Mensuration 2D - Study Mode
[#16] The minute hand of a clock is 20 cm long. Find the area on the face of the clock swept by the minute hand between 8 a.m. and 8:45 a.m.
Correct Answer
(A) $$frac{{6600}}{7},{ ext{c}}{{ ext{m}}^2}$$
Explanation
Solution: $$eqalign{
& { ext{Area}} = frac{{270}}{{360}}pi {r^2} cr
& = frac{3}{4} imes frac{{22}}{7} imes 20 imes 20 cr
& = frac{{6600}}{7},{ ext{c}}{{ ext{m}}^2} cr} $$
[#17] The area of the largest triangle that can be inscribed in a semicircle of radius 6 m is
Correct Answer
(A) 36 m 2
Explanation
Solution: $$eqalign{
& { ext{The area of largest }}Delta = frac{1}{2} imes b imes h cr
& = frac{1}{2} imes 12 imes 6 cr
& = 36{ ext{ }}{{ ext{m}}^2} cr} $$
[#18] In the given figure, PQRSTU is a regular hexagon of side 12 cm. What is the area (in cm 2 ) of triangle SQU?
Correct Answer
(C) 108√3
Explanation
Solution: $$eqalign{
& { ext{Area of }}Delta { ext{SQU}} = frac{1}{2}{ ext{ area of hexagon}} cr
& = frac{1}{2} imes left[ {6 imes frac{{sqrt 3 }}{4}{{left( {12}
ight)}^2}}
ight] cr
& = frac{1}{2} imes 216sqrt 3 cr
& = 108sqrt 3 { ext{ c}}{{ ext{m}}^2} cr} $$
[#19] In the given figure, two squares of sides 8 cm and 20 cm are given. What is the area (in cm 2 ) of the shaded part?
Correct Answer
(B) $$frac{{160}}{7}$$
Explanation
Solution: $$eqalign{
& Delta DHC sim Delta DFG cr
& { ext{So, }}frac{{DC}}{{DG}} = frac{{CH}}{{FG}} cr
& Rightarrow frac{8}{{28}} = frac{{CH}}{{20}} cr
& Rightarrow CH = frac{{40}}{7} cr
& { ext{So, area of }}Delta DCH cr
& = frac{1}{2} imes DC imes CH cr
& = frac{1}{2} imes 8 imes frac{{40}}{7} cr
& = frac{{160}}{7}{ ext{ c}}{{ ext{m}}^2} cr} $$
[#20] The area of an isosceles trapezium is 176 cm 2 and the height is $${frac{2}{{11}}^{{ ext{th}}}}$$ of the sum of its parallel sides. If the ratio of the length of the parallel sides is 4 : 7, then the length of a diagonal (in cm) is
Correct Answer
(A) $$2sqrt {137} $$
Explanation
Solution: Distance between two parallel line = $$frac{2}{{11}}$$ (sum of both parallel line) = $$frac{2}{{11}}$$ × (7x + 4x) = 2x Area = $$frac{1}{2}$$ (sum of parallel sides) × (distance between them) ⇒ $$frac{1}{2}$$ (7x + 4x) × 2x = 176 ⇒ 11x 2 = 176 ⇒ x 2 = 16 ⇒ x = 4 AB = 7 × 4 = 28 cm CD = 4 × 4 = 16 cm CM = 2 × 4 = 8 cm AM = AN + NM ⇒ AM = AN + 16 ⇒ AM = 6 + 16 ⇒ AM = 22 $$eqalign{
& left( {{ ext{AN}} = { ext{BM}} = frac{{12}}{2} = 6}
ight) cr
& { ext{A}}{{ ext{C}}^2} = { ext{C}}{{ ext{M}}^2} + { ext{A}}{{ ext{M}}^2} cr
& Rightarrow { ext{A}}{{ ext{C}}^2} = {8^2} + {22^2} cr
& Rightarrow { ext{AC}} = sqrt {64 + 484} cr
& Rightarrow { ext{AC}} = sqrt {548} cr
& Rightarrow { ext{AC}} = 2sqrt {137} cr} $$