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#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.

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#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.

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#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.

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#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 $AP:BP= 3:5$

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#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°.

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#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°

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#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.

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#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°

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

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#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°.

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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 .

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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.