Mensuration 2D - Study Mode

[#46] ABCD is a parallelogram. BC is produced to Q such that BC = CQ. Then
Correct Answer

(A) area (ΔABC) = area (ΔDCQ)

Explanation

Solution: In ΔABC & ΔDCQ ∠ABC = ∠DCQ ∠ACB = ∠DQC BC = CQ ΔABC ≅ ΔDCQ ar ΔABC = ar ΔDCQ

[#47] The perimeter of a sheet of paper in the shape of a quadrant of a circle is 75 cm. Its area would be $$left( {pi = frac{{22}}{7}}
ight)$$
Correct Answer

(B) 346.5 cm 2

Explanation

Solution: According to the figure, ⇒ Perimeter = r + r + $$l$$ ⇒ 75 cm = 2r + length of arc ⇒ 75 cm = 2r + $$frac{{2pi r}}{4}$$ ⇒ 75 cm = 2r + $$frac{{22 imes r}}{{7 imes 2}}$$ ⇒ r = 21 cm ⇒ Its area $$ = frac{1}{4}left[ {frac{{22}}{7} imes 21 imes 21}
ight] = 346.5{ ext{ c}}{{ ext{m}}^2}$$

[#48] Two adjacent sides of a parallelogram are of length 15 cm and 18 cm. If the distance between two smaller sides is 12 cm, then the distance between two bigger sides is
Correct Answer

(B) 10 cm

Explanation

Solution: Area of parallelogram = BC × FC = 15 × 12 = 180 cm 2 Area of parallelogram DC × AE = 180 18 × AE = 180 AE = 10 cm ∴ Distance between bigger sides = 10 cm

[#49] From a point in the interior of an equilateral triangle, the perpendicular distance of the sides are √3, cm 2√3 cm and 5√3 cm. The perimeter (in cm) of the triangle is
Correct Answer

(C) 48

Explanation

Solution: Let P be the point inside the equilateral ΔABC Let, PD = √3, PE = 2√3, PF = 5√3 and AB = BC = AC = $$x$$ $$eqalign{
& { ext{ar}}{ ext{.}},Delta { ext{ABC}} = { ext{ar}}{ ext{.}},Delta { ext{ABP}} + { ext{ar}}{ ext{.}},Delta { ext{ACP}} + { ext{ar}}{ ext{.}},Delta { ext{BCP}} cr
& frac{{sqrt 3 }}{4}{x^2} = frac{1}{2} imes x imes sqrt 3 + frac{1}{2} imes x imes 2sqrt 3 + frac{1}{2} imes x imes 5sqrt 3 cr
& sqrt 3 x = 2sqrt 3 + 4sqrt 3 + 10sqrt 3 cr
& x = 16 cr} $$ ∴ Perimeter of triangle = 3$$x$$ = 3 × 16 = 48 cm

[#50] At each corner of a triangular field of sides 26 m, 28 m and 30 m, a cow is tethered by a rope of length 7 m, the area (in m 2 ) ungrazed by the cows is
Correct Answer

(B) 259

Explanation

Solution: $$eqalign{
& { ext{Area Grazed by the cow}} cr
& = frac{{{A^ circ }}}{{360}}pi {left( 7
ight)^2} + frac{{{B^ circ }}}{{360}}pi {left( 7
ight)^2} + frac{{{C^ circ }}}{{360}}pi {left( 7
ight)^2} cr
& = pi {left( 7
ight)^2}left[ {frac{{{A^ circ } + {B^ circ } + {C^ circ }}}{{360}}}
ight] cr
& = pi {left( 7
ight)^2} imes frac{{180}}{{360}} cr
& = frac{1}{2}pi {left( 7
ight)^2} cr
& = 77{ ext{ }}{{ ext{m}}^2} cr
& s = frac{{26 + 30 + 28}}{2} = 42 cr
& { ext{Area of field}} = sqrt {sleft( {s - a}
ight)left( {s - b}
ight)left( {s - c}
ight)} cr
& = sqrt {42 imes 16 imes 14 imes 12} cr
& = 336{ ext{ }}{{ ext{m}}^2} cr
& Rightarrow { ext{Remaining area}} = 336 - 77 = 259{ ext{ }}{{ ext{m}}^2} cr} $$