Area - Study Mode

[#81] The length of a rectangle is decreased by r%, and the breadth is increased by (r + 5)%. Find r, if the area of the rectangle is unaltered :
Correct Answer

(D) 20

Explanation

Solution: Let original length = x and original breadth = y Then, Original area = xy New area : $$eqalign{
& = left[ {frac{{left( {100 - r}
ight)}}{{100}} imes x}
ight]left[ {frac{{left( {105 + r}
ight)}}{{100}} imes y}
ight] cr
& = left[ {left( {frac{{10500 - 5r - {r^2}}}{{10000}}}
ight)xy}
ight] cr
& herefore left( {frac{{10500 - 5r - {r^2}}}{{10000}}}
ight)xy = xy cr
& Rightarrow {r^2} + 5r - 500 = 0 cr
& Rightarrow left( {r + 25}
ight)left( {r - 20}
ight) = 0 cr
& Rightarrow r = 20 cr} $$

[#82] The cost of cultivating a square field at the rate of Rs. 685 per hector is Rs. 6165. The cost of putting a fence around it at the rate of Rs. 48.75 per metre would be :
Correct Answer

(C) Rs. 58500

Explanation

Solution: $$eqalign{
& { ext{Area}} = frac{{{ ext{Total cost}}}}{{{ ext{Rate}}}} cr
& ,,,,,,,,,,,,,, = left( {frac{{6165}}{{685}}}
ight){ ext{hectares}} cr
& ,,,,,,,,,,,,,, = left( {9 imes 10000}
ight){m^2} cr} $$ ∴ Side of the square : $$eqalign{
& = sqrt {90000} ,m cr
& = 300,m cr} $$ Perimeter of the field = (300 × 4) m = 1200 m Cost of fencing = Rs. (1200 × 48.75) = Rs. 58500

[#83] If the length of the diagonal of a square is 20 cm, then its perimeter must be :
Correct Answer

(C) $$40sqrt 2 ,$$ cm

Explanation

Solution: $$eqalign{
& d = sqrt 2 imes l cr
& Rightarrow l = frac{{20}}{{sqrt 2 }} cr} $$ ∴ Perimeter : $$eqalign{
& = left( {4l}
ight)cm cr
& = left( {frac{{4 imes 20}}{{sqrt 2 }} imes frac{{sqrt 2 }}{{sqrt 2 }}}
ight)cm cr
& = 40sqrt 2 ,cm cr} $$

[#84] The area of a square is twice that of a rectangle. The perimeter of the rectangle is 10 cm. If its length and breadth each is increased by 1 cm, the area of the rectangle become equal to the area of the square. The length of side of the square is :
Correct Answer

(A) $$2sqrt 3 $$ cm

Explanation

Solution: Let the length and breadth of the rectangle be $$l$$ cm and n cm respectively Then, $$eqalign{
& 2left( {l + b}
ight) = 10 cr
& Rightarrow l + b = 5 cr
& Rightarrow b = left( {5 - l}
ight)cm cr} $$ Area of the rectangle : $$eqalign{
& = lleft( {5 - l}
ight)c{m^2} cr
& = left( {5l - {l^2}}
ight)c{m^2} cr} $$ Area of the square : $$eqalign{
& = 2left( {5l - {l^2}}
ight)c{m^2} cr
& = left( {10l - 2{l^2}}
ight)c{m^2} cr} $$ $$eqalign{
& herefore left( {l + 1}
ight)left( {6 - l}
ight) = left( {10l - 2{l^2}}
ight) cr
& Rightarrow {l^2} - 5l + 6 = 0 cr
& Rightarrow left( {l - 3}
ight)left( {l - 2}
ight) = 0 cr
& Rightarrow l = 3 cr} $$ Area of the square : $$eqalign{
& = left( {10 imes 3 - 2 imes 9}
ight)c{m^2} cr
& = 12,c{m^2} cr} $$ ∴ Side of the square $$ = sqrt {12} ,cm = 2sqrt 3 $$

[#85] In ΔPQR, side PQ = 32 cm and side PR = 25 cm. What is the measure of side QR ?
Correct Answer

$$4sqrt {154} $$xa0 cm

Explanation

Solution: $$eqalign{
& QR = sqrt {{{left( {PR}
ight)}^2} - {{left( {PQ}
ight)}^2}} cr
& ,,,,,,,,,,,,, = sqrt {{{25}^2} + {3^2}} cr
& ,,,,,,,,,,,,, = sqrt {625 - 9} cr
& ,,,,,,,,,,,,, = sqrt {616} cr
& ,,,,,,,,,,,,, = 2sqrt {154} ,cm cr} $$