MCQ Questions on Geometry for Competitive Exams

QP and RS extended to M to make an equilateral ΔMQR.
∠SPM = 180° - 150° = 30°
In ΔMPS

PS = 9 (Given)
$$\eqalign{ & \tan {30^ \circ } = \frac{{{\text{MS}}}}{{{\text{PS}}}} \cr & \frac{1}{{\sqrt 3 }} = \frac{{{\text{MS}}}}{9} \cr & {\text{MS}} = \frac{9}{{\sqrt 3 }} \cr & {\text{MS}} = 3\sqrt 3 \cr} $$
Area of quadrilateral PQRS = Area of Equilateral Δ (MQR) - Area of ΔMSP
$$\eqalign{ & = \frac{{\sqrt 3 }}{4} \times {\left( {18} \right)^2} - \frac{1}{2} \times 9 \times 3\sqrt 3 \cr & = \frac{{\sqrt 3 }}{4} \times 18 \times 18 - \frac{{27\sqrt 3 }}{2} \cr & = \frac{{27\sqrt 3 }}{2}\left[ {6 - 1} \right] \cr & = \frac{{135\sqrt 3 }}{2}{\text{ c}}{{\text{m}}^2} \cr} $$

∠PQO = 2n° (by exterior angle theorem)
PO = OQ (Radius)
∠OPQ = ∠OQP = 2n° (by exterior angle theorem)
∠POS = ∠OPQ + ∠ORP
m = 2n° + n°
m = 3n°

According to question
Given:

OO' = 13 cm
OA = 7 cm
O'B = 2 cm
∴ Length of direct common tangent
$$\eqalign{ & {\text{AB}} = \sqrt {{{\left( {{\text{OO}}'} \right)}^2} - {{\left( {{\text{R}} - {\text{r}}} \right)}^2}} \cr & {\text{AB}} = \sqrt {{{\left( {13} \right)}^2} - {{\left( {7 - 2} \right)}^2}} \cr & {\text{AB}} = \sqrt {169 - 25} \cr & {\text{AB}} = \sqrt {144} \cr & {\text{AB}} = 12{\text{ cm}} \cr} $$

P - B = 17 (Given)
Area = $$\frac{1}{2}$$ × B × P
BP = 2 × 84
BP = 168 . . . . . . (i)
P - B = 17
Squaring both side
P² + B² - 2BP = 289
H² - 2(168) = 289 (∴ H² = P² + B²)
H² = 289 + 336
H = $$\sqrt {625} $$
H = 25 cm
(P + B)² = P2 + B2 + 2PB
(P + B)² = 625 + 2(168)
(P + B)² = 625 + 336
P + B = $$\sqrt {961} $$
P + B = 31
∴ H + P + B = 31 + 25 = 56

$$\eqalign{ & \angle {\text{B}} = \frac{4}{7}\angle {\text{A}} \cr & \frac{{\angle {\text{B}}}}{{\angle {\text{A}}}} = \frac{4}{7} \cr} $$
By exterior angle theorem
7x + 4x = 132°
11x = 132°
x = 12°
∠A = 7 × 12° = 84°

∵ OC = CD = radius
According to property of circle

Same arc angle
Make line AD

Angle BDA = 90° because of semicircle property
∠P = 90° - 30° = 60°

AB & AC ⇒ tangent
AB = AC
ΔAPQ
x + x + a + a = 30
x + a = 15
AB = x + a = 15

Let bigger circle radius = R
Let Smaller circle radius = r
Am = 15
Cm = 12
According to the question
R² - 15² = r² - 12²
[By Pythagoras theorem]
R² - r² = 15² - 12²
R² - r² = 81
$$\sqrt {{{\text{R}}^2} - {{\text{r}}^2}} $$   = 9
MN = 2NP = 9 × 2 = 18 cm

$$\eqalign{ & {\text{Area of }}\Delta ABC = \frac{1}{2}\sin \theta \times 12 \times 10 \cr & {\text{Let }}\frac{1}{2}\sin \theta = x \cr & {\text{Area of }}\Delta ABC = 120x \cr & {\text{Same as,}} \cr & {\text{Area of }}\Delta BEF = 10x \cr & {\text{Area of }}\Delta BDG = 9 \times 8x = 72x \cr & M = 72x - 10x = 62x \cr & N = 120x \cr & \frac{M}{N} = \frac{{62x}}{{120x}} = \frac{{31}}{{60}} \cr} $$