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.
ЁЯОУ 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:
- Draw a circle with a radius of 2.7 cm.
- Take a point P anywhere on the circle.
- 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.
- Take any point R on the major arc (the larger side of the circle) and join PR and QR to form \( \Delta PQR \).
- Place the compass at point R and draw an arc intersecting sides PR and QR. Keep the same radius.
- Place the compass point at P and draw a similar arc intersecting chord PQ. Do the same at point Q intersecting chord QP.
- Measure the distance of the arc drawn at angle R. Cut this distance on the arcs drawn at P and Q to replicate \( \angle PRQ \).
- Draw a line passing through P and the intersection point of the arcs. This is the required tangent at P.
- Draw a line passing through Q and the intersection point of the arcs. This is the required tangent at Q.
Visual Guide:
Fig 1: Rough Figure (Planning the Construction)
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
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’.
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 = 90┬║,
Q8. Construct any right angled triangle and draw the incircle of that triangle.
To solve this problem, we will first construct a right-angled triangle based on specific measurements, and then construct its incircle (a circle touching all three sides of the triangle internally).
Step 1: Triangle Construction Details
Let \(\Delta ABC\) be the required right-angled triangle with the following dimensions:
- Length of side \(AB = 6 \text{ cm}\)
- Length of side \(BC = 9 \text{ cm}\)
- Angle \(m\angle ABC = 90^\circ\)
Rough Figure (Analytical Diagram)
Figure 1: Rough Sketch
Step 2: Steps of Construction
- Draw a line segment \(BC\) of length \(9 \text{ cm}\).
- At point \(B\), construct an angle of \(90^\circ\).
- Cut off a segment \(AB\) of \(6 \text{ cm}\) from the ray of the \(90^\circ\) angle.
- Join points \(A\) and \(C\) to form \(\Delta ABC\).
- To find the incenter, draw the angle bisectors of \(\angle B\) and \(\angle C\).
- Mark the point of intersection of these bisectors as \(I\) (the Incenter).
- From \(I\), draw a perpendicular to side \(BC\) intersecting at point \(D\).
- With \(I\) as the center and \(ID\) as the radius, draw a circle. This is the required incircle.
Final Construction
Figure 2: Final Construction
8. Construct incircle of ∆SGN such that SG = 6.7 cm, ∠S = 70┬║, ∠G = 50┬║ and draw incircle of ∆ SGN.
8. Construct incircle of ∆SGN such that SG = 6.7 cm, ∠S = 70┬║, ∠G = 50┬║ and draw incircle of ∆ SGN.
7. Construct the incircle of ∆DEF in which DE = DF = 5.8 cm, ∠EDF = 65┬║.
Geometry Challenge: Constructing an Incircle
7. Construct the incircle of ∆DEF in which DE = DF = 5.8 cm, and ∠EDF = 65┬║.
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 = 120┬║, TU = 5 cm.
6. Construct the incircle of ∆ STU in which, ST = 7 cm, ∠T = 120┬║, 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 = 40┬║, ∠PRQ = 90┬║. Draw circumcircle of ∆ PQR.
4. Construct a right angled triangle ∆PQR where PQ = 6 cm, ∠QPR = 40┬║, ∠PRQ = 90┬║. Draw circumcircle of ∆ PQR.
3. Construct the circumcircle of ∆ KLM in which KM = 7 cm, ∠ K = 60┬║, ∠ M = 55┬║.
3. Construct the circumcircle of ∆ KLM in which KM = 7 cm, ∠ K = 60┬║, ∠ M = 55┬║.
2. Construct the circumcircle of ∆ SIM in which SI = 6.5 cm, ∠ I = 125┬║, IM = 4.4 cm.
2. Construct the circumcircle of ∆ SIM in which SI = 6.5 cm, ∠ I = 125┬║, IM = 4.4 cm.
1. Draw the circumcircle of ∆ PMT such that, PM = 5.4 cm, ∠ P = 60┬║, ∠ M = 70┬║.
1. Draw the circumcircle of ∆ PMT such that, PM = 5.4 cm, ∠ P = 60┬║, ∠ M = 70┬║.
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.
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 = 60┬║, ∠
M = 70┬║.
2. Construct the
circumcircle of ∆ SIM in which SI = 6.5
cm, ∠ I = 125┬║, IM = 4.4 cm.
3. Construct the circumcircle of ∆ KLM in which KM = 7 cm, ∠ K = 60┬║, ∠ M = 55┬║.
4. Construct a
right angled triangle ∆PQR where PQ = 6
cm, ∠QPR = 40┬║, ∠PRQ = 90┬║. Draw
circumcircle of ∆ PQR.
5. Construct the
incircle of ∆RST in which RS = 6 cm, ST = 7 cm and RT = 6.5 cm.
6. Construct the
incircle of ∆ STU in which, ST = 7 cm, ∠T = 120┬║, TU = 5
cm.
7. Construct the
incircle of ∆DEF in which DE = DF = 5.8
cm, ∠EDF = 65┬║.
8. Construct any
right angled triangle and draw incircle of that triangle.
9. Construct the
circumcircle and incircle of an equilateral ∆ XYZ with side 6.3 cm. [Ans.]
Ex. No. 3.1
1. Draw the circumcircle of ∆ PMT such that, PM = 5.4 cm, ∠ P = 60┬║, ∠
M = 70┬║.
2. Construct the
circumcircle of ∆ SIM in which SI = 6.5
cm, ∠ I = 125┬║, IM = 4.4 cm.
3. Construct the circumcircle of ∆ KLM in which KM = 7 cm, ∠ K = 60┬║, ∠ M = 55┬║.
4. Construct a
right angled triangle ∆PQR where PQ = 6
cm, ∠QPR = 40┬║, ∠PRQ = 90┬║. Draw
circumcircle of ∆ PQR.
5. Construct the
incircle of ∆RST in which RS = 6 cm, ST = 7 cm and RT = 6.5 cm.
6. Construct the
incircle of ∆ STU in which, ST = 7 cm, ∠T = 120┬║, TU = 5
cm.
7. Construct the
incircle of ∆DEF in which DE = DF = 5.8
cm, ∠EDF = 65┬║.
8. Construct any
right angled triangle and draw incircle of that triangle.
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’.
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.
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.
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.
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.
7. Draw a circle having radius 3 cm draw a chord XY = 5
cm. Draw tangents at point X and Y without using 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.
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’.
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.
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 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’.
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.
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.
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.
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.
7. Draw a circle having radius 3 cm draw a chord XY = 5
cm. Draw tangents at point X and Y without using 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.
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’.
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.
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
125┬║ and bisect it.
2. Draw
perpendicular bisector of seg AB of length 8.3 cm.
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 60┬║. 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 60┬║. 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 = 60┬║
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 30┬║ 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 30┬║ 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 = 30┬║
BC = 30 m
In right angled ∆DBC,
tan 30┬║ = side opposite to angle 30┬║/adjacent
side of 30┬║
∴ tan 30┬║ = 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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