MCQ Questions on Geometry for Competitive Exams

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

$$\eqalign{ & {\text{Area}}\left( {\Delta \,1} \right) = \frac{{30}}{{10 \times 10}} = \frac{3}{{10}} = \frac{{24}}{{80}} \cr & {\text{Area}}\left( {\Delta \,2} \right) = \frac{{10}}{{10 \times 8}} = \frac{{10}}{{80}} \cr & {\text{Area}}\left( {\Delta \,3} \right) = \frac{{12}}{{8 \times 10}} = \frac{{12}}{{80}} \cr & {\text{Area}}\left( {ABCD} \right) = 80 - \left( {24 + 10 + 12} \right) = 34 \cr} $$

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

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°

n² = 8 × 16
n² = 128
n = 8$$\sqrt 2 $$

b = (8$$\sqrt 2 $$ + a)
a² + (16)² = b²
a² + (16)² = (8$$\sqrt 2 $$ + a)²
a² + 256 = 128 + a² + 16$$\sqrt 2 $$ a
128 = 16$$\sqrt 2 $$ a
a = $$\frac{8}{{\sqrt 2 }}$$
a = 4$$\sqrt 2 $$

Now, Area of arc AC₁B $$ = \pi {r^2}.\frac{{90}}{{360}} = \frac{\pi }{4}{\left( 1 \right)^2} = \frac{\pi }{4}$$
And area of arc AC₂B = $$\frac{\pi }{4}$$
Area of square = (side)² = 1
Area of common portion = area of arc (AC₁B + AC₂B) - 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.

BC² = 48² + 14²
BC² = 2304 + 196
BC² = 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