Constructions

Step 1: Draw line segment AB = 7.6 cm

Step 2: Draw a ray AC making an acute angle with line segment AB .

Step 3: Along AC, divide (5+8) 13 equals points

A₁,A₂,A₃,……………….,A₁₃ on AX such that

AA₁=AA₂=A₂A₃ and so on .

Step 4: Join BA₁₃

Step 5: Through the point A₅, draw a line parallel to BA₁₃ (by making an angle equal to ∠AA₁₃B) at A₅ intersecting AB at point D.

C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8.

The lengths of AD and DB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.

Justification

The construction can be justified by proving that

By construction, we have A₅D∥A₁₃B. By applying Basic proportionality theorem for

From the figure, it can be observed that AA₅ and A₅A₁₃ contain 5 and 8 equal divisions of line segments respectively.

On comparing equations (1) and (2), we obtain

This justifies the construction.

Step 1: Draw a line segment AB of 5 cm.

Step 2: Taking A and B as Center, draw arcs of 6 cm and 5 cm radius respectively. Let these arcs intersect each other at point C. △ABC is the required triangle having length of sides as 5 cm , 6 cm , 7 cm respectively.

Step 3: Draw a ray AX making acute angle with line AB on the opposite side of vertex C.

Step 4: Locate 7 points, A₁,A₂,A₃,A_4, A₅,A₆,A₇( as 7 is greater between 5 and 7), on line AX such that AA₁=A1 A2=A2 A3=A3 A4=A4 A5=A5 A6=A6 A7

Step 5: Join BA₅ and draw a line through A₇ parallel to BA₅ to intersect extended line segment AB at point B′.

Step 6: Draw a line through B′ parallel to BC intersecting the extended line segment AC at C′.△AB’C’ is the required triangle.

Justification

∆AB' C'~∆ABCby AAA Similarity Condition

And

2. Join the point of intersection from A and B.

Hence, this gives the required triangle ABC

3. Dividing the base, draw a ray AX which is at an acute angle from AB


4. Plot seven points on AX such that:

AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.

5. Join A₅ to B.

6. Draw a line from point A₇that is parallel to A₅B and joins AB’ (AB extended to AB’).

7. Draw a line B’C’ || BC.

Hence, triangle AB’C’ is the required triangle.

△A′BC′ whose sides of the corresponding sides of can be drawn as follows.

Step 1 Draw a △ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = .

Step 2 Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3 Locate 4 points (as 4 is greater in 3 and 4), B1,B2,B3,B4 on line segment BX.

Step 4 Join B4C and draw a line through B3 , parallel to B4C intersecting BC at C′

Step 5 Draw a line through C' parallel to AC intersecting AB at A'.△A′BC′ is the required triangle.

Justification

∠A=60° (Common)

∠C=∠C'

By AA similarity condition

Also,

AB’=AB ,B’C’=BC and AC’=AC

Given in Δ ABC,

• Length of side BC = 6 cm.
• Length of side AB = 5 cm.
• ∠ ABC = 60°.

Steps of Construction:

1. Draw a line segment BC of length 6 cm.

Construction of 60° angle at B:
a. With B as centre and with some convenient radius draw an arc which cuts the line BC at D.


b. With D as radius and with same radius (in step a), draw another arc which cuts the previous arc at E.


c. Join BE. The line BE makes an angle 60° with BC.

3. Again with B as centre and with radius of 5 cm, draw an arc which intersects the line BE at point A.

5. Now, from B, draw a ray BX which makes an acute angle on the opposite side of the vertex A.

Steps of construction:

1. Draw a line segment BC = 7 cm.

2. Draw ∠ABC = 45° and ∠ACB = 30° i.e. ∠BAC = 105°.

We obtain ΔABC.

3. Draw a ray BX making an acute angle with BC. Mark four points B₁, B₂, B₃,B₄ at equal distances.

4. Through B₃ draw B₃C and through B₄ draw B₄C₁ parallel to B₃C.

Then draw A₁C₁ parallel to AC.

∴ A₁BC₁ is the required triangle.

Steps of construction:

1. Now in order to make a triangle, draw a line segment AB = 3 cm.

2. Make a right angle at point A and draw AC = 4 cm from this point.

3. Join points A and B to get the right triangle ABC.


4. Now, Dividing the base, draw a ray AX such at it forms an acute angle from AB.

AG = GH = HI = IJ = JK.

6. Join I to point line AB and Draw a line from K which is parallel to IB such that it meets AB at point M.

7. Draw MN || CB.


This is the required construction, thus forming AMN which have all the sides 5/3 times the sides of ABC

Triangle AMN is the required triangle.

Step1: Draw circle of radius 6cm with center A, mark point C at 10 cm from center

Step 2: find perpendicular bisector of AC

Step3: Take this point as center and draw a circle through A and C

Step4:Mark the point where this circle intersects our circle and draw tangents through C

Length of tangents = 8cm

AE is perpendicular to CE (tangent and radius relation)

In ΔACE

AC becomes hypotenuse

AC² = CE² + AE²

10² = CE² + 6²

CE² = 100-36

CE² = 64

CE = 8cm

Steps of construction:

i. Draw two concentric circles with radii 4 cm and 6 cm respectively.

ii. Now, draw the radius OP of the larger circle.

iii. Construct a perpendicular bisector of OP intersecting OP at point O’.

iv. Considering O’P as radius, draw another circle.

v. From point P, draw tangents PQ and PR(can see in the figure)

Justification: By applying Pythagoras theorem, we have;

PQ² = OP² – OQ²

= 6² – 4²

= 36 – 16 = 20

Or, PQ = 2√5 cm

Steps of construction:

i. At first draw a circle with radius 3 cm.

ii. Now, extend the diameter to A and B on both the sides.

iii. Then, draw the perpendicular bisectors of OA and OB.

Such that:

Perpendicular bisector of OA intersects it at point O’.

And,

Perpendicular bisector of OB intersects it at point O”.

iv. Considering O’A as radius, construct another circle.

v. considering O”B as radius, construct the third circle.

vi. From point A, draw tangents AP and AQ.

vii. From point B, draw tangents BR and BS.

Step1: Draw a line PQ=8cm. taking P and Q as a center draw circle of 3cm and 4cm.

Step2: Now bisect PQ. We get midpoint of PQ be T.

Now take T as a center , draw a circle of PT radius , this will intersect the circle at point A,B,C,D. Join PB,PD,AQ,QC.

Justification:

It can be justified by prove that PB,PD are tangents of circle (whose center is P and radius is 3cm) and AQ,QC are tangents of circle (whose center is Q and radius is 4cm)

Join PA, PC, QB, QD

∠PBQ=90° (Angle is on semicircle)

BQ⊥PB

Since, BQ is radius of circle, PB has to be a tangent. Similarly PD, QA ,QC are tangents.

Step1: Draw a circle from a bangle.

Step2: Take a point A outside the circle and two chords BC and DE.

Step3: Bisect chords. They intersect at point O.

Step4: join OA and bisect it. We get midpoint of OA is P. Taking P as a center draw a circle of OP radius which intersect circle at W and X.

Join AW and AX. These are tangent.

Justification:

We have to be proved that O is center of circle.

Join OW and OX.

∠AWO=90° (angle is on semicircle)

OW is perpendicular to AW.

Since, OW is radius of circle, AW has to be tangent. Similarly AX is tangent.