Q6 of 7 Page 223

Sujata drew three line segments of length 4cm, 6cm and 10 cm. by scale and compass Rahul bisects the 1st line segment, trisect the 2nd line segment divide 3rd line segment into five equal part Sabnam drew a triangle PQR taking of first line segment, of the 2nd line segment and th of the 3rd line segment. Let’s calssify the triangle on the basis of their sides.

Let the segments drawn be AB, CD and EF of length 4 cm, 6 cm and 10 cm respectively



Now we will divide the lines AB, CD and EF in the given ratios 1/2, and


Dividing AB in ratio 1/2 that is bisecting AB


Step1: Take any distance in compass approximately greater than half of AB. Keep the needle of compass on point A and mark arcs above and below AB.



Step2: Keeping the distance in compass same as that of in step1 keep the needle on point B and mark arcs intersecting arcs drawn in step1 and join the intersection points of both arcs. Mark the intersection point of line with AB as X



Dividing CD in 3 equal parts


Step3: Draw a ray at any angle below CD from C


Take any distance in compass and keep the needle of compass on point C and draw an arc intersecting ray drawn. Name the intersection point as X1. keeping the distance same in compass keep the needle on point X1 and mark an arc intersecting the ray at X2. Draw 3 such parts that is upto X3. By doing this we are making 3 equal parts on the ray



Step4: Join X3D and draw lines parallel to X3D from X2 and X1 intersecting CD at Y and Y1



Dividing EF in 5 parts


Step5: Repeat step3 and step4 for segment EF and divide EF in 5 parts by marking upto X5 on the ray from E as shown



Now we have to construct a ΔPQR in which the lengths are 1/2 of AB, of CD and of EF


Now from constructions in step2, step4 and step5


PQ = AX


QR = CY and


PR = EZ


Let us construct the ΔPQR


Step6: Draw any line and take a point P on it. Take distance AX in compass from step2. Keep the needle on point P and mark an arc intersecting the line at Q



Step7: Take distance CY in compass from step4 (QR = CY). Keep the needle of compass on point Q and mark an arc above PQ



Step8: Take distance EZ in compass from step5 (PR = EZ). Keep the needle of compass on point P and mark an arc intersecting arc drawn in step7 at R. Join PR and QR, ΔPQR is ready



Measure lengths PQ, QR and PR using scale each is 2 cm and as each side of ΔPQR is equal hence it is an equilateral triangle


More from this chapter

All 7 →
2

Ajij divided a line segment XY of length 12 cm in some parts in such a way that each part is of length 2.4 cm. First decide how many parts are there and then divide the line XY into that number of equal parts.

 3

Anoara drew a triangle ABC in her exercise book. We bisect side BC by compass and then the median AD. I trisect AD in AE, EF and FD. Now we join B and F by scale and extended it to X such that it intersect AC at X.

Measuring by scale we see,


AX= CX(Let’s put the number)


 4

Let’s divide a line segment of length 12.6 cm into 7 equals parts, by scale and compass. Using the above construction let’s draw an equilateral. Triangle of length 7.2 cm.

 5

Ramu pradhan drew a parallelogram ABCD where AB=6cm, BC=9 cm. and ∠ABC=60°. Let’s select two points P and Q on diagonal BD such that BP=PQ=QD.

Let’s write the nature of the quadrilateral APCQ joining the point A,P;P,C;C,Q and Q,A.