Constructions : Exercise 11.1 (Mathematics NCERT Class 10th)
Â Â Â Â Â Â In each of the following, give the justification of the construction also :

Class 10 : Maths + Science Video Lecture + Paper Solution+ TestMr. Kapil Sharma, GRS Academy, Mr. Neetin Agrawal
Sol. Â Â Â Steps of Construction :
Â Â Â Â Â Â Â Â 1. Draw a line segment AB = 7.6 cm
Â Â Â Â Â Â Â Â 2. Draw a ray AC making any acute angle with AB, as shown in the figure.
Â Â Â Â Â Â Â Â 3. On ray AC, starting from A, mark 5 + 8 = 13 equal line segments :
Â Â Â Â
Â Â Â Â Â Â Â Â 4. Join
Â Â Â Â Â Â Â Â 5. From , meeting AB at P.
Â Â Â Â Â Â Â Â 6. Thus, P divides AB in the ratio 5 : 8.
Â Â Â Â Â Â Â Â On measuring the two parts, we find AP = 2.9 cm and PB = 4.7 cm (approx).
Â Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Â In Â
Â Â Â Â Â Â Â Â Therefore, Â ~Â Â
Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â AP : PB = 5 : 8
Q.2 Â Â Â Construct a triangle of sides 4 cm, 5 cm and 6cm and then a triangle similar to it whose sides are of the corresponding sides of it.
Sol.Â Â Â Â Â Â Steps of Construction :
Â Â Â Â Â Â Â Â 1. Draw a line segment BC = 6 cm
Â Â Â Â Â Â Â Â 2. With B as centre and radius equal to 5 cm, draw an arc.
Â Â Â Â Â Â Â Â 3. With C as centre and radius equal to 4 cm, draw an arc intersecting the previously drawn arc at A.
Â Â Â Â Â Â Â Â 4. Join AB and AC, then ABC is the required triangle.
Â Â Â Â Â Â Â Â 5. Below BC, make an acute angle CBX.
Â Â Â Â Â Â Â Â 6. Along BX, mark off three points : such that .
Â Â Â Â Â Â Â Â 7. Join .
Â Â Â Â Â Â Â Â 8. From , meeting BC at D.
Â Â Â Â Â Â Â Â 9. From D, draw ED  AC, meeting BA at E. Then,
Â Â Â Â Â Â Â Â EBD is the required triangle whose sides are of the corresponding sides of ABC.
Â Â Â Â Â Â Â Â Justification : Since DE  CA
Â Â Â Â Â Â Â Â Therefore, Â Â ~ Â Â
Â Â Â Â Â Â Â Â and
Â Â Â Â Â Â Â Â Hence, we get the new triangle similar to the given triangle whose sides are equal to rd of the corresponding sides of ABC.Â
Q.3 Â Â Â 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.
Sol.Â Â Â Â Â Steps of construction :
Â Â Â Â Â Â Â Â 1. With the given data, construct ABC in which BC = 7 cm , CA = 6 cm and AB = 5 cm
Â Â Â Â Â Â Â Â 2. Below BC, make an acute .
Â Â Â Â Â Â Â Â 3. Along BX, mark off seven points : such that Â Â
Â Â Â Â Â Â Â Â 4. Join
Â Â Â Â Â Â Â Â 5. From , meeting BC produced at D.
Â Â Â Â Â Â Â Â 6. From D, draw DE  CA, meeting BA produced at E.
Â Â Â Â Â Â Â Â Then EBD is the required triangle whose sides are th of the corresponding sides of .
Â Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Â Since DE  CA
Â Â Â Â Â Â Â Â Therefore, Â
Â Â Â Â Â Â Â Hence, we get the new triangle similar to the given triangle whose sides are equal to the of the corresponding sides of .
Q.4 Â Â Â 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.
Sol.Â Â Â Â Steps of Construction :
Â Â Â Â Â Â Â 1. Draw BC = 8 cm
Â Â Â Â Â Â Â 2. Construct PQ the perpendicular bisector of line segment BC meeting BC at M.
Â Â Â Â Â Â Â 3. Along MP cut off MA = 4 cm
Â Â Â Â Â Â Â 4. Join BA and CA. Then so obtained is the required
Â Â Â Â Â Â Â 5. Extend BC to D, such that BD = 12 cm.
Â Â Â Â Â Â Â 6. Draw DE  CA, meeting BA produced at E.
Â Â Â Â Â Â Â Then is the required triangle.
Â Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Â Since DE  CA
Â Â Â Â Â Â Â Â Therefore, Â ~ Â
Â Â Â Â Â Â Â Â and
Â Â Â Â Â Â Â Â Hence, we get the new triangle similar to the given triangle whose sides are , i.e., times of the corresponding sides of the isosceles .
Q.5 Â Â Â Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and . Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.
Sol.Â Â Â Â Steps of Construction :
Â Â Â Â Â Â Â Â 1. With the given data, construct in which BC = 6 cm, and AB = 5 cm.
Â Â Â Â Â Â Â Â 2. Below BC, make an acute .
Â Â Â Â Â Â Â Â 3. Along BX, mark off 4 points : such that .
Â Â Â Â Â Â Â Â 4. Join .
Â Â Â Â Â Â Â Â 5. From D, draw ED  AC, meeting BA at E.Then, EBD is the required triangle whose sides are th of the corresponding sides of .
Â Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Â Since, DE  CA Therefore Â ~Â
Â Â Â Â Â Â Â Â and,
Â Â Â Â Â Â Â Â Hence, we get the new triangle similar to the given triangle whose sides are equal to th of the corresponding sides of
Q.6 Â Â Â Draw a triangle ABC with side BC = 7 cm, . Then construct a triangle whose sides are times the corresponding sides of .
Sol.Â Â Â Â Steps of Construction :
Â Â Â Â Â Â Â 1. With the given data, construct in which BC = 7 cm ,
Â Â Â Â Â Â Â
Â Â Â Â Â Â Â
Â Â Â Â Â Â Â
Â Â Â Â Â Â Â 2. Below BC, make an acute
Â Â Â Â Â Â Â 3. Along BX, mark off four points : such thatÂ
Â Â Â Â Â Â Â 4. Join
Â Â Â Â Â Â Â 5. From draw meeting BC produced at D.
Â Â Â Â Â Â Â 6. From D, draw ED  AC, meeting BA produced at E.
Â Â Â Â Â Â Â Then EBD is the required triangle whose sides are times of the corresponding sides ofÂ
Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Since, DE  CA Therefore, ~Â
Â Â Â Â Â Â Â and
Â Â Â Â Â Â Â Hence, we get the new triangle similar to the given triangle whose sides are equal to times of the corresponding sides of
Sol.Â Â Â Â Â Steps of Construction :
Â Â Â Â Â Â Â 1. With given data, construct in which BC = 4 cm and BA = 3 cm.
Â Â Â Â Â Â Â 2. Below BC, make an acute
Â Â Â Â Â Â Â 3. Along BX, mark off five points :
Â Â Â Â Â Â Â such thatÂ
Â Â Â Â Â Â Â 4. Join
Â Â Â Â Â Â Â 5. From draw meeting BC produced at D.
Â Â Â Â Â Â Â 6. From D, draw ED AC, meeting BA produced at E.
Â Â Â Â Â Â Â Then EBD is the required triangle whose sides are times of the corresponding sides of
Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Since, DE  CA Therefore, ~Â
Â Â Â Â Â Â Â and
Â Â Â Â Â Â Â Hence, we get the new triangle similar to the given triangle whose sides are equal to times of the corresponding sides of .
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