Mensuration Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

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°

Answer: (a)

Suppose the smaller and larger sides of a right triangle be x and y, respectively.

By given condition,

$x^2 + y^2 = (3 √{10})^2$

⇒ $x^2 + y^2$ = 90 ... (i)

and $9x^2 + 4y^2$ = 405 ... (ii)

On solving Eqs. (i) and (ii), we get

x = 3 units and y = 9 units

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: (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: (c)

In right triangle ABC,

AC = $√{(AB)^2 + (BC)^2} = √{(40)^2 + (9)^2}$ = 41 cm

Area of ΔABC = $1/2 × 9 × 40 = 180 cm^2$

∵ Area of quadrilateral ABCD

= 126 + 180 = 306 $cm^2$