Question

$\sf\red{Question :}$
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

don't spam ✖
___________​

Answer 1

Answer:

Question :-

• The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

Given :-

• The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months.

Find Out :-

• How much rent did it pay?

Solution :-

Let the sides of the triangle are a=122 m, b=22 m & c= 120 m.

✯ Semi Perimeter of the ∆,s = (a+b+c) /2

➸ s =(122 + 22 + 120) / 2

➸ s = 264/2

➸ s = 132m

◆ Using heron’s formula,

✯ Area of the wall = √s (s-a) (s-b) (s-c)

➸ √132(132 – 122) (132 – 22) (132 – 120)

➸ √132 × 10 × 110 × 12

➸ √11×12×10×11×10×12

➸ √11×11×12×12×10×10

➸ 11×12×10

➸ 1320m²

Given, that

• Earning on 1m² per year = ₹5000

➸ Earning on 1320 m² per year=1320×5000= ₹6600000

➸ Now, earning in 1320 m² in 12 months= ₹6600000

➸ Earning in 3 months = ₹ 6600000 ×3/12 = ₹ 1650000

Henceforth, the rent paid by the company for 3 months is ₹ 1650000.

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

Answer 2

Answer:

s=2a+b+c=2122+22+120

=132m

ar△ABC=s(s−a)(s−b)(s−c)=132(132−122)(132−23)(132−120)

=132×10×110×12=132×10×11×10×12=(132)2×(10)2

=132×10=1320 m2

Rent of 1 m2 per year = Rs. 5000

Rent of 1 m2 per month = Rs.125000

Rent of 1 m2 for 3 months=Rs.123×5000

Rent of 1320 m2 for 3 months=Rs. (123×5000×1320)

= Rs.16,500,000