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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 = 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. 3.1
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 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.
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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