mensuration area volumes Model Questions & Answers, Practice Test for ssc cgl tier 1 2023

Answer: (c)

In ΔCDE, ∠CDE = ∠EDC = 32°

∴ ∠DEC = 180° – ∠ECD – ∠EDC

= 180° – 32° – 32° = 116°

Again, ∠CED + ∠CEF = ∠DEF = 180°

116° + ∠CEF = 180°

∴ ∠CEF = 180° – 116° = 64° = ∠COF

Answer: (b)

x = 180 – 40 – 55

x = 85°

Answer: (d)

Let AD is diameter of circle of centre O. Find OP = q

AD = 34 cm, AO = OD = 17 cm

AB = 30 cm, AP = ${30}/2$ = 15 cm

OP = $√{(17)^2 – (15)^2} = √{64}$ = 8cm.

Answer: (c)

For inscribed square.

Diameter of circle = Diagonal of square using Pythagoras theorem,

⇒ $AB^2 + BC^2 = AC^2$

⇒ $2AB^2 = (2a)^2$ [∴ AB = BC]

$2AB^2 = 4a^2$

∴ $AB^2 = 2a^2$

∵ AB = $√2$a

Area of square ABCD = $(√2a)^2 = 2a^2$

For circumscribed circle,

Diameter of circle = Side of square = 2a

Area of square PQ RS = $(2a)^2 = 4a^2$

Difference between area of outer square and inner square

= $4a^2 – 2a^2 = 2a^2$

Answer: (d)

Area of sector = 72 π $cm^2$

⇒ ${π r^2 θ}/{360°}$ = 72 π

∴ θ = ${72 × 360}/{36 × 36}$ = 20°

Length of arc = ${πrθ}/{180°} = {π × 36 × 20}/{180}$ = 4 πcm