Question

please help me please​

Answer 1

Given:

• The triangular side walls of a flyover have been used for advertisements

• The sides of the wall are 122m,22m,120m

• The advertisements yield at Rupees 5000 per m² per year

• A company hires one of its walls for 3months

Full Solution:

${ \blue{ \underline{ \boxed{ \rm{(i)answer}}}}}$

Question:

• Find the semiperimeter of the triangular wall

Solution:

➼ We know,

${ \pink{ \star{ \underline{ \boxed{ \sf{semi \: perimeter = \frac{sum \: of \: all \: sides}{2} }}}}}}$

➱Here,

• 1st Side = 122m
• 2nd side = 22m
• 3rd side = 120m

Substituting we get,

${ : \implies} \rm \: semi \: perimeter_{(triangle)} = \frac{122 + 22 + 120m}{2} \\ \\ \\ { : \implies} \rm \: semi \: perimeter_{(triangle)} = \frac{114 + 120}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm \: semi \: perimeter_{(triangle)} = \frac{264}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm \: semi \: perimeter_{(triangle)} = { \purple{ \boxed{ \frak{132m}} \star}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Henceforth the semi perimeter of the triangular wall is option (b) 132m

${ \blue{ \underline{ \boxed{ \rm{(ii)answer}}}}}$

➱ We know,

• ↠ Semiperimeter of the wall is 132m

➼ Now,

• ↠ First side = 122m

So, the difference will be

★ 132 - 122m

★ 10m

_____________

• ↠ Second side = 22m

So, the difference will be

★ 132 - 22m

★ 110m

_________________

• ↠ Third side = 120m

So, the difference will be

★ 132 - 120m

★ 12m

• Henceforth the correct option is ( c ) 10m,110m,12m respectively

${ \blue{ \underline{ \boxed{ \rm{(iii)answer}}}}}$

➱ Now, we are asked to find the area

We know that,

${ \pink{ \star \: { \boxed{ \sf{ a = \sqrt{s(s - a)(s - b)(s - c)} }}}}}$

where,

• ↠ A stands for Area
• ↠S stand for Semiperimeter
• ↠ A,B,C are the sides

here,

• ↝Semiperimeter = 132m
• ↝Side A = 112m
• ↝Side B = 22m
• ↝ Side C = 120m

Substituting we get,

${ : \implies} \rm area _{(triange)} = \sqrt{s(s - a)(s - b)(s - b)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm area _{(triange)} = \sqrt{132 \: (132 -122 )(132 - 22)(132- 120)} \\ \\ \\ { : \implies} \rm area _{(triange)} = \sqrt{132(10)(110)(12)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm area _{(triange)} = \sqrt{132 \times 10 \times 110 \times 12} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm area _{(triange)} = \sqrt{1742400} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm area _{(triange)} = { \purple{ \boxed{ \frak{1320 {m}^{2} }} \star}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• Henceforth the area of the triangle is 1320m² option B

${ \blue{ \underline{ \boxed{ \rm{(iv)answer}}}}}$

We know that the company used the walls for 3 months

• The cost per the advertisements is rupees 5,000 per m²per year

So,

• Cost per month will be month per m² will be

★ 5000/12

★416 + ( 8/12 ) rupees

${ : \implies} \rm \: cost \: for \: 3 \: months = 416 \frac{8}{12} \times 3 \times 1320 \\ \\ \\ \\ { : \implies} \rm \: cost \: for \: 3 \: months = \frac{5000}{ \cancel{12}} \times \cancel3 \times 1320 \\ \\ \\ { : \implies} \rm \: cost \: for \: 3 \: months = \cancel \frac{5000}{4} \times 1320 \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \rm \: cost \: for \: 3 \: months = { \purple{ \boxed{ \frak{rs.1650000}} \star}} \: \: \: \:$

• Henceforth the cost for the company to advertise per 3monts is rs.165000 option A

${ \blue{ \underline{ \boxed{ \rm{(v)answer}}}}}$

We know,

• The payment yearly per m² = 5,000 rupees

${ : \implies} \rm \: cost_{(yearly)} = 1320 \times 5000 \\ \\ \\ { : \implies} \rm \: cost_{(yearly)} = { \purple{ \boxed{ \frak{ = rs.6600000}}}}$

• Hence forth the cost of advertisement per annum = Rupees 66,00000

Additional Info:

• The area of a triangle ( When hieght and base are give) = 1/2 × base×hieght

• Sum of interior angles in a triangle add up to 180°

• The sum of its exterior angles add upto 360°

• In an right angled triangle and square of its hypotenuse is equal to the sum of squares of the other two sides