Question

Please answer this questions only if you know the answer

The triangular sides walls of a fly over have been used for advertisements. The sides of the walls are 122m, 22m & 120m. The advertisements yield an earning of Rs.5000 per m² ar year. A company hired one its walls for 3 months. How much rent did it pay?​

Answer 1

Answer:

1650000rs

Step-by-step explanation:

semi-p = a+b+c/2 = 122+22+120/2= 264/2 = 132

Area = $\sqrt{s(s-a)(s-b)(s-c)}$

=$\sqrt{132(132-122)(132-22)(132-120)}$

= $\sqrt{132*10*110*12} \\\sqrt{12*11*10*10*11*12}\\\sqrt{12^{2} * 11^{2} * 10^{2}$

= 12x11x10 = 1320m

rent paid = 3/12 = 1/4

= 1320x5000x$\frac{1}{4}$ = 1650000

Answer 2

➤ Answer

• Rent paid by the company = ₹ 16,50,000.

➤ Given

• Sides of triangular walls are 122m, 22m, and 120m.
• Rate of advertising = ₹ 5000 per m² for one year.
• A company hired wall for 3 months.

➤ To Find

• The rent paid by the company.

➤ Step By Step Explanation

Given that

• Sides of triangular walls are 122m, 22m, and 120m.
• Rate of advertising = ₹ 5000 per m² for one year.
• A company hired wall for 3 months.

We need to find the rent paid by the company. So let's do it !!

First we need to find the area of the triangular walls using herons formula.

➠ Formula Used

$\underline{\boxed{\tt{\red{\cfrac{a + b + c}{2} = s}}}} \\ \\\underline{\boxed{ \tt{\green{ \sqrt{s \times (s - a) \times (s - b) \times (s - c)}}}}}$

➠ Substituting

$\longmapsto\tt \cfrac{122+22+120}{2}\\ \\ \longmapsto\tt 132 \\ \\ \longmapsto\tt \sqrt{132 \times (132 - 122) \times (132 - 22) \times (132 - 120)} \\ \\ \longmapsto\tt \sqrt{132\times 10 \times 110 \times 12} \\ \\ \longmapsto\tt \sqrt{1742400} \\ \\\longmapsto\tt 1320 {m}^{2}$

➠ Rent

Rent of 1m² of wall for 1 year = ₹ 5000

Wall is hired for 3 month or 3/12 => ¼

Now, Rent paid will be ⤵

$\longmapsto\tt\cfrac{1}{4} \times 1320 \times 5000 \\ \\ \longmapsto\tt \cfrac{6600000}{4} \\ \\ \longmapsto\tt 1650000 \: rupees$

Therefore, rent paid by the company = ₹ 16,50,000.

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