Question

please tell me amswer of this with solution and correct amswer will mark as brainlist

Answer 1

Answer :


Let sides of triangular side walls of fly-over be a,b and c metres

therefore,

a = 122 m
b = 22 m
c = 120 m

therefore,

$semiperimeter \: (s) = \frac{a + b + c}{2} \\ \\ = > \: s \: = \frac{122 + 22 + 120}{2} \\ \\ = > \: s = \frac{264}{2} \\ \\ = > \: s = 132 \: m$

By Heron's formula, area of triangular side of walls =>

$\sqrt{s(s - a)(s - b)(s - c)} \\ \\ = > \sqrt{132(132 - 122)(132 - 22)(132 - 120)} \\ \\ = > \: \sqrt{132 \times 10 \times 110 \times 12} \\ \\ = > \sqrt{11 \times 12 \times 10 \times 10 \times 11 \times 12} \\ \\ = > \sqrt{10 \times 10 \times 11 \times 11 \times 12 \times 12} \\ \\ = > 10 \times 11 \times 12 \\ \\ = > 1320 \: m {}^{2}$

$rent \: for \: advertisement \: on \: wall \\ \: for \: 1 \: year(i.e. \: 12 \: months) = rs. \: 5000 \: per \: {m}^{2} \\ \\ rent \: for \: advertisement \: on \: wall \\ for \:one \: month = rs. \: \frac{5000}{12} \: per \: {m}^{2} \\ \\ rent \: for \: advertisement \: on \: wall \\ for \: 3 \: months = rs. \: ( \frac{5000}{12} \times 3) \: per \: {m}^{2} \\ \\ now \: rent \: paid \: for \: advertisement \: per \: {m}^{2} = > \\ \\ = \: rs. \frac{5000}{4} \\ \\ rent \: paid \: for \: advertisement \: 1320 \: {m}^{2} \\ \\ = > \: rs .( \frac{5 000 }{4} \times 1320) \\ \\ = > \: rs.(5000 \times 330) \\ \\ rs. \: 1650000.$

Hence rent paid by company = ₹ 16,50,000


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