Chapter 3:Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progression

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 12:Area Related to Circles

Constructions

Step 1:Draw line AB of 7.6cm

Step 2:Draw a line AX with angle BAX

Step 3:Mark point A_{1},A_{2},A_{3},A_{4}............A_{13} On AX such that AA1 = A_{1}A_{2} = A_{2}A_{3}

Step 4:Join a line B and A_{13}

Step 5:Draw a parallel line A_{5} to BA_{13} abd makes an angle ∠AA_{13}B ,join a C

Step 6:now AP :PB = 5:8 and AP=2.9cm ,PB=4.7cm

Justification:

By construction, we have CA_{5} || B_{13} .

AC/CB =AA_{5}/AA_{5}AA_{13} = 5/8

AC/CB = 5/ 8

Step 1: Draw a line AB of 4 cm

Step 2: Using a point A draw an arc of radius 5 cm

Step 3: Similarly, Using a point B as its centre, and draw an arc of radius 6 cm.

Step 4: Now join C to A and B and form ΔABC triangle.

Step 5:Draw a line AX with angle BAX

Step 6: Make 3 points such as A_{1}, A_{2}, A_{3} on line AX such that AA1= A1A2 = A2A3.

Step 8: Join the BA_{3} through A_{2} which is parallel to the line BA_{3} meeting AB at point B’.

Step 9: Using point B’, draw a line parallel to the line BC meeting AC at C’.

Justification:

Here

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

By construction we get B’C’ || BC

∴ ∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’ (Common)

∴ ΔAB’C’ ∼ ΔABC (by AA similarity )

So AB’/AB = B’C’/BC= AC’/AC …. (1)

In ΔAAB’ and ΔAAB,

∠A_{2}AB’ =∠A_{3}AB (Common)

∠AA_{2}B’ =∠AA_{3}B (Corresponding angles)

∴ ΔAAB’ ∼ ΔAAB (by AA similarity )

AA_{2}B’ and AA_{3}B

So, AB’/AB = AA_{2}/AA_{3}

AB’/AB = 2/3 ……. (2)

From the eq (1) and (2)

AB’/AB=B’C’/BC = AC’/ AC = 2/3

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

Construction Step

Step 1: Draw a line AB =5 cm.

Step 2: Take point A and B as centre, and draw the arcs of radius 6 cm and 5 cm and join at C.

Step 3: Draw a line AX with angle BAX

Step 5: Mark 7 points such as A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7} , on line AX such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7}

Step 6: Join the points B to A_{5}

Step 7: A_{7}B' is draw a parallel to A_{5}B.

Step 8: Now,from B’ draw a line with point C’ that is parallel to the line BC.

Step 9: ΔAB’C’ is the triangle.
1
/
2

times the corresponding sides of the isosceles triangle.
Step 1: Draw a line of 8cm

Step 2:Make a perpendicular bisector of BC' at point D

Step 3: D as center deaw a arc of 4 cm intersect the perpendicular bisector at the point A

Step 4:meeting AB and AC

Step 5:Draw a line BX with angle line BC

Step 6:Mark 3 points B_{1}, B_{2} and B_{3} on the line BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}

Step 7:Join the points B_{2}C and draw a line from B_{3} parallel to the line B_{2}C,to intersects the extended line BC at point C’.

Step 8:Draw a line from C’ parallel to the line AC to intersect extended line AB at A’

3
/
4

of the corresponding sides of the triangle ABC.
Step 1:Draw a line BC = 6 cm,

Step 2:At point B draw ∠ABC = 60°.

Step 3: B as center make arc of 5cm and mark point A

Step 4:Join AB and AC

Step 5: Draw a line BX makes an acute angle with BC

Step 6:Mark 4 points such as B_{1}, B_{2}, B_{3}, B_{4}, on line BX.

Step 7:Join the points B_{4}C ,draw a line through B_{3}, parallel to B_{4}C intersecting the line BC at C’

Step 8: From C’ draw a line that is parallel to the line AC and intersects the line AB at A’.

4
/
3

times the corresponding sides of ∆ABC.
∠B = 45°, ∠A = 105°

We know that

Sum of angles in a triangle is 180°.

∠A+∠B +∠C = 180°

105°+45°+∠C = 180°

∠C = 180° − 150°

∠C = 30°

Step 1:Draw a line BC = 7 cm

Step 2:Draw ∠B =45 and ∠C =30

Step 3:Make a point A and join AB and AC

Step 4: Draw a line BX with acute angle with BC

Step 5: Mark 4 points such as B1, B2, B3, B4, on the ray BX.

Step 6: Join the points B3C

Step 7: from B4 draw a line parallel to B3C to intersects extended line BC at C’.

Step 8: From C’, draw a line parallel to the line AC to intersects extended line at C’.

5
/
3f

times the corresponding sides of the given triangle.
Solution:
Step 1: Draw a line BC of 7cm

Step 2:At B draw ∠B =90

Step 3:B as center draw arc with 4 cm radius

Step 4:join AC

Step 5:ABC Triangle

Step 6: Draw a line BX with acute angle with BC .

Step 7: Mark 5 point B^{1}, B^{2}, B^{3}, B^{4}, on the line BX such that BB^{1} = B^{1}B^{2} = B^{2}B^{3}= B^{3}B^{4} = B^{4}B^{5}

Step 8: Join the points B^{3}C.

Step 9: Draw a line from B^{5} parallel to B^{3}C intersects the extended line BC at C’.

9. From C’, draw a line parallel to AC intersects the extended line AB at A’.