Geometry Solved Examples- 2 - CSAT Preparation - UPSC PDF Download

Question 11: What is the ratio of longest diagonal to the shortest diagonal in a regular octagon? 
A. √3 : 1
B. 2 : 1
C. 2 : √3
D. √2 : 1

Answer. √2 : 1
Explanation.

Consider a regular octagon with centre O and circumradius R.
The longest diagonal (for example AE) joins two opposite vertices and subtends 180° at the centre, so its length is the diameter:
AE = 2R.
The shortest non-side diagonal (for example AC) joins vertices two steps apart and subtends 90° at the centre, so its length is the chord for 90°:
AC = 2R sin(90°/2) = 2R sin(45°) = 2R · (√2/2) = R√2.
Therefore,
AE / AC = (2R) / (R√2) = 2 / √2 = √2.
So the ratio of the longest diagonal to the shortest diagonal is √2 : 1.

Hence, the answer is √2 : 1

Choice D is the correct answer.

Question 12: Find the altitude to side AC of triangle with side AB = 20 cm, AC = 20 cm, BC = 30 cm. 
A. 10√7
B. 8√7
C. 7.5√7
D. 15√7

Answer. 7.5√7

Explanation.

Given triangle ABC with AB = AC = 20 cm and base BC = 30 cm, the triangle is isosceles with apex A.
Drop altitude AD from A to BC. This altitude bisects BC because the triangle is isosceles.
So DC = DB = 15 cm.
Apply Pythagoras in triangle ADC:
AD² + DC² = AC²
AD² + 15² = 20²
AD² = 400 - 225 = 175
AD = 5√7 (this is the altitude to BC).
Area of ΔABC = (1/2) × BC × AD = (1/2) × 30 × 5√7 = 75√7.
The altitude from B to side AC (call it h) satisfies Area = (1/2) × AC × h.
So (1/2) × 20 × h = 75√7
h = (75√7 × 2) / 20 = 150√7 / 20 = 7.5√7.

Hence, the answer is 7.5√7.

Choice C is the correct answer.

Question 13: ABCDEF is a regular hexagon inscribed inside a circle. If the shortest diagonal of the hexagon is of length 3 units, what is the area of the shaded region. 

A. 1/6(3π - (9√3)/2)
B. 1/6(2π - (6√3)/2)
C. 1/6(3π - (8√3)/2)
D. 1/6(6π - (15√3)/2)

Answer. 1/6(3π - (9√3)/2)

Explanation.

Let the side of the regular hexagon be a.
In a regular hexagon each pair of vertices separated by one vertex (two steps) are at distance √3·a (this is the shorter diagonal across one vertex).
Given the shortest diagonal = 3, so √3·a = 3 ⇒ a = 3/√3 = √3.
For a regular hexagon, the circumradius R equals the side length, so R = a = √3.
Area of the circumscribed circle = πR² = π(√3)² = 3π.
Area of the regular hexagon = (3√3/2) a² = (3√3/2) · (√3)² = (3√3/2) · 3 = (9√3)/2.
Area between the circle and the hexagon = 3π - (9√3)/2.
The shaded region (as shown) is one-sixth of this difference, so its area is
(1/6) [3π - (9√3)/2] = 1/6(3π - (9√3)/2).

Hence, the answer is 1/6(3π - (9√3)/2)

Choice A is the correct answer.

Question 14: A circle of radius 5 cm has chord RS at a distance of 3 units from it. Chord PQ intersects with chord RS at T such that TS = 1/3 of RT. Find minimum value of PQ. 
A. 6√3
B. 4√3
C. 8√3
D. 2√3

Answer. 4√3

Explanation.

Let O be the centre and OM be perpendicular distance to chord RS where OM = 3 and radius OS = 5.
Half-length of RS is MS = √(OS² - OM²) = √(25 - 9) = √16 = 4.
Hence RS = 8 cm.
Let RT = x and TS = (1/3) x. Then RT + TS = RS gives x + x/3 = 8 ⇒ (4/3) x = 8 ⇒ x = 6, TS = 2.
By the intersecting-chords theorem, RT · TS = PT · TQ.
So PT · TQ = 6 × 2 = 12.
For fixed product, the sum PT + TQ is minimized when PT = TQ (AM-GM).
Therefore PQ = PT + TQ ≥ 2√(PT·TQ) = 2√12 = 4√3.
Minimum possible PQ = 4√3.

Hence, the answer is 4√3.

Choice B is the correct answer.

Question 15: Triangle has perimeter of 6 + 2√3 . One of the angles in the triangle is equal to the exterior angle of a regular hexagon another angle is equal to the exterior angle of a regular 12-sided polygon. Find area of the triangle. 
A. 2√3
B. √3
C. √3/2
D. 3

Answer. 2√3

Explanation.

Exterior angle of a regular hexagon = 360°/6 = 60°.
Exterior angle of a regular 12-gon = 360°/12 = 30°.
Hence two angles of the triangle are 60° and 30°, so the third angle is 90° and the triangle is right-angled at the third vertex.
In a 30°-60°-90° triangle the sides are in ratio x : √3x : 2x (opposite 30°, 60°, 90° respectively).
Perimeter = x + √3x + 2x = x(3 + √3) = 6 + 2√3.
Solve for x: x = (6 + 2√3)/(3 + √3) = 2.
Thus the sides are 2 (opposite 30°), 2√3 (opposite 60°) and 4 (hypotenuse).
Area = (1/2) × (two perpendicular sides) = (1/2) × 2 × 2√3 = 2√3.

Hence, the answer is 2√3.

Choice A is the correct answer.

Question 16: Area of a Rhombus of perimeter 56 cms is 100 sq cms. Find the sum of the lengths of its diagonals. 
A. 33.40
B. 34.40
C. 31.20
D. 32.30

Answer. 34.40

Explanation.

Let each side be a. Given 4a = 56 ⇒ a = 14.
Let the diagonals be d1 and d2. Area of rhombus = (1/2) d1 d2 = 100 ⇒ d1 d2 = 200.
Diagonals of a rhombus are perpendicular and their half-lengths form right triangles with side a, so
(d1/2)² + (d2/2)² = a² ⇒ (d1² + d2²)/4 = a² ⇒ d1² + d2² = 4a² = 4·196 = 784.
(d1 + d2)² = d1² + d2² + 2d1 d2 = 784 + 2·200 = 1184.
d1 + d2 = √1184 = 34.40 (approx.).
Therefore the sum of the diagonals is 34.40 cm.

Hence, the answer is 34.40.

Choice B is the correct answer.

Question 17: Rhombus has a perimeter of 12 and one angle = 120°. Find its area. 
A. 9 * (√3)/2
B. 3 * (√3)/2
C. 9 * √3
D. 18 * √3

Answer. 9 * (√3)/2

Explanation.

Let the side be a. Given perimeter 4a = 12 ⇒ a = 3.
Area of a rhombus = a² sin θ, where θ is any interior angle.
Here θ = 120°, so sin 120° = sin 60° = √3/2.
Area = a² sin θ = 9 · (√3/2) = 9 · (√3/2).

Hence, the answer is 9 * (√3/2).

Choice A is the correct answer.

Question 18: Circle with center O and radius 25 cms has a chord AB of length of 14 cms in it. Find the area of triangle AOB 
A. 144 cm²
B. 121 cm²
C. 156 cm²
D. 168 cm² 

Answer. 168 cm² 

Explanation. 

Triangle AOB is isosceles with OA = OB = 25 cm and base AB = 14 cm.
For an isosceles triangle with equal sides a and base b, the altitude to the base is √(a² - (b/2)²).
Here a = 25, b = 14, so altitude from O to AB equals √(25² - 7²) = √(625 - 49) = √576 = 24.
Area of ΔAOB = (1/2) × base × altitude = (1/2) × 14 × 24 = 7 × 24 = 168 cm².

Choice D is the correct answer.

Question 19: Two mutually perpendicular chords AB and CD intersect at P. AP = 4, PB = 6, CP = 3. Find radius of the circle. 
A. 31.25^(1/2)
B. 37.5^(1/2)
C. 26^(1/2)
D. 52^(1/2)

Answer. 31.25^(1/2)

Explanation.

By intersecting chords theorem, AP·PB = CP·PD.
So 4 × 6 = 3 × PD ⇒ PD = 24/3 = 8.
Hence AB = AP + PB = 4 + 6 = 10 and CD = CP + PD = 3 + 8 = 11.
Place P at the origin and align AB on the x-axis and CD on the y-axis: then A(-4,0), B(6,0), C(0,3), D(0,-8).
Midpoint of AB is X = ((-4+6)/2, 0) = (1,0).
Midpoint of CD is Y = (0, (3-8)/2) = (0, -2.5).
Perpendicular bisectors of AB and CD are the vertical line x = 1 and the horizontal line y = -2.5 respectively, so the centre O is at their intersection O(1, -2.5).
Radius = distance OB = √[(6 - 1)² + (0 + 2.5)²] = √[5² + 2.5²] = √[25 + 6.25] = √31.25.
Thus the radius is 31.25^1/2.

Hence, the answer is 31.25^(1/2).

Choice A is the correct answer.

Question 20: Triangle ABC has angles A = 60^° and B = 70^°. The incenter of this triangle is at I. Find angle BIC. 
A. 90^°
B. 130^°
C. 80^°
D. 120^°

Answer. 120^°

Explanation.

Angles of ΔABC sum to 180°, so C = 180° - (A + B) = 180° - 130° = 50°.
The incenter I is intersection of angle bisectors. The angles at I in triangle BIC are formed by the bisectors of ∠B and ∠C.
∠BIC = 180° - (½∠B + ½∠C) = 180° - (35° + 25°) = 180° - 60° = 120°.
(Equivalently, known formula: ∠BIC = 90° + ½∠A = 90° + 30° = 120°.)

Hence, the answer is 120^°.

Choice D is the correct answer.