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Question-1

Draw a circle of radius 3 cm. From a point 10 cm away from its centre. Construct the pair of tangents to the circle.

Solution:
Given: A circle with centre O and radius 3 cm.
Required: To construct the pair of tangents.



 


Steps of Construction:

(i) Draw a circle of radius 3 cm.

(ii) Take an external point P which is 10 cm away from its centre. Join OP.
(i) Bisect the line segment OP = 10 cm. Let the point of bisection be M.
(ii) Taking M as centre and OM as radius, draw a circle. Let it intersect the given circle at the points Q and R.
(iii) Join PQ and PR.
These are the required tangents.

Question-2

Construct a triangle ABC whose sides are 7.5 cm, 7 cm and 6.5 cm. Construct another triangle similar to Δ ABC and with sides of the corresponding sides of triangle ABC.

Solution:
Given: Δ ABC, AB = 7.5 cm, BC = 7 cm and CA = 6.5 cm.

Required: To construct a Δ A’BC’ in which A’B =
 AB, A’C’ = AC and BC’ = BC.


              
 

Steps of construction:
(i) Divide the base BC into three equal parts. Let C’ be the point on BC such that BC’ = BC.

 

Steps of construction
1. Draw a line segment BC = 7 cm, 
 AB = 7.5 cm and CA = 6.5 cm.

2. Below BC, make an acute angle  CBP

3. Divide the base BC into three equal parts. Let C’ be the point on BC such that BC’ = BC.

4. Along BP, mark off three points X1, X2, X3 such that XX1 = X1X2 = X2X3

5. Join X3C

6. Draw a line C’A’ || CA intersecting BA at A’.
Then A’BC’ is the required triangle.


Question-3

Construct a triangle similar to a given triangle with sides 5 cm, 12 cm and 13 cm and whose sides are of the corresponding sides of the given triangle.

Solution:
Given: Δ ABC, AB = 5 cm, BC = 12 cm and CA = 13 cm.
Required: To construct a Δ A’BC’ in which A’B =AB, A’C’ =AC and BC’ =BC.
         

                                                
Steps of construction:
1. Draw a line segment BC = 12 cm

2. With B as centre and with radius 5 cm, draw an arc. 
3. With C as centre and with radius 13 cm, draw another arc, intersecting the previously drawn arc at A. 
4. Join AB and AC. Then, Δ ABC is the required triangle. 
5. Below BC, make an acute angle  CBP. 
6. Along BP, mark off seven points X1, X2, X3…..X7 such that XX1 = X1X2…… X6X7 
7. Join X5 to C and draw a line through X3 parallel to X5 C, intersecting the extended line segment BC at C’. 
8. Draw a line through C’ parallel to CA intersecting the line segment BA at A’. Then A’BC’ is the required triangle. 
  

Question-4

Construct a triangle similar to a given triangle with sides 6 cm, 7 cm and 8 cm and whose sides are th of the corresponding sides of the given triangle.

Solution:
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Steps of construction:  
1. Draw a line segment BC = 7 cm 
2. With B as centre and with radius 6 cm, draw an arc. 
3. With C as centre and with radius 8 cm, draw another arc, intersecting the previously drawn arc at A. 
4. Join AB and AC. Then, Δ ABC is the required triangle. 
5. Below BC, make an acute angle  CBX. 
6. Along BX, mark off seven points X1, X2, X3…..X7 such that XX1 = X1X2…… X6X7 
7. Join X7 to C and draw a line through X5 parallel to X7C, intersecting the extended line segment BC at C’. 
8. Draw a line through C’ parallel to CA intersecting the line segment BA at A’. Then A’BC’ is the required triangle. 
  

Question-5

Divide a line segment of 7 cm length externally in the ratio of 3 : 5.

Solution:
Given: AB is a line segment of 7 cm length.

Required: To divide a line segment of 7 cm length externally in the ratio of 3 : 5.


               

Steps of Construction:

1. Draw the line segment AB = 7 cm.

2. Draw ray BX making an acute ABX .

3. Along BX, mark off five points B1,, B2, B3,, B4 and B5. Join B2 to A.

4. Through B5 draw B5P || B2A , intersecting BA produced at P.

5. The point P so obtained is the required point which divides AB externally in the ratio
3 : 5.

Proof: In Δs ABB2 and PBB5,

B5P || B2A ABB2 PBB5

=(Property of similarty).


Question-6

Construct a triangle similar to a given Δ ABC such that each of its sides is rd of the corresponding sides of the ΔABC. Given AB = 4.2 cm, BC = 5 cm and AC = 6.2 cm.

Solution:
Given: In Δ ABC, AB = 4.2 cm, BC = 5 cm and AC = 6.2 cm.
Required: To construct Δ AB’C’ such that each of its sides is rd of the corresponding sides of the Δ ABC.

Steps of Construction:

1. Draw a line segment AB = 4.2 cm.

2. With A as centre and radius = AC = 6.2 cm, draw an arc.

3. With B as centre and radius = BC = 5 cm, draw another arc, intersecting the previous arc at C.

4. Join AC and BC to obtain Δ ABC.

5. Below AB, make an acute angle BAX.

6. Along AX, mark off three points A1 , A2 , A3 such that AA1 = A1A2 = A2A3

7. Join A3B.

8. Draw A2B’ || A3B, meeting AB at B’.

9. From B’, draw B’C’ || BC meeting AC at C’.

AB’C’ is the required Δ .

Proof: Since B’C’ || BC , Δ ABC Δ AB’C’.

B’C’/BC = AC’/AC = AB’/AB = 2/3.


Question-7

Construct a triangle similar to a Δ XYZ with its sides equal to ()th of the corresponding sides of ΔXYZ. It is given that XY = 6 cm, XZ = 5 cm and ZY = 4 cm. Write the steps of construction.

Solution:
Given: D XYZ in which XY = 6 cm, XZ = 5 cm and ZY = 4 cm
Required: To construct a Δ XY'Z' in which XY’ = (3/4)XY, Y’Z’ = (3/4)ZY and
XZ’ = (3/4)XZ.



Steps of construction:
(i) Draw a ray XP.
(ii) Construct a Δ XYZ in which XY = 6 cm, XZ = 5 cm and ZY = 4 cm.
(iii) Draw any ray XP inclined at certain angle with X.
(iv) Starting from X, cut off seven equal line – segment XX1, X1X2, X2X3, X3X4 on XQ.
(v) Join YX4 and draw a line – segment X3Y’ parallel to X4Y to intersect XP at Y’.
Draw a line Y’Z’ parallel to YZ which intersects XP in Y’
Then XY’Z’ is the required quadrilateral.


Question-8

Draw a Δ ABC in which AB = 5 cm, BC = 4.6 cm, and AC = 5.8 cm. Construct a triangle similar to Δ ABC such that each of its sides is 2/3rd of the corresponding sides of Δ ABC.

Solution:
Given: In Δ ABC, in which AB = 5 cm, BC = 4.6 cm, and AC = 5.8 cm.

Required: To construct a triangle similar to Δ ABC such that each of its sides is two-third of the corresponding sides of Δ ABC.

   
 

Steps of Construction:

(i) Draw BC = 4.6 cm.

(ii) With B as centre and radius equal to 5 cm draw an arc and with C as centre and radius equal
to 5.8 cm draw another arc to cut the previous arc at A.

(iii) Join AB and AC.

(iv) Make an acute angle CBE.

(v) Set off three equal distances along BE at B1, B2 and B3.

(vi) Join B3C.

(vii) From B2 draw B2C’ || B3C, meeting BC at C’.

(viii) Join AC’.

Then, ABC’ is the required triangle.





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