Geometry — Part 2 MCQs with Explanations

$$\frac{{AB}}{{PQ}} = \frac{2}{3}\left( {{\text{Given}}} \right)$$ $$\eqalign{ & \frac{{{\text{area of }}\Delta ABC}}{{{\text{area of }}\Delta PQR}} = {\left( {\frac{{AB}}{{PQ}}} \right)^2} = \frac{4}{9} \cr & \left[ {{\text{Ratio of similar }}\Delta } \right] \cr & \frac{{{\text{area of }}\Delta ABD}}{{{\text{area of }}\Delta PQS}} = \frac{{\frac{1}{2} \times {\text{area of }}\Delta ABC}}{{\frac{1}{2} \times {\text{area of }}\Delta PQR}} \cr & = \frac{4}{9} = \frac{{B{D^2}}}{{Q{S^2}}} \cr & \left[ {\therefore AD\,\& \,PS\,{\text{are median}}} \right] \cr} $$

$$\frac{{{\text{AD}}}}{{{\text{AB}}}} = \frac{{{\text{AE}}}}{{{\text{AC}}}}$$ (It's mean ΔADE ∽ ΔABC) ∠B = ∠D, ∠C = ∠E ∠D = 78° ∠ADE = ∠ACB + 30° 78° = ∠ACB + 30° ∠ACB = 48° ∠A + ∠D + ∠E = 180° ∠A + 78 + 48 = 180° ∠A = 54°

Let, ∠CAB = x ∠ABC = 2x ∠ACD = 3x = 126° (by exterior angle) 3x = 126° x = 42° ∠CAB = 42°

We know that AB + CD = BC + AD 6.5 + 5.3 = 5.4 + AD 11.8 = 5.4 + AD AD = 11.8 - 5.4 AD = 6.4 cm Note:- AB + CD = (AB + PB) + (DR + RC) . . . . . . (i) The length of tangent drawn from an external point to a circle are equal. AB = CD, AS = BQ, CQ = RC, DR = DS From equation (i) AB + CD = AS + BQ + DS + CQ = (AS + DS) + (BQ + BC) $$\boxed{{\text{AB}} + {\text{CD}} = {\text{AD}} + {\text{BC}}}$$ If a circle is inscribed inside a quadrilateral then the sum of a pair opposite side equal to the sum of another pair of opposite sides. i.e. AB + CD = AD + BC

Given that AP = 22 cm and PQ = 8 cm Made a triangle such that B, is on side PQ and BQ = 2 cm And C is on RQ such that QC = 16 cm Because all the vertices are on sides of PQRA. Now, PQRA is a rectangle so all the angle will be of 90°. ∠ARQ = 90° and RC = RQ - CQ = 22 - 16 cm = 6 cm In right angle ΔARC AC 2 = AR 2 + RC 2 = 8 2 + 6 2 $$\boxed{{\text{AC}} = 10}$$ Now in ΔABC, AC is 10 cm and M, N are the mid point of ΔABC. So, MN = $$\frac{{10}}{2}$$ = 5 cm

Height will be same for both triangles In triangles ADB, if base = 4x and area = 60 cm 2 Area of ΔADB = 60 cm 2 $$\frac{1}{2}$$ × base × height = 60 ⇒ $$\frac{1}{2}$$ × 4x × height = 60 ⇒ height = $$\frac{{60}}{{2x}}$$ ⇒ height = $$\frac{{30}}{x}$$ cm ⇒ Therefore using height Area of ΔADC will be = $$\frac{1}{2}$$ × base × height = $$\frac{1}{2}$$ × 5x × $$\frac{{30}}{x}$$ = 75 cm 2

$$\eqalign{ & \left( {2 + \sqrt 2 } \right){\text{unit}} \to 8\left( {\sqrt 2 + 1} \right) \cr & 1\,{\text{unit}} \to \frac{{8\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}} = 4\sqrt 2 \cr & {\text{AC}} = 4\sqrt 2 \times \sqrt 2 = 8{\text{ cm}} \cr} $$

According to figure ⇒ AD || BC ⇒ ∠ABC = 75° Then ⇒ ∠ABC + ∠ADC = 180° ⇒ 75° + ∠ADC = 180° ⇒ ∠ADC = 180° - 75° ⇒ ∠ADC = 105° ⇒ As we know in a cyclic trapezium ∠ADC + ∠DCB = 180° (AD || BC, corresponding angle) ⇒ 105° + ∠DCB = 180° ⇒ ∠DCB = 75°

Hypotenuses = 26 cm Perimeter = 60 cm Using by pythagoras triplet - 5, 12, 13 13 unit → 26 cm 1 unit → 2 cm Area of triangle = $$\frac{1}{2}$$ × 10 × 24 = 120 cm 2

∠PDQ = 45° + 45° = 90°

AM = 12 AB = 2 × 12 = 24 cm

In ΔAPO 2 unit → 12 1 unit → 6 PT = 6 PQ = 2 × PT = 2 × 6 = 12 cm

∠APO = 60° ∠POA = 30° 1 unit → 12.5 2 unit → 12.5 × 2 = 25 PO = 25 cm

ΔPO 1 A is similar to ΔPO 2 B AO 1 : BO 2 = 2 : 3 O 1 O 2 = 10 = 3 - 2 = 1 unit 1 unit → 10 PO 2 = 3 units → 30 ΔPO 3 C is similar to ΔPO 2 B. BO 2 : CO 3 = 2 : 3 2 units → 30 1 unit → 15 PO 3 = 45 CO 3 = 9 $$\eqalign{ & {\text{PC}} = \sqrt {{{45}^2} - {9^2}} \cr & = \sqrt {2025 - 81} \cr & = \sqrt {1944} \cr & = 18\sqrt 6 \cr} $$

∠QPS = $$\frac{1}{2}$$∠PSR = x (Let) ∠PSR = 2x ∠PQS + ∠QPS = 2x ∠QPR = 78° and ∠PRS = 44° In ΔPQR ∠P + ∠Q + ∠R = 180° 78° + x + 44° = 180° x = 180° - 122° x = 58° ∠PSQ = 180° - 2x = 180° - 2 × 58° = 180° - 116° = 64°

PQ = 12 cm r = 5 cm PQ 2 = PA × PB 12 2 = x(x + 10) 144 = x(x + 10) x 2 + 10x - 144 = 0 x 2 + (18x - 8x) -144 = 0 x(x + 18) - 8(x + 18) = 0 x = 8, x = -18 PA = 8 cm

AD : BD = 6 : 1 ΔADE : trapezium BCDE = 36 : 13

$$\eqalign{ & {\text{Let the length of }}CQ = x\,\& \,\angle PQC = \theta \cr & {\text{In }}\Delta ABQ, \cr & \tan \theta = \frac{{15}}{{24 + x}}.\,.\,.\,.\,.\,\left( {\text{i}} \right) \cr & {\text{Again in }}\Delta PCQ \cr & \tan \theta = \frac{5}{x}\,.\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr & {\text{From equation}}\left( {\text{i}} \right)\,\& \,\left( {{\text{ii}}} \right) \cr & \frac{{15}}{{24 + x}} = \frac{5}{x} \cr & 3x = 24 + x \cr & 2x + 24 \cr & x = 12 \cr & {\text{In }}\Delta ABQ, \cr & {\text{By pythagoras theorem}} \cr & A{Q^2} = A{B^2} + B{Q^2} \cr & = {15^2} + {36^2} \cr & = 225 + 1296 \cr & = 1521 \cr & AQ = 39 \cr} $$

Since, AB 2 = AD × AC AB 2 = 6 × 11 = 66 cm And, BD 2 = AD × DC = 6 × 5 = 30 ∴ AB 2 + BD 2 = 66 + 30 = 96

PA × PB = PT 2 5(5 + x) = 100 x = 15 In right angle triangle PTB, TB 2 = 20 2 - 10 2 TB 2 = $$10\sqrt 3 $$ So, OT $$ = \frac{{10\sqrt 3 }}{2} = 5\sqrt 3 $$

Let ABCD is a square of x unit side Then ∠AOD = 90° Then OD = $$\frac{{\text{x}}}{{\sqrt 2 }}$$ Diagonal of square ABCD = $$\sqrt 2 $$ x Line MB || OD i.e OD = MB = $$\frac{{\text{x}}}{{\sqrt 2 }}$$ ⇒ Then MBOD will be a rectangle become MB || OD, MB = OD = $$\frac{{\text{x}}}{{\sqrt 2 }}$$ BD || MO, MO = BD = $$\sqrt 2 $$ x $${\text{R}} = \sqrt {{{\left( {\frac{{\text{x}}}{{\sqrt 2 }}} \right)}^2} + {{\left( {\sqrt 2 {\text{x}}} \right)}^2}} = \frac{{\sqrt 5 {\text{x}}}}{{\sqrt 2 }}{\text{ Ans}}{\text{.}}$$

Let the side of BC & CA are 3x and 3y QC = 2x QB = x PC = 2y AP = y In triangle AQC AQ 2 = (3y) 2 + (2x) 2 = 9y 2 + 4x 2 BP 2 = (3x) 2 + (2y) 2 = 9x 2 + 4y 2 AB 2 = (3x) 2 + (3y) 2 = 9x 2 + 9y 2 = 9(x 2 + y 2 ) $${\text{Now, }}\frac{{{\text{A}}{{\text{Q}}^2} + {\text{B}}{{\text{P}}^2}}}{{{\text{A}}{{\text{B}}^2}}} = \frac{{13\left( {{x^2} + {y^2}} \right)}}{{9\left( {{x^2} + {y^2}} \right)}} = \frac{{13}}{9}$$

∠P + ∠R = 180° (∵ PQRS is a cyclic quadrilateral) 110° + ∠R = 180° ∠R = 70° ∠R + ∠Q = 180° (supplementary angle) 70° + ∠Q = 180° ∠PQR = 110° Note:- In any cyclic quadrilateral if two sides are parallel, then non-parallel sides are also equal in length. This type of quadrilateral is called an isosceles trapezium. If AB || CD Then AD = BC and ∠A = ∠B, ∠C = ∠D

From figure ∠BOC = 120 ∴ ∠AOC = 180 - 120 = 60 So, ∠ADC = $$\frac{1}{2}$$∠AOC (Angle made on circumference is half of the angle made on centre) = $$\frac{1}{2}$$ × 60 So, ∠ADC = 30°

$$\angle {\text{OAB}} = \frac{{{{180}^ \circ } - {{38}^ \circ }}}{2} = {71^ \circ }$$

$$\eqalign{ & AD{\text{ is angle bisector of }}\angle A \cr & \therefore \frac{{AB}}{{AC}} = \frac{{BD}}{{DC}} \cr & \Rightarrow \frac{6}{{2x - 3}} = \frac{{x - 2}}{x} \cr & \Rightarrow 6x = 2{x^2} - 4x - 3x + 6 \cr & \Rightarrow 2{x^2} - 13x + 6 = 0 \cr & \Rightarrow 2{x^2} - 12x - x + 6 = 0 \cr & \Rightarrow 2x\left( {x - 6} \right) - 1\left( {x - 6} \right) = 0 \cr & \Rightarrow \left( {x - 6} \right)\left( {2x - 1} \right) = 0 \cr & x - 6 = 0 \cr & x = 6 \cr & 2x - 1 = 0 \cr & x = \frac{1}{2}\left( {{\text{not valied}}} \right) \cr & \therefore BD = x - 2 = 6 - 2 = 4 \cr} $$

$$\eqalign{ & \angle QXY = {120^ \circ } \cr & \angle PXY = {60^ \circ } \cr & \Delta PXY \sim \Delta PRQ \cr & \therefore \frac{{PY}}{{PQ}} = \frac{{XY}}{{QR}} \cr & \frac{{18}}{{PQ}} = \frac{9}{{14}} \cr & PQ = \frac{{18 \times 14}}{9} = 28\,{\text{cm}} \cr} $$

AB = 35 cm, BC = 12 cm and AC = 37 cm 12, 35 and 37 are triplet ABC is a right angle triangle at ∠B = 90° Use appoloneous theorem $$\eqalign{ & A{B^2} + B{C^2} = 2\left( {A{D^2} + B{D^2}} \right) \cr & {35^2} + {12^2} = 2 \times {\left( {\frac{{AC}}{2}} \right)^2} + 2B{D^2} \cr & {37^2} = 2 \times {\left( {\frac{{37}}{2}} \right)^2} + 2B{D^2} \cr & 2B{D^2} = {37^2}\left( {1 - \frac{1}{2}} \right) \cr & 2B{D^2} = 1369 \times \frac{1}{2} \cr & B{D^2} = \frac{{1369}}{4} \cr & BD = \frac{{37}}{2} \cr & BG = BD \times \frac{2}{3} \cr & BG = \frac{{37}}{2} \times \frac{2}{3} \cr & BG = \frac{{37}}{3} = 12.3{\text{ cm}} \cr} $$

In ΔACD ∠A + ∠C + ∠D = 180° θ + θ + 52° = 180° 2θ + 52° = 180° θ = $$\frac{{{{128}^ \circ }}}{2}$$ θ = 64°

Let, Angle ABC = x Then, Angle AOC = 2x In Quadrilateral ABCO, 30° + 50° + x + 360° - 2x = 360° ⇒ x = 80° So, 2x = 160°

$$\eqalign{ & \frac{{PQ}}{{PR}} = \frac{2}{3} \cr & \therefore \frac{{{\text{ar }}\Delta PSR}}{{{\text{ar }}\Delta PQR}} = \frac{3}{{2 + 3}} = \frac{3}{5} \cr & 5{\text{ units}} = 180 \cr & 3{\text{ units}} = \frac{{180}}{5} \times 3 = 108 \cr & {\text{Area of }}\Delta PSR = 108{\text{ c}}{{\text{m}}^2} \cr} $$

$$\eqalign{ & PQ \times QR = QS \times PR \cr & 35 \times QR = 28 \times \sqrt {{{\left( {35} \right)}^2} + Q{R^2}} \cr & {\left( {\frac{5}{4}QR} \right)^2} = {\left( {35} \right)^2} + Q{R^2} \cr & \frac{{25}}{{16}}Q{R^2} - Q{R^2} = 1225 \cr & \frac{9}{{16}}Q{R^2} = 1225 \cr & Q{R^2} = \frac{{1225 \times 16}}{9} \cr & QR = \frac{{35 \times 4}}{3} \cr & QR = \frac{{140}}{3} \cr & S{R^2} = Q{R^2} - Q{S^2} \cr & S{R^2} = {\left( {\frac{{140}}{3}} \right)^2} - {\left( {28} \right)^2} \cr & S{R^2} = \frac{{{{14}^2}\left( {{{10}^2} - 9 \times 4} \right)}}{9} \cr & S{R^2} = \frac{{{{14}^2} \times 64}}{9} \cr & SR = \frac{{14 \times 8}}{3} \cr & SR = \frac{{112}}{3} \cr & SR = 37.33{\text{ cm}} \cr} $$

∠A + ∠B + ∠C = 180° . . . . . . . (i) ∠A - ∠B = 33° ∠A = 33° + ∠B ∠B - ∠C = 18° ∠C = ∠B - 18° From equation (i) 33° + ∠B + ∠B + ∠B - 18° = 180° 3°∠B + 15° = 180° ∠B = 55° ∠A = 88° ∠C = 37° The sum of largest and smallest angles of the triangle = 88° + 37° = 125°

(PQ) 2 + (PR) 2 = 2(QD 2 + PD 2 ) (30) 2 + (50) 2 = 2[(18) 2 + PD 2 ] 900 + 2500 = 2[324 + PD 2 ] 3400 = 2[324 + PD 2 ] 1700 = 324 + PD 2 PD 2 = 1376 PD = $$4\sqrt {86} $$ CD = $$\frac{1}{3}$$PD CD = $$\frac{{4\sqrt {86} }}{3}$$

2(x + y + z) = 32 x + y + z = 16 AB - AC = x - y x - y = 2 . . . . . . . (i) AB - BC = z - y z - y = 4 . . . . . . . (ii) Add equation (i) + equation (ii) x + z - 2y = 6 $$\frac{{{\text{x}} + {\text{z}} - 6}}{2}$$ = y x + y + z = 16 x + z + $$\frac{{{\text{x}} + {\text{z}} - 6}}{2}$$ = 16 2(x + z) + x + z - 6 = 32 3(x + z) = 38 x + z = $$\frac{{38}}{3}$$ PB + AR = $$\frac{{38}}{3}$$

∠PMO = 90° (Angle made on tangent) ∠QOR = 90° - $$\frac{{\angle {\text{P}}}}{2}$$ ∠QOR = 90° - 22.5° ∠QOR = 67.5°

ABCD is cyclic quadrilateral ∴ 3p + 2p = 180° 5p = 180° p = 36° ∠ADC = 3p = 3 × 36 = 108° ∠AOC = $$\frac{1}{2}$$∠ADC = $$\frac{1}{2}$$ × 108° = 54° Reflex of ∠AOC = 360° - 54° = 206°

∠ADB = 180° - 62° - 62° = 56°

According to question Given: ∠BDC = 120°, ∠ABP =? ∴ ∠CDP = 180° - ∠BDC ∠CDP = 180° - 120° ∠CDP = 60° CD || AB ∴ ∠CDP = ∠ABP = 60°

$$\eqalign{ & R{T^2} = {25^2} - {20^2} \cr & RT = 15 \cr & {\text{Inradius }}\left( {\text{r}} \right) = \frac{{20 + 15 - 25}}{2} = \frac{{10}}{2} = 5 \cr & {\text{In }}\Delta {O_1}A{O_2} \cr & {O_1}O_2^2 = {O_1}{A^2} + AO_2^2 \cr & = {\left( 5 \right)^2} + {\left( {15} \right)^2} \cr & = 25 + 225 \cr & = 250 \cr & {O_1}{O_2} = 5\sqrt {10} {\text{ cm}} \cr} $$