Math, asked by Anonymous, 7 months ago

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?


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Answers

Answered by khandelwalsudha012
1

Answer:

In △ABC

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

s=

2

a+b+c

=

2

122+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 m

2

Rent of 1 m

2

per year = Rs. 5000

Rent of 1 m

2

per month = Rs.

12

5000

Rent of 1 m

2

for 3 months=Rs.

12

3×5000

Rent of 1320 m

2

for 3 months=Rs. (

12

3×5000×1320

)

= Rs.16,500,000

Answered by Anonymous
2

Answer:

apun bola

tu meri laila

voh boli

chal phektha hai saala

me jab bhi

sach bolta

usko kyu Sab jhoot lagta

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