Geometry — Part 1 MCQs with Explanations

$$\eqalign{ & ABCD{\text{ is a square}} \cr & AC = \sqrt 2 AB \cr & AO = OC = \frac{{AC}}{2} = \frac{{\sqrt 2 AB}}{2} = \frac{{AB}}{{\sqrt 2 }} \cr & \Delta AOF \sim \Delta ABG \cr & \left[ {{\text{By }}AA{\text{ property}}} \right] \cr & \frac{{AO}}{{AB}} = \frac{{OF}}{{BG}} \cr & \frac{{\frac{{AB}}{{\sqrt 2 }}}}{{AB}} = \frac{{OF}}{{BG}} \cr & \frac{1}{{\sqrt 2 }} = \frac{{OF}}{{BG}} \cr & BG = \sqrt 2 OF{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & AG{\text{ is angle bisector of }}\Delta ABC \cr & \frac{{AB}}{{AC}} = \frac{{BG}}{{GC}} = \frac{1}{{\sqrt 2 }} \cr & \left[ {{\text{angle bisector theorem}}} \right] \cr & BG = \frac{1}{{\sqrt 2 }}GC{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & {\text{Compare }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \sqrt 2 OF = \frac{1}{{\sqrt 2 }}GC \cr & OF:CG = 1:2 \cr} $$

$$\eqalign{ & {\text{Area}}\left( {\Delta \,1} \right) = \frac{{30}}{{10 \times 10}} = \frac{3}{{10}} = \frac{{24}}{{80}} \cr & {\text{Area}}\left( {\Delta \,2} \right) = \frac{{10}}{{10 \times 8}} = \frac{{10}}{{80}} \cr & {\text{Area}}\left( {\Delta \,3} \right) = \frac{{12}}{{8 \times 10}} = \frac{{12}}{{80}} \cr & {\text{Area}}\left( {ABCD} \right) = 80 - \left( {24 + 10 + 12} \right) = 34 \cr} $$ $$\eqalign{ & P{M^2} = {10^2} - {4^2} = 84 \cr & PM = \sqrt {84} \cr & {\text{Area}}\left( {PQR} \right) = \frac{1}{2} \times 8 \times \sqrt {84} = 4\sqrt {84} \cr & {\text{Area of quadrilateral }}ABCD \cr & = \frac{{34}}{{80}} \times {\text{ar}}\left( {\Delta PQR} \right) \cr & = \frac{{34}}{{80}} \times 4\sqrt {84} \cr & = \frac{{17\sqrt {21} }}{5}\,{\text{c}}{{\text{m}}^2} \cr} $$

In ΔABC ∠B = ∠A = 50° ∠ACD = 50° + 50° = 100° ∠ACD is the external angle or ΔABC ∠ACD + ∠CAD + ∠ADC = 180° ∠CAD = ∠ADC ∵ AC = CD 2∠CAD = 180° - 100° ∠CAD = 40° ∠BAD = 50° + 40° = 90°

According to the figure. OM = OY = ON ∴ In ΔOMY ∠OMY = ∠OYM = 15° ∴ ∠MOY = 180° - 15° - 15° ∠MOY = 150° In ΔONY ∠ONY = ∠OYN = 50° ∴ ∠NOY = 180° - 50° - 50° ∠NOY = 80° ∴ ∠MON = 150° - 80° ∠MON = 70°

n 2 = 8 × 16 n 2 = 128 n = 8$$\sqrt 2 $$ b = (8$$\sqrt 2 $$ + a) a 2 + (16) 2 = b 2 a 2 + (16) 2 = (8$$\sqrt 2 $$ + a) 2 a 2 + 256 = 128 + a 2 + 16$$\sqrt 2 $$ a 128 = 16$$\sqrt 2 $$ a a = $$\frac{8}{{\sqrt 2 }}$$ a = 4$$\sqrt 2 $$

Now, Area of arc AC 1 B $$ = \pi {r^2}.\frac{{90}}{{360}} = \frac{\pi }{4}{\left( 1 \right)^2} = \frac{\pi }{4}$$ And area of arc AC 2 B = $$\frac{\pi }{4}$$ Area of square = (side) 2 = 1 Area of common portion = area of arc (AC 1 B + AC 2 B) - Area of square $$\eqalign{ & = \frac{\pi }{4} + \frac{\pi }{4} - 1 \cr & = \frac{\pi }{2} - 1\,{\text{sq}}{\text{. m}} \cr} $$

In ΔBAD and ΔBDC ∠BDA = ∠BDC (Each 90°) If ∠C = 30° Then ∠A = 60° Also ∠ABD = 30° ∠BCD = ∠ABD ΔBAD ∼ ΔBDC . . . . . (i) In ΔBAD and ΔCAB ∠BDA = ∠ABC (Each 90°) ∠BAD = ∠BAC (common) ΔBAD ∼ ΔCAB . . . . . (ii) Similarly ΔCBD ∼ ΔCAB . . . . . (iii) All conditions are true.

BC 2 = 48 2 + 14 2 BC 2 = 2304 + 196 BC 2 = 2500 BC = 50 cm In radius r = $$\frac{{14 + 48 - 50}}{2}$$ = 6 cm

RL × LS = PL × LQ 6 × RL= 9 × 4 RL = 6 cm

$$\eqalign{ & AD = 7{\text{ cm}} \cr & CE = 4\sqrt 6 {\text{ cm}} \cr & {\text{Since, }}4\left( {A{D^2} + C{E^2}} \right) = 5A{C^2} \cr & 4\left( {{{\left( 7 \right)}^2} + {{\left( {4\sqrt 6 } \right)}^2}} \right) = 5A{C^2} \cr & 4\left( {49 + 96} \right) = 5A{C^2} \cr & 4\left( {145} \right) = 5A{C^2} \cr & A{C^2} = \frac{{4 \times 145}}{5} \cr & AC = \sqrt {4 \times 29} \cr & AC = 2\sqrt {29} {\text{ cm}} \cr} $$

Sides of triangle are 6 cm, 10 cm, x cm. As we know in a triangle sum of two sides is always greater than the 3 rd side. So minimum integral value of x is 5. Hence option (D) is correct answer.

As D and E are mid-points of AF and AG, ∴ Length ⇒ DE = $$\frac{1}{2}$$ FG ⇒ FG = 2 × DE ⇒ FG = 2 × 2.4 ⇒ FG = 4.8 cm As F and G are the mid-points of AB and AC ∴ Length ⇒ FG = $$\frac{1}{2}$$ BC ⇒ BC = 2 × FG ⇒ BC = 2 × 4.8 ⇒ BC = 9.6 cm

In ΔODS 1 unit → 9 cm √3 units → 9√3 cm So, D is the mid point of SR So DS = 9√3 SR = 18√3 In ΔPCO √3 units → 9 unit 1 unit → $$\frac{9}{{\sqrt 3 }}$$ = 3√3 = PC ⇒ PQ = 6√3 Sum of pair of 2 opposite side is equal to sum of pair of other two opposite sides. So, PQ + SR = SP + QR Perimeter = 2(18√3 + 6√3) = 48√3

ΔABC is a right angle triangle. In which ∠C = 90° and D is a point on AB such that D is perpendicular on AB. Let AC = BC = a ∴ AB 2 = AC 2 + BC 2 = a 2 + a 2 $$\eqalign{ & \boxed{AB = a\sqrt 2 } \cr & \therefore BD = AD = \frac{{a\sqrt 2 }}{2} = \frac{a}{{\sqrt 2 }} \cr & {\text{Now in }}\Delta ACD \cr & = A{C^2} = C{D^2} + A{D^2} \cr & {a^2} = C{D^2} + \frac{{{a^2}}}{2} \cr & {a^2} - \frac{{{a^2}}}{2} = C{D^2} \cr & \boxed{\frac{{{a^2}}}{2} = C{D^2}} \cr & C{D^2} = \frac{{{a^2}}}{2} \cr & 2C{D^2} = {a^2} \cr & {\text{and }}A{D^2} + B{D^2} = {\left( {\frac{a}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{a}{{\sqrt 2 }}} \right)^2} \cr & = \frac{a}{{\sqrt 2 }} + \frac{a}{{\sqrt 2 }} \cr & = {a^2} \cr & \therefore \boxed{2C{D^2} = A{D^2} + B{D^2}} \cr} $$

Distance = 8 + 8 = 16

$$\eqalign{ & {\text{Let }}\angle PUQ = \theta \cr & {\text{Then }}\angle SUT = 180 - \theta \cr & {\text{And}}\,{\text{let }}\angle RTP = \alpha \cr & {\text{Then }}\angle STU = 180 - \alpha \cr & \therefore {\text{Ratio of area of }}\frac{{\Delta PUQ}}{{\Delta SUT}} \cr & = \frac{{\frac{1}{2} \times 18 \times 24 \times \sin \theta }}{{\frac{1}{2} \times 12 \times 9 \times \sin \left( {180 - \theta } \right)}} = \frac{{4\sin \theta }}{{\sin \theta }} \cr & \frac{{\Delta PUQ}}{{\Delta SUT}} = \frac{4}{1} \cr & {\text{Now, ratio of area of }}\frac{{\Delta PTR}}{{\Delta SUT}} \cr & = \frac{{\frac{1}{2} \times 27 \times 24 \times \sin \alpha }}{{\frac{1}{2} \times 9 \times 8 \times \sin \left( {180 - \alpha } \right)}} = \frac{{9\sin \alpha }}{{\sin \alpha }} \cr & \frac{{\Delta PTR}}{{\Delta SUT}} = \frac{9}{1} \cr & \therefore {\text{Ratio of area of }}\frac{{\Delta PQU}}{{\Delta PTR}} = \frac{4}{9} \cr} $$

PO × OX = QO × OY = RO × OZ 6 × 2 = 4 × OY OY = 3 ∴ Then QY = QO + OY = 4 + 3 = 7 cm

$$\eqalign{ & AB = \sqrt {4{r_1}.{r_2}} \cr & BC = \sqrt {4{r_2}.{r_3}} \cr & AC = \sqrt {{{\left( {{r_1} + 2{r_2} + {r_3}} \right)}^2} - {{\left( {{r_3} - {r_1}} \right)}^2}} \cr & AB + BC = AC \cr & \sqrt {4{r_1}.{r_2}} + \sqrt {4{r_2}.{r_3}} = \sqrt {{{\left( {{r_1} + 2{r_2} + {r_3}} \right)}^2} - {{\left( {{r_3} - {r_1}} \right)}^2}} \cr & {\text{Squaring both sides}} \cr & \Rightarrow 4{r_1}.{r_2} + 4{r_2}.{r_3} + 2\sqrt {16{r_1}.{r_2}^2.{r_3}} = {\left( {{r_1} + 2{r_2} + {r_3}} \right)^2} - {\left( {{r_3} - {r_1}} \right)^2} \cr & \Rightarrow 4{r_1}.{r_2} + 4{r_2}.{r_3} + 8{r_2}\sqrt {{r_1}.{r_3}} = {r_1}^2 + 4{r_2}^2 + {r_3}^2 + 4{r_1}.{r_2} + 4{r_2}.{r_3} + 2{r_1}.{r_3} - {r_3}^2 - {r_1}^2 + 2{r_1}.{r_3} \cr & \Rightarrow 8{r_2}\sqrt {{r_1}.{r_3}} = 4{r_2}^2 + 4{r_1}.{r_3} \cr & \Rightarrow 2{r_2}\sqrt {{r_1}.{r_3}} = {r_2}^2 + {r_1}.{r_3} \cr & {\text{Squaring both sides}} \cr & \Rightarrow 4{r_2}^2.{r_1}.{r_3} = {r_2}^4 + {r_1}^2.{r_3}^2 + 2{r_2}^2.{r_1}.{r_3} \cr & \Rightarrow {\left( {{r_2}^2 - {r_1}.{r_3}} \right)^2} = 0 \cr & \Rightarrow {r_2}^2 = {r_1}.{r_3} \cr & \Rightarrow {r_2} = \sqrt {{r_1}.{r_3}} \cr} $$

∠DBC = 28° (Given) BD = DC, (Given) ∠BCD = 28° So, ∠BDC = 180° - (28° + 28°) = 124° ∠BAC = 180° - 124° = 56° (Cyclic $$\square $$) ∠BOC = 2 × ∠BAC = 56° × 2 = 112° Answer.

2x = x + 60° x = 60° This triangle is a equilateral triangle.

1 unit → 3 cm 3 unit → 9 cm YZ = 2LM = 2 × 9 = 18 cm

$$\eqalign{ & s = \frac{{25 + 34 + 39}}{2} = 49 \cr & {\text{Area of }}\Delta ABC = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr & = \sqrt {49 \times 15 \times 10 \times 24} \cr & = \sqrt {{s^2} \times 3 \times 2 \times {2^3} \times 3} \cr & = 7 \times 5 \times 3 \times 4 \cr & = 420 \cr} $$ So, area of square is 21 more than area of Δ = (i.e.) = 420 + 21 = 441 square unit $$\eqalign{ & {{\text{A}}^2} = 441 \cr & {\text{A}} = {\left[ {{{\left( {21} \right)}^2}} \right]^{\frac{1}{2}}} = 21 \cr} $$

According to question QR is the common tangent and NO is also the common tangent. ∴ QN = NP = NR In ΔQPN ∠NQP = ∠NPQ ∠NRP = ∠NPR In ΔPQR ∠P + ∠Q + ∠R = 180° x + y + x + y = 180° 2x + 2y = 180° x + y = 90° As shown in the figure x + y = ∠P = 90°

Maximum 4 tangents can be drawn to two non-intersecting circles.

Perimeter of triangle ABC = 36 cm By Pythagorean triplets 6, 8, 10 Now, a = 8 cm x = 10 cm Area of triangle ABC = $$\frac{1}{2}$$ × 16 × 6 = 48 cm 2

So, OO' = $$6 + \sqrt {161} $$

AC 2 + BD 2 = 2(AB 2 + BC 2 ) AC 2 + BD 2 = 2(10 2 + 3 2 ) = 218

P : Q : R = 3 : 2 : 1 3 + 2 + 1 = 6u → 180° 1u → 30° P = 90° ∠PRS = 90° Q = 60° R = 30° ∠P + ∠Q = ∠PRT 90° + 60° = ∠PRS + ∠TRS 150° = 90° + ∠TRS ∠TRS = 60°

$$\eqalign{ & {\text{In }}\Delta {O^1}CP \cr & {O^1}P = \sqrt {{{20}^2} + {{15}^2}} = 25{\text{ cm}} \cr & OP = 25 - 15 - r = 10 - r \cr & {\text{In }}\Delta OPD\,\& \,\Delta {O^1}CP \cr & \frac{{OP}}{{{O^1}P}} = \frac{{OD}}{{{O^1}C}} \cr & \frac{{10 - r}}{{25}} = \frac{r}{{15}} \cr & 150 - 15r = 25r \cr & r = \frac{{150}}{{40}} \cr & r = 3.75{\text{ cm}} \cr} $$

XY = 9 cm, TX = 15 cm ⇒ We know Length of tangents drawn from a point to the circle are equal Therefore XY = XZ = 9 cm, TZ = RT TX = 15 cm XZ + ZT = 15 ZT = 15 - 9 = 6 RT = ZT = 6 cm

According to question AB = AD = DB = r ∴ ΔADB is a equilateral triangle ∠DBA = 60° Similar in ΔABC ∠ABC = 60° ∠DBC = 60° + 60° ∴ ∠DBC = 120°

Let ∠DAC = x ∠ADB = 2x ∠ACD + ∠DAC = ∠ADB ∠ACD + x = 2x ∠ACD = 2x - x = x In ΔABC ∠A + ∠B + ∠C = 180° 70° + 56° + ∠C = 180° ∠C = 54° ∠ADC = 180° - 2x = 180° - 2 × 54° = 180° - 108° = 72°

Let, PA = x PA × PB = PC × PD x × (x + 4) = 4 × 15 x × (x + 4) = 60 x 2 + 4x - 60 = 0 x 2 + 10x - 6x - 60 = 0 x(x + 10) - 6(x + 10) = 0 x = 6 PB = x + 4 = 6 + 4 = 10 cm

If AC || BD, 2a + b + a = 180° . . . . . . (1) 2a = 38° a = 19° From equation (1) 2 × 19° + b + 19° = 180° b = 123° Now, ∠ABC + 5∠CBD = 123° + 5 × 19° = 218°

$${\text{R}} = \frac{{{\text{Hypotenuse}}}}{2} = \frac{{14}}{2} = 7$$

∠AOB = 64° Then, ∠BPA = 32° ∠PAQ = 90° [∴ PQ is a diameter] So, ∠PXA = 90 - 32 = 58°

$$\eqalign{ & {\text{In }}\Delta APQ\,\& \,\Delta ABC \cr & \frac{{AP}}{{AB}} = \frac{{AQ}}{{AC}} \cr & \frac{{AP}}{{20}} = \frac{{4\sqrt 2 }}{{10\sqrt 2 }} \cr & AP = \frac{{20 \times 2}}{5} = 8{\text{ cm}} \cr & PB = AB - AP = 20 - 8 = 12{\text{ cm}} \cr} $$

$$\eqalign{ & CD:DB = 3:2 \cr & CD = 5\sqrt {13} \times \frac{3}{5} \cr & CD = 3\sqrt {13} \cr & \sin \theta = \frac{{3\sqrt {13} }}{{DE}} = \frac{{5\sqrt {13} }}{{10}} \cr & DE = 6{\text{ cm}} \cr} $$

ΔABC ≅ ΔDBC y + 15 = 63° y = 48° 2x - 4 = 29° 2x = 33° 2x + 5y = 33° + 5 × 48° = 33° + 240° = 273°

Area of ΔABC : Area of ΔBDC = a : b = ak : bk $$\eqalign{ & ak = x \cr & k = \frac{x}{a} \cr & bk = b \times \frac{x}{a} = \frac{{bx}}{a} \cr} $$