Chapter 10. Constructions                   

 Class10 Construction 3 or 4 Marks Questions and Solution :

Question : Construct a triangle similar to a given triangle ABC with its sides equal to of the corresponding sides of the triangle ABC . [SEBA 2015 , 2018 , 2021]

Solution :  Given a triangle ABC, we are required to construct another triangle whose sides are of the corresponding sides of the triangle ABC.

Steps of Construction :

     

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

(ii) Along BX mark off 4 points  and such that  .

(iii) Join  .

(iv)  Start from B and reach to point  on BX . Draw  parallel to which meets AB at .

(v) From  draw  meeting AB at.

Then,  is the required triangle .

Question : Construct a triangle similar to a given triangle ABC with its sides equal to of the corresponding sides of the triangle ABC . [SEBA 2020,2023]

Solution: Given a triangle ABC, we are required to construct a triangle whose sides are of the corresponding sides of  .

Steps of Construction :

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

(ii) Along BX mark off 5 points  and such that  .

(iii) Join  .

(iv)  Start from B and reach to point  on BX . Draw  parallel to which meets BA at .

(v) From  draw meeting BA at.

Then,  is the required triangle .

Question: 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 .[SEBA 2019]

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 :

(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 C.

(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 .

[ Note : Justification :  Since   (By construction)

  

 But ,

So ,    Verified .]

Question : 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  , , , , , andon 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 .

[ Note : Justification :  Since   [By construction]

  But ,   

So ,      Verified . ]

Question : 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 .[SEBA 2017]

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  , 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 .

[ Note : Justification : Since,    [ by construction ]

   So, 

    But, 

  Verified. ]

Question : 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 sides of .

Step of construction :

        

(i) Draw 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 .

[ Note : Justification :  Since   [By construction]

   

 But , 

So ,    Verified . ]

Question : 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 .

 Now , = 180° – (105° + 45°) = 180° – 150° = 30° 

Step of construction :

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

(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 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 .

[ Note :  Justification :      [ By construction]

 

verified ]

Question : 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 , such that ∠ABC=90°  , 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 .

[ Note : Justification : Since,    [ By construction ]

Therefore ,

But ,

    Verified . ]

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

Solution :  Given a line segment AB is 7.6 cm , we want to divide it in the ratio 5 : 8 ,where 5 and 8 are positive integers .  

Steps of construction : 

  (i) We draw the line segment AB of length 7.6 cm .

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

                     

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

(iv) Locate the points   on AX and  on BY such that   .

(v)  We join    and this line intersect AB at P . 

 Then     

 

Measurement :        (by construction)

 cm

and  cm