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle .

Step of construction : Step of construction :

(i) Draw a line segment AB = 6 cm . With A and B as the centres and radius 4 cm .

(ii) Now we draw two arcs intersecting each other at C and join AC = 5 cm and BC = 4 cm .  

(iii) Draw any ray AX making an acute angle with AB on the side opposite to the vertex A.

(iv) Locate 3 points  ,  and  on AX , so that  .

 (v) Join  and draw a line through  parallel to  to intersect AB at   .

(vi) Draw a line through parallel to the line BC to intersect AC at  . Then,  is the required triangle .

Justification :  Since,   [By construction]

 But,

 So,     .

3. Construct a triangle with sides 5 cm , 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle .

Solution:  Given, a triangle of the sides 5 cm, 6 cm and 7 cm , we are required to construct another triangle whose sides are of the corresponding sides of the first triangle .

 Step of construction : 

(i) Draw a  with PQ = 7 cm , QR = 5 cm and PR = 6 cm .

 (ii) Draw an acute angle QPX below PQ at point P .

(iii) Locate 7 points  , , , , , and on PX , so that   .

 (iv) Join  and draw a line through  parallel to  to intersecting the extended line segment PQ at Q’ .

(v) Draw a line through Q’ parallel to the line RQ to intersecting the extended line segment PR at R’ .

Then, is the required triangle .

Justification: Since,   [By construction]

  But,

 So ,        Verified .

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are  times the corresponding sides of the isosceles triangle .

Solution:  Given, an isosceles triangle whose base is 8 cm and altitude 4 cm , we are required to construct another triangle whose sides are  times the corresponding sides of the isosceles triangle.

Steps of Construction : 

(i) Construct an isosceles triangle PQR in which PR = 8 cm and Altitude AD = 4 cm .

(ii) Draw a ray PX , making an acute angle with PR .

(iii) Locate 3 points  , and on PX so that  . Join  .

(iv) Through R , draw a line  parallel to  , meeting produced line PR at  .

(v) Through R , draw a line  parallel to QR , meeting the produced line PQ at  .

Thus,  is the required isosceles triangle .

Justification :     [ by construction ]

So, 

But, 

   Verified.

5. Draw a triangle ABC with side   and . Then construct a triangle whose sides are  of the corresponding sides of the triangle ABC .

Solution: Given, a triangle ABC with side BC = 6 cm , AB = 5 cm and . Then, we are required to construct a triangle whose sides are of the corresponding sidesof ABC .

 Step of construction :

 (i) Draw ∆ABC with side BC = 6 cm , AB = 5 cm and   .

(ii) Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

(iii) Locate 4 points  , , and on BX , so that   .

 (iv) Join  and draw a line through  parallel to  to intersect BC and   .

(v) Draw a line through   parallel to the line CA to intersect BA at  . Then,  is the required triangle .

Justification :

Since,   [By construction]

 But, 

 So,     Verified .

6. Draw a triangle ABC with side  and  . Then, construct a triangle whose sides are  times the corresponding sides of  .

Solution: Given, A triangle ABC with side BC = 7 cm ,  and  . Then , we are required to construct  a triangle whose sides are time the corresponding  sides of ABC .

 Now ,  

                   = 180° – (105° + 45°)

                   = 180° – 150° = 30° 

Step of construction :

(i) Draw  with side BC = 7 cm , and   

(ii) Draw a ray BX making an acute angle with BC on opposite side of vertex A .

(iii) Locate 4 points  ,  ,  ,  on BX , so that  .

(iv) Join C and draw a line through   parallel to  , intersecting the extended line segment BC at  .

(v) Draw a line through  parallel to CA intersecting the extended  line segment BA at  .

Then,   is the required triangle .

Justification:      [ By construction]

Now, 

   Verified .

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm . Then construct another triangle whose sides are  times the corresponding sides of the given triangle .

Solution:  Given, a right triangle in which the side ( other than hypotenuse) are of lengths  4 cm and 3 cm . We are required to construct another triangle whose sides are times the corresponding sides of the given triangle .

  Steps of construction :

 (i) Draw a ∆ABC , such that  , BC = 4 cm  and AC = 3 cm .

(ii) Draw a ray BX making an acute angle with BC .

(iii) Locate 5 points ,  , , and on BX , so that  .

(iv)  Join  and draw a line through  parallel to  , intersecting the extended line segment BC at  .

(v) Draw a line through   parallel to CA intersecting the extended line segment BA at .Then   is the required triangle .

Justification : Since,   [ By construction ]

Therefore , 

But ,

   Verified .

Class 10 Maths Chapter 11. CONSTRUCTIONS Exercise 11.2 Solutions

1. Draw a circle of radius 6 cm . From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths .

Solution:  Step of construction:

(i) Take a point O as a centre and draw a circle of radius 6 cm .

(ii) We draw a point P at the distance 10 cm from the centre O .

(iii) Taking P as centre and OP as radius , Draw a circle . let it intersect the given circle at the points B and C .

(iv) We join AB and AC .

Then AB and AC are the required two tangents .

Justification:  We join OB .

Then we find  .

So, ABC is a right triangle .

Here,  OB = 6 cm  and OA = 10 cm

In  , we have

2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measures its length . Also verify the measurement by actual calculation .

Solution:    Step of construction :

(i) Take a point O and draw a concentric circle of radius 4 cm and 6 cm respectively .

(ii) let P be the mid-point of OC .

(iii) Taking P as centre and OP = CP as radius, draw a circle . Let it intersect the given circle at the points A and B .

(iv) Join AC and BC . Then AC and BC are the required two tangents .

Justification : Justification:  We join OA .

Then we find .

So, AOC is a right triangle .

Here,  OA = 4 cm  and OC = 6 cm

In  , we have

 

3. Draw a circle of radius 3 cm . Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre . Draw tangents to the circle from these two points P and Q .

Solution:  Step of construction:

(i) Take a point O as centre , draw a circle of radius 3 cm .

(ii) Take two point P and Q at the distance 7 cm (I.e., OP = OQ = 7 cm) from the centre O.

(iii)   M and N are the bisector of OP and OQ . Let M and N be the mid-point of OP and OQ .

iv) Taking M as centre and OM as radius , draw a circle . Let it intersect the given circle at the points A and B . Join AP and BP .

(v) Taking N as centre and ON as radius , draw a circle .

Let it intersect the given circle at the points C and D . Join CQ and DQ .

Then AP, BP , CQ and DQ are required two tangents .

4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60° .

Solution:   Step of construction:

(i) Take a point O as centre and draw a circle of the radius OA = OC = 5 cm .

(ii) We draw  a angle between the radius OA and OC is 120° ,i.e.,  .

(iii) We draw and  . Now , AB and BC intersecting at the point B .

Then, AB and BC are the required two tangents .

Justification:   Here,  , and

Since, OABC is a cyclic quadrilateral .

 

 

 

 

 

5. Draw a line segment AB of length 8 cm . Taking A as centre , draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm . Construct tangents to each circle from the centre of the other circle .

Solution: Step of construction:

(i) We draw a segment  AB = 8 cm .

(ii) Taking A as centre , draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm .

(iii) Let P be the mid-point of the segment AB .

(iv) Taking P as centre and AP = BP as radius , draw a circle . let it intersect the given circle at the point C , D , E and F .

(v) Join AE , AF , BC and BD .

Then  AE , AF , BC and BD are the required two tangents .

6. Let ABC be a right triangle in which  and . BD is the perpendicular from B on AC . The circle through B , C , D is drawn . Construction the tangents from A to this circle .

Solution: Step of construction: 

(i) Draw ABC is a right triangle with base AB = 6 cm  and    .

(ii) Taking O as centre and draw a circle with diameter BC = 8 cm . This circle passes through D .

(iii) let O be the mid-point of BC and join OA .

(iv)   Draw a circle with diameter OA . This circle cut the given circle at B and E . Join AE .

 Then AB and AE are the required two tangents .

7. Draw a circle with the help of a bangle . Take a point outside the circle . Construct the pair of tangents from this point to the circle .

Solution:  Step of construction:

(i) Draw a circle with help of a bangle .

(ii) Draw BD and BE are two non-parallel chords .

(iii) We draw the perpendicular bisector of BD and BE intersecting at O (say) .

(iv)  Join OC and bisect it . let P be the mid -point of OC .

(v) Taking P as centre and OP as radius, draw a circle . Let it intersect the given circle at the point A and B .

(vi) Join AC and BC .

Then, AC and BC are the required two tangents .