Geometry Solved Examples- 1 - CSAT Preparation - UPSC PDF Download

Q1: O is a point in the interior of ΔABC such that OA = 12 cm, OC = 9 cm, ∠AOB = ∠BOC = ∠COA and ∠ABC = 60°. What is the length (in cm) of OB?
(a) 6√2
(b) 4√6
(c) 6√3
(d) 4√3
Ans: (c)
Given:
O is a point in the interior of ΔABC
OA = 12 cm
OC = 9 cm
∠AOB = ∠BOC = ∠COA
∠ABC = 60°
Concept used:
An angle around a point will always make a 360°
The sum of all the angles of a triangle is 180°
Calculation:

Explanation:
The three equal angles around O imply each central angle is 360°/3 = 120°.
Let ∠OBA = θ. In ΔAOB, the angles give:
θ + ∠BAO + 120° = 180° ⇒ ∠BAO = 60° − θ.
In ΔBOC, ∠OBC = ∠ABC − ∠OBA = 60° − θ. Triangle BOC then gives:
(60° − θ) + 120° + ∠OCB = 180° ⇒ ∠OCB = θ.
Thus ∠BAO = ∠OBC and ∠OBA = ∠OCB. By AA, ΔAOB ∼ ΔBOC.
From similarity, OA/OB = OB/OC ⇒ OB^2 = OA × OC.
So OB^2 = 12 × 9 = 108 ⇒ OB = √108 = 6√3 cm.
∴ The required value of OB is 6√3 cm.

Q2: O is the centre of the circle. A tangent is drawn which touches the circle at C. If ∠AOC = 80°, then what is the value (in degrees) of ∠BCX?

(a) 80
(b) 30
(c) 40
(d) 50
Ans: (d)
Given:
∠AOC = 80°
Concept used:
The angle subtended by a chord at the centre is twice the angle subtended by it at any point on the circle.
Calculation:

Explanation:
Chord AC subtends ∠AOC = 80° at the centre, so the angle subtended by AC at any point on the circumference is half of this:
∠ABC = 1/2 × ∠AOC = 40°.
In isosceles triangle OBC (OB = OC = radius), base angles are equal, hence ∠OBC = ∠OCB = 40°.
The tangent at C is perpendicular to radius OC, so ∠XCO = 90°. The required angle is the difference between ∠XCO and ∠OCB:
∠BCX = ∠XCO − ∠OCB = 90° − 40° = 50°.
∴ The value of ∠BCX is 50°.

Q3: In ΔPQR, ∠Q = 85° and ∠R = 65°. Points S and T are on the sides PQ and PR, respectively such that ∠STR = 95° and the ratio of the QR and ST is 9 : 5. If PQ = 21.6 cm, then the length of PT is:
(a) 12 cm
(b) 10.5 cm
(c) 9 cm
(d) 9.6 cm
Ans: (a)
Given:
QR : ST = 9 : 5, PQ = 21.6 cm,
∠Q = 85°, ∠R = 65 and ∠STR = 95°
Concept Used:
Two triangles are similar if two angles are equal (AA similarity).
Calculation:

Explanation:
From the figure, ΔPTS ∼ ΔPQR by angle-angle correspondence (the given angles match).
Thus QR/TS = PQ/PT.
Substitute the given ratio 9/5 and PQ = 21.6 cm:
9/5 = 21.6 / PT ⇒ PT = (21.6 × 5) / 9 = 12 cm.
∴ The length of PT is 12 cm.

Q4: Chords AB and CD of a circle, when produced, meet at a point P outside the circle, If AB = 6 cm, CD = 3 cm and PD = 5 cm, then PB is equal to∶
(a) 6 cm
(b) 4 cm
(c) 5 cm
(d) 6.25 cm
Ans: (b)

Explanation:
Use the external chord (power of a point) theorem: PB × PA = PD × PC.
Let PB = x, then PA = x + AB = x + 6 and PC = PD + CD = 5 + 3 = 8.
So x(x + 6) = 5 × 8 = 40.
⇒ x^2 + 6x − 40 = 0.
Factorise: (x + 10)(x − 4) = 0 ⇒ x = −10 (discard) or x = 4.
Hence PB = 4 cm.

Q5: In the given figure, B and C are the centres of the two circles. ADE is the common tangent to the two circles. If the ratio of the radius of both the circles is 3 : 5 and AC = 40, then what is the value of DE?

(a) 3√15
(b) 5√15
(c) 6√15
(d) 4√15
Ans: (d)
Concept Used:
Tangents are always perpendicular to the radius at the point of contact.
Calculation:

Explanation:
Let the radii be 3x and 5x. Then BC = 3x + 5x = 8x and given AC = 40. From similarity of right triangles (or using given ratios) we get AB : AC = 3 : 5, so BC = 2/5 × 40 = 16.
Thus 8x = 16 ⇒ x = 2. So BD = 3x = 6 and EC = 5x = 10.
In right triangle AEC, AE^2 = AC^2 − EC^2 = 40^2 − 10^2 = 1600 − 100 = 1500 ⇒ AE = 10√15.
Since AD : DE = 3 : 2 (from similar triangles or radii ratio), DE = (2/5) × AE = (2/5) × 10√15 = 4√15.
∴ DE = 4√15.

Q6: If D and E are points on the sides AB and AC respectively of a triangle ABC such that DE||BC. If AD = x cm, DB = (x - 3) cm, AE = (x + 3) cm and EC = (x - 2) cm, then what is the value (in cm) of x?
(a) 3
(b) 3.5
(c) 4
(d) 4.5
Ans: (d)
The given triangle is shown below,

Considering similar triangles ∆ABC and ∆ADE,
⇒ AB/AD = AC/AE
⇒ (AD + DB)/AD = (AE + EC)/AE
⇒ (x + x - 3)/x = (x + 3 + x - 2)/(x + 3)
⇒ (2x - 3)/x = (2x + 1)/(x + 3)
⇒ (2x - 3)(x + 3) = x(2x + 1)
⇒ 2x² + 6x - 3x - 9 = 2x² + x
⇒ 2x = 9
∴ x = 9/2 = 4.5

Explanation:
Because DE ∥ BC, triangles ADE and ABC are similar. Write the ratio of corresponding sides and solve:
(AD + DB)/AD = (AE + EC)/AE ⇒ (2x − 3)/x = (2x + 1)/(x + 3).
Cross-multiply and simplify:
(2x − 3)(x + 3) = x(2x + 1) ⇒ 2x^2 + 3x − 9 = 2x^2 + x ⇒ 2x = 9 ⇒ x = 4.5 cm.

Q7: Circumcentre of ΔABC is O. If ∠BAC = 75° and ∠BCA = 80°, then what is the value (in degrees) of ∠OAC?
(a) 45
(b) 65
(c) 90
(d) 95
Ans: (b)
Given: 
In ΔABC,
⇒ ∠BAC + ∠BCA + ∠ABC = 180°
⇒ ∠ABC = 180° - 75° - 80°
⇒ ∠ABC = 25°
Since O is the circumcentre
⇒ 2 × ∠ABC = ∠AOC
⇒ ∠AOC = 50°
In ΔAOC, AO = OC (radius of circle)
Hence, ΔAOC is an isosceles Δ
⇒ ∠OAC + ∠ACO + ∠AOC = 180°
⇒ ∠OAC + ∠ACO = 180° - 50°
⇒ 2 × ∠OAC = 130°
⇒ ∠OAC = 65°
∴ the correct option is 2)

Explanation:
Compute the third angle: ∠ABC = 180° − 75° − 80° = 25°. The central angle subtending arc AC is twice the inscribed angle at B, so ∠AOC = 2×25° = 50°. Triangle AOC is isosceles (AO = OC), so the base angles are equal and 2×∠OAC = 180° − 50° = 130°, giving ∠OAC = 65°.

Q8: In the given figure, ABC is a triangle. The bisectors of internal DB and external DC interest at D. If ∠BDC = 48°, then what is the value (in degrees) of ∠A?

(a) 48
(b) 96
(c) 100
(d) 114
Ans: (b)
Given:
The bisectors of internal DB and external DC intersect at D.
∠BDC = 48°
Calculation:

Explanation:
Let x and y denote the indicated adjacent angles so that exterior-angle relations apply. From the exterior angle property in ΔBDC we have y + 48 = x, hence x − y = 48° ......(i).
In ΔABC, angle at C equals 180° − 2y − ∠A ......(ii).
From ΔBCD, angle at C is 180° − y − x − 48 ......(iii).
Equate (ii) and (iii): 180° − 2y − ∠A = 180° − y − x − 48 ⇒ ∠A = x − y + 48°. Using (i), ∠A = 48° + 48° = 96°.
∴ ∠A = 96°.

Q9: Points P, Q, R, S and T lie in this order on a circle with centre O. If chord TS is parallel to diameter PR and ∠RQT= 58°, then find the measure (in degrees) of ∠RTS.
(a) 45
(b) 29
(c) 32
(d) 58
Ans: (c)
Given:
Points P, Q, R, S and T lie in this order on a circle with centre O
chord TS is parallel to diameter PR
∠RQT = 58°

Calculation:
PR is a diameter
∠PTR = 90° .....(angle inscribe in a semicircle)
also ∠TPR = 58°   .....(angle formed on same chord)
so, ∠PRT = 180° - ∠PTR - ∠TPR
⇒ ∠PRT = 180° - 90° - 58°
⇒ ∠PRT = 32°
∠PRT = ∠RTS = 32°   (Alternate interior angle as chord TS is parallel to diameter PR)
∴ The measure of ∠RTS is 32.

Explanation:
PR is a diameter so ∠PTR (inscribed angle on semicircle) = 90°. The angle subtended by chord TR at P equals 58° (given by ∠TPR). Hence ∠PRT = 180° − 90° − 58° = 32°. Because TS ∥ PR, angle ∠RTS equals alternate interior angle ∠PRT = 32°.

Q10: In the given figure, a circle inscribed in ∆PQR touches its sides PQ, QR and RP at points S, T and U, respectively. If PQ = 15 cm, QR = 10 cm, and RP = 12 cm, then find the lengths of PS, QT and RU?

(a) PS = 6.5 cm, QT = 8.5 cm and RU = 3.5 cm
(b) PS = 3.5 cm, QT = 6.5 cm and RU = 8.5 cm
(c) PS = 8.5 cm, QT = 6.5 cm and RU = 3.5 cm
(d) PS = 8.5 cm, QT = 3.5 cm and RU = 6.5 cm
Ans: (c)
Let PS be x cm, then QS = (15 - x) cm
PS = PU, QS = QT, RT = RU [tangents]
⇒ QT = (15 - x) cm
⇒ RT = 10 - (15 - x) = x - 5
⇒ RU = (x - 5)
⇒ PU = 12 - x + 5 = 17 - x
⇒ 17 - x = x
⇒ 2x = 17
⇒ x = 17/2
⇒ x = 8.5 cm
⇒ PS = 8.5
⇒ QT = 15 - 8.5 = 6.5
⇒ RU = 8.5 - 5 = 3.5

Explanation:
Tangents from a point to a circle are equal. Set PS = PU = x. Then QS = QT = 15 − x. From side QR = 10, RT = 10 − QT = 10 − (15 − x) = x − 5, so RU = x − 5. The side RP = 12 gives PU + RU = x + (x − 5) = 12 ⇒ 2x − 5 = 12 ⇒ 2x = 17 ⇒ x = 8.5 cm.
Thus PS = 8.5 cm, QT = 15 − 8.5 = 6.5 cm and RU = 8.5 − 5 = 3.5 cm.

Q11: ΔABC, BE ⊥ AC, CD ⊥ AB and BE and CD intersect each other at O. The bisectors of ∠OBC and ∠OCB meet At P. If ∠BPC = 148°, then what is the measure of ∠A?
(a) 28°
(b) 32°
(c) 64°
(d) 56°
Ans: (c)

As we know,
∠BPC = 90° + ∠BOC/2
⇒ 148° = 90° + ∠BOC/2
⇒ ∠BOC/2 = 148° - 90° = 58°
⇒ ∠BOC = 58° × 2 = 116°
⇒ ∠BOC = ∠DOE = 116° [opposite angle]
In quadrilateral ADOE
∠DAE + ∠ADO + ∠DOE + ∠OEA = 360
⇒ ∠DAE + 90° + 116° + 90° = 360°
⇒ ∠DAE = 360° - 296° = 64°

Explanation:
Point P is the intersection of internal bisectors of ∠OBC and ∠OCB, so angle at P satisfies ∠BPC = 90° + (1/2)∠BOC. Given ∠BPC = 148°, therefore ∠BOC = 2(148° − 90°) = 116°.
∠BOC is vertically opposite ∠DOE, so in quadrilateral ADOE the sum of angles is 360°: ∠DAE + 90° + 116° + 90° = 360°. Hence ∠DAE = 64°, which is ∠A of ΔABC.

Q12: ABCD is a cyclic quadrilateral whose diagonals intersect at P. If AB = BC, ∠DBC = 70° and ∠BAC = 30°, then the measure of ∠PCD is:
(a) 50°
(b) 35°
(c) 55°
(d) 30°
Ans: (a)

In ΔABC,
If AB = BC
then ∠BAC = ∠BCA = 30°
⇒ ∠BAC + ∠BCA + ∠ABC = 180°
⇒ ∠ABC = 180° - 30° - 30° = 120°
⇒ ∠ABC = ∠ABD + ∠DBC
⇒ 120° = ∠ABD + ∠DBC
⇒ ∠ABD = 50°
As we know,
⇒ ∠ABD = ∠ACD = 50° (angles drawn from the same base to the circumference of circle)
and we can write ∠ACD = ∠PCD
or ∠PCD = 50°

Explanation:
Since AB = BC, triangle ABC is isosceles with base AC, so ∠BAC = ∠BCA = 30°. Therefore ∠ABC = 180° − 30° − 30° = 120°. Given ∠DBC = 70°, we get ∠ABD = 120° − 70° = 50°. In the cyclic quadrilateral, angles subtended by the same chord are equal, so ∠ABD = ∠ACD = 50°. As P lies on diagonal AC, ∠PCD = ∠ACD = 50°.

Q13: An equilateral triangle of area 300 cm2 is cut from its three vertices to form a regular hexagon. Area of hexagon is what percent of the area of triangle?(Each side of the regular hexagon is 1/3 rd the original side of equilateral triangle)
(a) 66.66%
(b) 33.33%
(c) 83.33%
(d) 56.41%
Ans: (a)
Each side of the regular hexagon is 1/3 rd the original side of equilateral triangle
⇒ Area of regular hexagon = (3√3/2) × (side)²
⇒ (3√3/2) × (side of triangle/3)²
⇒ (√3/6) × side of triangle²
⇒ (2/3) × (√3/4) × side of triangle²
⇒ 2/3 × Area of equilateral triangle = (2/3) × 100 = 66.66%
∴ Required percentage is 66.66%

Explanation:
Let the triangle side be s. Area of the equilateral triangle = (√3/4) s^2 = 300.
Side of the hexagon = s/3. Area of a regular hexagon = (3√3/2) × (side_hex)^2 = (3√3/2) × (s/3)^2 = (√3/6) s^2.
Ratio (hexagon area)/(triangle area) = (√3/6) s^2 ÷ (√3/4) s^2 = (1/6)/(1/4) = 4/6 = 2/3 = 66.66%.

Q14: ΔXYZ is similar to ΔPQR. If ratio of Perimeter of ΔXYZ and Perimeter of ΔPQR is 4 : 9 and if PQ = 27 cm, then what is the length of XY (in cm)?
(a) 9
(b) 12
(c) 16
(d) 15
Ans: (b)
Given that ΔXYZ is similar to ΔPQR
Since they are similar we know,
(Perimeter of ∆XYZ)/(perimeter of ∆PQR) = (length of XY)/(length of PQ)
Ratio of Perimeter of ΔXYZ and Perimeter of ΔPQR is 4 : 9
⇒ 4/9 = (length of XY)/27
⇒ Length of XY = 4/9 × 27= 12 cm
∴ Length of XY is 12 cm

Explanation:
Similarity implies corresponding linear measures are in the same ratio as perimeters. So XY/PQ = 4/9. Given PQ = 27 cm, XY = (4/9) × 27 = 12 cm.

Q15: In the given figure, triangle ABC is drawn such that AB is tangent to a circle at A whose radius is 10cm and BC passes through centre of the circle. Point C lies on the circle. If BC = 36cm and AB = 24cm, then what is the area (in cm2) of triangle ABC?

(a) 134.5
(b) 148
(c) 166.15
(d) 180
Ans: (c)
Join OA where OA = 10 cm
Here, AO is perpendicular to AB
⇒ Area of ∆ABC = Area of ∆OAB + Area of ∆AOC
⇒ Area of ∆ABC = 1/2(OA)(AB) + 1/2(OC)(OA)sin(∠AOC)
⇒ Area of ∆ABC = 1/2(10)(24) + 1/2(10)(10) sin(∠AOC)
⇒ Area of ∆ABC = 120 + 50 sin(∠AOC)
In right angled ∆OAB
⇒ tan(∠AOB) = 24/10 = 12/5
⇒ ∠AOB = 67.38°
So, ∠AOC = 180° - 67.38° = 112.62°
Hence, Area of ∆ABC = 120 + 50 (0.93) = 120 + 46.15 = 166.15
∴ The area (in cm²) of triangle ABC is 166.15.

Explanation:
Let O be the circle centre and OA = 10 cm. Since AB is tangent at A, OA ⟂ AB, so area of ΔOAB = (1/2) × OA × AB = (1/2) × 10 × 24 = 120 cm2.
Also, triangle AOC has sides OA = OC = 10 cm and included angle ∠AOC. Note that ∠AOB in right triangle OAB satisfies tan ∠AOB = AB/OA = 24/10 = 12/5, so ∠AOB ≈ 67.38°. Then ∠AOC = 180° − ∠AOB ≈ 112.62°.
Area of ΔAOC = (1/2) × OA × OC × sin ∠AOC = (1/2) × 10 × 10 × sin 112.62° ≈ 50 × 0.9231 ≈ 46.15 cm2.
Total area ≈ 120 + 46.15 = 166.15 cm2.
∴ The area of triangle ABC is approximately 166.15 cm2.