Dividing A Line Segment Into Three Or Five Equal Parts
Question 1. I with the help of pencil, scale and compass, trisect a straight line 9 cm long and will measure the length of each part by scale.
Solution:
A 9 cm long straight line AB is drawn. A On point A of any measure ∠BAC is drawn. At AB on point B equal to ∠BAC, ∠ABX is drawn.
From straight line AC taking same radius two equal parts AD and DE are cut.
From straight line BD taking same radius two equal parts BF and FG cut. E, F and D, G are joined. EF and DG both straight lines cut AB straight line respectively points P and Q points.
AB straight line is divided into three equal parts on points P and Q.
i.e., AP = PQ = QB = \(\frac{1}{3}\) AB = \(\frac{1}{3}\) × 9 cm = 3 cm
Length of each part is 3 cm.
Read and Learn More WBBSE Solutions For Class 8 Maths
Dividing A Line Segment Exercise 23.1
Question 1. Rihana draw a straight line PQ of length 10 cm. I divided the segment PQ into five equal parts by scale and compass. Let’s verify by scale that each part of the segment is 2 cm or not.
Solution:
Given
Rihana has drawn in her copy a 10 cm long straight line PQ. I divided the segment PQ into five equal parts by scale and compass.
At PQ on point P an acute angle ∠RPQ is drawn and on PQ straight line on the opposite side of ∠RPQ equal to that angle ∠PQS is drawn.
Now from straight line PR taking the same radius four equal parts PA, AB, BC and CD are cut.
Similarly the straight line QS, taking same radius, is cut into four equal parts QE, EF, FG and GH. A,H; B,G; C,H and D,E are joined which cut PQ straight line at points I, J, K and L respectively.
PQ straight line is divided by I, J, K and L points into five equal parts and length of each part 2 is cm,
i.e., PI = IJ=JK=KL=LQ= \(\frac{1}{5}\) PQ = \(\frac{1}{5}\)×10 cm
= 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.
Solution:
Given
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.
To divide 12 long straight line into parts of length 2.4, it should be cut into 5 equal parts.
∴ Ajij will cut XY straight line into 5 equal parts.
At point X on XY an acute angle ∠ZXY is drawn and on XY straight line on the opposite side of ∠ZXY, equal to that angle ∠XYP is drawn.
Now from straight line XY taking same radius, it is cut into 4 equal parts XA, AB, BC and CD.
Similarly taking the same radius YP straight line is cut into 4 equal parts YE, EF, FG and GH. A, H; B,G; C,H and D,E are joined by which XY straight line is divided into five equal parts at points l, J, K and L; and length of each part is 2.4 cm, i.e.,
XI = IJ = JK= KL= LY = \(\frac{1}{5}\) XY = \(\frac{1}{5}\)×12 cm
=2.4 cm.
Question 3. Anoara drew a triangle ABC in her exercise book. We bisect side BC by compass and then draw the median AD. I trisect AD in AE, EF and FD. Now we join B and F by scale and extend it to X such that it intersects AC at X.
Solution:
Given
Anoara drew a triangle ABC in her exercise book. We bisect side BC by compass and then draw the median AD.
I trisect AD in AE, EF and FD. Now we join B and F by scale and extend it to X such that it intersects AC at X.
Measuring by scale we see,
AX = CX [Let’s put the number]
In ΔABC side BC is bisected with the help of pencil and compass and AD median is drawn. Median AD is divided into three parts AE, EF and FD with help of scale and compass.
Now, with help of scale, I join two points B and F and extend them which cut straight line AC at point X.
Measuring by scale we see,
AX = CX, i.e., AX= 1 CX
Question 4. Let’s divide a line segment of length 12.6 cm into 7 equal parts, by scale and compass. Using the above construction, let’s draw an equilateral triangle of length 7.2 cm.
Solution:
A 12.5 cm long straight line AB is drawn. At point A on AB an acute angle ∠XAB is drawn and on AB on the opposite side of ∠XAB, equal to that angle, ∠ABY is drawn.
Now from AX straight line taking same radius six equal parts AC, CD, DE, EF, FG and GH are cut off.
Similarly by taking same radius BY straight line is cut into six equal parts Bl, IJ, JK, KL, LM and MO. C,0 ; D,M; E,L; F,K; G,J and H,l are joined by which AB which straight line is divided into 7 equal parts respectively P, Q, R, S, T and U points and length of each part is 1.8 cm. i.e.,
AP = PQ = QR = RS = ST = TU = UB = 1.8 cm
AS = 7.2 cm. Now taking radius equal to AS and taking centre A and S two arcs are drawn which cut each other at W point. A, W and S, W are joined.
ASW equilateral triangle is formed whose length of each side is 7.2 cm.
Question 5. Rama pradhan drew a parallelogram ABCD where AB = 6 cm, BC = 9 cm and ABC = 60°. Let’s select two points P and Q on the diagonal BD such that BP = PQ = QD. Joining the points A, P; P, C; C, Q and Q, A, let’s write the nature of quadrilateral APCQ.
Solution:
Given
Rama pradhan drew a parallelogram ABCD where AB = 6 cm, BC = 9 cm and ABC = 60°.
Let’s select two points P and Q on the diagonal BD such that BP = PQ = QD. Joining the points A, P; P, C; C, Q and Q, A,
ABCD is a parallelogram whose sides AB = CD = 6 cm and BC = AD = 9 cm and ABC = 60°. I choose with help of pencil, scale and compass, on BD diagonal of the parallelogram ABCD two points at P and Q such that BP = PQ = QD.
Now joining A,P; P,C; C,Q and Q, A, APCQ quadrilateral is formed. In it ∠AQC = 130°, ∠PAQ = ∠PCQ = 50° and AP = CQ and AQ = CP.
∴ It is a parallelogram.
Question 6. Sujata drew three line segments of length 4 cm, 6 cm and 10 cm. By scale and compass, Rahul bisects the 1st line segment, trisects the 2nd line segment and divides the 3rd line segment into five equal parts. Sabnarh drew a triangle PQR taking \(\frac{1}{3}\) th of 1 1 the first line segment, \(\frac{1}{5}\) th of the 2nd line segment and \(\frac{1}{2}\) th of the 3rd line segment. Let’s classify the triangles on the basis of their sides.
Solution:
Given
Sujata drew three line segments of length 4 cm, 6 cm and 10 cm.
By scale and compass, Rahul bisects the 1st line segment, trisects the 2nd line segment and divides the 3rd line segment into five equal parts.
Sabnarh drew a triangle PQR taking \(\frac{1}{3}\) th of 1 1 the first line segment, \(\frac{1}{5}\) th of the 2nd line segment and \(\frac{1}{2}\) th of the 3rd line segment.
Half of the length of the first straight line = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) × 4 cm = 2 cm
= \(\frac{1}{3}\) rd of the length of the 2nd straight line = \(\frac{1}{3}\) PQ = \(\frac{1}{3}\) × 6 cm = 2 cm
= \(\frac{1}{5}\) th of length of the third straight line = \(\frac{1}{5}\) XY = \(\frac{1}{5}\) × 10 cm = 2 cm
∵ Half of the length of the first straight line, \(\frac{1}{3}\) rd of that of the second and \(\frac{1}{5}\) th of that of the third, are equal.
∴ PQR is an equilateral triangle whose length of each side is 2 cm.
∴ Shabnam has drawn equilateral triangles.
