Chapter-9, CIRCLES

Explanation:

Provided, two concentric circles have radii of 4 and 5 cm.

Each chord of a circle that is tangent to another circle needs to have its length determined.

Think of two circles that are in close proximity.

OA = 4 cm

OB = 5 cm

The figure shows that OA BC.

Triangle OAB has a right triangle shape with A at a right angle.

Pythagorean Theorem states that OB² = OA² + AB² (5)² = (4)²5 = 2 + AB² 16 + AB² AB² = 25 - 16

AB² = 9

Square root equals AB = 3 cm.

Knowing that BC = 2AB

Therefore, BC = 2(3) BC = 6 cm.

Consequently, each chord is 6 cm long.

Explanation:

Since AOB = 125°, we must determine the size of angle COD.

We can see from the figure that ABCD is a quadrilateral enclosing the circle.

We are aware that the opposing sides of a quadrilateral enclosing a circle subtend additional angles there.

AB and CD are the opposing sides.

<AOB + <COD = 180° + 125° +< COD = 180° +< COD = 180° <COD = 180° - 125° <COD = 55°

Consequently, 55° is the angle COD measurement.

Explanation:

Because AB is a circle chord.

The diameter of the circle is AOC.

Moreover, ACB = 50°

The circle's tangent is AT.

The magnitude of the angle BAT must be determined.

We are aware that a right angle exists in a semicircle.

People are aware that the sum of a triangle's three interior angles is always 180 degrees.

Triangular AOB

So, ∠CBA = 90°

∠CAB + ∠CBA + ∠ACB = 180°

∠CAB + 90° + 50° = 180°

∠CAB + 140° = 180°

∠CAB = 180° - 140°

∠CAB = 40° ------------------- (1)

We are aware that a circle's radius at the point of contact is perpendicular to the tangent.

The figure shows,

OA ⟂ OT

So, ∠CAB + ∠BAT = 90°

 For (1) in the preceding expression,

40° + ∠BAT = 90°

∠BAT = 90° - 40°

∠BAT = 50°

Consequently, the angle BAT is 50° in size.

Explanation:

The right response is (A) 60 cm2.

Draw an O-centered circle with a radius of 5 cm first. P is a point that is 13 cm away from O. PQ and PR are depicted as a pair of tangents.

Quadrilateral PQOR is subsequently produced.

QP is a tangent line because.

∴OQ⊥QP

OP2OQ2+QP2132

=52+QP2xQP2=169-25=144

QP=12 cm Now, the area of OQP=1/2QPxQO

=1/2xQP12x5=30 cm2 in right-angled PQO∴Quadrilateral area QORP=2 and OQP=230=60 cm2.

Explanation:

Because AB is a circle's diameter

The circle's radius is 5 cm.

Tangent XAY is drawn at the circle's end A.

The chord CD's length parallel to XY and 8 cm away from A must be determined.

From the figure, the chord is represented by XAY, which is the circle's tangent.

Chord CD is 8 cm from A, and its radius is AO = OC = 5 cm.

AE = AO + OE 8 = 5 + OE 

OE = 8 - 5

OE = 3 cm, as we are aware.

It is clear that OE is parallel to the chord CD.

So, ∠ OEC = 90°

We are aware that the chord is divided by the perpendicular taken from its center.

that is, CE = DE

OED is a right triangle having a right angle at E in the triangle OED.

According to the Pythagorean Theorem, OC² = OE² + CE²,

(5)² = (3)² + CE²,

and CE² = 25 - 9 CE².

CE equals 4 cm when squared.

Chord length CE² = 2(CE) = 2(4) = 8 cm

Consequently, the chord's length is 8.

Explanation:

Assuming that AT is tangent to the circle with origin O

The OT is 4 cm long.

Moreover, OTA = 30°

We must determine AT's length.

O, please join A.

The circle's radius will be OA.

We are aware that a circle's radius at the point of contact is perpendicular to the tangent.

So, ∠OAT = 90°

AOT is a right triangle with A at its right angle in the triangle AOT.

Pythagorean Theorem states that cos 30° = AT/OT.

Cos 30° = 3/2 utilizing trigonometric ratios of angles.

So, √3/2 = AT/4

When multiplied by themselves, 43/2 equals AT-AT = 23 cm

The AT thus has a 23 cm measurement.

Explanation:

O obviously sits at the heart of a circle.

As a chord, PQ

A tangent at P, PR, forms a 50° angle with PQ.

We must ascertain the POQ angle's measurement.

We are aware that at the point of contact, the radius of a circle is perpendicular to the tangent.

i.e., ∠OPR = 90°

Assuming RPQ = 50°

We are aware that 90° = OPQ + 50° and OPR = OPQ + RPQ.

∠OPQ = 90° - 50°

∠OPQ = 40°

We are aware that if two sides are equal, then two angles must also be equal.

The figure shows that OP = OQ = Radius.

Therefore, OPQ = OQP = 40°

Triangle POQ teaches us that the sum of a triangle's three interior angles is always 180 degrees.

The formula is 180°, 40°, and 40°.+ POQ = 180°

POQ = 180° - 80° + POQ = 180°

Consequently, POQ = 100°.

Explanation:

Since the circle with center O has tangents PA and PB.

Moreover, <APB = 50°

We must determine the size of the OAB angle.

We are aware that the tangents to a circle through an outside point are equal.

Tangents, then PA = PB

Triangular PAB

PAB is an isosceles triangle because PA = PB.

Two sides and two angles are equal in an isosceles triangle.

that is, PA = PB

Moreover, <PAB = <PBA —------------------ (1)

We are aware that the sum of a triangle's three interior angles is always 180 degrees.

Now, from (1), 50° + PAB + PAB = 180°. 

APB + PAB + PBA = 180°.

2∠PAB = 180° - 50°

2∠PAB = 130°

Since PAB = PBA and PBA = 65°, PAB = 130°/2.

We are aware that a circle's radius is parallel.

Explanation:

 A circle with a radius of 3 cm and an angle of 60° has two tangents drawn to it. We must determine the length of each tangent.

The figure shows,

Let PA and PC represent the circle's tangents.

PA and PC are 60° inclined.

So, ∠APC = 60°

We are aware that the tangents to a circle through an outside point are equal.

So, PA = PC

Triangle OAP and OCP have the formula PA = PC.

OA = OC = circle radius

OP stands for the common side.

Triangles OAP and OCP are similar according to the SSS criterion since we know that a circle's radius is perpendicular to the tangent at its point of contact.

<OAP = <OCP = 90°, therefore.

APC = <OAP + <OCP since OA = OC = radius and <OAP = <OCP.

<OAP = 60°/2 <OAP = 30°, hence 2 <OAP = 60°.

OAP is a right triangle with A at a right angle in the triangle OAP.

30° tan = OA/AP

By comparing angles using trigonometry.

Explanation:

Assuming that PQR is the tangent at Q to a circle with O as its center.

A chord similar to PR is AB.

Moreover,< BQR = 70°

We must determine the AQB angle's measurement.

We are aware that at the point of contact, the radius of a circle is perpendicular to the tangent.

<OQP = 90°, thus <OQR = <OQP.

According to the graph, <OQR = <OQB + <BQR 90° =< OQB + 70°

∠OQB = 90° - 70°

∠OQB = 20°

Allow the line from O to intersect AB at C.

In light of triangles AQM and BQM, we know that the chord is divided in half by the perpendicular running from the circle's centre.

Therefore, AC = BC, <QMA = 90°, <QMB = common side, and QM = QM.

The triangles AQM and BQM are comparable using the SAS criterion.

Therefore, <MQA = <MQB = 20°

From the equation, "<AQB = <MQA + <MQB," 

"<AQB = 20° + 20°,"

and <"AQB = 40°."

Consequently, the angle AQB is measured at 40°.

Explanation:

Since a chord, AB subtends an angle of 60 at a circle's center, i.e., "<AOB=60," and since OA = OB = the circle's radius, "<OAB="<OBA=60,"

Drawn are the tangents at positions A and B. which come together at C.

We are aware. OA⊥ AC, OB ⊥ BC.

∴ ∠OAB=90∘,∠OBC=90∘

⇒ ∠OAB+∠BAC=90∘

And ∠OBA+∠ABC=90∘

⇒ ∠ABC=90∘−60∘=30∘

In △ABC∠BAC+∠CBA+∠ACB=180∘

[Because a triangle's internal angles add up to 180],

⇒ ∠ACB=180∘−(30∘+30∘)=120∘.

Explanation:

Let P be a point outside the circle that is moving away from O.

Create a tangent on the circle PA such that PA > r.

Let's now have a look at point P1 on the tangent AP that is close to contact point A of the tangent PA, P1A AO.

PA and P1A's tangent lengths are longer and shorter than OA's radius, respectively.

as a result, the radius may or may not be bigger than the length of the tangent.

The claim is untrue as a result.

Explanation:

true

Because the hypotenuse of a right-angled triangle (PQO) formed by the tangent and the circle at point Q is the greatest in a triangle, the length of PQ is always less than PO.

14. The angle between two tangents to a circle may be 0°. 

Explanation:

Untrue

Only one tangent can be traced from the place where we are on the circle.

Only when two tangents coincide will the angle between them be 0 degrees. it is unattainable.

Therefore, a circle's two tangents cannot have an angle of 0 degrees.

Explanation:

Untrue

Only one tangent can be traced from the place where we are on the circle.

Only when two tangents coincide will the angle between them be 0 degrees. it is unattainable.

Therefore, a circle's two tangents cannot have an angle of 0 degrees.

Explanation:

Consider a circle with a radius of and tangents PT and PR with an angle between them of 90 degrees.

In OTP and ORP, TO = OR [same circle radius]

OP = OP [common]

TP = PR (circles have equal tangents through external points)

ORP [By SSS Criterion] OTP

<OPR [c.p.c.t] = <TPO (1) Given that <TPR = 90°

<TPO + <OPR equals 90°

<TPO plus <TPO equals 90° [from 1]

∠TPO = 45°

OT⏊ TP< OTP = 90° POT is a right-angled triangle because the tangent at any point on the circle is perpendicular to the radius through the point of contact.

We are aware that sin is the perpendicular and hypotenuse sin. TPO = a/OP + OT/OP

45° of sin = a/OP

1/√2 = a/OP

OP = a√2

Thus, that is accurate.

Explanation:

Un true

Two tangents, PT and PR, are drawn starting at point P.

Given:Radius = an

Line OP also divides the <RPT <TPO=RPO=30

OTPT, too.

OTP with a right angle.

Sin30∘=OTOP 

⇒12=aOP 

⇒OP=2a.

Explanation:

Assume a circle in which ABC is an isosceles triangle within the circle, where AB = AC and EF is a tangent that passes through point A on the circle.

To Establish: EF || BC

Building: Join OA, OB, and OC

Evidence: AB = AC [Given]

Angles that are perpendicular to equal sides are equal (<ACB = <ABC) [1]

Since we are aware that the angle formed by the chord in the alternate segment equals the angle generated by the tangent, we can conclude that <EAB = <ACB --- [2]

<EAB equals <ACB from [1] and [2].

If the alternative interior angles of two lines are equal, then they are parallel EF || BC.

The assertion is accurate as a result.

Explanation:

Consider n circles with centres C1, C2, C3,..., Cn as S1, S2, S3,..., Sn.

At point A, PQ is a common tangent to all the circles, and it is shared by all circles.

We are aware that the radius across the point of contact is perpendicular to the tangent at any point on the circle.

C1A ⏊ PQ

C2A ⏊ PQ

C3A ⏊ PQ

CnA ⏊ PQ

As PA may or may not be equal to AQ, C1 C2 C3... Cn does not lie on the perpendicular bisector in this case, but rather on the perpendicular line to PQ. The claim is untrue as a result.

Explanation:

True

With the endpoints P and Q of a line segment PQ as our starting point, we formed two circles with centers C1 and C2.

We are aware that the center of a circle is always traversed by the perpendicular bisector of a circle's chord.

Therefore, C1 and C2 are included in the perpendicular bisector of PQ.

Similar to this, the centers of all circles that pass through PQ will lie on their perpendicular bisectors.

Explanation:

Arc BC subtends <DOC at the circle's middle and <BAC at the remaining portions of the circle after first joining OC <BCD = BAC = 30° (angles in the alternate segment).

∠BOC = 2 ∠BAC = 2 × 30° = 60°

In triangle OCD

∠BOC = ∠DOC = 60°

∠OCD = 90°

OC is perpendicular to CD

∠DOC + ∠ODC = 90°

60° + ∠ODC = 90°

∠ODC = 90° - 60° = 30°

In triangle BCD

∠ODC = ∠BDC = ∠BCD = 30°

So BC = BD

The assertion is accurate as a result.

Explanation:

According to the inquiry,

The larger circle's radius is 5 cm. 

and the 8 cm tangent AC to the smaller circle.

We are aware that the midpoint of the tangent from the figure is M, and that the tangent is perpendicular to the radius through the point of tangency.

therefore AM=4 cm

AMO is the right-angled triangle as a result, with OM being the radius of the smaller circle.

In light of the Pythagorean theorem,

OA2=OM2+AM2

=> 52=OM2+42

=> OM=3

Hence, the smaller circle's radius is 3 cm.

Explanation:

Provided that, PQ and PR are the two tangents created by connecting an outside point to an O-centered circle.

We must demonstrate QORP's status as a cyclic quadrilateral.

O is the circle's center according to the illustration.

The circle radius is equal to OR and OQ.

The two tangents to the circle from an outside point P are PR and PQ.

We are aware that at the point of contact, the radius of a circle is perpendicular to the tangent.

Consequently, OR ⟂ PR, OQ ⟂ PQ, and <OQP = 90°

We are aware that a quadrilateral's total interior angles are always equal to 360 degrees.

With regard to quadrilateral PQOR,

∠OQP + ∠QOR + ∠ORP + ∠RPQ = 360°

90° + ∠QOR + 90° + ∠RPQ = 360°

180° + ∠QOR + ∠RPQ = 360°

∠QOR + ∠RPQ = 360° - 180°

So, ∠O + ∠P = 180°

Here, opposing angles complement one another.

PQOR is a cyclic quadrilateral as a result.

Explanation:

Assumedly, BC and BD are the tangents taken from point B on the outside of an O-centered circle.

Moreover, <DBC = 120°

We must demonstrate that BO = 2BC and that BC + BD = BO.

We are aware that at the point of contact, the circle's radius is perpendicular to the tangent.

Therefore, OC⟂  BC and OD⟂  BD

Triangles OCB and ODB are represented by OB = OB = common side.

We are aware that all tangents to a circle drawn via an outside point are equal.

Thus, BC = BD.

In addition,< OCB = <ODB = 90°

The triangles OCB and ODB are comparable using RHS.

Angles OBC and OBD are equal thanks to comparable triangular portions.

Therefore, <OBC = <OBD = 60°

Triangle OCB has C at a right angle, making it a right triangle.

Pythagorean theorem states that cos 60° = BC/OB.

Cos 60° = 1/2 according to the trigonometric ratio of angles.

then 1/2 = BC/OB

OB = 2BC

As a result, it is established that OB = 2BC.

Explanation:

Two tangents are given, and PQ and PR are drawn from an exterior point P to a circle with O as its center.

To demonstrate: A circle's middle rests on the angle formed by the intersection of two lines.

Construction Combined OR and OQ.

In ∠POR and ∠POQ

∠ PRO = ∠ PQO = 90∘

Tangents are perpendicular to the radius across the point of contact at any point on a circle.

Radii of some circle: OR = OQ

since OP is widespread.

PQO≅ [RHS] PRO

<RPO hence equals <QPO [by CPCT]

O thus lies on the angle that divides PR and PQ.

So it was proved.

Explanation:

Because AB and CD are tangents to two circles with different radii,

the case

Create AB and CD so they meet at P during construction.

Proof: PC > PA

[Drawn tangents from an internal point to a circle have identical lengths]

PB = PD as well.

[Drawn tangents from an internal point to a circle have identical lengths]

AB = CD; PA - PB = PC - PD

So it was proved.

Explanation:

Given AB and CD are tangents to two equal-radius circles.

showing that AB = CD

Construction Come to OA, OC, O' B, and O'D.

Proof Here: <OAB=90∘

A circle's tangents are always parallel to the radius via the point of contact.

Consequently, AC is a line.

Also. ∠OAB+∠OCD=180∘ ∴AB∥CD

BD is also a straight line.

Moreover, <O′BA=O′DC=90∘.

Additionally, AC = BD (two circles' radii are equal).

ABCD is a quadrilateral. <A=<B=<C=D=90∘ and AC = BD afterwards. The rectangle's opposite sides are equal when AB=CD.

Explanation:

We know that the tangents of two circles AB and CD are similar.

At E, the tangents come together.

We must demonstrate that AB equals CD.

E is the external point in the illustration.

EA, EC, EB, and ED are the tangents that are traced through external point E.

We are aware that all tangents to a circle drawn via an outside point are equal.

EA equals EC, therefore. ---------------- (1) EB = ED ---------------------- (2) By combining (1) and (2), EA + EB equals EC + ED.

The figure shows that AB = EA + EB.

CD equals EC plus ED.

Thus, AB = CD.

As a result, it is established that AB = CD.

Explanation:

Given are a circle, a chord (PQ), and a tangent (MRN) at R, such that (QPMRN).

To demonstrate: R bisects the PRQ arc.

Build by combining RP and RQ.

In order to prove that 1=2, the chord RP subtends 1 with tangent MN and 2 in alternate segments of the circle.

MRN∣∣PQ

[Alternate Interior Angles] <1 =< 3

⇒∠2=∠3 

PR=RQ [Sides against equaling s in RPQ]

In a circle, equal chords subtend equal arcs, so

RQ=arc PR

or the arc PRQ is bisected by R. So it was proved.

Explanation:

We must demonstrate that the tangents formed at a circle's chord's endpoints form equal angles with the chord.

O is the circle's center according to the figure.

A chord of the circle is AB.

The tangents that are drawn at the extremities of a circle's chord are PA and PB.

At P, the tangents come together.

Come to OA and OB.

The formula OA = OB = circle radius.

Triangle OAB has the formula: OA = OB = circle radius.

OAB is an isosceles triangle since its two sides are equal.

We are aware that a triangle's equal sides are opposing equal angles.

Thus, <OAB =< OBA.

i.e., ∠1 = ∠2 ------------- (1) We are aware that at the point of contact, the circle's radius is perpendicular to the tangent.

As a result, OA ⟂PA and OB⟂  PB <OAP = <OBP = 90°, or

<2 +< 3 =< 1 + 4------------- (2) By replacing (1) in (2), <1 +< 3 = <1 +< 4

< 1 + <3 - <1 = <4 

<3 = <4 

resulting in< PAB = <PBA.

Explanation:

The right answer is option A, where l is the tangent to the circle with center O at B, AB is the diameter of the circle, and Q is the location of chord CD.

Because: CQ=DQ

AB⊥l AND I CD because l is tangent to the circle.

OQ⊥CD, which is a chord of the circle, and CD

OQ divides CD. 

Consequently, AB bisects CD.

Hence, CQ=QD.

Therefore, a circle's diameter AB cuts through all chords that are perpendicular to the tangent at point A.

Therefore, the assertion is accurate.

Explanation:

Provided, a circle is bounded by a hexagon ABCDEF.

That AB + CD + EF = BC + DE + FA must be demonstrated.

We are aware that a circle's tangents through an outside point are equal.

As seen in the picture, the circle's tangent through point A is represented by AR and AM.

Thus, AM = AR. -------------------- (1)

In a similar manner, the tangents through point B are BM = BN. ------------------------ (2)

The tangents at point C are CO = CN. ------------------------ (3)

The tangents at point D are DO = DP. ------------------------ (4)

The tangents passing through E EQ = EP ------------------------ The tangents across point F FQ = FR are (5). ------------------------ (6)

AM + BM + CO + DO + EQ + FQ = AR + BN + CN + DP + EP + FR after adding (1) to (6).

(AM + BM) + (CO + DO) + (EQ + FQ) rearranged into (AR + FR) + (BN + CN) + (EP + DP)

The figure shows,

AM + BM = AB

CO + DO = CD

EQ + FQ = EF

AR + FR = AF

BN + CN = BC

EP + DP = ED

So, AB + CD + EF = AF + BC + ED

Therefore, AB + CD + EF = BC + DE + FA.

Explanation:

Provided, s represents a triangle ABC's semiperimeter.

Triangle sides BC = a, CA = b, and AB = c.

At D, E, and F, a circle intersects sides BC, CA, and AB.

We must demonstrate that BD = s - b.

Half the sum of the triangle's sides makes up the semiperimeter.

The formula for a triangle's semiperimeter is s = (a + b + c)/2.

We are aware that a circle's tangents through an outside point are equal.

AF and AE are the tangents through point A.

AF = AE, therefore. -------------------- (1) Point B's tangents are BF and BD.

BF = BD, then. -------------------- (2) CD and CE are the tangents through point C.

Thus, CD equals CD. -------------------- (3)

Semiperimeter, defined as s = (AB + BC + CA)/2

AB + BC + CA = 2s

According to the diagram, AB = AF + FB, BC = BD + DC

CE + EA = CA

2s = AF, FB, BD, DC, CE, and EA

2s = AE + AE + CE + CE + BD + BD from (1, 2) and (3)

2AE = 2CE + 2BD

AE + CE + BD = 2s

S = AE + CE + BD when the common word is eliminated.

The figure shows that AC = AE + CE.

Thus, s = AC + BD.

Assuming AC = b

s = b + BD

BD = s - b

Thus, it is established that BD = s - b.

Explanation:

Given that a circle with an exterior point P and a center O has two tangents drawn to it at PA and PB, respectively.

A tangent is made at point E on the circle, which intersects PA and PB at C and D.

Assuming PA = 10 cm

The triangle PCD's perimeter must be determined.

We are aware that any tangents to a circle that passes through an outside point are equal.

The tangents drawn through point P in the figure are PA and PB.

Thus, PA = PB. (1) The tangents to the circle passing through C are CA and CE.

Thus, CA = CE. -------------- (2) The circle's tangents through D are DB = DE

Thus, DB = DE. -------------- (3)

In light of triangular PCD,

Perimeter equals PC + PD + CD

The figure shows that CD = CE + DE.

Perimeter then equals PC + PD + CE + DE.

Perimeter = PC + PD + CA + DB from (2) and (3)

Perimeter = PC + CA + PD + DB upon rearrangement

The figure shows that PC + CA equals PA.

PB = PD + DB

PA = PB by (1)

Therefore, the perimeter is equal to PA plus PB.

The perimeter is equal to two PAs.

Assuming PA = 10 cm

So, 2PA = 2(10) = 20 cm

Consequently, the triangle PCD's circumference is 20 cm.

Explanation:

Given that AB is a circular chord with O as its center.

AOC stands for "circle diameter."

At point A, AT is the tangent.

We must demonstrate that BAT = ACB.

We are aware that a semicircle's angle is always equal to 90 degrees.

So, ∠CBA = 90°

With respect to triangle CBA, we are aware that the total of a triangle's three inner angles is always 180 degrees.

∠CBA + ∠BAC + ∠ACB = 180°

90° + ∠BAC + ∠ACB = 180°

∠BAC + ∠ACB = 180° - 90°

∠BAC + ∠ACB = 90°

∠ACB = 90° - ∠BAC ----------------- (1)

They are aware that at the place of contact, the circle's radius is perpendicular to the tangent.

So, OA ⟂ AT

∠OAT + ∠CAT = 90°

From the figure,

∠CAT = ∠BAT + ∠BAC

90° = ∠BAT + ∠BAC

∠BAT = 90° - ∠CAB ---------------- (2)

Comparing (1) and (2), it can be shown that since the RHS are the same, <ACB =< BAT.

Explanation:

Provided are two circles with radii of 3 cm and 4 cm and centers of O and O.

P and Q are the intersection points of two circles.

The tangents to the two circles are OP and O'P.

The common chord PQ's length must be determined.

We are aware that at the point of contact, the radius of a circle is perpendicular to the tangent.

So, ∠OPO' = 90°

Triangle OPO' has P at a right angle, making it a right triangle.

Using Pythagoras's Theorem

(0OP)² + (O'P)² = (OP')²

The figure shows that OP = 3 cm for the circle's radius.

O'P = 4 cm (OO') (other circle's radius).² = (3)² + (4)² (OO’)² = 9 + 16 (OO’)² = 25

Square root equals OO' = 5 cm.

Assume ON = x cm.

Thus, O'N = 5 - x cm.

triangular ONP

(OP)2 equals (ON) according to Pythagoras' Theorem.² + (PN)² (3)² = (x)² + (PN)²

9 = x² + (PN)² PN² = 9 - x² -------------------------- (1)

O'P = O'N in the triangle O'NP.² + (PN)² (4)² = (5 - x)² + PN² PN² = 16 - (5 - x)²

Algebraic identity demonstrates that (a - b)2 = a2 - 2ab + b2.

PN² = 16 - (25 -10x + x²)

PN² = 16 - 25 + 10x - a²

PN² = -x² + 10x - 9 ----------------- (2) When (1) and (2) are compared, 9 - x² = -x² + 10x - 9

9 = 10x - 9

10x = 9 + 9 10x = 18 x = 18/10 x = 1.8

Change the value of x.

in (1), PN² = 9 - (1.8)² PN² = 9 - 3.24

PN² = 5.76

Square root equals PN = 2.4 cm.

Knowing that PQ = 2PN

PQ = 2(2.4)

PQ = 4.8 cm

Consequently, the chord PQ is 4.8 cm long.

Explanation:

Since B is at a right angle, ABC is a right triangle.

Draw a circle with diameter AB that intersects the hypotenuse AC at point P.

We must demonstrate that BC is divided in half by the circle's tangent at P.

We are aware that a semicircle's angle is always equal to 90 degrees.

So, ∠APB = 90°

When two lines come together, they generate a pair of linear angles.

<BPC = 90° by the linear pair of angles.

So, ∠3 + ∠4 = 90° -------------------- (1)

Since B = 90°

We are aware that the total of a triangle's three interior angles is always 180 degrees, as in triangle ABC.

ACB + BAC + ABC = 180°

∠1 + 90° + ∠5 = 180°

∠1 + ∠5 = 180° - 90°

∠1 + ∠5 = 90° -------------------------- (2) We are aware that the angle formed by the chord in the alternate segment is equal to the angle formed by the tangent and chord of a circle.

So, ∠1 = ∠3 ------------------ (3) Replace (3) in (2), where< 3 +< 5 = 90° --------------- (3) Comparing (1) and (4), we find that <3 + <4 = <3 +< 5

Explanation:

Provided, the tangents drawn to a circle are PQ and PR.

Moreover, RPQ = 30°

The tangent PQ is paralleled by the chord RS.

The size of the angle RQS must be determined.

We are aware that all tangents to a circle drawn via an outside point are equal.

PQ = PR, so.

We are aware that a triangle has equal-sized angles on either side of its equal sides.

Consequently,< PRQ = <PQR ------------ (1)

We know that the sum of a triangle's three inner angles is equal when we consider the triangle PQR.

<PQR + <RPQ +< PRQ = 180° 

<PQR + 30° +< PRQ = 180°

<PQR + 30° +< PRQ = 180° - 30° = 150°

<PQR +< PQR equals 150° based on (1).

<PQR = 150°/2 and <PQR = 75° respectively.

PQ = PR, hence <PQR = <PRQ = 75°, indicating that the angle

The angle formed by the chord in the alternate segment is identical to the angle between a circle's tangent and chord.

Therefore, <PQR = <RSQ = 75°. ------------------- (2)

The opposite angles are equal since RS || PQ.

For example, <PQR = <QRS = 75° ------------------ (3)

With respect to (2) and (3),< RSQ = <QRS = 75°

We are aware that equal angles have equal sides.

Thus, SQ = RQ.

Inferring an isosceles triangle from this, QRS.

By the angle sum property, in the triangle <QRS, <RSQ + <QRS + <RQS = 180° 

75° + 75° + 180°

150° +< RQS = 180°

 <RQS = 180° - 150°.

Consequently,< RQS = 30°.

Explanation:

Because AB is the circle's center's diameter A circle's chord is O AC.

Moreover, <BAC = 30°

At point D, the tangent at C crosses extended AB.

We must demonstrate that BC equals BD.

Using the alternative segment theorem,

We are aware that the angle formed by the chord in the alternate segment is equivalent to the angle formed by the tangent and chord of a circle.

Thus, <BCD = <BAC.

Assuming <BAC = 30°

So, <BCD = 30°. ------------------- (1)

We are aware that a semicircle's angle is always exactly 90 degrees.

∠ACB = 90°

ABC is a triangle.

∠CAB + ∠CBA + ∠ACB = 180°

30° + ∠CBA + 90° = 180°

120° + ∠CBA = 180°

∠CBA = 180° - 120°

∠CBA = 60°

The figure shows,

a pair of parallel angles,

∠CBA + ∠CBD = 180°

60° + ∠CBD = 180°

∠CBD = 180° - 60°

∠CBD = 120°

Thinking of triangle CBD,

We are aware that the sum of a triangle's three interior angles is always 180 degrees.

∠CBD + ∠BCD + ∠BDC = 180°

120° + 30° + ∠BDC = 180°

150° + ∠BDC = 180°

∠BDC = 180° - 150°

∠BDC = 30° ----------------------------- (2)

Comparing (1) and (2),

∠BCD = ∠BDC = 30°

We are aware that equal angles have equal sides.

Thus, BC = BD.

Thus, it is established that BC = BD.

Explanation:

Provided, a circle's arc's centre is marked with a tangent.

We must demonstrate that the chord connecting the endpoints of the arc and the tangent drawn at the centre of the circle's arc are parallel to one another.

The circle's centre can be seen as O in the figure.

P is the centre of the arc APB.

PT is a circle's tangent at P.

We are aware that at the point of contact, the circle's radius is perpendicular to the tangent.

Therefore, OP⟂  PT 

OPT = 90°

Arc AP = Arc APB because P is the midway of arc APB. BP

<AOP equals <BOP

<AOM equals <BOM

When examining the triangles AOM and BOM, OA = OB = circle radius.

OM stands for the common side.

Moreover, <AOM = <BOM.

The triangles AOM and BOM are comparable using the SAS criterion.

The relationship between the sides and angles of two congruent triangles is referred to as "corresponding parts of congruent triangles," or cpct.

Using cpct, <AMO = <BMO —------------------ (1) The pair of linear angles <AMO + <BMO= 180°

Using (1), <AMO + <AMO equals 180°.

AMO = 90° and 180° for two.

Therefore, <AMO = <BMO = 90°

The relevant angles are equivalent, as far as we are aware.

The figure shows that <BMO = <OPT = 90°.

It suggests AB || PT.

As a result, the chord connecting the arc's endpoints and the tangent at the arc's midway are parallel.

Explanation:

Given that two circles with centers O and O' share the tangents AB and CD,

At E, the common tangents AB and CD come together.

The intersection of points O, E, and O must be demonstrated.

Join O'D, O'B, and OA.

Triangles AOE and EOC show that OE = OE = common side.

OA = OC = circle radius

We are aware that at the point of contact, the radius of a circle is perpendicular to the tangent.

EA and EC are tangents to the exterior point E.

Thus, AE = EC.

The triangles <AOE and <EOC are comparable using the SAS criterion.

The relationship between the sides and angles of two congruent triangles is referred to as "corresponding parts of congruent triangles," or cpct.

According to cpct, <CEO=A<EO ------------------ (1)

According to the figure,< AEC = <AEO + <CEO From (1),< AEC = 2 <AEO ------------------- (2) CD appears in the figure as a straight line.

The angle combination <AED + <AEC = 180° is linear.

2<AEO + 2<AED = 180° when deriving from (2).

AEO + (1/2)AED = 90° by 2.

∠AEO = 90° - (1/2)∠AED --------------- (3)

Triangles O'ED and O'EB show that O'B = O'D = radius of the circle.

"O'E" stands for "common side."

We are aware that any tangents to a circle that passes through an outside point are equal.

Thus, EB = ED.

The triangles O'ED and O'EB are comparable using the SSS criterion.

<O'ED = <O'EB via cpct. ------------ (4) The figure shows that DEB = O'EB + O'ED.

Deriving from (4), <DEB = <O'ED + <O'ED, and <DEB = 2<O'ED. ------------- (5)

The figure shows that AB is a straight line.

The angle combination <AED +< DEB = 180° is a linear pair.

AED + 2<O'ED = 180° according to (5).

When divided by 2, (1/2)<AED + <O'ED equals 90°.

<O'ED equals 90° - (1/2)<AED -------------- (6) By combining (3) and (6), <AEO +< AED + <O'ED = 90° - (1/2)<AED + <AED + 90° - (1/2)<AED = 180° - <AED + <AED = 180°

Therefore, <AEO + <AED + <O'ED = 180°.

Explanation:

Given that O is the center of a circle with a 5 cm radius.

T is a location outside the circle where OT equals 13 centimeters.

OT intersects the circle at E.

If AB is the tangent to the circle at E, we must determine its length.

Tangents to a circle through an outside point are equal, as far as we are aware.

The tangents to the circle from point A are AP and AE.

Thus, AP = AE.

Let x be the value of AP = AE.

We are aware that at the point of contact, the radius of a circle is perpendicular to the tangent.

Therefore, OP ⟂PT and OQ ⟂QT

In addition,< OPT = <OQT = 90°

Triangle OPT implies that P = 90°.

With P at a right angle, OPT is a right triangle.

Using Pythagoras's Theorem

OT² = OP² + PT²

(13)² = (5)² + PT²

169 = 25 + PT²

PT² = 169 - 25

PT² = 144

Square root equals PT = 12 cm.

Triangle AET has an angle of 90°, so E =

With E at a right angle, AET is a right triangle.

Using Pythagoras's Theorem

AT² = AE² + ET²

The figure shows,

PT = PA + AT
12 = x + AT

AT = 12 - x

(12 - x)² = x² + ET²

ET² = (12 - x)² - x²

Using algebraic identity,

(a - b)² = a² - 2ab + b²

So, ET² = 144 - 24x + x² - x²

ET² = 144 - 24x

The figure shows,

OT = OE + ET

13 = 5 + ET

ET = 13 - 5

ET = 8 cm

So, (8)² = 144 - 24x

64 = 144 - 24x

24x = 144 - 64

24x = 80

x = 80/24

x = 20/6

x = 10/3 cm

Given that AB = 2(AE),

AB = 2(10/3)

AB = 20/3

AB = 6.6 cm

Consequently, AB is 6.6 cm long.

Explanation:

O obviously sits at the heart of a circle.

The circle's diameter is AB.

A tangent to the circle at point C is formed by PC and the extension of AB to P.Assuming <PCA = 110o

We need to locate <CBA.

We are aware that a semicircle's angle is always equal to 90 degrees.

So, ∠BCA = 90°

We are aware that at the point of contact, the radius of a circle is perpendicular to the tangent.

that is, OC ⟂ PC

So, ∠OCP = 90°

According to the diagram, <PCA = <BCA + <PCB + 110° = 90° +< PCB + <PCB = 110° - 90° + 20°

The angle between a circle's tangent and chord is equal to the angle made by the chord in the alternate segment, according to the alternate segment theorem.

<CAB = <PCB

So, ∠CAB = 20°

With respect to triangle ABC, we are aware that the sum of a triangle's three internal angles is 180°.

<BCA +< CBA +< CAB equals 180° 

90° +< CBA + 20° equals 180° 

110° + <CBA equals 180°

CBA equals 180° - 110°

Consequently, CBA.

Explanation:

ABC is a triangle with an isosceles angle.

The distance between sides AB = AC is 6 cm.

A 9 cm-diameter circle contains ABC.

We must calculate the triangle's area.

O is the circle's center according to the illustration.

Get with OB and OC.

Let M represent the center of BC.

Thus, OM ⟂  BC

The median from the vertex of an isosceles triangle is parallel to the base.

since M is the middle of BC and ABC is an isosceles triangle.

AM ⟂ BC

Suppose AM = x and MB = CM = y.

Triangle AMB has M at a right angle, making it a right triangle.

Pythagoras's Theorem states that AB² = AM² + BM²(6)² = x² + y² x² + y² = 36 ---------------------------- (1)

Triangle OMB has M at a right angle, making it a right triangle.

OB² = OM² + BM² and OB = OC = 9 (circle radius).

According to the diagram, AM + MO = AO AO = circle's radius of x + OM = 9 OM = 9 - xSo, (9)² = (9 - x)² + y²

By the rule of algebra, (a - b)² = a² - 2ab + b².

81 is now equal to 81 - 18x + x² + y² x²+ y² = 18x. ----------------------------- (2)

Using (1) and (2) as a comparison, 36 = 18x x = 36/18 x = 2 cm

AM = 2 cm, therefore.

Add x = 2 to (1) and (2).² + y² = 36

 4 + y² = 36

 y² = 36 - 4

 y² = 32

Square rooting gives us y = 42 cm.

Area of a triangle equals base height (1/2)

Triangle ABC has an area of (1/2) BC AM BC = BM + CM = 4√2 + 4√2 BC = 8√2 cm.

Triangle ABC's surface area is equal to (1/2) 8√  2√  8 x2 square centimeters.

As a result, the triangle has an area of 8√2 square centimeters.

Explanation:

Given that O is the center of a circle with a radius of 5 cm, the distance between points A and O is 13 cm.

The tangents to the circle at P and Q are AP and AQ.

To intersect AP at B and AQ at C, a tangent BC is drawn at point R located on the minor arc PQ.

The triangle ABC's perimeter must be determined.

The figure shows,

Circle radius = OP = OQ = 5 cm

Assuming OA = 13 cm

We are aware that any tangents to a circle that passes through an outside point are equal.

Given that tangent to the circle from an outside point, A are AP and AQ.

Thus, AP = AQ. ------------------- (1)

We are aware that at the point of contact, the circle's radius is perpendicular to the tangent.

Therefore, OP ⟂PA and OQ ⟂AQ

In addition, <OPA = <OQA = 90°

Triangle OPA has P at a right angle, making OPA a right triangle.

Pythagoras's Theorem states that OA² = OP² + PA²

 (13)² = (5)².

PA²=144 

- 169 - 25 PA² = 2 + PA² 169 = 25 + PA²

Square root equals PA = 12 cm.

The total of a triangle's three sides equals its perimeter.

the triangular.

Triangle ABC's perimeter equals AB + BC + AC

BC = BR + CR according to the diagram.

Triangle ABC's perimeter is calculated as AB + BR + CR + AC.

We are aware that any tangents to a circle that pass through an outside point are equal.

BP and BR are tangents to the outside point B.

Thus, BP = BR.

CQ and CR are tangents to the exterior point C.

CQ = CR, therefore.

Triangle ABC's perimeter is calculated as AB + BR + CR + AC = AB + BP + CQ + AC.

Knowing that AP = AB + BP

So, Triangle ABC's perimeter equals AP + CQ + AC

The figure shows that AQ = AC + CQ.

Now, the ABC triangle's perimeter equals AP plus AQ.

Because of (1), AP plus AP = 2 AP.

Triangle ABC's perimeter is 2AP = 2(12) = 24 cm.

Triangle ABC's perimeter is 24 cm as a result.