PE 2 - Circles | Geometry - Circles

Answer: Option B

Explanation :

Tangents make a right angle with the radius ⇒ ∠APO = ∠AQO = 90°.

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In right ∆APO
AP² + PO² = AO²
⇒ AP² + 5² = 13²
⇒ AP = 12 cm.

Similarly, AQ = AP = 12 cm.

Now AD = AO - radius = 13 - 5 = 8 cm.

Let BP = x, hence, BD = x [Tangents drawn from a point are equal in length]
and, AB = 12 - x

In right ∆BDA, ∠BDA = 90°
AD² + DB² = AB²
⇒ 8² + x² = (12 - x)²
64 + x² = 144 - 24x + x²
⇒ 24x = 80
⇒ x = 10/3 cm

Similalry, BD = CD = 10/3

⇒ AB = BD + CD = 20/3

Hence, option (b).

Answer: Option D

Explanation :

Let the radius of the largest circle be R and that of the two smaller circles be r₁ and r₂.

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When two chords of a circle intersect, product of two parts of both these chords is same.

⇒ PC × CQ = AC × BC
⇒ 2r₁ × 2r₂ = 16 × 16
⇒ r₁ × r₂ = 64

Also, 2r₁ + 2r₂ = 2R
⇒ R = r₁ + r₂
⇒ R² = r₁² + r₂² + 2 × r₁ × r₂
⇒ R² - (r₁² + r₂²) = 2 × r₁ × r₂ = 2 × 64 = 128

Area of the shaded part = πR² - (πr₁² + πr₂²)
= π[R² - (r₁² + r₂²)]
= π × 128
= 128π

Hence, option (d).

Answer: Option D

Explanation :

∠AOB = 60° and AO = BO = radius of the circle.
⇒ Triangle AOB is an equilateral triangle.

Since CE || OA, DE || BO and ∠AOB = 60°, it implies that ∠CED is also equal to 60°.
Also, CE = ED, hence, triangle CED is also an equilateral triangle.

Height of triangle CED = radius of the circle = √3

Height of an equilateral triangle = √3 × side/2 = √3
⇒ Side of triangle CED = 2 cm

Hence, option (d).

Answer: 19

Explanation :

Let PO = QO = x (radius of the semicircle)

∆POB ≈ ∆QCO [∠BPO = ∠OQC = 90° and ∠POB = ∠QCO]

⇒ $\frac{\mathrm{PO}}{\mathrm{QC}}$ = $\frac{32}{24}$ 

⇒ $\frac{x}{\mathrm{QC}}$ = $\frac{4}{3}$

QC = $\frac{3x}{4}$ 

Now, in ∆QCO

⇒ OC² = QO² + QC²
⇒ 24² = x² + (3x/4)²
⇒ 24² × 16 = 25x²
⇒ 24 × 4 = 5x
⇒ x = 19.2

Hence, 19.

Answer: Option B

Explanation :

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Let ∠ACD = x

Now, ∠ACD = ∠ABD = x (angle subtended by same chord AD)

∠ACB = 90° (angle subtended by diameter AB on circle will be right angle)

⇒ ∠DCB = 90 - x

Now, In triangle DCB, DC = DB
⇒ ∠DCB = ∠DBC = 90 - x
⇒ ∠ABC = ∠DBC - ∠DBO = 90 - x - x = 90 - 2x

Now, in triangle OCB, 
external ∠AOD = ∠OCB + ∠OBC
⇒ 30 = (90 - x) + (90 - 2x)
⇒ 3x = 150
⇒ x = 50°

∠ADC = ∠ABC = 90 - x = 40°

Hence, option (b).

Answer: Option A

Explanation :

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In the figure drawn above, XR = OM = 5 cm (radius)
∴ PX = 9 - 5 = 4 cm.

Now, In right triangle PXO, PX = 4, PO = 5
⇒ OX = 3 cm (pythagorean triplet)

Now, PR || OM || ST and RQ and RQ are two common transversals, hence
RM : MT = PO : OS = 1 : 1
⇒ RM = MT
⇒ OX = SY = 3 cm.

Now, In right triangle OYS, SY = 3, SO = 5
⇒ OY = 4 cm (pythagorean triplet)

Now, YM = OM - OY = 5 - 4 = 1 cm.

Also, YM = ST = 1 cm.

Hence, option (a).

Answer: Option B

Explanation :

Let us join QR and SO

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OP is the diameter of smaller circle, hence ∠OSP = 90°
QP is the diameter of larger circle, hence ∠QRP = 90°

Now, ∆PQR ≈ ∆POS [∠OSP = ∠QRP = 90° and ∠OPS is common angle]

⇒ PR : PS = PQ : PO = 2 : 1

⇒ 34 : PS = 2 : 1

∴ PS = 17 cm.

Hence, option (b).

Answer: Option B

Explanation :

Let us join BD.

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∠ABD = 90° (angle subtended by diameter on cirlce is right angle)

In ∆ABD and ∆AEC
∠ABD = ∠AEC = 90°
∠BDA = ∠BCA (angle subtened by same chord AB)

∴ ∆ABD ≈ ∆AEC (by AA rule)

⇒ AB : AE = AD : AC
⇒ 10 : 4 = AD : 6
∴ AD = 15 cm

∴ Radius = 15/2 = 7.5 cm.

Hence, option (b).

Answer: Option C

Explanation :

∆ABC is a right triangle, hence AB = 25 cm (15, 20 and 25 is a pythagorean triplet)

∴ CD = $\frac{\mathrm{AC}\times \mathrm{BC}}{\mathrm{AB}}$ = $\frac{15\times 20}{25}$ = 12 cm

In right ∆ACD
AD = 9 cm (9, 12 and 15 is a pythagorean triplet)

Inradius = $\frac{\mathrm{AD}+\mathrm{CD}-\mathrm{AC}}{2}$ = $\frac{9+12-15}{2}$ = 3 cm

In right ∆ACD
BD = 25 - 9 = 16 cm.

Inradius = $\frac{\mathrm{CD}+\mathrm{BD}-\mathrm{BC}}{2}$ = $\frac{12+16-20}{2}$ = 4 cm

In ∆PQR
PR = 3 + 4 = 7
QR = 4 - 3 = 1

PQ² = PR² + QR²
⇒ PQ = √50 = 5√2

Hence, option (c).

Answer: Option B

Explanation :

Let us draw EP perpendicular to AB and join EA.

Let AP = x

In ∆BEP, ∠EBP = ∠BEP = 45°
∴ BP = EP = 3 + x

In ∆AEP
⇒ AE² = AP² + PE²
⇒ 6² = (x)² + (3 + x)²
⇒ 36 = x² + 9 + x² +  6x
⇒ 2x² + 6x - 27 = 0
⇒ x = $\frac{-6+6\sqrt{7}}{4}$ = $\frac{3\sqrt{7}-3}{2}$

⇒ EB = √2(3 + x) = $\sqrt{2}\left(3+\frac{3\sqrt{7}-3}{2}\right)$ = $\frac{3\sqrt{7}+3}{\sqrt{2}}$

⇒ ED = EB - DB =  $\frac{3\sqrt{7}+3}{\sqrt{2}}$ - 3 = $\frac{3\sqrt{7}-3\sqrt{2}+3}{\sqrt{2}}$

Hence, option (b).