MCQ: Geometry - 3 - SSC CGL MCQ

In ΔsADE and ΔABC
∠A = ∠A [common]
∠ADE = ∠ACB = x°(Given)
∴ ΔADE ∼ ΔACB (AA Similarly)



Hence BD = AB - AD = 19.5 - 6 = 13.5 cm.

In ΔACB and PCQ
∠C = ∠C (common)
∠ABC = ∠PQC (each 90°)
∴ ΔACB ∼ ΔPC (AA Similarity)

As per the given in question , we draw a figure of a quadrilateral ABCD inscribed in a circle with centre O

Given , ∠COD = 120°
∠BAC = 30°


∴ ∠BAD = 90°
∴ ∠BCD = 180° - ∠BAD
∴ ∠BCD = 180° – 90° = 90°

According to question , we can draw a figure

Hence , option C is correct answer .

As we know that Sum of two supplementary angles = 180°
∴ (5y + 62°) + (22° + y) = 180°
⇒ 6y + 84° = 180°
⇒ 6y = 180° – 84° = 96°

According to question , we draw a figure of a circle with centre O,

Here , BO = OC = 15 cm and OD = 9 cm.
From ∆ BDO,

∴ BC = 2 × 12 = 24 cm.

On the basis of question we draw a figure of a circle with centre O ,

Given , ∠AOD = 100°
We know that the angle subtended at the centre is twice to that of angle at the circumference by the same arc.

Again, ∠BOC = 70°

∴ ∠APC = 180° – 50° – 35° = 95°

On the basis of question we draw a figure of a circle with centre O ,

Let OE = y cm
then OF = (y +1) cm
OA = OC = r cm
AE = 4 cm; CF = 3 cm
From ∆ OAE,
OA² = AE² + OE²
⇒ r² = 16 + y²
⇒ y² = r² – 16 ......(i)
From ∆OCF,
(y + 1)² = r² – 9 ..... (ii)
By equation (ii) – (i),
(y + 1)² – y² = r² – 9 – r² + 16
⇒ 2y + 1 = 7
⇒ y = 3 cm
∴ From equation (i),
9 = r² – 16
⇒ r² = 25
⇒ r = 5 cm

As we know that the orthocentre of an obtuse angled triangle lies outside the triangle. Hence , correct answer is option B.

Given , The sum of the interior angles of a regular polygon = 1080°
Sum of the interior angles of a regular polygon of n sides = (2n – 4) × 90°
∴ (2n – 4) × 90° = 1080°
⇒ 2n – 4 = 1080 ÷ 90 = 12
⇒ 2n = 12 + 4 = 16
∴ n = 8

As we know the centroid of a triangle is the point where the point of intersection of medians of a triangle is called centroid.

Firstly, We draw a figure of an isosceles triangle ABC,

Given that , AB = AC and ∠B = 35°
⇒ ∠ABC = ∠ACB = 35°
Now, ∠ADB = 90°
In ∆ADB , ∠ BAD + ∠ ABD + ∠ ADB = 180°
⇒ ∠ BAD + 90° + 35° = 180°
∴ ∠BAD = 55°

We know that in a triangle, if three altitudes are equal, then the triangle is called equilateral.

As per the given in question , we draw a figure of an isosceles triangle ABC,

In isosceles triangle ,
AC = BC = 5 cm

A circle is touching the side AQ and AR,

Then AR = AQ = 6 cm
Now AQ + AR = AB + BQ + AC + CR
⇒ 6 + 6 = AB + BQ + AC + CR
⇒ 6 + 6 = AB + BQ + AC + CR 
[here BQ = BP and CR = CP]
12 = AB + AC + BP + PC
12 = AB + AC + BC
Hence, perimeter of a triangle ABC is 12 cm.