Saturday, May 31, 2014

10. Draw a tangent to the circle with centre O and radius 3.3 cm from a point A such that d (O, A) = 7.5 cm. Measure the length of tangent segment.

10. Draw a tangent to the circle with centre O and radius 3.3 cm from a point A such that d (O, A) = 7.5 cm. Measure the length of tangent segment.




9. Draw a tangent to the circle from the point L with radius 2.8 cm. Point ‘L’ is at a distance 5 cm from the centre ‘M’.

9. Draw a tangent to the circle from the point L with radius 2.8 cm. Point ‘L’ is at a distance 5 cm from the centre ‘M’.




8. Draw a tangent to the circle from the point B, having radius 3.6 cm and centre ‘C’. Point B is at a distance 7.2 cm from the centre.

8. Draw a tangent to the circle from the point B, having radius 3.6 cm and centre ‘C’. Point B is at a distance 7.2 cm from the centre.




7. Draw a circle having radius 3 cm draw a chord XY = 5 cm. Draw tangents at point X and Y without using centre.

7. Draw a circle having radius 3 cm draw a chord XY = 5 cm. Draw tangents at point X and Y without using centre.





6. Draw a circle of radius 2.7 cm and draw chord PQ of length 4.5 cm. Draw tangents at P and Q without using centre.

Q6. Draw a circle of radius 2.7 cm and draw chord PQ of length 4.5 cm. Draw tangents at P and Q without using the centre.
🎓 Concept: Tangent-Secant Theorem (Alternate Segment Theorem)

To draw tangents without using the centre, we use the property that the angle between a tangent and a chord is equal to the angle in the alternate segment.

If we construct a triangle \( \Delta PQR \) inside the circle, the angle required for the tangent at point P will be equal to \( \angle R \), and similarly for point Q.

Steps of Construction:

  1. Draw a circle with a radius of 2.7 cm.
  2. Take a point P anywhere on the circle.
  3. Using a compass, take a distance of 4.5 cm, place the metal point on P, and cut an arc on the circle to mark point Q. Join chord PQ.
  4. Take any point R on the major arc (the larger side of the circle) and join PR and QR to form \( \Delta PQR \).
  5. Place the compass at point R and draw an arc intersecting sides PR and QR. Keep the same radius.
  6. Place the compass point at P and draw a similar arc intersecting chord PQ. Do the same at point Q intersecting chord QP.
  7. Measure the distance of the arc drawn at angle R. Cut this distance on the arcs drawn at P and Q to replicate \( \angle PRQ \).
  8. Draw a line passing through P and the intersection point of the arcs. This is the required tangent at P.
  9. Draw a line passing through Q and the intersection point of the arcs. This is the required tangent at Q.

Visual Guide:

Rough Figure - Tangent Construction Fig 1: Rough Figure (Planning the Construction)
Final Construction - Tangents without Centre Fig 2: Final Construction (Fair Figure)

5. Draw a circle of radius 3.5 cm. Take any point K on it. Draw a tangent to the circle at K without using the centre of the circle.

5. Draw a circle of radius 3.5 cm. Take any point K on it. Draw a tangent to the circle at K without using the centre of the circle. 



4. Draw a circle with centre P and radius 3.1 cm. Draw a chord MN of length 3.8 cm. Draw tangent to the circle through points M and N.

4. Draw a circle with centre P and radius 3.1 cm. Draw a chord MN of length 3.8 cm. Draw tangent to the circle through points M and N. 





3. Draw a circle of radius 2.6 cm. Draw tangent to the circle from any point on the circle using centre of the circle.

3. Draw a circle of radius 2.6 cm. Draw tangent to the circle from any point on the circle using centre of the circle.

Steps of Construction:

  • Take a point O as the center and draw a circle with radius 2.6 cm.
  • Take any point P on the circle.
  • Draw ray OP.
  • Construct a line perpendicular to ray OP at point P. This line is the required tangent to the circle.
Analytical Figure

Analytical Figure

Geometric Construction

Final Construction

2. Draw a tangent at any point R on the circle of radius 3.4 cm and centre ‘P’.

2. Draw a tangent at any point R on the circle of radius 3.4 cm and centre ‘P’.




1. Draw a tangent at any point ‘M’ on the circle of radius 2.9 cm and centre ‘O’.


1. Draw a tangent at any point ‘M’ on the circle of radius 2.9 cm and centre ‘O’.




Thursday, May 29, 2014

9. Construct the circumcircle and incircle of an equilateral ∆ XYZ with side 6.3 cm.


9. Construct the circumcircle and incircle of an equilateral ∆ XYZ with side 6.3 cm.




8. Construct any right angled triangle and draw incircle of that triangle. ∆ ABC is the required right angled triangle. Such that AB = 6 cm, BC = 9 cm and m ∠ABC = 90º,


8. Construct any right angled triangle and draw incircle of that triangle.

∆ ABC is the required right angled triangle.

Such that AB = 6 cm, BC = 9 cm and m ABC = 90º,




8. Construct incircle of ∆SGN such that SG = 6.7 cm, ∠S = 70º, ∠G = 50º and draw incircle of ∆ SGN.

8. Construct incircle of  ∆SGN such that SG = 6.7 cm, S = 70º, G = 50º and draw incircle of  ∆ SGN.




7. Construct the incircle of ∆DEF in which DE = DF = 5.8 cm, ∠EDF = 65º.

Geometry Challenge: Constructing an Incircle

7. Construct the incircle of ∆DEF in which DE = DF = 5.8 cm, and ∠EDF = 65º.

Glossary of Terms
Incircle
The largest possible circle that can be drawn inside a triangle, touching all three sides without crossing them.
Triangle (∆)
A fundamental shape in geometry consisting of three straight sides and three angles.
Isosceles Triangle
A triangle that has two sides of equal length. In this problem, ∆DEF is isosceles because DE = DF.
Angle Bisector
A line or ray that divides an angle into two smaller, equal angles. The construction of an incircle relies on finding where these bisectors meet.
Incenter
The center point of a triangle's incircle. It is found at the intersection of the triangle's three angle bisectors.

6. Construct the incircle of ∆ STU in which, ST = 7 cm, ∠T = 120º, TU = 5 cm.

6. Construct the incircle of  ∆ STU in which, ST = 7 cm,  ∠T = 120º, TU = 5 cm.




5. Construct the incircle of ∆RST in which RS = 6 cm, ST = 7 cm and RT = 6.5 cm.

5. Construct the incircle of ∆RST in which RS = 6 cm, ST = 7 cm and RT = 6.5 cm.




4. Construct a right angled triangle ∆PQR where PQ = 6 cm, ∠QPR = 40º, ∠PRQ = 90º. Draw circumcircle of ∆ PQR.

4. Construct a right angled triangle  ∆PQR where PQ = 6 cm,  QPR = 40º, PRQ = 90º. Draw circumcircle of ∆ PQR.






3. Construct the circumcircle of ∆ KLM in which KM = 7 cm, ∠ K = 60º, ∠ M = 55º.

3.  Construct the circumcircle of  ∆ KLM in which KM = 7 cm, K = 60º, M = 55º.




2. Construct the circumcircle of ∆ SIM in which SI = 6.5 cm, ∠ I = 125º, IM = 4.4 cm.

2. Construct the circumcircle of  ∆ SIM in which SI = 6.5 cm,   I = 125º, IM = 4.4 cm.




1. Draw the circumcircle of ∆ PMT such that, PM = 5.4 cm, ∠ P = 60º, ∠ M = 70º.

1. Draw the circumcircle of  ∆ PMT such that, PM = 5.4 cm, P = 60º,   M = 70º.


IMPORTANT POINTS TO REMEMBER FOR CONSTRUCTING CIRCUMCIRCLE OF TRIANGLE.



1. A circle passing through the vertices of the triangle is called the circumcircle of a triangle.

2. Circumcentre can be obtained by drawing perpendicular bisectors of any two sides of a triangle.

3. The point of intersection of the perpendicular bisectors is called
circumcentre and it is equidistant from the vertices of the triangle.

The position of circumcentre depends upon the type of a triangle.

(i) If the triangle is an obtuse angled triangle, the circumcentre lies outside the triangle.

(ii) If the triangle is an acute angled triangle, the circumcentre lies inside the triangle.

(iii) If the triangle is a right angled triangle, the circumcentre lies on the midpoint of the hypotenuse.

Wednesday, May 28, 2014

Draw an angle of 125º and bisect it.


Draw perpendicular bisector of seg AB of length 8.3 cm.


CONSTRUCTION

BASIC CONSTRUCTION


To draw a perpendicular bisector of a given line segment.


To draw an angle bisector of a given angle.


To draw a perpendicular to a line at a given point on it.


To draw a perpendicular to a given line from a point outside it.


To draw an angle congruent to a given angle.


To draw a line parallel to a given line through a point outside it.



Ex. No. 3.1 


1. Draw the circumcircle of  ∆ PMT such that, PM = 5.4 cm, P = 60º,   M = 70º.  [Ans.]

2. Construct the circumcircle of  ∆ SIM in which SI = 6.5 cm,  ∠ I = 125º, IM = 4.4 cm.  [Ans.]

3.  Construct the circumcircle of  ∆ KLM in which KM = 7 cm, ∠ K = 60º, ∠ M = 55º. [Ans.]

4. Construct a right angled triangle  ∆PQR where PQ = 6 cm,  ∠QPR = 40º, ∠PRQ = 90º. Draw circumcircle of ∆ PQR.  [Ans.]

5. Construct the incircle of ∆RST in which RS = 6 cm, ST = 7 cm and RT = 6.5 cm. [Ans.]

6. Construct the incircle of  ∆ STU in which, ST = 7 cm,  ∠T = 120º, TU = 5 cm. [Ans.]

7. Construct the incircle of  ∆DEF in which DE = DF = 5.8 cm, ∠EDF = 65º. [Ans.]


8. Construct any right angled triangle and draw incircle of that triangle. [Ans.]



9. Construct the circumcircle and incircle of an equilateral ∆ XYZ with side 6.3 cm. [Ans.]


Ex. No. 3.2

1. Draw a tangent at any point ‘M’ on the circle of radius 2.9 cm and centre ‘O’.[Ans.]

2. Draw a tangent at any point R on the circle of radius 3.4 cm and centre ‘P’. [Ans.]

3. Draw a circle of radius 2.6 cm. Draw tangent to the circle from any point on the circle using centre of the circle. [Ans.]

4. Draw a circle with centre P and radius 3.1 cm. Draw a chord MN of length 3.8 cm. Draw tangent to the circle through points M and N. [Ans.]

5. Draw a circle of radius 3.5 cm. Take any point K on it. Draw a tangent to the circle at K without using the centre of the circle. [Ans.]

6. Draw a circle of radius 2.7 cm and draw chord PQ of length 4.5 cm. Draw tangents at P and Q without using centre. [Ans.]

7. Draw a circle having radius 3 cm draw a chord XY = 5 cm. Draw tangents at point X and Y without using centre.[Ans.]

8. Draw a tangent to the circle from the point B, having radius 3.6 cm and centre ‘C’. Point B is at a distance 7.2 cm from the centre. [Ans.]

9. Draw a tangent to the circle from the point L with radius 2.8 cm. Point ‘L’ is at a distance 5 cm from the centre ‘M’. [Ans.]


10. Draw a tangent to the circle with centre O and radius 3.3 cm from a point A such that d (O, A) = 7.5 cm. Measure the length of tangent segment. [Ans.]


Exercise No. 3.3. 

1. Sum No. 1. [video]

5. Sum No. 5 [video]

6. Sum No. 6 [Video]



Problem Set No. 3

1. Draw an angle of 125º and bisect it.


2. Draw perpendicular bisector of seg AB of length 8.3 cm.



To draw a line parallel to a given line through a point outside it.


To draw an angle congruent to a given angle.


To draw a perpendicular to a given line from a point outside it.


To draw a perpendicular to a line at a given point on it.


To draw an angle bisector of a given angle.


To draw a perpendicular bisector of a given line segment.


A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to the ground. The inclination of the string with the ground is 60º. Find the length if the string, assuming that there is no slack in the string. ( √3 = 1.73)

6. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to the ground. The  inclination of the string with the ground is 60º. Find the length  if the string, assuming that there is no slack in the string. ( √3 = 1.73)





Sol.
seg AB represents the distance of a kite from ground.
∴ AB = 60 m
seg AC represents the length of the string
m ∠ ACB = 60º
In right angled ∆ ABC,
sin 600 = side opposite to 600 /Hypotenuse
∴ sin 600 = AB/AC
∴ √3/2 = 60/AC
∴ AC = 120/√3
∴ AC = (120/√3)× (√3/√3)
∴ AC = 40√3 m
∴ AC = 40 × 1.73
∴  AC = 69.2 m

∴  The length of the string, assuming that there is no slack in the string is 69.2 m.

A tree is broken by the wind. The top struck the ground at an angle of 30º and at a distance of 30 m from the root. Find the whole height of the tree. ( √3 = 1.73)

5. A tree is broken by the wind. The top struck the ground at an angle of 30º and at a distance of 30 m from the root. Find the whole height of the tree. ( √3 = 1.73)





Sol. seg AB represents the height of the tree

The tree breaks at point D

∴ seg AD is the broken part of tree which then takes the position of DC
∴  AD = DC
m∠ DCB = 30º
BC = 30 m
In right angled ∆DBC,
tan 30º = side opposite to angle 30º/adjacent side of 30º
∴ tan 30º = BD/BC
∴ 1/√3 = BD/30
∴ BD = 30/√3
∴ BD = (30/√3)×(√3/√3)
∴ BD = 10√3 m
cos 300 = adjacent side of angle 300/Hypotenuse

∴ cos 300 = BC/DC
∴ cos 300 = 30/DC
∴ √3/2  = 30/DC
∴ DC = (30×2)/√3
∴ DC = (60/√3)×(√3/√3)
∴ DC = 20√3 m
AD = DC = 20 √3 m
AB = AD + DB [∵ A - D - B]
∴  AB = 20 3 + 10 3
∴  AB = 30 3 m
∴  AB = 30 × 1.73
∴  AB = 51.9 m

 The height of tree is 51.9 m.

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