Q4 of 20 Page 11

Construct a ΔABC in which AB = 5 cm. ∠B=60° altitude CD = 3 cm. Construct a ΔAQR similar to ΔABC such that side of ΔAQR is 1.5 times that of the corresponding sides of ΔABC.

Steps of construction


1. Construct a line segment AB of length 5 cm.



2. From B construct an angle of 60° using compass and name it ABX.



3. Construct a line EF parallel to line segment AB at a height of 3 cm such that it intersects BX at point C. Join AC.



4, Construct CO perpendicular to AB. CO is the altitude of the triangle ABC of height 3 cm.



5. Construct AL making an acute angle with AB, on the side opposite to vertex C and mark 3 points (corresponding to the greater of 2 or 3) X1,X2 and X3 on AL Such that AX1 =X1X2 =X2X3



6. Join X2B and construct the line passing through X3 and parallel to X2B.Let this line intersect the extended line AB at R.


3.



7.Now construct a line passing through R and parallel to BC which intersect the extended line AC at point Q.


AQR is the triangle.



More from this chapter

All 20 →
2

Draw a right ∆ABC in which BC=12 cm, AB=5 cm and ∠B=90°. Construct a triangle similar to it and of scale factor .

 3

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

 5

Draw a triangle ABC in which AB=5 cm, BC=6 cm and ∠ABC= 60°. Construct another triangle similar to ΔABC with scale factor .

 6

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are times the corresponding sides of the given triangle.