Area - Study Mode

[#116] A hall 50 m long and 45 m broad is to be paved with square tiles. Find the largest tile as well as its number in the given options so that the tiles exactly fit in the hall :
Correct Answer

(C) 25 sq. m and 90 tiles

Explanation

Solution: Length of hall = 50 m Breadth of hall = 45 m Area of hall = (50 × 45) m Maximum length of a square tiles = HCF of 50 m and 45 m = 5 metres Area of tiles = 5 × 5 = 25 sq. m ∴ Number of tiles : $$eqalign{
& = frac{{50 imes 45}}{{25}} cr
& = 90{ ext{ tiles}} cr} $$

[#117] A rectangular farm has to be fenced on one long side, one short side and the diagonal. If the cost of fencing is Rs. 100 per m, the area of the farm is 1200 m 2 and the short side is 30 m long, how much would the job cost ?
Correct Answer

(B) Rs. 12000

Explanation

Solution: Length : $$eqalign{
& = left( {frac{{1200}}{{30}}}
ight)m cr
& = 40,m cr} $$ Diagonal : $$eqalign{
& = left( {sqrt {{{left( {40}
ight)}^2} + {{left( {30}
ight)}^2}} }
ight)m cr
& = 50,m cr} $$ Length to be fenced : $$eqalign{
& = left( {40 + 30 + 50}
ight)m cr
& = 120,m cr} $$ ∴ Cost of fencing : $$eqalign{
& = { ext{Rs}}{ ext{.}}left( {120 imes 100}
ight) cr
& { ext{ = Rs}}.12000 cr} $$

[#118] Twenty-nine times the area of a square is one square metre less than six times the area of the second square and nine times its side exceeds the perimeter of other square by 1 metre. The difference in the sides of these squares is :
Correct Answer

(C) 6 m

Explanation

Solution: Let the sides of the two squares be x metres and y metres respectively Then, $$ Rightarrow 29{x^2} = 6{y^2} - 1.....(i)$$ And, $$eqalign{
& Rightarrow 9x - 4y = 1 cr
& Rightarrow 4y = 9x - 1 cr
& Rightarrow y = frac{{9x - 1}}{4}.....(ii) cr} $$ From (i) and (ii), we get : $$eqalign{
& Rightarrow 29{x^2} = 6{left( {frac{{9x - 1}}{4}}
ight)^2} - 1 cr
& Rightarrow 29{x^2} = 6left( {frac{{81{x^2} + 1 - 18x}}{{16}}}
ight) - 1 cr
& Rightarrow 243{x^2} + 3 - 54x - 8 = 232{x^2} cr
& Rightarrow 11{x^2} - 54x - 5 = 0 cr
& Rightarrow left( {x - 5}
ight)left( {11x + 1}
ight) = 0 cr
& Rightarrow x = 5,,m cr
& herefore y = frac{{9x - 1}}{4} = frac{{9 imes 5,,m - 1}}{4} = 11,,m cr
& ext{Required difference} cr
& = left(11-5
ight) m cr
& = 6,m } $$

[#119] In ΔPQR, side PQ = 3 cm and side PR = 25 cm. What is the area of ΔPQR ?
Correct Answer

(B) $$3sqrt {154} { ext{ sq}}{ ext{.cm}}$$

Explanation

Solution: $$eqalign{
& { ext{Area of }}Delta { ext{ }}PQR cr
& = frac{1}{2} imes QR imes PQ cr
& { ext{Given}},,PQ = 3,cm cr
& QR = sqrt {{{left( {PR}
ight)}^2} - {{left( {PQ}
ight)}^2}} cr
& ,,,,,,,,,, = sqrt {{{left( {25}
ight)}^2} - {3^2}} ,cm cr
& ,,,,,,,,,, = sqrt {625 - 9} ,cm cr
& ,,,,,,,,,, = sqrt {616} ,cm cr
& ,,,,,,,,,, = 2sqrt {154} ,cm cr
& herefore { ext{Area}},{ ext{of}},Delta ,PQR cr
& = left( {frac{1}{2} imes 2sqrt {154} imes 3}
ight)c{m^2} cr
& = 3sqrt {154} {mkern 1mu} {mkern 1mu} c{m^2} cr} $$

[#120] In an isosceles triangle, the measure of each of the equal sides is 10 cm and the angle between them is 45° . The area of the triangle is :
Correct Answer

(C) $$25sqrt 2 ,c{m^2}$$

Explanation

Solution: $$eqalign{
& { ext{Area of the traingle :}} cr
& = frac{1}{2}absin heta cr
& = left( {frac{1}{2} imes 10 imes 10 imes sin {{45}^ circ }}
ight)c{m^2} cr
& = left( {frac{1}{2} imes 10 imes 10 imes frac{1}{{sqrt 2 }}}
ight)c{m^2} cr
& = left( {frac{{50}}{{sqrt 2 }} imes frac{{sqrt 2 }}{{sqrt 2 }}}
ight)c{m^2} cr
& = 25sqrt 2 ,c{m^2} cr} $$