NCERT Exemplar Solutions Class 10 Maths Chapter 10

Strategic angle. Spot the right triangle first. The instant you read ``chord of one circle tangent to the other'', draw the radius to the point of contact; it is perpendicular, so a right triangle with hypotenuse equal to the outer radius appears. The numbers 4,5 are two members of the famous 3,4,5 triple, so the half-chord is 3 without any heavy arithmetic.

1. Right triangle: legs 4 and AB, hypotenuse 5.
2. 3,4,5 triple gives the missing leg AB=3 cm at sight.
3. Chord =2× 3=6 cm.

• Trap to avoid: Watch the trap option (A) 3 cm: that is only the half-chord. The question asks for the full chord, so you must double it.
• Why it works: Recognising the Pythagorean triple turns a two-line calculation into a one-glance answer, which is exactly the time saving boards reward in the one-mark MCQ block.

Quick reading. The supplementary-angle property for the two pairs of opposite sides of a tangential quadrilateral is the whole question. Each vertex gives two equal tangent segments, and joining a vertex to the centre bisects the vertex angle; summing the four half-angles around O forces the opposite central angles to add to 180^∘.

1. Opposite sides AB,CD give ∠ AOB+∠ COD=180^∘.
2. ∠ COD=180^∘-125^∘=55^∘.

• Exam habit: The distractor (A) 62.5^∘ is exactly half of 125^∘, which is what you would write if you wrongly halved the given angle instead of subtracting from 180^∘. Keeping the supplementary rule in mind blocks that slip.
• Quick check: This pairing of opposite central angles recurs whenever a circle is inscribed in a polygon.

Picture-first. The cleanest route is the tangent-chord (alternate segment) theorem: the angle between tangent AT and chord AB equals the inscribed angle ∠ ACB in the alternate segment. That gives the answer in one line, and the diameter-and-right-angle method confirms it.

1. Alternate segment: ∠ BAT=∠ ACB=50^∘.
2. Cross-check via the semicircle: ∠ BAC=40^∘, and 90^∘-40^∘=50^∘.

• Common slip: The two methods agreeing is a strong self-check under exam pressure.
• Key insight: Students who only learn the triangle method spend longer and risk an angle-chase slip; the alternate-segment theorem is worth memorising because it collapses many tangent problems to a single equality.

Strategic angle. The kite PQOR splits along the diagonal OP into two right triangles that are mirror images. Computing one and doubling is faster and safer than treating the kite as a single shape. The distractor (C) 30 cm² is exactly one triangle, the classic ``forgot to double'' answer.

1. Tangent PQ=12 cm via the 5,12,13 triple.
2. One triangle =12(5)(12)=30 cm².
3. Two triangles =60 cm².

• Cross-check: A clean habit here is to write the doubling step explicitly so you never lose the factor of two.
• Watch out: The same kite-area-by-two-triangles idea reappears in many tangent-pair questions, so internalise it now.

Picture-first. Sketch the diameter vertical with A at the bottom and the tangent horizontal through A. A chord parallel to that tangent is horizontal, so the vertical diameter cuts it at right angles and halves it. The only care needed is converting ``8 cm from A'' into ``3 cm from O''.

1. Distance from centre: 8-5=3 cm.
2. Half-chord: √25-9=4 cm via the 3,4,5 triple.
3. Full chord: 8 cm.

• Time saver: The neat coincidence that CD equals the diameter ×45 is incidental; what matters is the reflex of always referencing distances to the centre when a radius is the hypotenuse.
• Deeper reason: That single habit prevents the most common error in chord problems.

Picture-first. Mark the right angle at A before touching trigonometry. With the right angle placed, OT is clearly the hypotenuse and AT sits next to the 30^∘ angle, so cosine is forced. The distractor (B) 2 cm comes from using sin 30^∘ by mistake.

1. Right angle at A; OT=4 is the hypotenuse.
2. AT=OTcos 30^∘=4·32=23 cm.

• Sanity test: A quick reasonableness test: 23≈ 3.46 cm, which is less than the hypotenuse 4 cm, as a leg must be. That sanity check catches a sine-cosine swap in seconds.
• Reusable idea: Always confirm a leg is shorter than the hypotenuse before committing.

Strategic angle. Two clean routes agree. The isosceles-triangle method uses tangent ⊥ radius then the angle sum; the alternate-segment method doubles the tangent-chord angle directly. Choosing the doubling route is fastest once you trust the theorem.

1. ∠ OPQ=90^∘-50^∘=40^∘.
2. Base angles equal, so ∠ POQ=180^∘-2(40^∘)=100^∘.

• Spot this: The distractor (B) 80^∘ appears if you stop at 2× 40^∘ instead of completing the triangle, a common half-finished answer.
• Careful here: Carrying the calculation to the angle sum, or trusting the doubling theorem, keeps you on 100^∘.

Quick reading. Recognise the shortcut ∠ OAB=12∠ APB. It comes from the radius bisecting the picture: ∠ OAP=90^∘ and the isosceles base angle is 90^∘-12∠ APB, so ∠ OAB is the leftover 12∠ APB.

1. Base angle ∠ PAB=65^∘.
2. ∠ OAB=90^∘-65^∘=25^∘, i.e. 12(50^∘).

• Big picture: The distractor (D) 50^∘ is simply ∠ APB copied without working, a tempting but wrong shortcut.
• Takeaway: Knowing the genuine half-relation keeps you from falling for it and confirms the answer instantly.

Strategic angle. The whole problem reduces to one right triangle with a 30^∘ angle at P and the radius 3 opposite it. Then PA=3cot 30^∘=33. Keeping the radius as the side opposite the half-angle makes the trig ratio unambiguous.

1. Half-angle at P is 30^∘, radius opposite is 3.
2. PA=3cot 30^∘=33 cm.

• Distractor: The distractor (C) 3 cm matches the radius and tempts students who guess the tangent equals the radius; it does not, since cot 30^∘=3≠ 1.
• Confirm it: Anchoring on the right triangle with the correct half-angle removes that guesswork.

Picture-first. The parallel chord turns the diagram symmetric about the diameter through Q. That symmetry hands you ∠ AQB=180^∘-2(70^∘) once you see both base angles are 70^∘ via alternate angles.

1. Alternate angle: ∠ ABQ=70^∘.
2. Isosceles: ∠ QAB=70^∘, so ∠ AQB=180^∘-140^∘=40^∘.

• Method note: The distractor (A) 20^∘ comes from mistakenly halving 40^∘ once more.
• Why chosen: Tracking exactly which angle the question wants, the apex and not a base angle, keeps the answer at 40^∘.

Strategic angle. Lock onto the supplementary rule ∠ ACB=180^∘-∠ AOB. It comes straight from the two right angles in the kite OACB. Plugging ∠ AOB=60^∘ gives 120^∘ at once, exposing the claim as false.

1. ∠ ACB=180^∘-∠ AOB.
2. =180^∘-60^∘=120^∘≠ 60^∘.

• Generalises: The statement would be true only if ∠ AOB were 90^∘, the lone self-supplementary value.
• Mark grabber: Recognising that single fixed point stops you from assuming the two angles are generally equal, which is the trap this true/false question sets.

Quick reading. The tangent length grows smoothly from 0 (point on the circle) upward as P moves away. So it passes through every positive value, including values below r. One nearby point settles the matter.

1. For P near the circle, OP→ r⁺ and → 0.
2. Such an is below r, so ``always greater'' fails.

• Pitfall: The honest statement is that the tangent length can be anything positive; it equals the radius only at one special distance OP=r2.
• One-line proof: Treating ``always'' words sceptically is a reliable instinct for these reasoning questions.

Strategic angle. The contrast with the previous question is instructive: the tangent is compared to OP, not to the radius. Against OP it is always shorter, because OP is the hypotenuse of the contact right triangle. The comparison target decides the verdict.

1. Right triangle OAP with hypotenuse OP.
2. Leg PA always.

• Trap to avoid: This is a strict inequality: PA can never equal OP, since that would force the radius OA to be zero.
• Why it works: Reading the comparison carefully, tangent versus OP here, versus radius in the last item, is the skill these paired questions test.

Picture-first. Visualise the two tangents as a moving pair. As the external point recedes, the angle between the tangents shrinks; in the limit, the point goes to infinity and the tangents become the parallel pair at the diameter ends, where the angle is 0^∘.

1. Tangents at opposite ends of a diameter are both ⊥ to it.
2. Both ⊥ to the same line, hence parallel, hence 0^∘.

• Exam habit: The subtlety is that no finite external point gives 0^∘; the parallel tangents meet only at infinity. But the angle they make is genuinely 0^∘, so ``may be 0<sup>∘</sup>'' is a true statement.
• Quick check: Distinguishing ``meets'' from ``angle between'' resolves the apparent paradox.

Strategic angle. The fastest mental model is the square OAPB of side a. When the tangents meet at 90^∘, the figure is a square, and OP is its diagonal, so OP=a2 immediately.

1. ∠ APB=90^∘⇒ OAPB is a square of side a.
2. Diagonal OP=a2.

• Common slip: Either route, the 45^∘ right triangle or the square diagonal, lands on a2.
• Key insight: Carrying both interpretations gives a built-in cross-check, which is reassuring when a numerical-looking claim must be judged true or false.

Quick reading. The claim swaps the trig ratio. With the radius opposite the 30^∘ half-angle, sine governs and OP=a/sin 30^∘=2a. The value a3 would come from a cosine, which applies to the tangent length, not OP.

1. Half-angle 30^∘; sin 30^∘=a/OP.
2. OP=a/12=2a≠ a3.

• Cross-check: A useful tie-in: the tangent length here is acot 30^∘=a3, which is exactly the number the false statement mis-assigns to OP.
• Watch out: Spotting that the claim attached the right number to the wrong segment nails the verdict.

Strategic angle. The proof hinges on one equality chain: tangent-chord angle equals an inscribed base angle, which equals the other base angle by isosceles symmetry. That makes alternate angles equal, so the lines are parallel.

1. ∠ XAB=∠ ACB (alternate segment).
2. ∠ ACB=∠ ABC (isosceles), so ∠ XAB=∠ ABC, giving XAY∥ BC.

• Time saver: Symmetry offers an even quicker view: the perpendicular bisector of BC passes through A and the centre, and the tangent at A is perpendicular to that line, just as BC is.
• Deeper reason: Two lines perpendicular to the same line are parallel, confirming the result without angle chasing.

Quick reading. The contact point A is pinned, so every radius to A is perpendicular to PQ there. All centres line up on that single perpendicular at A. Unless A is the midpoint, this is not the perpendicular bisector.

1. Tangent at A forces centre on the perpendicular at A.
2. That line is the bisector only when A is the midpoint, so the claim fails in general.

• Sanity test: Contrast this deliberately with the next item, where circles pass through both P and Q and the centres really do lie on the perpendicular bisector.
• Reusable idea: Holding the two scenarios side by side prevents the wording from misleading you.

Strategic angle. Reduce the circle language to a distance condition: passing through P and Q means OP=OQ. That single equality is the definition of the perpendicular bisector, so the conclusion is immediate.

1. Through P,Q⇒ OP=OQ.
2. OP=OQ⇒ O on the perpendicular bisector of PQ.

• Spot this: This is the true counterpart to the previous false item: there the contact point was fixed; here the two endpoints are fixed. Endpoints fixed gives the perpendicular bisector; a single contact point gives a perpendicular at that point.
• Careful here: Keeping the distinction sharp earns easy marks.

Strategic angle. The proof flows through three standard facts in order: semicircle right angle, alternate-segment equality, and isosceles converse. Each is one line, and together they pin BC=BD cleanly.

1. Semicircle: ∠ ACB=90^∘, so ∠ ABC=60^∘.
2. Alternate segment: ∠ BCD=30^∘; angle sum in BCD gives ∠ BDC=30^∘.
3. Equal angles ⇒ BC=BD.

• Big picture: The crucial move is reading the exterior angle ∠ DBC=120^∘ at B, since D lies on AB produced. Getting that exterior angle right is what makes the triangle BCD resolve to two 30^∘ angles.
• Takeaway: This is a recurring board proof, so the route is worth rehearsing.

Strategic angle. Treat the inner radius as the unknown leg of a right triangle whose hypotenuse is the outer radius and whose other leg is the half-chord. The tangent condition is what makes that triangle right-angled.

1. Half-chord AM=4 cm, outer radius OA=5 cm.
2. Inner radius r=√25-16=3 cm.

• Distractor: The structure mirrors Exercise 10.1 Question 1, only the unknown has moved from the chord to the radius. Recognising the same right triangle across questions builds speed; here the answer drops out the moment you halve the chord and apply Pythagoras.
• Confirm it: A common error is to forget halving the chord and to use the full 8 cm as a leg, which would give a meaningless negative under the square root.
• Method note: Always halve a chord before feeding it into a centre-perpendicular right triangle, and confirm the inner radius is smaller than the outer one as a final sanity check.

Strategic angle. The proof needs only the two contact right angles. Because they sit at opposite vertices Q and R, their sum is automatically 180^∘, which is the cyclic condition. No circle through the four points needs to be constructed explicitly.

1. Two contact right angles at Q and R.
2. Opposite angles sum to 180^∘⇒ cyclic.

• Why chosen: A vivid way to see it: OP becomes the diameter of the circle through Q,O,R,P, since ∠ OQP and ∠ ORP are angles in a semicircle on OP.
• Generalises: That viewpoint also explains why Q and R always lie on a circle with OP as diameter.

Strategic angle. Bisect the 120^∘ to land a 60^∘ right triangle, where the adjacent-over-hypotenuse cosine is exactly 12. That single value delivers OB=2BC, and equal tangents finish the addition.

1. Half-angle ∠ OBC=60^∘.
2. cos 60^∘=12=BC/OB⇒ OB=2BC.
3. BD=BC, so BC+BD=2BC=OB.

• Mark grabber: The geometry is engineered so the half-angle is precisely 60^∘, the one angle whose cosine is 12.
• Pitfall: Any other tangent-angle would not produce the tidy ``BO=2BC'' statement, which is a hint that the 120^∘ in the question was chosen deliberately.

Strategic angle. Convert ``touches both lines'' into ``equidistant from both lines'', then invoke the bisector locus. The RHS congruence makes the equidistance rigorous, since both perpendicular distances equal the radius.

1. Perpendicular distances to both lines equal r.
2. Equidistant from two lines ⇒ on the angle bisector.

• One-line proof: This is the principle behind the incircle of a triangle: its centre, the incentre, is the meeting point of the angle bisectors precisely because it is equidistant from all three sides. The same equidistance argument scales from two lines to three.
• Trap to avoid: Note also that there are two bisectors at an intersection, the internal and external ones; a circle nestled inside the angle has its centre on the internal bisector, while a circle straddling the vertex on the far side would sit on the external bisector.
• Why it works: The question's phrasing ``touching two intersecting lines'' covers the internal-bisector case, which is the one drawn in routine board figures.

Picture-first. Extend the tangents until they cross at P. That one point gives two equal-tangent pairs, one per circle. The common-tangent segment AB is the difference PA-PB, and CD is the matching difference PC-PD, so they are equal.

1. PA=PC (larger circle), PB=PD (smaller circle).
2. AB=PA-PB=PC-PD=CD.

• Exam habit: The unequal radii are a deliberate distraction: the proof never uses the actual radii, only the equal-tangents property at the shared external point.
• Quick check: Noticing that the radii are irrelevant is a sign you have found the intended, economical argument.

Strategic angle. With equal radii, abandon the intersection-point method of the previous part; the tangents are now parallel and never meet. Instead, read each tangent as a side of a rectangle whose opposite side is the line of centres.

1. Each tangent and the two equal radii build a rectangle.
2. AB=O₁O₂ and CD=O₁O₂, so AB=CD.

• Common slip: The contrast with Question 25 is the teaching point: unequal radii give converging tangents handled by equal-tangent subtraction, while equal radii give parallel tangents handled by a rectangle.
• Key insight: Matching the method to the radius condition is the skill being tested.

Picture-first. The crossing point E is the only tool needed. It is external to both circles, so it yields one equal-tangent pair per circle. Because E sits inside each tangent segment, the parts add to the wholes AB and CD.

1. EA=EC, EB=ED.
2. AB=EA+EB=EC+ED=CD.

• Cross-check: This transverse common-tangent configuration is the mirror image of the direct-tangent case: same equal-tangent property, opposite combination.
• Watch out: Reading where E lies relative to the contact points tells you instantly whether to add or subtract, which is the heart of both proofs.

Strategic angle. Chain two angle facts: tangent-chord equality and parallel-line alternate angles. They meet at a common angle, forcing the triangle RPQ to be isosceles, which gives equal chords and hence equal arcs.

1. ∠ XRP=∠ RQP (tangent-chord) and ∠ XRP=∠ RPQ (parallel).
2. So RP=RQ, equal arcs, R bisects arc PRQ.

• Time saver: A symmetry shortcut confirms it: the diameter through R is perpendicular to the tangent at R, hence perpendicular to the parallel chord PQ, so it bisects PQ and the arc.
• Deeper reason: The angle proof and the symmetry proof agree, a comforting double-check.

Strategic angle. Anchor on the isosceles triangle OAB. Its equal base angles, each subtracted from the 90^∘ tangent-radius angle, yield equal tangent-chord angles directly, with no need for the alternate-segment theorem.

1. Base angles ∠ OAB=∠ OBA (isosceles).
2. Each tangent angle =90^∘- base angle, so they are equal.

• Sanity test: The figure is symmetric about the perpendicular bisector of AB, which also passes through the centre. That mirror symmetry is the deep reason the two tangent angles match, and it is worth visualising even when the algebraic proof is the one you write down.
• Reusable idea: The same symmetry argument proves the equal-tangents property from an external point and the fact that the tangents at the ends of a diameter are parallel, so investing in the mirror picture pays off across the whole chapter.
• Spot this: Whenever a circle problem has a chord or diameter as an axis of symmetry, expect a pair of equal angles or equal lengths to fall out.

Strategic angle. Track one property, perpendicularity to AB, as it passes from the tangent to every parallel chord. Once a chord is perpendicular to the diameter, the centre-perpendicular-bisects-chord rule finishes the job.

1. Tangent ⊥ AB, so parallel chords ⊥ AB.
2. Diameter ⊥ chord ⇒ bisects it.

• Careful here: The word ``all'' is justified because the argument never names a particular chord; any chord parallel to the tangent inherits the perpendicularity and hence the bisection. Recognising that the proof is independent of which parallel chord you pick is what makes the universal claim valid.
• Big picture: To write a fully rigorous version in an exam, state ``let CD be an arbitrary chord parallel to the tangent at A'' at the start; the word ``arbitrary'' signals that nothing special was assumed, which is exactly what a universal statement needs.
• Takeaway: The converse is also worth knowing: a chord that the diameter bisects must be parallel to the tangent at A.

Strategic angle. List the six equal-tangent pairs first, then split every side at its contact point. The two target sums collapse to the same six tangent lengths, so the equality is bookkeeping once the labels are set.

1. Six equal pairs, one per vertex.
2. Both sums reduce to the same collection of tangent segments.

• Distractor: The cleanest way to avoid label errors is to walk around the hexagon in order, naming contact points as you go.
• Confirm it: Keeping the cyclic order straight is the only place this proof can go wrong; the algebra itself is just substitution.

Strategic angle. Set up the x,y,z tangent labels, sum to find x+y+z=s, then isolate the one you want. BD=y, and y=s-(x+z)=s-b because x+z is exactly side b.

1. a=y+z, b=z+x, c=x+y, so x+y+z=s.
2. BD=y=s-(x+z)=s-b.

• Method note: This identity BD=s-b is a building block for the incircle and for Heron-style results, and the symmetric labels make the analogous formulas CD=s-c and so on fall out for free.
• Why chosen: Memorising the labelling, not the final formula, is what transfers to new problems.

Strategic angle. Slide the contact point E along the tangent CD and use CA=CE, DB=DE to fold the broken path P→ C→ E→ D→ P back onto the two straight tangents PA and PB.

1. CE=CA, DE=DB.
2. Perimeter =PA+PB=2(10)=20 cm.

• Generalises: The striking feature is independence from E: the answer never uses the position of the tangent CD.
• Mark grabber: That invariance is the heart of the problem, and recognising it tells you the radius and the exact place of E are deliberately withheld because they do not matter.

Strategic angle. Show both angles equal the same expression 90^∘-∠ BAC. One comes from the triangle's angle sum after the semicircle gives 90^∘; the other from the tangent being perpendicular to the diameter.

1. ∠ ACB=90^∘-∠ BAC (semicircle).
2. ∠ BAT=90^∘-∠ BAC (tangent ⊥ diameter).

• Pitfall: This question is a guided derivation of the alternate-segment theorem for the special case where the chord and a diameter share the endpoint A.
• One-line proof: Seeing the theorem emerge from elementary facts makes it far easier to trust and reuse later.

Strategic angle. The mutual-tangent condition is what makes ∠ OPO'=90^∘. With a right triangle in hand, the half-chord is simply the altitude to the hypotenuse, 345, and doubling gives the chord.

1. Right triangle legs 3,4, hypotenuse 5.
2. Half-chord =345=2.4; chord PQ=4.8 cm.

• Trap to avoid: The 3,4,5 triple keeps the arithmetic exact and clean. Recognising the common chord as twice the foot-of-altitude segment is the move that avoids a longer coordinate computation, which is why this set-up rewards knowing the altitude formula.
• Why it works: If you forget the altitude shortcut, you can fall back to areas: the area of right triangle OPO' is 12(3)(4)=6, and also 12(OO')(PM)=12(5)(PM), so PM=12/5=2.4 cm gives the same half-chord.
• Exam habit: Two independent derivations of 2.4 confirm the chord 4.8 cm without doubt.

Strategic angle. Spot that BC is itself a tangent (at B), because the diameter AB is perpendicular to it. Then the point M where the tangent at P crosses BC has two equal tangents MB,MP, and the right angle at P converts that into MB=MC.

1. BC tangent at B; MB=MP (equal tangents from M).
2. ∠ BPC=90^∘ gives MP=MC, so MB=MC.

• Quick check: The hidden tangent BC is the insight; once you see that the hypotenuse-adjacent leg is tangent at B, the equal-tangent machinery applies immediately. Students who miss it try coordinate geometry and take far longer for the same midpoint result.
• Common slip: The reason MP=MC deserves a word: in right triangle BPC, the angle ∠ MPC and the angle ∠ MCP are both complements of the equal angles ∠ MPB=∠ MBP, so they are equal, and equal angles give the equal sides MP=MC.
• Key insight: That little angle-chase is the bridge from ``M is equidistant to the two tangent points'' to ``M is the midpoint of BC''.

Strategic angle. Anchor on the isosceles base angle 75^∘, push it into the alternate segment via the tangent-chord theorem, then use RS∥ PQ to make triangle QRS isosceles with two 75^∘ angles. The apex ∠ RQS is the leftover 30^∘.

1. Base angle =75^∘; alternate segment gives ∠ QSR=75^∘.
2. Parallel chord gives ∠ QRS=75^∘, so ∠ RQS=30^∘.

• Cross-check: Notice the answer equals the original ∠ RPQ=30^∘.
• Watch out: That neat coincidence is a quick sanity flag: the apex angle of the inscribed triangle matches the angle between the tangents, which you can use to confirm you have not dropped a factor along the way.

Strategic angle. Build the central angle ∠ BOC=60^∘ from the isosceles triangle OAC, then use the tangent-radius right angle to read ∠ ODC=30^∘. The tangent-chord angle ∠ BCD=30^∘ closes the loop and makes BCD isosceles.

1. ∠ BOC=60^∘; tangent ⊥ OC gives ∠ ODC=30^∘.
2. ∠ BCD=30^∘ (alternate segment), so BC=BD.

• Time saver: The specific value 30^∘ is chosen so the two base angles of BCD coincide; a different ∠ BAC would not give BC=BD. Recognising that the question's data is tuned for this equality helps you anticipate the isosceles conclusion before the algebra confirms it.
• Deeper reason: A general way to see it: the tangent-chord angle ∠ BCD always equals ∠ BAC, and the angle ∠ BDC comes out as 90^∘-∠ ABC from the right angle at C; these two coincide exactly when ∠ BAC=30^∘, since then both equal 30^∘.
• Sanity test: Spotting which special angle makes a figure isosceles is a recurring board skill.

Strategic angle. Route everything through the radius OR to the arc midpoint. The arc midpoint sits on the chord's perpendicular bisector, so OR⊥ PQ; the tangent is ⊥ OR by definition. Two perpendiculars to OR are parallel.

1. Arc midpoint ⇒ OR⊥ PQ.
2. Tangent ⊥ OR, so tangent ∥ PQ.

• Reusable idea: This is the converse companion to Question 28, where a chord parallel to the tangent forced the contact point to bisect the arc. Here the arc midpoint forces the parallel tangent.
• Spot this: Seeing the two statements as a matched pair makes both easier to recall under exam conditions. The single shared fact behind both is that the diameter through the arc midpoint is the axis of symmetry of the chord and its arc: it is perpendicular to the chord, perpendicular to the tangent at the midpoint, and bisects both the chord and the arc.
• Careful here: Naming that diameter early turns either direction of the proof into one or two lines.

Strategic angle. The key realisation is that AB and CD are common tangents, so they form one and the same angle at E for both circles. Each centre lies on the bisector of that single angle, and a single angle has only one bisector.

1. EO bisects ∠ at E (first circle).
2. EO' bisects the same ∠ at E (second circle); one bisector, so collinear.

• Big picture: The proof never needs the radii or the distance between centres, only that the tangents are shared.
• Takeaway: Identifying that the two angle-bisector claims refer to the identical angle is the whole argument, and it is a clean reminder that ``common tangent'' is a strong shared constraint.

Strategic angle. Find TP=12 via the 5,12,13 triple, locate E at ET=8, then use the shared angle at T to set up the proportion AE/ET=OP/TP. Doubling the half-tangent AE gives AB.

1. TP=12 (5,12,13); ET=13-5=8.
2. AE=8·512=103; AB=203 cm.

• Distractor: The similar-triangle proportion is the engine: it converts the radius-and-tangent of the big triangle into the small tangent segment AE. Keeping the matching of opposite-to-adjacent consistent across the two triangles is the one place to be careful, and it lands the exact fraction 203.
• Confirm it: A quick reasonableness test: AB=203≈ 6.67 cm is comfortably less than the chord of contact PQ of the big tangent pair, which fits the picture of AB sitting closer to the circle than PQ.
• Method note: Carrying the answer as an exact fraction rather than a rounded decimal is also what board schemes expect here.

Strategic angle. Strip the 90^∘ tangent-radius angle out of ∠ PCA to get ∠ OCA=20^∘, bounce it to ∠ CAB=20^∘ by the isosceles radii, then use the semicircle's 90^∘ to finish the triangle.

1. ∠ OCA=110^∘-90^∘=20^∘=∠ CAB.
2. ∠ ACB=90^∘ (semicircle), so ∠ CBA=70^∘.

• Why chosen: A fast cross-check uses the tangent-chord theorem directly: the angle between tangent PC and chord CB equals ∠ CAB=20^∘, and since ∠ PCA=110^∘, the chord CB splits it as 110^∘-20^∘=90^∘ at C, agreeing with the semicircle right angle.
• Generalises: Two independent paths to 70^∘ give confidence.

Strategic angle. The whole problem turns on one unknown, the altitude h=AM, and one shared quantity, the half-base BM. Write BM² two different ways: once from the radius using right triangle OMB with hypotenuse OB=9, and once from the equal side using right triangle AMB with hypotenuse AB=6. Setting the two expressions equal eliminates BM entirely and leaves a single linear equation in h.

1. From the radius: BM²=9²-(h-9)². From the equal side: BM²=6²-h².
2. Equate: 81-(h-9)²=36-h². The h² terms cancel, giving 18h=36, so h=2 cm.
3. Back-substitute: BM²=36-4=32, so BM=42 and base BC=82 cm.
4. Area =12× BC× AM=12(82)(2)=82 cm².

• Mark grabber: A qualitative check seals the work: the altitude h=2 is tiny against the radius 9, so the triangle is short and wide and sits low in the circle, exactly as the diagram shows. That mental picture guards against the most common slip, getting the sign of OM=h-9 wrong.
• Pitfall: Had you written OM=9-h instead, the same cancellation still gives h=2 here because the term is squared, but in problems where the apex lies on the far side of the centre the sign genuinely matters, so always read the figure before committing.
• One-line proof: The clean surd 82 is a reassuring sign that the radius and side lengths were chosen to produce an exact answer.

Strategic angle. Find AP=12 from the 5,12,13 triple, then fold the broken boundary A→ B→ R→ C→ A onto the two straight tangents AP and AQ using BR=BP and CR=CQ.

1. AP=AQ=12 cm (5,12,13).
2. Perimeter =AP+AQ=24 cm.

• Trap to avoid: As in Question 33, the position of the contact point R never enters the answer; the perimeter is locked at twice the tangent length. Spotting this invariance saves time and tells you the ``minor arc'' detail is there to test understanding, not to change the number.
• Why it works: The reason R must lie on the minor arc PQ is simply that only then does the tangent at R cross both AP and AQ between A and the contact points, keeping the small triangle ABC well defined.
• Exam habit: Once that condition holds, the broken path folds onto AP+AQ exactly, and 2× 12=24 cm follows.