Get Answers to all your Questions

Filter By

Clear Filter

All Questions

#10.3

Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre

Solution
Given : Radius = 4cm

Steps of construction
1. Draw a circle at radius r = 4 cm at point O.
2. Take a point P at a distance 6cm from point O and join PO
3. Draw a perpendicular bisector of line PO M is the mid-point of PO.
4. Taking M as centre draw another circle of radius equal to MO and it intersect the given circle at point Q and R

5. Now, join PQ and PR.
Then PQ and PR one the required two tangents.

View Full Answer(1)

Posted by

infoexpert27

#10.3

Draw a triangle ABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm.Construct a triangle similar to it and of scale factor $\frac{5}{3}$

Solution
Given : BC = 6cm, CA = 5cm, AB = 4cm
Scale factor = 5/3
Let m = 5 and n = 3

Steps of construction

Draw line BC = 6cm
Taking B and C as centre mark arcs of length 4cm and 5cm respectively which intersecting each other at point A.
Join BA and CA
Draw a ray BX making acute angle with BC.
Locate 5 Points on BX in equidistant as B₁, B₂, B₃, B₄, B₅.
Join B₃ to C and draw a line through B5 parallel to B₃C to intersect at BC extended at ${C}'$ .

7. Draw a line through ${C}'$ parallel to AC intersect AB extended at ${A}'$ .

Now ${A}'B{C}'$ is required triangle.

View Full Answer(1)

Posted by

infoexpert27

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

#10.3

Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and $\angle B= 90^{\circ}$ .Construct a triangle similar to it and of scale factor $\frac{2}{3}$ . Is the new triangle also a right triangle ?

Solution
Given BC = 12cm, AB = 5 cm, $\angle B= 90^{\circ}$

Steps of construction
1. Draw a line BC = 12 cm
2.   From B draw AB = 5 cm which makes an angle of $90^{\circ}$ at B.
3.   Join AC
4.   Make an acute angle at B as $< CBX$
5.   On BX mark 3 point at equal distance at $X_{1},X_{2},X_{3}$ .
6.   Join $X_{3}C$
7.   From $X_{2}$ draw $X_{2}{C}'\parallel X_{3}C$ intersect AB at ${C}'$
8.   From point ${C}'$ draw ${C}'{A}'\parallel CA$ intersect AB ${A}'$
Now   $\bigtriangleup {A}'B{C}'$ is the required triangle and   $\bigtriangleup {A}'B{C}'$ is also a right triangle.

View Full Answer(1)

Posted by

infoexpert27

#10.3

Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio $3:5$ .

Solution
Given: AB = 7cm
The required ratio is $3:5$
Let m = 3, n = 5
$m+n= 3+5= 8$

Steps of constructio
1.   Draw a line segment AB = 7cm.
2. Draw a ray AX making acute angle with AB
3.   Locate 8 points $A_{1},A_{2},A_{3}\cdots A_{8}$ on AX on equidistant.         (because $m+n= 8$ )
4.   Join $BA_{8}$
5.   Through the point $A_{3}$ draw a line parallel to $BA_{8}$ which intersect line AB at P..
Here triangle AA₃P is similar to triangle $AA_{8}B$
   $AA_{3}/AA_{8}= AP/AB= 3/5$         (by construction)
Therefore $AP:BP= 3:5$

View Full Answer(1)

Posted by

infoexpert27

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

#10.3

In Fig. 10.16, $\angle$ OAB = 30º and $\angle$ OCB = 57º. Find $\angle$ BOC and $\angle$ AOC.

$\angle$ AOC = 54° and $\angle$ BOC = 66°

Solution:

Given: $\angle$ OAB = 30°, $\angle$ OCB = 57°C

In $\triangle$ OAB,

AO = BO (radius of the same circle)

$\angle$ OAB = $\angle$ OBA = 30° (Angles opposite to equal sides are equal)

In $\triangle$ AOB, the sum of all angles is 180°.

$\Rightarrow$ $\angle$ OAB + $\angle$ OBA + $\angle$ AOB = 180°.

$\Rightarrow$ 30° + 30° + $\angle$ AOB = 180°

$\Rightarrow$ $\angle$ AOB = 180° – 30° – 30°

$\Rightarrow$ $\angle$ AOB = 120° … (i)

Now, in $\triangle$ OBC,

OC = OB (radius of the same circle)

$\angle$ OBC = $\angle$ OCB = 57° (Angles Opposite to equal sides are equal).

In $\triangle$ OBC, the sum of all angles is 180°.

$\angle$ OBC + $\angle$ OCB + $\angle$ BOC = 180°

$\Rightarrow$ 57° + 57° + $\angle$ BOC = 180°

$\angle$ BOC = 180° – 57° – 57°

$\Rightarrow$ $\angle$ BOC = 66° … (ii)

Now, form equation (i) we have

$\Rightarrow$ $\angle$ AOB = 120°.

$\Rightarrow$ $\angle$ AOC + $\angle$ COB = 120°

$\Rightarrow$ $\angle$ AOC + 66° = 120° (from ii)

$\Rightarrow$ $\angle$ AOC = 120° – 66°

$\Rightarrow$ $\angle$ AOC = 54° and $\angle$ BOC = 66°.

View Full Answer(1)

Posted by

infoexpert24

#10.3

In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of $\angle$ ACD + $\angle$ BED.

270°

Solution:

Join AE

ACDE is a cyclic quadrilateral and sum of opposite angles in a cyclic quadrilateral is 180°

$\therefore$ $\angle$ ACD + $\angle$ AED = 180° … (i)

Now, we know that the angle in a semi-circle is 90°

So, $\angle$ AEB = 90° … (ii)

An adding equation (i) & (ii), we get.

$\angle$ ACD + $\angle$ AED + $\angle$ AEB = 180° + 90°

$\angle$ ACD + $\angle$ BED = 270°

Hence the value of $\angle$ ACD + $\angle$ BED is 270°

View Full Answer(1)

Posted by

infoexpert24

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

#10.3

Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO′ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2 OO′.

Two circles with centres O and O′ intersect at two points A and B.

PQ || OO'

To Prove that: PQ = 2OO'

Proof:

Draw OM perpendicular to PB and O'N perpendicular to BQ

From the figure, we have:

OP = OB (radius of same circle)

O'B = O'Q (radius of the same circle)

In $\triangle$ OPB,

BM = MP … (i) (perpendicular from the centre of the circle bisects the chord)

Similarly in $\triangle$ O'BQ,

BN = NQ … (ii) (perpendicular from the centre of the circle bisects the chord)

Adding (i) and (ii),

BM + BN = PM + NQ

Adding BM + BN to both the sides.

BM + BN + BM + BN = BM + PM + NQ + BN.

2BM + 2BN = PQ

2(BM + BN) = PQ … (iii)

OO' = MN (As OO' NM is a rectangle)

OO' = BM + BN ... (iv)

Using equation (iv) in (iii)

2 OO'= PQ

Hence proved.

View Full Answer(1)

Posted by

infoexpert24

#10.3

A quadrilateral ABCD is inscribed in a circle such that AB is diameter and $\angle$ ADC = 130º. Find $\angle$ BAC.

40°

Solution:

Given: Quadrilateral ABCD inscribed in a circle

$\angle$ ADC = 130º

$\therefore$ Quadrilateral ABCD is a cyclic Quadrilateral. Sum of opposite angles in a cyclic quadrilateral is 180°

$\Rightarrow$ $\angle$ ADC + $\angle$ ABC = 180°

$\Rightarrow$ 130° + $\angle$ ABC = 180°

$\Rightarrow$ $\angle$ ABC = 180° – 130°

$\Rightarrow$ $\angle$ ABC = 50° …(i)

AB is a diameter of a circle (given)

$\therefore$ ACB = 90° (angle in a semicircle is 90º)

In $\triangle$ ABC,

$\Rightarrow$ $\angle$ ABC + $\angle$ ACB + $\angle$ BAC = 180°(angle sum property of a triangle)

Putting all the values,

$\Rightarrow$ 50° + 90° + $\angle$ BAC = 180°

$\Rightarrow$ $\angle$ BAC = 180° –50° – 90°

$\therefore$ $\angle$ BAC = 40°

View Full Answer(1)

Posted by

infoexpert24

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

#10.3

In Fig.10.14, $\angle$ ACB = 40º. Find $\angle$ OAB.

50°

Solution:

Given, $\angle$ ACB = 40°

As we know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle

$\\\Rightarrow \angle AOB = 2 \angle ACB.\\ \\ \Rightarrow \angle ACB = \frac{ \angle AOB}{2}\\ \\ \Rightarrow 40^{\circ} =\frac{1}{2} \angle AOB\\ \\ \Rightarrow \angle AOB = 80^{\circ}$

In $\triangle$ AOB, AO = BO (Both are the radius of the same circle)

$\angle$ OBA = $\angle$ OAB (angles opposite to equal sides in a triangle are equal)

As we know, the sum of all three angles in a triangle AOB is 180°.

In $\triangle$ AOB
$\\ \angle AOB + \angle OBA + \angle OAB = 180^{\circ}\\ \Rightarrow 80^0 + \angle OAB + \angle OAB = 180^{\circ}\\ \Rightarrow 2\angle OAB = 180^{\circ} -80^{\circ}\\ \Rightarrow 2\angle OAB = 100^0\\\\ \Rightarrow \angle OAB = \frac{100^{\circ}}{2}=50^{\circ}$

View Full Answer(1)

Posted by

infoexpert24

#10.3

In Fig.10.13, $\angle$ ADC = 130° and chord BC = chord BE. Find $\angle$ CBE.

100°

Solution:

Consider $\triangle$ BCO and $\triangle$ BEO

$\Rightarrow$ BC = BE as given.

$\angle$ BCO = $\angle$ BEO (angles opposite to equal sides in a triangle are equal)

BO = BO (common sides)

$\Rightarrow$ $\triangle$ BCO $\cong$ $\triangle$ BEO (SAS congruence)

$\angle$ CBO = $\angle$ OBE (CPCT) …(i)

Now we know that ABCD is a Cyclic quadrilateral,

So, $\angle$ ADC + $\angle$ ABC = 180°

$\Rightarrow$ 130° + $\angle$ ABC = 180°

$\Rightarrow$ $\angle$ ABC = 180°-130° = 50°

$\therefore$ $\angle$ OBE = 50°

From (i)

$\angle$ CBO = $\angle$ OBE = 50°

$\Rightarrow$ $\angle$ CBE = $\angle$ CBO + $\angle$ OBE = 50° + 50° = 100°.

View Full Answer(1)

Posted by

infoexpert24

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Resources

Diploma Colleges

Other Exams

Exams

Resources

Upcoming Events

Exams

Top Schools

Products & Resources

NCERT Study Material

Top Countries

Student Visas

Exams

Colleges

Exams

Colleges

Upcoming Events

Resources

Exams

Colleges & Courses

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors & E-Books

Exams

Predictors & Articles

Colleges

Resources

Exams

Colleges

Resources

Top Courses & Careers

Colleges

Exams

Resources

Get Answers to all your Questions

All Questions

Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre

Posted by

infoexpert27

Draw a triangle ABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm.Construct a triangle similar to it and of scale factor

Posted by

infoexpert27

Crack CUET with india's "Best Teachers"

Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and .Construct a triangle similar to it and of scale factor . Is the new triangle also a right triangle ?

Posted by

infoexpert27

Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio .

Posted by

infoexpert27

JEE Main high-scoring chapters and topics

In Fig. 10.16, OAB = 30º and OCB = 57º. Find BOC and AOC.

Draw a triangle ABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm.Construct a triangle similar to it and of scale factor $\frac{5}{3}$

Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and $\angle B= 90^{\circ}$ .Construct a triangle similar to it and of scale factor $\frac{2}{3}$ . Is the new triangle also a right triangle ?

Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio $3:5$ .

In Fig. 10.16, $\angle$ OAB = 30º and $\angle$ OCB = 57º. Find $\angle$ BOC and $\angle$ AOC.

In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of $\angle$ ACD + $\angle$ BED.

A quadrilateral ABCD is inscribed in a circle such that AB is diameter and $\angle$ ADC = 130º. Find $\angle$ BAC.

In Fig.10.14, $\angle$ ACB = 40º. Find $\angle$ OAB.

In Fig.10.13, $\angle$ ADC = 130° and chord BC = chord BE. Find $\angle$ CBE.