Q1 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 Rs. 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?
Q2 There is a slide in a park. One of its side walls has been painted in some colour with a message "KEEP THE PARK GREEN AND CLEAN" (see Fig. 12.10). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
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Verified answer
Step-by-step explanation:
[tex]\large\underline{\sf{Solution-1}}[/tex]
Given that, triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m.
Let assume that a = 122 m, b = 22 m and c = 120 m
Now, Consider
[tex]\sf\: {b}^{2} + {c}^{2} \\ [/tex]
[tex]\sf\: = \: {(22)}^{2} + {(120)}^{2} \\ [/tex]
[tex]\sf\: = \:484 + 14400 \\ [/tex]
[tex]\sf\: = \: 14884 \\ [/tex]
[tex]\sf\: = \: {(122)}^{2} \\ [/tex]
[tex]\sf\: = \: {a}^{2} \\ [/tex]
[tex]\implies\bf\: {b}^{2} + {c}^{2} = {a}^{2} \\ [/tex]
So, it means, Triangle is right angle triangle.
[tex]\sf\: Area\:of\:triangular\:wall = \dfrac{1}{2} \times b \times c \\ [/tex]
[tex]\sf\: Area\:of\:triangular\:wall = \dfrac{1}{2} \times 22 \times 120 \\ [/tex]
[tex]\sf\: Area\:of\:triangular\:wall = 22 \times 60 \\ [/tex]
[tex]\implies\bf\:Area\:of\:triangular\:wall = 1320 \: {m}^{2} \\ [/tex]
Further given that, the advertisements yield an earning of Rs. 5000 per m² per year.
Thus,
[tex]\sf\: Rent\:paid\:for\:3\:months \\ [/tex]
[tex]\sf\: = \: 1320 \times \dfrac{5000}{12} \times 3 \\ [/tex]
[tex]\sf\: = \: 1320 \times \dfrac{5000}{4} \\ [/tex]
[tex]\sf\: = \: 330 \times 5000 \\ [/tex]
[tex]\sf\: = \: 1650000 \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:Rent\:paid\:for\:3\:months \: is \: Rs \: 1650000 \: }\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex]\large\underline{\sf{Solution-2}}[/tex]
Let assume that a = 25 m, b = 11 m and c = 6 m
Now,
[tex]\sf\: Semi-perimeter, s = \dfrac{a + b + c}{2} = \dfrac{15 + 11 + 6}{2} = \dfrac{32}{2} = 16 \: m[/tex]
Now,
[tex]\sf\: Area\:of\:wall \: to \: be \: painted \\ [/tex]
[tex]\sf\: = \: \sqrt{s(s - a)(s - b)(s - c)} \\ [/tex]
[tex]\sf\: = \: \sqrt{16(16 - 15)(16 - 11)(16 - 6)} \\ [/tex]
[tex]\sf\: = \: \sqrt{16(1)(5)(10)} \\ [/tex]
[tex]\sf\: = \: 4 \times 5 \sqrt{2} \\ [/tex]
[tex]\sf\: = \: 20 \sqrt{2} \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:Area\:of\:wall \: to \: be \: painted \: is \: 20 \sqrt{2} \: {m}^{2} \: } \\ [/tex]