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## All Questions

#### Write True or False and give reasons for your answer in each of the following:A pair of tangents can be constructed to a circle inclined at an angle of $170^{\circ}$.

Solution

Tangent : - it is a straight line that touches the curve but not cross it. A pair of tangents can be constructed to a circle inclines at an angle greater than $0^{\circ}$ but less than $180^{\circ}$.Here, the inclination angle is $170^{\circ}$ Hence it is possible.

#### Write True or False and give reasons for your answer in each of the following:A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of 3 cm from the center.

Solution
Point distance from center = 3 cm
But, $r> d\Rightarrow 3.5 cm> 3cm$
So, the point P lies inside the circle.
So, pair of tangents cannot be drawn to point P to a circle.

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#### Write True or False and give reasons for your answer in each of the following: To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides $\frac{7}{3}$ of the corresponding sides of $\bigtriangleup ABC$, draw a ray BX making acute angle with BCand X lies on the opposite side of A with respect to BC. The points B1, B2, ....,B7 are located at equal distances on BX, B3 is joined to C and then a linesegment $B_6{C}'$ is drawn parallel to B3C where ${C}'$ lies on BC produced. Finally,line segment ${A}'{C}'$ is drawn parallel to AC.

Solution
According to question:-

To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides $\frac{7}{3}$ of the corresponding sides of $\bigtriangleup ABC$

1. Draw a line segment BC
2. Taking B and C as centres draw two arcs of suitable radii intersecting each other at A.
3. Join BA and CA. ?ABC is the required triangle.
4. From B draw any ray BX downwards making an acute angle CBX.
5. Locate seven points B1, B2, b3, …. B7 on BX such that BB1 = B1B2 = B1B3 = B3B4 = B4B5 = B5B6 = B6B7.
6. Join B3C and from B7 draw a line B7C’ ? B3C intersecting the extended line segment BC at C’.
7. From point C’ draw C’A’ ? CA intersecting the extended line segment BA at A

But as given if we join B3C and from B6 draw a line B6C’ ? B3C intersecting the extended line segment BC at C’.

$BB_{3}/BB_{6}={BC}/{{BC}'} = 3/6$
${BC}/{{BC}'} = 1/2$
$BC:{BC}'= 1:2$

Hence the sides are not in the ratio of 7:3
So, the required triangle can not be constructed in this way.
Hence the given statement is false.

#### Write True or False and give reasons for your answer in each of the following:By geometrical construction, it is possible to divide a line segment in the ratio $\sqrt{3}:\frac{1}{\sqrt{3}}$

Solution
To divide a line segment in the ratio, we need both positive integer.
So, we can be simplified it by multiply both the terms by  .$\sqrt{3}$
We obtain  $\sqrt{3}\times \sqrt{3} : \sqrt{3}\times \frac{1}{\sqrt{3}}$
$\Rightarrow 3:1$
So, the required ratio is $3:1$
Geometrical construction, is possible to divide a line segment in the ratio $3:1$

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#### Write True or False and justify your answer in each of the following: In Fig. 10.10, if AOB is a diameter and $\angle$ADC = 120°, then $\angle$CAB = 30°.

True

Solution:

Given, AOB is a diameter and $\angle$ADC = 120°

Join CA & CB

$\angle$ADC + $\angle$CBA = 180°  (Sum of opposite angles of a cyclic quadrilateral is 180°)

120° + $\angle$CBA = 180°

$\angle$CBA = 60°

In $\triangle$ACB,

$\angle$CAB + $\angle$CBA +$\angle$ACB = 180°  (angle sum property of a triangle)

$\angle$ACB = 90° (Since, $\angle$ACB is an angle in a semi-circle)

$\angle$CAB + 60° + 90° = 180°

$\Rightarrow$$\angle$CAB = 180° – 150° = 30°

Therefore the given statement is true.

#### Write True or False and justify your answer in each of the following: If A, B, C and D are four points such that $\angle BAC=45^{\circ}$ and $\angle BDC=45^{\circ}$, then A, B, C, D are concyclic.

True

Solution:

Definition of concyclic points: The points that lie on the same circle.

We know that angles in the same segment of a circle are equal

Since $\angle BAC=45^{\circ}=\angle BDC$

We know that these angles are in the same segment of a circle.

This can be represented as shown below:

BC is the chord of the circle and hence A, B, C & D are concyclic.

Therefore the given statement is true.

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#### Write True or False and justify your answer in each of the following: If A, B, C, D are four points such that $\angle BAC = 30^{\circ} \;\; and \;\;\angle BDC = 60^{\circ}$, then D is the centre of the circle through A, B and C.

True

Solution:

Given, ∠BAC = 30° and ∠BDC = 60°

We know that,

The angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle.

Considering BC,

At centre, $\angle BDC = 60^{\circ}$

At any other point on the circle,

$\angle BAC = 30^{\circ}=\frac{1}{2} (60^{\circ})$

Hence the above rule is justified.

Therefore the given statement is true.

#### Write True or False and justify your answer in each of the following: ABCD is a cyclic quadrilateral such that $\angle A = 90^{\circ}, \angle B = 70^{\circ}, \angle C = 95^{\circ} \; and \; \angle D = 105^{\circ}$.

False.

Solution:

In a cyclic Quadrilateral, the sum of opposite pairs of the angles in the given quadrilateral must be equal to 180°.

Here    $\angle A + \angle C = 90 ^{\circ} + 95 ^{\circ} = 185 ^{\circ}$

$\angle B + \angle D = 70 ^{\circ} + 105 ^{\circ} = 175 ^{\circ}$

Hence, it is not a cyclic Quadrilateral because the sum of the measures of opposite angles is not equal to 180°

Therefore the given statement is false.

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#### Write True or False and justify your answer in each of the following: If AOB is a diameter of a circle and C is a point on the circle, then AC2 + BC2 = AB2.

True

Solution:

Given: AOB is a diameter of a circle

$\Rightarrow$ AOB is a straight line

$\Rightarrow$ $\angle$AOB = 180°

Now, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle

$\\\angle ACB =\left ( \frac{1}{2} \right ) \angle AOB\\ \angle ACB =\left ( \frac{1}{2} \right ) 180^{\circ}\\ \angle ACB = 90^{\circ}$

$\Rightarrow$ ACB is a Right angle

So$\angle ABC$ follows Pythagoras theorem,

i.e., AB2 = AC2 + BC2

Therefore the given statement is true.

#### Write True or False and justify your answer in each of the following: A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.

True

Solution:

Yes, we can draw a circle of radius 3 cm.

The points A & B are such that AB = 6 cm.

So AB will become the diameter of the required circle as AB = 2 (Radius)

Hence, 2 × 3 = 6 cm

Therefore the given statement is true.

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