Mensuration 2D - Study Mode
[#81] A rectangular lawn whose length is twice of its breadth is extended by having four semicircular portions on its sides. What is the total cost (in Rs.) of levelling the entire lawn at the rate of Rs. 100 per square metre, it the smaller side of the rectangular lawn is 12 m? (Take π = 3.14)
Correct Answer
(A) 85,320
Explanation
Solution: Breadth = 12 m Length = 24 m Cost = 100 Rs./m 2 r = 6 R = 12 π × 144 + π × 36 + 288 = 3.14 × 144 + 3.14 × 36 + 288 = 853.2 ∴ Total cost = 853.2 × 100 = Rs. 85320
[#82] ABCD is a trapezium with AD and BC parallel sides. The ratio of the area of ABCD to that of ΔAED is
Correct Answer
(D) $$frac{{{ ext{AD}} + { ext{BC}}}}{{{ ext{AD}}}}$$
Explanation
Solution: $$eqalign{
& { ext{Let EN }} x08ot ,{ ext{AD}} cr
& { ext{Area of }}Delta { ext{AED}} = frac{1}{2} imes { ext{EN}} imes { ext{AD}} cr
& { ext{Area of trapezium ABCD}} cr
& = frac{1}{2}left( {{ ext{AD}} + { ext{BC}}}
ight) imes { ext{EN}} cr
& frac{{{ ext{ar}}left( {{ ext{ABCD}}}
ight)}}{{{ ext{ar}}left( {{ ext{AED}}}
ight)}} cr
& = frac{{frac{1}{2}left( {{ ext{AD}} + { ext{BC}}}
ight) imes { ext{EN}}}}{{frac{1}{2} imes { ext{EN}} imes { ext{AD}}}} cr
& = frac{{{ ext{AD}} + { ext{BC}}}}{{{ ext{AD}}}} cr} $$
[#83] a and b are two sides adjacent to the right angle of a right angled triangle and p is the perpendicular drawn to the hypotenuse from the opposite vertex. Then p 2 is equal to
Correct Answer
(C) $$frac{{{{ ext{a}}^2}{{ ext{b}}^2}}}{{{{ ext{a}}^2} + {{ ext{b}}^2}}}$$
Explanation
Solution: Length of perpendicular drawn from the right angle to hypotenuse, $$eqalign{
& { ext{P}} = frac{{{ ext{a}} imes { ext{b}}}}{{ ext{H}}} cr
& {{ ext{P}}^2} = frac{{{{ ext{a}}^2}{{ ext{b}}^2}}}{{{{ ext{H}}^2}}} cr
& {{ ext{P}}^2} = frac{{{{ ext{a}}^2}{{ ext{b}}^2}}}{{{{ ext{a}}^2} + {{ ext{b}}^2}}},,,,left[ {x08ecause {{ ext{H}}^2} = {{ ext{a}}^2} + {{ ext{b}}^2}}
ight] cr} $$
[#84] A took 15 sec to cross a rectangular field diagonally walking at the rate of speed 52 m/min and B took the same time to cross the same field along its sides walking at the rate speed 68 m/min. The area of the field is:
Correct Answer
(D) 60 m 2
Explanation
Solution: BD = length of diagonal = speed × time $$eqalign{
& = frac{{52}}{{60}} imes 15 cr
& = 13{ ext{ m}} cr
& { ext{BD}} = sqrt {{l^2} + {b^2}} cr
& Rightarrow {l^2} + {b^2} = {13^2} cr
& Rightarrow {l^2} + {b^2} = 169 cr
& { ext{Again, }}l + b = frac{{68}}{{60}} imes 15 = 17 cr
& {left( {l + b}
ight)^2} = {l^2} + {b^2} + 2lb cr
& {17^2} = 169 + 2lb cr
& lb = frac{{120}}{2} = 60{ ext{ }}{{ ext{m}}^2} cr} $$
[#85] The area of the shaded region in the figure given below is
Correct Answer
(C) $${a^2}left( {frac{pi }{2} - 1}
ight){ ext{ sq}}{ ext{. units}}$$
Explanation
Solution: Area of shaded region = Area of semicircle - Area of triangle $$eqalign{
& = frac{{pi {{left( a
ight)}^2}}}{2} - frac{1}{2} imes a imes 2a cr
& = frac{{pi {a^2}}}{2} - {a^2} cr
& = {a^2}left( {frac{pi }{2} - 1}
ight){ ext{ sq}}{ ext{. units}} cr} $$