Dividing A Line Segment Into Three Or Five Equal Parts
Class 8th Mathematics West Bengal Board Solution
- Rihana drew a straight-line segment PQ of length 10 cm I divided the line segment PQ in…
- Ajij divided a line segment XY of length 12 cm in some parts in such a way that each…
- Anoara drew a triangle ABC in her exercise book. We bisect side BC by compass and then…
- Let’s divide a line segment of length 12.6 cm into 7 equals parts, by scale and…
- Ramu pradhan drew a parallelogram ABCD where AB=6cm, BC=9 cm. and ∠ABC=60°. Let’s…
- Sujata drew three line segments of length 4cm, 6cm and 10 cm. by scale and compass…
Lets Do 23
Question 1.Let's divide a line segment of length 9 cm in three equal parts by scale and pencil compassed and let's measure the length of the parts by scale.
Answer:
Given: Length of a line segment = 9cm
Steps of Construction
Step1: Draw a line segment AB = 9cm
Step 2: From point A, draw a line segment at any angle.
Step 3: Now, take the compass and keep the needle on point A, and set its width to a bit less than one third of the length of AX and mark the three arcs of same length.
Step 4: Label the last arc C.
Step 5: Take the compass and keep the needle on point C and take the width CB, draw an arc from A just below it.
Step 6: Now, set the width of AC in compass, draw an arc from B intersecting the arc drawn in step 5. This intersection point is D. Join D to B
Step 7: Using the same compass width as used in step 3 along AC, step the compass from D along DB and make 3 arcs across the line.
Step 8: Join C to B and draw the lines between the corresponding points along AC and DB.
Hence, the lines divide the given line segment into 3 equal parts and each part is of 3cm
Lets Work Out 23
Question 1.Rihana drew a straight-line segment PQ of length 10 cm I divided the line segment PQ in five equal parts by scale and compass. Let’s verify by scale which each part of segment is 2 cm or not.
Answer:
Rihana’s construction is a segment PQ of length 10 cm
Step1: Using scale draw a segment of length 10 cm and mark the endpoints as P and Q
Now we have to divide this segment in 5 equal parts
Step2: Draw a ray at any angle below PQ from point P using a scale
Step3: Take any distance in compass and keep the needle of compass on point P and draw an arc intersecting the ray drawn in step2. 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 5 such parts that is upto X5. By doing this we are making 5 equal parts on the ray
Step4: Join X5 and Q
Now we will draw lines parallel to X5Q from X4, X3, X2 and X1
Step5: Take any distance in compass keep the needle of compass on point X5 and mark an arc intersecting X5Q and ray drawn in step2 at A1 and B1 respectively. Keeping the distance in compass same keep the needle on point X4 and mark arc intersecting ray at B2.
Step6: Take distance A1B1 in compass keep the needle on point B2 mark an arc intersecting the arc drawn using X4 as centre at A2
Draw line passing through X4A2 which cuts PQ at P4
Step7: Repeat step5 and step6 to draw lines parallel to X5Q from points X3, X2 and X1
Using scale measure each part that is PP1, P1P2, P2P3, P3P4 and P4Q. Each part is 2 cm
Question 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.
Answer:
The segment XY is of length 12 cm and let it be divided in n parts such that each part is 2.4 cm
Hence n × 2.4 = 12
⇒ n = 5
Hence the segment XY is divided in 5 parts
Let us begin the construction
Step1: Using scale draw a segment of length 12 cm and mark the endpoints as X and Y
Now we have to divide this segment in 5 equal parts
Step2: Draw a ray at any angle below XY from point X using a scale
Step3: Take any distance in compass and keep the needle of compass on point X and draw an arc intersecting the ray drawn in step2. 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 5 such parts that is up to X5. By doing this we are making 5 equal parts on the ray
Step4: Join X5 and Y
Now we will draw lines parallel to X5Y from X4, X3, X2 and X1
Step5: Take any distance in compass keep the needle of compass on point X5 and mark an arc intersecting X5Y and ray drawn in step2 at A1 and B1 respectively. Keeping the distance in compass same keep the needle on point X4 and mark arc intersecting ray at B2.
Step6: Take distance A1B1 in compass keep the needle on point B2 mark an arc intersecting the arc drawn using X4 as centre at A2
Draw line passing through X4A2 which cuts XY at P4
Step7: Repeat step5 and step6 to draw lines parallel to X5Y from points X3, X2 and X1
Question 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)
Answer:
Step1: Draw a ΔABC of any length with scale
Now we will bisect BC and mark the centre as D
Step2: Take any distance in compass approximately greater than half of BC. Keep the needle of compass on point B and mark arcs above and below BC
Step3: Keeping the distance in compass same as that of in step2 keep the needle of compass on point C and intersect arcs drawn in step2. Join there intersection points. Mark the intersection of line with BC as D and join AD. AD is the median
Now we have to trisect the median AD that is we have to divide the median in 3 equal parts
The first step when dividing a line in equal parts we draw a ray at any angle to the given line
Here the line which we have to divide is AD. Let AB be the ray
Step4: Take any distance in compass and keep the needle of compass on point A and draw an arc intersecting AB. Name the intersection point as X1. keeping the distance same in compass keep the needle on point X1 and mark an arc intersecting AB at X2. Draw 3 such parts that is upto X3. By doing this we are making 3 equal parts on the AB
Step5: Join X3 and D and draw parallels to DX3 from X2 and X1 which intersects AD at F and E respectively.
Thus we have divided AD in 3 equal parts AE, EF and FD
Step6: Draw a line passing through points B and F and let it intersect AC at X
Measure AX and XC using scale both are equal
AX = CX
Hence the box should be filled with 1
⇒ AX = 1 × CX
Question 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.
Answer:
Step1: Draw a segment AB of length 12.6 cm
Step2: Draw a ray below AB at any angle from point A
Step3: Take any distance in compass and keep the needle of compass on point A and draw an arc intersecting ray drawn in step2. 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 7 such parts that is upto X7. By doing this we are making 7 equal parts on the ray
Step4: Join X7B and draw lines parallel to X7B from X1, X2, X3, X4, X5 and X6 which intersect AB at A1, A2, A3, A4, A5 and A6
Measuring each part it is of length 1.8 cm
So length of 4 parts combined will be 7.2 cm
Using this we will construct the equilateral triangle of length 7.2 cm
Step5: Take distance AA4 in compass. Keep the needle of compass on point A and mark an arc above AA4
Step6: Keeping the distance in compass same as that of in step5 which is AA4 keep the needle of compass on A4 and mark arc intersecting arc drawn in step5 at C. Join AC and A4C equilateral ΔAA4C of side 7.2 cm is ready
Question 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.
Answer:
First let us construct the parallelogram ABCD
Step1: Draw segment BC of length 9 cm
Step2: Draw a line segment BA of length 6 cm at 60° to BC from point B
Now we will draw line parallel to AB from point C
Step3: Take any distance in compass keep the needle on point B and mark an arc intersecting AB and BC at X and Y. Keeping the distance in the compass same keep the needle on point C and mark an arc intersecting BC extended at S
Step4: Take distance XY in compass keep the needle on point S and mark arc intersecting the arc with centre C at R. Draw a line CD of length 6 cm passing through R. Join AD. Join BD
Now we have to take points P and Q on BD such that BP = PQ = QD which means we have to divide BD in 3 equal parts
We first draw a ray at any angle to the given line which is to be divided. Here the line to be divided is BD let AD be the line which will act as the ray for BD.
Step5: Take any distance in compass and keep the needle of compass on point D and draw an arc intersecting AD. Name the intersection point as X1. Keeping the distance same in compass keep the needle on point X1 and mark an arc intersecting AD at X2. Draw 3 such parts that is upto X3. By doing this we are making 3 equal parts on AD
Step6: Join BX3 and draw lines parallel to BX3 from X2 and X1 intersecting BD at P and Q respectively
Step7: Join AP, AQ, CP and CQ. Using scale measure lengths AP, AQ, CP and CQ
As the lengths AP = QC and AQ = PC which are opposite sides of quadrilateral APCQ.
As opposite sides of quadrilateral are equal quadrilateral APCQ is a parallelogram
Question 6.
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.
Answer:
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