Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to with scale factor . Justify the construction. Are the two triangles congruent? Note that all the three angles and two sides of the two triangles are equal
Solution
Steps of construction
Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is . Also justify the
construction. Measure the distance between the centre of the circle and the point of intersection of tangents
Solution
Steps of construction
1. Draw a circle of radius OA = 4 cm with centre O
2. Produce OA to P such that
3. Draw a perpendicular bisector of OP=8cm
4.Now taking A as Centre draw circle of radius AP = OA = 4 cm
5.Which intersect the circle at x and y
6.Join PX and PY
7.PX and PY is the tangent of the circle
Justification
In we have
(Radius)
(Radius of circle with centre A)
is equilateral triangle
In we have
Hence
View Full Answer(1)
Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and .Construct a triangle similar to ABC with scale factor . Justify the construction
Solution
Given : AB = 5 cm, BC = 6 cm
Steps of construction
Justification: Here, [by construction]
Now
Also,
Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.
Solution
Steps of construction
1. Draw line BC = 5 cm
2. Taking B and C as centres, draw two arcs of equal radius 6 cm intersecting each other at point A.
3. Join AB and AC DABC is required isosceles triangle
4 From B draw ray with an acute angle
6. draw at with equal distance
7. Join and from draw line , intersect extended segment BC at point D.
8. From point D draw meting BA produced at E.
Then EBD is required triangle. We can name it PQR.
Justification
Study 40% syllabus and score up to 100% marks in JEE
Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.
Solution
Steps of construction
1. Draw two concentric circles with center O and radii 3 cm and 5 cm
2. Taking any point P on outer circle, Join P and O
3. Draw perpendicular bisector of OP let M be the mid point of OP
4. Taking M as centre and OM as radius draw a circle which cuts inner circle at Q and R
5. Join PQ and PR. Thus PQ and PR are required tangents
On measuring PQ and PR we find that PQ = PR = 4 cm
Calculations
[using pythagoras theorem]
Hence the length of both tangents is 4 cm.
View Full Answer(1)
Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and , divide it into triangles BCD and ABD
by the diagonal BD. Construct the triangle BD'C' similar to DBDC with scale factor . Draw the line segment D'A' parallel to DA where A' lies on extended side BA. Is A'BC'D' a parallelogram?
Solution
Steps of construction
1. Draw, A line AB = 3 cm
2 Draw a ray by making
3. Taking B as centre and radius equal to 5 cm. Draw an arc which cut BP at point C
4. Again draw ray AX making
5. With A as centre and radius equal to 5 cm draw an arc which cut AX at point D
6. Join C and D Here ABCD is a parallelogram
7. Join BD , BD is a diagonal of parallelogram ABCD
8. From B draw a ray BQ with any acute angle at point B i.e., is acute angle
9. Locate 4 points on BQ with equal distance.
10. Join and from parallel to which intersect at point
11. From point draw line which is parallel to CD
12. Now draw a line segment parallel to DA
Note : Here and are the extended sides.
13. is a parallelogram in which and and divide it into triangles and by the diagonal
View Full Answer(1)
Two line segments AB and AC include an angle of where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC, respectively such that and . Join P and Q and measure the length PQ.
Solution
Given :
AB = 5 cm and AC = 7 cm
From equation 1
P is any point on B
scale of a line segment AB is
Steps of construction
1. Draw line segment AB = 5 cm
2. Now draw ray AO which makes an angle i.e.,
3. Which A as center and radius equal to 7 cm draw an arc cutting line AO at C
4. Draw ray AP with acute angle BAP
5. Along AP make 4 points with equal distance.
6. Join
7. From draw which is parallel to which meet AB at point P.
Then P is point which divides AB in ratio 3 : 1
AP : PB = 3 : 1
8. Now draw ray AQ, with an acute angle CAQ.
9. Along AQ mark 4 points with equal distance.
10. Join
11. From draw which is parallel to which meet AC at point Q.
Then Q is point which divides AC in ratio 1 : 3
AQ : QC = 1 : 3
12. Finally join PQ and its measurement is 3.25 cm.
In Fig. 10.21, O is the centre of the circle, BD = OD and CD AB. Find CAB.
30°
Solution:
Given: In the figure BD = OD, CD AB
In OBD,
BD = OD (given)
OD = OB (radius of the same circle)
OB = OD = BD
Hence the triangle is equilateral
BOD = OBD = ODB = 60°
Consider MBC and MBD
MB = MB (common)
CMB = BMD = 90° (given)
CM = MD (perpendicular from the centre on the chord bisects the chord]
MBC =MBD (SAS rule)
MBC =MBD (CPCT)
MBC = OBD = 60° (OBD = 60°)
Since AB is the diameter of the circle
ACB = 90° (angle is a semi-circle)
CAB +CBA + ACB = 180° (By angle sum property of a triangle)
Putting the values,
CAB + 60° + 90° = 180°
CAB = 180° – (60° + 90°) = 30°
View Full Answer(1)
In Fig. 10.20, O is the centre of the circle, BCO = 30°. Find x and y.
x = 30° and y = 15°
Solution:
Given: O is the centre of the circle
BCO = 30°
Join OB and AC.
In BOC,
CO = BO (Radius of the same circle)
OBC =OCB = 30° (angles opposite to equal sides in a triangle are equal)
In OBC,
OBC + OCB + BOC = 180° (angle sum property of a triangle)
BOC = 180° – (30° + 30°) = 120°
BOC = 2BAC (Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)
Consider AEB, AEC
AE = AE (common)
AEB =AEC = 90°
BE = EC (given)
AEB AEC (SAS congruence)
BAE =CAE (CPCT)
BAE + CAE = BAC = 60°
2BAE = 60°
BAE = 30° = x
Now, OD is perpendicular to AE
and, CE is perpendicular to AE
Two lines perpendicular to the same line are parallel to each other
So, OD || CE and OC is transversal
DOC = ECO = 30° (alternate interior angles)
Now, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle.
Consider the arc CD,
DOC = 2DBC = 2y
30° = y
y = 15°
Hence, x = 30° and y = 15°
View Full Answer(1)AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q^{2} = p^{2} + 3r^{2}
AB and AC are two chords of a circle of radius r such that AB = 2AC.
The distance of AB and AC from the centre are p and q, respectively
To Prove: 4q^{2 }= p^{2} + 3r^{2}
Proof:
Let AB = 2x then AC = x (Given, AB = 2AC)
Draw ON perpendicular to AB and OM perpendicular to AC
AM = MC = x/2 (perpendicular from the centre bisects the chord)
AN = NB = x (perpendicular from the centre bisects the chord)
In DOAM, applying Pythagoras theorem
AO^{2} = AM^{2 }+ MO^{2}
AO^{2} = (x/2)^{2} + q^{2} ….(i)
InOAN, applying Pythagoras theorem
AO^{2} = (AN)^{2} + (NO)^{2}
AO^{2} = (x)^{2} + p^{2} ….(ii)
From equation (i) and (ii)
x^{2} + 4q^{2} = 4x^{2} + 4p^{2 } [Multiply both sides by 4]
4q^{2} = 3x^{2} + 4p^{2}
4q^{2 }= p^{2} + 3 (x^{2} + p^{2})
4q^{2 }= p^{2 }+ 3r^{2} (In right angle OAN, r^{2} = x^{2} + p^{2})
Hence Proved.
View Full Answer(1)Study 40% syllabus and score up to 100% marks in JEE
CBSE 9 Class
CBSE 12 Class
CBSE 7 Class
Class 12
Class 10
Class 7
Class 9
Maths
Mathematics Part II Textbook for Class XII
Mathematics Textbook for Class IX
Mathematics Textbook for Class VII
Exemplar Maths for Class 9
Exemplar Maths for Class 10
Circles
Constructions
Vector Algebra
Practical Geometry