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Here, O is the centre of circle.

PQ and PT are tangents to the circle from a point P

R is any point on the circle. RT and RQ are joined.

∠TPQ = 70°

Now,

Join TO and QO.

∠TOQ = 180° – 70° = 110°

Here, OQ and OT are perpendicular on QP and TP.

∠TOQ is on the centre and ∠TRQ is on the rest part.

∠TRQ = 1/2∠TOQ = ½( 110°) = 55°

Therefore, ∠TRQ = 55°.

Correct answer is (c) 2√3 cm

OA ⊥ AT

So, \(\frac{AT}{OT}=\) cos 30°

\(\Rightarrow\) \(\frac{AT}{4}=\frac{\sqrt{3}}{2}\)

\(\Rightarrow\) AT = \((\frac{\sqrt{3}}{2}\times4)\)

\(\Rightarrow\) AT = \(2\sqrt{3}\)

A) obtuse angle

Correct option is: (A) Minor segment

The shaded portion is the region bounded between the chord and minor are of the circle which is known as minor segment.

Correct option is: (A) Minor segment

Option : (B)

The minor segment in a circle always forms an obtuse angle.

Correct answer is (d) 7.6 cm

We know that tangent segments to a circle from the same external point are congruent. Therefore, we have

PT = PO = 3.8 cm and PT = PR 3.8 cm

∴ QR = QP + PR = 3.8 + 3.8 = 7.6 cm

Correct answer is (a) 9 cm

Tangents drawn from an external point to a circle are equal.

So, AQ = AP = cm

CR = CS = 3cm

And BR = (BC - CR)

\(\Rightarrow\) BR = (7 - 3)cm

\(\Rightarrow\) BR = 4cm

BQ = BR = 4cm

∴ AB = (AQ + BQ)

\(\Rightarrow\) AB = (5 + 4)cm

\(\Rightarrow\) AB = 9cm

Answer : (c) AB = 4AP

Using the tangent-secant theorem, we have 

AB × AP = AD² = \(\big(\frac{AC}{2}\big)^2\)  ( ∵ AD = DC) 

⇒ AB × AP = \(\frac{1}{4}\) AC² = \(\frac{1}{4}\) AB² ( ∵ AB = AC) 

⇒ AB = 4 AP

Correct answer is (D) 8 cm

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

So we get

∠AOB = 2 ∠ACB

It is given that ∠ACB = 50^o

By substituting

∠AOB = 2 × 50^o

By multiplication

∠AOB = 100^o …… (1)

Consider △ OAB

We know that the radius of the circle are equal

OA = OB

Base angles of an isosceles triangle are equal

So we get

∠OAB = ∠OBA ……. (2)

Using the angle sum property

∠AOB + ∠OAB + ∠OBA = 180^o

Using equations (1) and (2) we get

100^o + 2 ∠OAB = 180^o

By subtraction

2 ∠OAB = 180^o – 100^o

So we get

2 ∠OAB = 80^o

By division

∠OAB = 40^o

We know that BD is the diameter of the circle

Angle in a semicircle is a right angle

∠BAD = 90^o

Consider △ BAD

Using the angle sum property

∠ADB + ∠BAD + ∠ABD = 180^o

By substituting the values

∠ADB + 90^o + 35^o = 180^o

On further calculation

∠ADB = 180^o – 90^o – 35^o

By subtraction

∠ADB = 180^o – 125^o

So we get

∠ADB = 55^o

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

∠ACB = ∠ADB = 55^o

So we get

∠ACB = 55^o

Therefore, ∠ACB = 55^o.

Option : (B)

∠BDC = 42° 

∠ABC = 90° 

(Angle in a semi-circle) 

In ABC, 

∠ABC + ∠BAC = 42° 

(Angles on the same segment) 

90o + 42° + ∠ACB = 180° 

∠ACB = 48°

Given,

ABCD is a square of side 7 cm. 

Area of the shaded region = Area of ABCD – Area of two semicircles with radius 7/2 = 3.5 cm 

APD and BPC are semicircles. 

= 7 × 7 – 2 × \(\frac{1}{2}\) × \(\frac{22}{7}\) × 3.5 × 3.5 

= 49 – 38.5 

= 10.5 cm² 

∴ Area of shaded region = 10.5 cm

Correct answer is (c) 16 cm

We know that the radius and tangent are perpendicular at their point of contact In right triangle AOP

AO² = OP² + PA²

\(\Rightarrow\) 10² = 6² + PA²

\(\Rightarrow\) PA² = 64

\(\Rightarrow\) PA = 8 cm

Since, the perpendicular drawn from the center bisect the chord

∴ PA = PB = 8 cm

Now, AB = AP + PB = 8 + 8 = 16 cm

Since,

AB is a diameter 

Then, 

∠ADB = 90° ...(i) 

(Angle in semi-circle) 

Since, 

AC is a diameter 

Then, 

∠ADC = 90° ... (ii) 

(Angle in semi-circle) 

Adding (i) and (ii), we get 

∠ADB + ∠ADC = 90° + 90° 

∠BDC = 180° 

Then, 

BDC is a line 

Hence, 

The circles on any two sides intersect each other on the third side.

Given that, 

ABCD is a cyclic quadrilateral in which 

(i) Since, 

EA = ED 

Then, 

∠EAD = ∠EDA  ...(i) 

(Opposite angles to equal sides) 

Since, 

ABCD is a cyclic quadrilateral 

Then, 

∠ABC + ∠ADC = 180° 

But, 

∠ABC + ∠EBC = 180° 

(Linear pair) 

Then, 

∠ADC = ∠EBC  ...(ii) 

Compare (i) and (ii), we get 

∠EAD = ∠EBC ...(iii) 

Since, 

corresponding angles are equal 

Then, 

BC ǁ AD 

(ii) From (iii), we have 

∠EAD = ∠EBC 

Similarly, 

∠EDA = ∠ECB .... (iv) 

Compare equations (i), (iii) and (iv), we get 

∠EBC = ∠ECB 

EB = EC 

(Opposite angles to equal sides)

We know that ABCD is a cyclic quadrilateral

So we get

∠ABC + ∠ADC = 180^o

By substituting the values

92^o + ∠ADC = 180^o

On further calculation

∠ADC = 180^o – 92^o

By subtraction

∠ADC = 88^o

We know that AE || CD

From the figure we know that

∠EAD = ∠ADC = 88^o

We know that the exterior angle of a cyclic quadrilateral = interior opposite angle

So we get

∠BCD = ∠DAF

We know that

∠BCD = ∠EAD + ∠EAF

It is given that ∠FAE = 20^o

By substituting the values

∠BCD = 88^o + 20^o

By addition

∠BCD = 108^o

Therefore, ∠BCD = 108^o.

∠DBC = 70°, ∠BAC = 30°, then ∠BCD =? 

AB = BC, then ∠ECD = ? 

∠DAC and ∠DBC are angles in same segment. 

∴ ∠DAC = ∠DBC = 70° 

∴ ∠DAC = 70° 

ABCD is a cyclic quadrilateral. 

∴ Sum of opposite angles is 180°. 

∠DAB + ∠DCB = 180 

100 + ∠DCB = 180 

[∵ ∠DAC + ∠BAC = ∠DAB 70 + 30 = 100] 

∠DCB = 180 – 100 

∴ ∠DCB = 80 

∠DCB = ∠BCD = 80 

∴ ∠BCD = 80 

In ∆ABC, AB = AC, 

∴ ∠BAC = ∠BCA = 30° 

∠BCA = 30° 

∠ECD = ∠BCD – ∠BCA = 80 – 30 

∴ ∠ECD = 50°.

Given that,

ABCD is cyclic quadrilateral in which AB = DC 

To prove : 

AC = BD 

Proof : 

In ΔPAB and ΔPDC, 

AB = DC (Given) 

∠BAP = ∠CDP 

(Angles in the same segment) 

∠PBA = ∠PCD 

(Angles in the same segment) 

Then,

 ΔPAB = ΔPDC ... (i)

(By c.p.c.t)

PC = PB ... (ii)

(By c.p.c.t)

Adding (i) and (ii), we get 

PA + PC = PD + PB 

AC = BD

(i) We know that the angles in the same segment are equal

So we get

∠BDC = ∠BAC = 40^o

Consider △ BCD

Using the angle sum property

∠BCD + ∠BDC + ∠DBC = 180^o

By substituting the values

∠BCD + 40^o + 60^o = 180^o

On further calculation

∠BCD = 180^o – 40^o – 60^o

By subtraction

∠BCD = 180^o – 100^o

So we get

∠BCD = 80^o

(ii) We know that the angles in the same segment are equal

So we get

∠CAD = ∠CBD

So ∠CAD = 60^o

Answer is A. 30 cm

Given: 

AQ = 4 cm 

BR = 6 cm 

PC = 5 cm 

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

By the above property, 

AR = AQ = 4 cm (tangent from A) 

BR = BP = 6 cm (tangent from B) 

CP = CQ = 5 cm (tangent from C) 

Now,

Perimeter of ∆ABC = AB + BC + CA 

⇒ Perimeter of ∆ABC = AR + RB + BP + PC + CQ + QA 

[∵ AB = AR + RB 

BC = BP + PC 

CA = CQ + QA] 

⇒ Perimeter of ∆ABC = 4 cm + 6 cm + 6 cm + 5 cm + 5 cm + 4 cm 

⇒ Perimeter of ∆ABC = 30 cm 

Hence, Perimeter of ∆ABC = 30 cm

∠OBC + ∠CBA = ∠OBA 

∠OBC + 30° = 50° 

∠OBC = 20° 

∠OBC + ∠BAC = ∠OBC + ∠CAB 

= 20° + 110° 

= 130°

ΔABC is an equilateral triangle. (Given) 

Each angle of an equilateral triangle is 60 degrees. 

In quadrilateral ABEC: 

∠BAC + ∠BEC = 180° (Opposite angles of quadrilateral) 

60° + ∠BEC = 180° 

∠BEC = 180° – 60° 

∠BEC = 120°

Since, 

Triangle ABC is an equilateral triangle 

∠BAC = 60° 

∠BAC + ∠BEC = 180°

(Opposite angles of quadrilateral) 

60° + ∠BEC = 180° 

∠BEC = 120°

We have, 

∠PQR = 35° 

∠PQR + ∠PRQ = 35° 

(Angle opposite to equal sides) 

In triangle PQR, 

By angle sum property,

∠P + ∠Q + ∠R = 180° 

∠P + 35° + 35° = 180° 

∠P = 110° 

Now, 

∠QSR + ∠QTR = 180° 

110° + ∠QTR = 180° 

∠QTR = 70°

Given : A circle with center O and AC as a diameter and AB and BC as two chords also AT is a tangent at point A

To Prove : ∠BAT = ∠ACB

Proof :

∠ABC = 90° [Angle in a semicircle is a right angle]

In △ABC By angle sum property of triangle

∠ABC + ∠ BAC + ∠ACB = 180 °

∠ACB + 90° = 180° - ∠BAC

∠ACB = 90 - ∠BAC [1]

Now,

OA ⏊ AT [Tangent at a point on the circle is perpendicular to the radius through point of contact ]

∠OAT = ∠CAT = 90°

∠BAC + ∠BAT = 90°

∠BAT = 90° - ∠BAC [2]

From [1] and [2]

∠BAT = ∠ACB [Proved]

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

So we get

∠AOB = 2 ∠ACB

From the figure we know that

∠ACB = ∠DCB

It can be written as

∠AOB = 2 ∠DCB

We also know that

∠DCB = ½ ∠AOB

By substituting the values

∠DCB = 40^o/2

By division

∠DCB = 20^o

In △ DBC

Using the angle sum property

∠BDC + ∠DCB + ∠DBC = 180^o

By substituting the values we get

100^o + 20^o + ∠DBC = 180^o

On further calculation

∠DBC = 180^o – 100^o – 20^o

By subtraction

∠DBC = 180^o – 120^o

So we get

∠DBC = 60^o

From the figure we know that

∠OBC = ∠DBC = 60^o

So we get

∠OBC = 60^o

Therefore, ∠OBC = 60^o.

It is given that △ ABC is equilateral

We know that

∠BAC = ∠ABC = ∠ACB = 60^o

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

∠BDC = ∠BAC = 60^o

So we get

∠BDC = 60^o

(ii) We know that the opposite angles of a cyclic quadrilateral are supplementary

So we get

∠BAC + ∠BEC = 180^o

By substituting the values

60^o + ∠BEC = 180^o

On further calculation

∠BEC = 180^o – 60^o

By subtraction

∠BEC = 120^o

It is given that O is the centre of a circle and ∠BOD =150^o

We know that

Reflex ∠BOD = (360^o – ∠BOD)

By substituting the values

Reflex ∠BOD = (360^o – 150^o)

By subtraction

Reflex ∠BOD = 210^o

Consider x = ½ (reflex ∠BOD)

By substituting the value

x = 210/2

So we get

x = 105^o

We know that

x + y = 180^o

By substituting the values

105^o + y = 180^o

On further calculation

y = 180^o – 105^o

By subtraction

y = 75^o

Therefore, the value of x is 105^o and y is 75^o.

It is given that ABCD is a cyclic quadrilateral

We know that the opposite angles of a cyclic quadrilateral are supplementary

So we get

∠A + ∠C = 180^o

By substituting the values

∠A + 100^o = 180^o

On further calculation

∠A = 180^o – 100^o

By subtraction

∠A = 80^o

Consider △ ABD

Using the angle sum property

∠A + ∠ABD + ∠ADB = 180^o

By substituting the values

80^o + 50^o + ∠ADB = 180^o

On further calculation

∠ADB = 180^o – 80^o – 50^o

By subtraction

∠ADB = 180^o – 130^o

So we get

∠ADB = 50^o

Therefore, ∠ADB = 50^o.

∠DBA = ∠DCA = 58° …(1) [Angles in same segment] 

ABCD is a cyclic quadrilateral : 

Sum of opposite angles = 180 degrees 

∠A +∠C = 180° 

75° + ∠C = 180° 

∠C = 105° 

Again, 

∠ACB + ∠ACD = 105° 

∠ACB + 58° = 105° 

or ∠ACB = 47° …(2) 

Now, 

∠ACB = ∠ADB = 47° [Angles in same segment] 

Also, ∠D = 77° (Given) 

Again From figure, ∠BDC + ∠ADB = 77° 

∠BDC + 47° = 77° 

∠BDC = 30° 

In triangle DPC 

∠PDC + ∠DCP + ∠DPC = 180° 

30° + 58° + ∠DPC = 180° 

or ∠DPC = 92° . Answer!!

∠AO'B = 50° 

Since, 

both the triangle are congruent so their corresponding angles are equal. 

∠AOB = AO'B = 50°

Now,

∠APB = \(\frac{AOB}{2}\)

∠APB = \(\frac{50}{2}\)

= 25°

∠DBA = ∠DCA = 58° 

(Angles on the same segment) 

In triangle DCA,

∠DCA + ∠CDA + ∠DAC = 180° 

58° + 77° + ∠DAC = 180° 

∠DAC = 45° 

∠DPC = 180° - 58° - 30° 

= 92°

The length of tangent from an external point P on a circle with centre O is always less than OP.

True

Correct answer is (A) 60 cm²

If angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°, then OP = a√2 .

True