Question

two crossroads each of width 4 m,run at right angles through the centre of a park of length 72 m and bredth 42 m and parallel to its sides.find the area of the roads.also find the cost of constructing the roads at the rate of rupees 700per sq m.

Answer 1

$\underline {\mathfrak {\huge{The\:Question:}}}$

Two crossroads each of width 4 m, run at right angles through the centre of a park of length 72 m and breadth 42 m , parallel to its sides. Find the area of the roads, also, find the cost of constructing the roads at the rate of rupees 700 per sq m.

$\underline{\mathfrak{\huge{Your\:Answer:}}}$

Lengths of the two roads = 42 m and 72 m

Breadth of the two roads = 4 m each

Area of one road = 42 m × 4 m

Area of one road = $\tt{168 m^{2}}$

Area of another road = 72 m × 4 m

Area of another road = $\tt{288 m^{2}}$

$\underline{\bold{Note :}}$ Area of the small square is included twice. So, we need to subtract it once.

Area of the small square = 4 m × 4 m

Area of the small square = $\tt{16 m^{2}}$

Area of the roads = $\tt{168 m^{2} + 288 m^{2} - 16 m^{2}}$

Area of the roads = $\tt{440 m^{2}}$

Cost of one $m^{2}$ = Rs. 700

Cost of 440 $m^{2}$ = Rs. 700 × 440

Cost = $\tt{Rs. 308000}$

Answer 2

Area of EFHG = EF × FH

AS FH is parallel to AC in rectangle

So FH = AC= 72

Area of EFHG = 4× 72 = 288 M^2


Area of IJLK = IJ × IK

= 4× 42 = 168 m^2


Now area of square formed by intersection of two roads= 4 × 4 = 16 m^2


So area of road = Area of EFGH + Area of IJLK - area of square

= 288 +168 -16 = 456 -16 = 440 m^2


Cost of constructing is 700 per m^2

So cost of constructing 440 m^2 is 440 × 700

= Rs 308000