Circles: Class 10 Case Study Previous Year Questions with Answers

AN = 5√3 cm.

OA ⟂ AN, since a radius is perpendicular to the tangent at its point of contact.

In right △OAN, tan 30° = $\frac{\text{(}\text{OA}\text{)/(}}{\text{AN}\text{)}}$ = $\frac{\text{(}\text{5}\text{)/(}}{\text{AN}\text{)}}$.

$\frac{\text{(}\text{1}\text{)/(}}{\text{√3}\text{)}}$ = $\frac{\text{(}\text{5}\text{)/(}}{\text{AN}\text{)}}$, so AN = 5√3 cm.

∠AOB = 120°.

In right triangles OAN and OBN, ON is common and OA = OB (radii).

Thus △OAN ≅ △OBN by RHS, so ∠BNO = ∠ANO = 30°.

∠ANB = 30° + 30° = 60°.

In quadrilateral OANB, ∠OAN = ∠OBN = 90°.

∠AOB = 360° − 90° − 90° − 60° = 120°.

NA + NB + AB = 15√3 cm.

NA = NB, since tangents from the same external point are equal.

∠ANB = 60°, as found above.

The base angles of isosceles △ANB are each (180° − 60°)/2 = 60°.

Thus △ANB is equilateral, so AB = NA = NB = 5√3 cm.

Total length = 3 × 5√3 = 15√3 cm.

OANB is a square.

Use the new angle ∠ANO = 45° for this alternative.

By RHS congruence, ∠BNO = ∠ANO = 45°, so ∠ANB = 90°.

The angles at A and B are 90°, since the radii are perpendicular to the tangents.

The angle at O is also 360° − 90° − 90° − 90° = 90°.

Thus OANB is a rectangle.

OA = OB because they are radii; these are adjacent sides of the rectangle.

A rectangle with equal adjacent sides is a square.

AC = 40 m.

B is the midpoint of AC, so BC = BA = 20 m.

AC = AB + BC = 20 + 20 = 40 m.

Shortest distance OB = 15 m.

OB ⟂ AB, since the road is tangent to the circle at B.

The perpendicular OB gives the shortest distance from O to the road.

By Pythagoras in right △OBA, OB² = OA² − AB².

OB² = 25² − 20² = 225.

OB = 15 m.

Circumference = 30π m.

The radius is OB = 15 m.

Circumference = 2πr = 2π × 15 = 30π m.

If π is taken as 22/7, this is $\frac{\text{(}\text{660}\text{)/(}}{\text{7}\text{)}}$ m, as also accepted in the marking scheme.

Area = 225π m².

The radius is OB = 15 m.

Area = πr² = π × 15² = 225π m².

If π is taken as 22/7, this is $\frac{\text{(}\text{4950}\text{)/(}}{\text{7}\text{)}}$ m², as also accepted in the marking scheme.

AB = 75√3 cm.

OA ⟂ AB, since OA is the radius to the point of contact.

In right △OAB, tan 30° = $\frac{\text{(}\text{OA}\text{)/(}}{\text{AB}\text{)}}$ = $\frac{\text{(}\text{75}\text{)/(}}{\text{AB}\text{)}}$.

AB = 75√3 cm.

OB = 150 cm.

In right △OAB, sin 30° = $\frac{\text{(}\text{OA}\text{)/(}}{\text{OB}\text{)}}$ = $\frac{\text{(}\text{75}\text{)/(}}{\text{OB}\text{)}}$.

$\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ = $\frac{\text{(}\text{75}\text{)/(}}{\text{OB}\text{)}}$, so OB = 150 cm.

AP = $\frac{\text{(}\text{75√3}\text{)/(}}{\text{2}\text{)}}$ cm.

OQ = OA = 75 cm, since both are radii.

QB = OB − OQ = 150 − 75 = 75 cm.

Thus Q is the midpoint of OB.

Since QP ∥ OA, the midpoint theorem in △BOA gives P as the midpoint of BA.

AP = $\frac{\text{(}\text{AB}\text{)/(}}{\text{2}\text{)}}$ = $\frac{\text{(}\text{75√3}\text{)/(}}{\text{2}\text{)}}$ cm.

PQ = $\frac{\text{(}\text{75}\text{)/(}}{\text{2}\text{)}}$ cm = 37.5 cm.

OQ = OA = 75 cm, so BQ = 150 − 75 = 75 cm.

Since PQ ∥ OA, △BQP ∼ △BOA by AA similarity.

$\frac{\text{(}\text{PQ}\text{)/(}}{\text{OA}\text{)}}$ = $\frac{\text{(}\text{BQ}\text{)/(}}{\text{BO}\text{)}}$ = $\frac{\text{(}\text{75}\text{)/(}}{\text{150}\text{)}}$ = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$.

PQ = $\frac{\text{(}\text{75}\text{)/(}}{\text{2}\text{)}}$ cm = 37.5 cm.

∠OSA = 90°.

OS is the radius to contact point S, and SA lies along the tangent AD.

Therefore OS ⟂ SA, so ∠OSA = 90°.

ABCD is a kite.

Tangents from A, B, C and D are equal in pairs: AP = AS, BP = BQ, CQ = CR and DR = DS.

Adding these equalities gives AB + CD = AD + BC.

Given AB = AD, subtracting them gives CD = BC.

Thus AB = AD and BC = CD: ABCD has two pairs of equal adjacent sides, so it is a kite.

AP = 4 cm.

DS = DR = 7 cm, since tangents from D are equal.

AS = AD − DS = 11 − 7 = 4 cm.

AP = AS, since tangents from A are equal.

Hence AP = 4 cm.

∠QOR = 120°.

OQ ⟂ CQ and OR ⟂ CR, since each radius is perpendicular to its tangent.

In quadrilateral OQCR, ∠OQC = ∠ORC = 90°.

∠QOR = 360° − 90° − 90° − 60° = 120°.

PQ = 30 cm.

AP = AQ because tangents from the same external point A are equal.

Since ∠PAQ = 60°, the other angles of isosceles △APQ are each 60°.

Thus △APQ is equilateral, so PQ = AP = 30 cm.

∠POQ = 120°.

OP ⟂ AP and OQ ⟂ AQ, by the radius–tangent theorem.

In quadrilateral OPAQ, ∠POQ = 360° − 90° − 90° − 60° = 120°.

OA = 20√3 cm.

In right triangles OPA and OQA, OA is common and OP = OQ (radii).

Thus △OPA ≅ △OQA by RHS; hence OA bisects ∠PAQ.

∠PAO = 60°/2 = 30°.

In right △OPA, cos 30° = $\frac{\text{(}\text{AP}\text{)/(}}{\text{OA}\text{)}}$ = $\frac{\text{(}\text{30}\text{)/(}}{\text{OA}\text{)}}$.

OA = $\frac{\text{(}\text{60}\text{)/(}}{\text{√3}\text{)}}$ = 20√3 cm.

Radius OP = 10√3 cm.

By RHS congruence of △OPA and △OQA, ∠PAO = 30°.

In right △OPA, tan 30° = $\frac{\text{(}\text{OP}\text{)/(}}{\text{AP}\text{)}}$ = $\frac{\text{(}\text{OP}\text{)/(}}{\text{30}\text{)}}$.

OP = $\frac{\text{(}\text{30}\text{)/(}}{\text{√3}\text{)}}$ = 10√3 cm.

∠PAO = 90°.

OA is the radius to the point where tangent PA touches the circle.

Therefore OA ⟂ PA, so ∠PAO = 90°.

PA = 3√3 cm.

In right triangles OAP and OBP, OP is common and OA = OB (radii).

Thus △OAP ≅ △OBP by RHS, so OP bisects ∠APB.

∠APO = 30°.

In right △OAP, tan 30° = $\frac{\text{(}\text{OA}\text{)/(}}{\text{PA}\text{)}}$ = $\frac{\text{(}\text{3}\text{)/(}}{\text{PA}\text{)}}$.

PA = 3√3 cm.

∠AOB = 120°.

∠OAP = ∠OBP = 90°, since radii are perpendicular to tangents.

In quadrilateral OAPB, ∠AOB = 360° − 90° − 90° − 60° = 120°.

∠OAB = 30°.

PA = PB because tangents from P are equal.

With ∠APB = 60°, △PAB is equilateral, so ∠PAB = 60°.

OA ⟂ PA, so ∠OAP = 90°.

∠OAB = ∠OAP − ∠BAP = 90° − 60° = 30°.

AR = x m.

AP and AR are tangents from the same external point A.

Therefore AR = AP = x m.

BQOR is a square.

OQ ⟂ BQ and OR ⟂ BR, since the radii meet tangents at right angles.

Also ∠QBR = 90°, because AB ⟂ BC.

The fourth angle is 90°, so BQOR is a rectangle.

OQ = OR because they are radii; these are adjacent sides of the rectangle.

Thus BQOR is a square.

PC = (8 + x) m; x = $\frac{\text{(}\text{√274 − 8}\text{)/(}}{\text{2}\text{)}}$ m ≈ 4.28 m.

AR = x, so BR = AB − AR = 7 − x.

BQ = BR = 7 − x, since tangents from B are equal.

CQ = BC − BQ = 15 − (7 − x) = 8 + x.

CP = CQ = 8 + x, since tangents from C are equal.

AC = AP + PC = x + (8 + x) = 8 + 2x.

By Pythagoras, AC² = AB² + BC² = 7² + 15² = 274.

8 + 2x = √274, since AC is positive.

x = $\frac{\text{(}\text{√274 − 8}\text{)/(}}{\text{2}\text{)}}$ m ≈ 4.28 m.

x = $\frac{\text{(}\text{√274 − 8}\text{)/(}}{\text{2}\text{)}}$ m; r = $\frac{\text{(}\text{22 − √274}\text{)/(}}{\text{2}\text{)}}$ m ≈ 2.72 m.

Let AR = AP = x m.

BR = BQ = 7 − x m, and CP = CQ = 15 − (7 − x) = 8 + x m.

AC = AP + PC = 8 + 2x m.

By Pythagoras, (8 + 2x)² = 7² + 15² = 274.

Thus x = $\frac{\text{(}\text{√274 − 8}\text{)/(}}{\text{2}\text{)}}$ m.

Since BQOR is a square, r = OR = BR = 7 − x.

r = 7 − $\frac{\text{(}\text{√274 − 8}\text{)/(}}{\text{2}\text{)}}$ = $\frac{\text{(}\text{22 − √274}\text{)/(}}{\text{2}\text{)}}$ m ≈ 2.72 m.

OB = 8 cm.

OD ⟂ BD because OD is the radius to tangent BC.

In right △ODB, sin 30° = $\frac{\text{(}\text{OD}\text{)/(}}{\text{OB}\text{)}}$ = $\frac{\text{(}\text{4}\text{)/(}}{\text{OB}\text{)}}$.

OB = 8 cm.

Yes, DE ∥ CA.

A, O and D lie on one line and OD ⟂ BC, so AD is an altitude.

In an equilateral triangle, an altitude is also a median, so D is the midpoint of BC.

Tangents from B are equal, so BE = BD = $\frac{\text{(}\text{BC}\text{)/(}}{\text{2}\text{)}}$.

Since AB = BC, BE = $\frac{\text{(}\text{AB}\text{)/(}}{\text{2}\text{)}}$; hence E is the midpoint of AB.

The segment joining the midpoints of AB and BC is parallel to AC, by the midpoint theorem.

Therefore DE ∥ CA.

∠OEB = 90°, ∠EBD = 60°, ∠BDO = 90°, ∠DOE = 120°; OEBD is cyclic.

OE ⟂ BE and OD ⟂ BD, since radii are perpendicular to tangents.

Thus ∠OEB = ∠BDO = 90°.

∠EBD = ∠ABC = 60°, because ABC is equilateral.

The angle sum of quadrilateral OEBD gives ∠DOE = 360° − 90° − 60° − 90° = 120°.

Opposite angles ∠OEB + ∠BDO = 90° + 90° = 180°.

Therefore OEBD is cyclic, by the converse of the cyclic-quadrilateral angle theorem.

Perimeter of △ABC = 41.52 cm.

OB = 8 cm, as found in part (i).

In right △OBD, BD = OB cos 30° = 8 × $\frac{\text{(}\text{√3}\text{)/(}}{\text{2}\text{)}}$ = 4√3 cm.

D is the midpoint of BC, so BC = 2BD = 8√3 cm.

Since ABC is equilateral, perimeter = 3BC = 24√3 cm.

Using √3 = 1.73, perimeter = 24 × 1.73 = 41.52 cm.

∠AOP = 60°.

The passage does not explicitly call AP and BP tangents. The 430/5/1 marking scheme (printed p. 17) treats them as tangents; this solution follows that reading.

OA ⟂ AP, so △OAP is right-angled at A.

OA = 7 m and OP = 14 m.

cos ∠AOP = $\frac{\text{(}\text{OA}\text{)/(}}{\text{OP}\text{)}}$ = $\frac{\text{(}\text{7}\text{)/(}}{\text{14}\text{)}}$ = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$.

Hence ∠AOP = 60°.

△OAP ≅ △OBP by RHS congruence.

∠OAP = ∠OBP = 90°, since radii are perpendicular to the tangents.

OP = OP, the common hypotenuse.

OA = OB, radii of the same circle.

Therefore △OAP ≅ △OBP by RHS congruence.

Length of fencing = 38.22 m.

In right △OAP, AP² = OP² − OA² = 14² − 7² = 147.

AP = 7√3 m, and BP = AP because tangents from P are equal.

Total fencing = OA + AP + PB + BO.

Total fencing = 7 + 7√3 + 7√3 + 7 = 14 + 14√3 m.

Using √3 = 1.73, fencing = 14 + 14 × 1.73 = 38.22 m.

Area of OAPB = 84.77 m².

AP = BP = 7√3 m, and OA = OB = 7 m.

OP divides OAPB into right triangles OAP and OBP.

Area = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ × OA × AP + $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ × OB × BP.

Area = 2 × $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ × 7 × 7√3 = 49√3 m².

Using √3 = 1.73, area = 49 × 1.73 = 84.77 m².

Height of the gate = 7 feet, using π = 22/7.

The full circular gate has height equal to its diameter, 2r.

Circumference = 2πr = 22 feet.

Using π = $\frac{\text{(}\text{22}\text{)/(}}{\text{7}\text{)}}$, as in the marking scheme, 2r = 22 × $\frac{\text{(}\text{7}\text{)/(}}{\text{22}\text{)}}$ = 7 feet.

Thus the gate’s height is 7 feet; the rectangular frame’s height is only r = 3.5 feet.

∠HOC = 135°.

Four evenly spaced diameters give eight equal central angles.

Each central angle = 360°/8 = 45°.

The non-reflex angle from OH to OC spans three such angles in the diagram.

∠HOC = 3 × 45° = 135°.

OHAB is a square.

HA and AB are tangents to the circle at H and B.

Thus OH ⟂ HA and OB ⟂ AB.

∠HAB = 90°, because HACD is a rectangle.

The fourth angle ∠HOB is 90°, by the angle sum of a quadrilateral.

Hence OHAB is a rectangle.

OH = OB because they are radii, and these are adjacent sides of the rectangle.

Therefore OHAB is a square.

Area of iron sheet = $\frac{\text{(}\text{49}\text{)/(}}{\text{2}\text{)}}$ square feet = 24.5 square feet.

Using π = 22/7, the circle has radius r = 3.5 feet.

The rectangle’s width AC = HD = 2r = 7 feet.

Its height CD = OB = r = 3.5 feet, from the diagram.

Area = AC × CD = 7 × 3.5 = 24.5 square feet.

The sheet covers rectangle HACD, not the full circular gate.

Diameter = 4 m.

The rope gives the radius r = 2 m.

Diameter = 2r = 2 × 2 = 4 m.

XY = 2√3 m.

Join OX and OY.

OA ⟂ XA and OY ⟂ XY because the radii are perpendicular to the tangents.

In right triangles OAX and OYX, OX is common and OA = OY (radii).

Thus △OAX ≅ △OYX by RHS, so OX bisects ∠AXY = 60°.

∠OXY = 30°.

In right △OYX, tan 30° = $\frac{\text{(}\text{OY}\text{)/(}}{\text{XY}\text{)}}$ = $\frac{\text{(}\text{2}\text{)/(}}{\text{XY}\text{)}}$.

XY = 2√3 m.

CD = 7 m.

Use the side lengths stated in this subpart.

Let the contact points on AB, BC, CD and DA be P, Q, R and S respectively.

AP = AS, BP = BQ, CQ = CR and DR = DS, since tangents from each external point are equal.

Adding and regrouping gives AB + CD = AD + BC.

3 + CD = 5 + 5.

CD = 7 m.

AC = 13 m; BC = 15 m.

Use this alternative’s separate circle of radius 4 m.

AF = AD = 6 m and BE = BD = 8 m, by equal tangents.

Let CF = CE = x m.

AB = 6 + 8 = 14 m, AC = 6 + x m and BC = 8 + x m.

Join O to A, B and C. The perpendicular distance from O to every side is 4 m.

Area of △ABC = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$ × 4 × [14 + (6 + x) + (8 + x)].

84 = 2(28 + 2x) = 56 + 4x.

x = 7 m.

AC = 6 + 7 = 13 m and BC = 8 + 7 = 15 m.

∠OAP = 90°.

OA is the radius to the point of contact A of tangent PA.

Therefore OA ⟂ PA, so ∠OAP = 90°.

PA = 9√3 m.

OA ⟂ PA, so △OAP is right-angled at A.

By Pythagoras, PA² = OP² − OA² = 18² − 9² = 243.

PA = √243 = 9√3 m.

∠AOB = 120°.

Join OA, OB and OP.

In right △OAP, cos ∠AOP = $\frac{\text{(}\text{OA}\text{)/(}}{\text{OP}\text{)}}$ = $\frac{\text{(}\text{9}\text{)/(}}{\text{18}\text{)}}$ = $\frac{\text{(}\text{1}\text{)/(}}{\text{2}\text{)}}$.

Hence ∠AOP = 60°.

In right triangles OAP and OBP, OP is common and OA = OB (radii).

Thus △OAP ≅ △OBP by RHS, so ∠POB = ∠AOP = 60°.

∠AOB = ∠AOP + ∠POB = 120°.

PA = PB = 9√3 m.

By RHS congruence of △OAP and △OBP, OP bisects ∠APB.

∠APO = 60°/2 = 30°.

In right △OAP, tan 30° = $\frac{\text{(}\text{OA}\text{)/(}}{\text{PA}\text{)}}$ = $\frac{\text{(}\text{9}\text{)/(}}{\text{PA}\text{)}}$.

PA = 9√3 m.

PB = PA = 9√3 m, since tangents from the same external point P are equal.