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#### 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.

#### 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

1. Draw line BC = 6cm
2. Taking B and C as centre mark arcs of length 4cm and 5cm respectively which intersecting each other at point A.
3. Join BA and CA
4. Draw a ray BX making acute angle with BC.
5. Locate 5 Points on BX in equidistant as B1, B2, B3, B4, B5.
6. 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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#### 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.

#### 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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#### 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°.

#### 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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#### 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.

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

#### 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°.