RD Sharma 2020 solution class 9 chapter 15 Circles VSAQS

VSAQS

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Question 1:

In the given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.
 

Answer 1:

Consider the smaller circle whose centre is given as ‘O’.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

Hence ,the measure of is.

Question 2:

In the given figure, two congruent circles with centres O and O' intersect at A and B. If ∠AOB = 50°, then find ∠APB.
 

Answer 2:

Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’.

So, from the given figure we have,

Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus.

One of the properties of a rhombus is that the opposite angles are equal to each other.

So, since it is given that, we can say that the angle opposite it, that is to say that should also have the same value.

Hence we get

Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is.

A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

This means that,

Hence the measure of is

Question 3:

In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.

Answer 3:

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to .

Here we have a cyclic quadrilateral ABCD. The centre of this circle is given as ‘O’.

Since in a cyclic quadrilateral the opposite angles are supplementary, here

Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment. The angle that the chord forms with any point on the circumference of a particular segment is always the same.

Here, ‘CD’ is a chord and ‘A’ and ‘B’ are two points along the circumference on the major segment formed by the chord ‘CD’.

So,

Now,

In any triangle the sum of the interior angles need to be equal to .

Consider the triangle ΔABP,

$$\begin{gathered} \angle PAB+\angle ABP+\angle APB=180° \\ \Rightarrow \angle APB=180°-30°-58° \\ \Rightarrow \angle APB=92° \end{gathered}$$

From the figure, since ‘AC’ and ‘BD’ intersect at ‘P’ we have,

Hence the measure of is.

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Question 4:

In the given figure, if ∠AOB = 80° and ∠ABC = 30°, then find ∠CAO.

Answer 4:

Consider the given circle with the centre ‘O’. Let the radius of this circle be ‘r’. ‘AB’ forms a chord and it subtends an angle of 80° with its centre, that is.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

In any triangle the sum of the interior angles need to be equal to 180°.

Consider the triangle

Since,  , we have. So the above equation now changes to

Considering the triangle ΔABC now,

Hence, the measure of is.

Question 5:

In the given figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD : ∠ABE.
 

Answer 5:

It is given that ‘ABCD’ is a parallelogram. But since ‘A’ is the centre of the circle, the lengths of ‘AB’ and ‘AD’ will both be equal to the radius of the circle.

So, we have.

Whenever a parallelogram has two adjacent sides equal then it is a rhombus.

So ‘ABCD’ is a rhombus.

Let.

We know that in a circle the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

By this property we have

In a rhombus the opposite angles are always equal to each other.

So,

Since the sum of all the internal angles in any triangle sums up to in triangle , we have

In the rhombus ‘ABCD’ since one pair of opposite angles are ‘’ the other pair of opposite angles have to be

From the figure we see that,

So now we can write the required ratio as,

Hence the ratio between the given two angles is .

Question 6:

In the given figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.
 

Answer 6:

Let us first consider the triangle ΔABQ.

It is known that in a triangle the sum of all the interior angles add up to 180°.

So here in our triangle ΔABQ we have,

By a property of the circle we know that an angle formed in a semi-circle will be 90°..

In the given circle since ‘AB’ is the diameter of the circle the angle which is formed in a semi-circle will have to be 90°.

So, we have

Now considering the triangle we have,

From the given figure it can be seen that,

Now, we can also say that,

Hence the measure of the angle is 115°.

Question 7:

In the given figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =
 

Answer 7:

Consider the circle with the centre ‘P’.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

Since ‘ACD’ is a straight line, we have

Now let us consider the circle with centre ‘Q’. Here let ‘E’ be any point on the circumference along the major arc ‘BD’. Now ‘CBED’ forms a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

So here,

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, now we have

Hence, the measure of is .

Question 8:

In the given figure, if O is the circumcentre of ∠ABC, then find the value of ∠OBC + ∠BAC.

Answer 8:

Since, O is the circumcentre of $△ABC$, So, O would be centre of the circle passing through points A, B and C.

$\angle ABC$= 90°             (Angle in the semicircle is 90°.)

$\Rightarrow \angle OAB+\angle OBC=90°$                .....(1)

As OA = OB    (Radii of the same circle)

$$\begin{gathered} \therefore \angle OAB=\angle OBA \left(\mathrm{Angle} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{sides} \mathrm{are} \mathrm{equal}\right) \\ \mathrm{or}, \angle BAC= \angle OBA \\ \mathrm{From} \left(1\right) \\ \mathit{\angle }BAC+\angle OBC=90° \end{gathered}$$

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Question 9:

If the given figure, AOC is a diameter of the circle and arc AXB = $\frac{1}{2}$ arc BYC. Find ∠BOC.

Answer 9:

We need to find

  
 

$$\begin{gathered} \mathrm{arc} AXB=\frac{1}{2}\mathrm{arc} BYC, \\ \angle AOB=\frac{1}{2}\angle BOC \\ \mathrm{Also} \angle AOB+\angle BOC=180° \\ \mathrm{Therefore}, \frac{1}{2}\angle BOC+\angle BOC=180° \\ \Rightarrow \angle BOC=\frac{2}{3}\times 180°=120° \end{gathered}$$

Question 10:

In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.

Answer 10:

We are given ABCD is a quadrilateral with center O, ∠ADE = 95° and ∠OBA = 30°

We need to find ∠OAC

We are given the following figure

Since ∠ADE = 95°

⇒ ∠ADC = 180 ° − 95° = 85°

Since squo;ABCD is cyclic quadrilateral

This means

∠ABC + ∠ADC = 180°

Since OB = OC (radius)

⇒ ∠OBC = ∠OCB = 65°

In ΔOBC

Since ÐBAC and ÐBOC are formed on the same base which is chord.

So

Consider ΔBOA which is isosceles triangle.

∠OAB = 30°