Mensuration Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

P does not lie on the circle at x = 6.5cm and

y = 6.5 cm because, as <APB = 90°

then, $AB^2 ∼ ∼ = AP^2 ∼ ∼ + BP^2$

$(13)^2 ≠ (6.5)^2 ∼ ∼ + (6.5)^2$

Hence, point P does not lies on the circle ∼

Answer: (a)

From figure.

BC || EF || AD

∴ x° = z° = 50° (corresponding interior angle

∴ θ + z° = 180° (linear pair)

∴ ∼ θ = 180° – 50° = 130°

In quadrilateral

AQFD, x° + y° + 120° + θ = 360°

50° + y° + 120° + 130° = 360°

y = 360° – 300° = 60°

Answer: (c)

${\text"Area of Δ DAE "}/{\text"Area of Δ DEC"} = {1/2 × DE × AE}/{1/2 × DE × CE}$

= ${AE}/{CE} = {(AD)^2}/{(DC)^2} = (6/8)^2 = {9/{16}}$

Similarly, in ΔABC,

${\text"Area of ΔBCF"}/{\text"Area of ΔBFA"} = 9/{16}$

∴ The area of shaded to unshaded region = ${16}/9$

Answer: (c)

ΔEAB is equilateral

ΔEDC is also equilateral

Area of trapezium ABCD

= $(1/2 × DB × OA) + 1/2 (DB × OC)$

= $1/2$ × DB × AC

Let AO = OB = x and DO = OC = y

Area (ABCD) = $1/2 (x + y)^2$ = 16(given)

⇒ x + y = 4 $√2$ ... (i)

ΔAOB is a right angled isosceles triangle.

So, AB = $√{x^2 + x^2} = √2x$

Similarly, DC = $√2$y

Now, FG = EF – EG

⇒ FG = AB sin 60° – DC sin 60°

= $√3/2 (AB - DC) = √6/2$ (x - y) ... (ii)

Area of trapezium

= Area ΔEAB – Area ΔEDC

= $√3/4 (AB^2 - DC^2)$

= $√3/4 [(x√2)^2 - (y√2)^2]$

⇒ Area = $√3/2$ (x + y) (x – y)

Now, $√3/2$ (x + y)(x - y) = 16

⇒ x – y = ${32}/{√3 (x + y)} ⇒ x - y = 8/√6 (∵ x + y = 4√2)$

Height = $√6/2 (x - y) = √6/2 × 8/√6$ = 4 cm

Answer: (d)

The tangents drawn from an outer point on a circle are always equal = ∠CBA.

Therefore, ∠CAB = ∠CBA

∴ 45° + x + x = 180°

⇒ 2x = 180° – 45°

⇒ x = 67 ${1°}/2$

∠AQP = ∠x = ∠BQP

= 67 ${1°}/2$

(alternate interior segments properties)

⇒ ∠AQB = ∠AQP + ∠BQP

= $67 {1°}/2 + 67{1°}/2$ = 135°