MCQ Questions on Geometry for Competitive Exams

$$\eqalign{ & \frac{1}{2} \times 10 \times 10 = \frac{1}{2} \times 10\sqrt 2 \times {\text{QS}} \cr & {\text{QS}} = 5\sqrt 2 \cr} $$

∠C = 90°
∠B = 180° - 118° = 62°
∠BAC = 180° - (62° + 90°) = 28°

x + y = 180° - 64°
x + y = 116°
2(x + y) = 232°
∠C + ∠D = 360° - 232° = 128°

According to question
Given:

DE || BC, AB = 17 cm, AC = ?
∠DAB = ∠ACB
(By alternate segment theorem)
∠DAB = ∠ABC
(Alternate angle)
∴ ∠ABC = ∠ACB
AB = AC = 17 cm

We know when a new triangle is formed by using mid points of big triangle.
⇒ In this case Area of 4 triangle is same

⇒ i.e. Area of ΔAFE = ΔFBD
= ΔFDE = ΔDEC = 1
⇒ Parallelogram
DEFB = ΔBFD + ΔDFE = 1 + 1
⇒ Area of Parallelogram
DEFB = 2 . . . . . . (i)
⇒ Again trapezium CAFD
= ΔAFE + ΔFED + ΔDCE = 1 + 1 + 1
Area of Trapezium
CAFD = 3 . . . . . . (ii)
Required Ratio will be = 2 : 3

At least two pairs of consecutive angles are congruent in a isosceles trapezium.

According to question

Let the radius of the circle be = $$r$$
∴ DO = OC = $$\frac{r}{2}$$
In right angle ΔAOD
By using Pythagoras theorem
$$\eqalign{ & A{D^2} = O{D^2} + A{O^2} \cr & {r^2} = \frac{{{r^2}}}{4} + A{O^2} \cr & A{O^2} = {r^2} - \frac{{{r^2}}}{4} \cr & A{O^2} = \frac{{3{r^2}}}{4} \cr & AO = \frac{{\sqrt 3 r}}{2}\,\,\,\left( {\because AB = 2 \times AO} \right) \cr & AB = \frac{{\sqrt 3 r}}{2} \times 2 \cr & AB = \sqrt 3 r{\text{ units}} \cr} $$

OCD = 30°
OD = 2
OC = 4
O₁E = R
O₁O = R + 2
ΔOCD Similar to ΔO₁CE

$$\eqalign{ & \frac{{{\text{OC}}}}{{{{\text{O}}_1}{\text{C}}}} = \frac{{{\text{OD}}}}{{{{\text{O}}_1}{\text{E}}}} \cr & \frac{4}{{{\text{R}} + 2 + 4}} = \frac{2}{{\text{R}}} \cr} $$
4R = 2R + 12
2R = 12
R = 6