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?​

Answer 1

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= 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, 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

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

Answer 2

Answer:

In △ABC

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

s= a+b+c/2 =122+22+120/2=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 m ²

Rent per year = Rs. 5000

Rent of 1 m²per month = Rs. 5000/12

Rent of 1 m ²for 3 months=Rs. (3×5000)/12

Rent of 1320 m

2

for 3 months=Rs. ( 3×5000×1320)/2

= Rs.16,500,000

Step-by-step explanation:

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