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

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)

Area will be maximum when P and B will be same

So $P^2 + P^2 = H^2 ⇒ P^2 = H^2/2$

⇒ P = $H/√2$

Area = $1/2 BP = 1/2 P^2 = 1/2. H^2/2 = H^2/4$

= ${100}/4 = 25 cm^2$

Answer: (a)

Let radius of cirlce is ‘r’ and side of the square is a ∼ then, from question

$π r^2 = a^2 ⇒ r/a = 1/√π$

Now ∼ ${2πr}/{4a} = {2π}/{4√π} = √π/2$

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: (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: (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°