WBBSE Solutions For Class 8 Maths Geometry Chapter 8 Construction

Geometry Chapter 8 Construction Examples

1. Construction of Triangles measurements.
2. Construction of Parallel lines.
3. Dividing a line segment into three or five equal parts.

⇔ Construction of Triangles measurements.

Example 1. Draw a triangle, whose two angles are 45° and 30° and length of the opposite side of 30° angle is 6 cm.

Solution:

Step 1.

⇔ At first, I draw a line segment of length 6 cm with a scale.

Read and Learn More WBBSE Solutions For Class 8 Maths

⇒ Then I draw two angles of measurement 45° and 30° with pencil compass.

Step 2.

⇒ The line segment BC of length 6 cm is cut off from the ray BX.

Step 3.

⇒ Then I draw two angles ∠YBC and ∠ZCX of measurements 45° each at B and C respectively.

Step 4.

⇒ Then I draw an angle ∠PCZ of measurement 30° with the help of pencil compass at C on the same side of the line segment CZ where BY lies.

PC and BY meet at point A.

∴ ΔABC is the required triangle where BC = 6 cm.

⇒ ∠ABC = 45° ∠BAC = ∠ACZ = 30°

Proof: ∠ABC= ∠ZCX= 45° [By construction] [Corresponding angles are equal] BC is transversal.

∴ BA || CZ

⇒ Again BA || CZ and AC is transversal.

∴ ∠BAC = ∠ACZ [Alternate angles]= 30°

∴ In ΔABC, ABC = 45° and ∠BAC = 30° and the opposite sides of ∠BAC is BC = 5 cm.

Example 2. Draw a triangle length of whose two sides are 5 cm, 3.5 cm and the measurement of the angle opposite to the side of length 3.5 cm is 40°.

Solution:

Step 1.

⇒ I draw two line segments of lengths 5 cm and 3.5 cm and an angle of measurement 40° with the help of protractor.

Step 2.

⇒ I draw a ray BX and I draw an angle ∠XBY of measurement 40° with the compass at B on BX.

Step 3.

⇒ I cut off a line segment AB of length 5 cm from the ray BX.

Step 4. Taking A as center I draw an arc of radius 3.5 cm which intersects the rays BX at C and D.

⇒ Joining the points A, C and A, D by a scale I get the two triangles ΔABC and ΔABD.

∴ ΔABC is a required triangle, whose AB = 5 cm, AC = 3.5 cm and ∠ABC= 40° which is opposite to the side AC.

⇒ ΔABD is also a required triangle whose AB = 5 cm, AD = 3.5 and ABD = 40° which is opposite to the side AD.

⇒ [Sometimes we are getting one triangle, sometimes two triangle and sometimes no triangle.]

Case 1. If a = b we will unable to construct one triangle.

Case 2. If h < b < a we will able to draw two triangles.

Case 3. If a < b we will able to draw one triangle.

Case 4. b = h, then we will able to draw one triangle, B.

⇔ Construction of Parallel lines:

Example 3. Draw a straight line PQ and let consider a point R outside the line segment PQ. Draw a line, parallel to the line PQ that passes through R by a scale and a compass.

Solution:

Method 1.

Step 1.

⇒ I draw a straight line PQ and take a point R outside the line PQ.

Step 2.

⇒ Now I take a point S on the straight line PQ and I join R, S by a scale.

As a result an angle ∠RSQ is formed.

Step 3.

⇒ Now with a scale and compass we draw ∠TRS at R equal to opposite side of ∠RSQ and it is in the opposite side of ∠RSQ.

⇒ I join T and R by a scale and producing the line TR on both the sides we get a straight line AB.

Proof: ∠ARS = ∠RSQ but they are alternate angles.

∴ AB || PQ.

∴ We draw a straight line AB parallel to the straight line PQ passing through the point R which is outside the line PQ.

Method 2.

Step 1.

⇒ I take a point S on the straight line PQ.

Step 2.

⇒ I join two points R and S by a scale and produced the line segment SR to M.

As a result, ∠MSQ is formed.

Step 3.

⇒ I draw an angle ∠MRT at R on the line segment RM equal to the ∠RSQ and ∠MRT is drawn on the same side of the line segment MS where ∠RSQ lies.

Step 4.

⇒ Now I produce the line segment TR on both side and the straight line AB is form.

∴ Proof: ∠MRB = ∠MSQ.

⇒ But they are corresponding angles.

∴ AB || PQ.

∴ AB is the required straight line through R which is parallel to line PQ.

Method 3.

Step 1.

⇒ I draw a straight line PQ and take a point R outside the straight line PQ.

Step 2.

⇒ Let I take a point S on the straight line PQ.

⇒ I join R, S and I got a line segment RS.

Step 3.

⇒ I cut off a line segment ST from ray SQ.

Step 4.

⇒ Then taking R as the center and the length of the line segment ST as radius; I draw an arc with a compass.

Step 5.

⇒ Then taking T as the center and the length of the line segment RS as radius, I draw a another arc by a compass which intersects the previous arc at M.

Step 6.

⇒ I join R, M and then produce RM on both side where AB straight line is form.

Proof: I join S, M and M, T.

In ΔRMS and ΔMTS, RM = ST, RS = MT and SM is common side.

∴ ΔRMS ≅ ΔMTS [By SSS congruency]

∴ ∠RMS = ∠MST [Corresponding angles of congruent triangles]

But they are alternate angle.

∴ RM || ST i.e. AB || PQ.]

∴ I got the straight line AB through R which is parallel to PQ.

Dividing a line segment into three or five equal parts.

Example 4. Draw a straight line segment PQ of length 12 cm and divided the line segment PQ in five equal parts by scale and compass. Verify by scale which each part of the segment is 2.4 cm or not.

Solution:

Step 1.

⇒ We draw a line segment PQ ∅ of length 10 cm.

Step 2.

⇒ Then I cut off the line segment PQ of length 10 cm from the ray PX.

Step 3.

⇒ I draw an angle ∠XPY at P on PX.

Step 4.

⇒ I draw ∠PQR equal to ∠YPQ on the opposite side of the line segment PQ, such that it lies below ∠QPY.

Step 5.

⇒ Then I cut four equal line segments PA₁, A₁B₁, B₁C₁ and C₁D₁ from PY by pencil compass.

⇒ Again I cut another four equal line segments QD₂, D₂C₂, C₂B₂ and B₂A₂ from QZ by pencil compass with same radius.

⇒ I join (A₁, A₂); (B₁, B₂); (C₁, C₂) and (D₁, D₂) by scale.

⇒ The line segment A₁A₂, B₁B₂, C₁C₂ and D₁D₂ intersects PQ at points A, B, C and D respectively.

⇒ So the line segment PQ is divided at A, B, C, and D with equal length.

⇒ So PA = AB = BC = CD = DQ = 2.4 cm.