KnowledgeBoat Logo
|

Mathematics

Case Study:
In most of the Indian metropolitan cities, government has made many flyovers to regulate the traffic flow. One of such flyover is shown below. The triangular side walls of the flyover have been used for advertisements. One of such side walls has lengths 122 m, 120 m and 22 m. The rate of advertisement on this wall is ₹ 1,000 per m2 per year. A private company hired this wall for 4 months.

In most of the Indian metropolitan cities, government has made many flyovers to regulate the traffic flow. One of such flyover is shown below. ARC Properties of Circle, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Based on the above information, answer the following questions:

  1. Semi-perimeter of the triangular wall is :
    (a) 130 m
    (b) 132 m
    (c) 134 m
    (d) 135 m

  2. Area of the triangular wall is :
    (a) 1400 m2
    (b) 1350 m2
    (c) 1340 m2
    (d) 1320 m2

  3. Total amount paid by the company as rent for the wall is:
    (a) ₹4,00,000
    (b) ₹4,20,000
    (c) ₹4,40,000
    (d) ₹13,20,000

  4. Another company paid ₹5,50,000 as a rent for the same wall. For how many months did the company hire the wall?
    (a) 5
    (b) 6
    (c) 7
    (d) 8

  5. Which of the following is the correct formula for finding the area of a triangle of sides x, y, z and semi perimeter p?
    (a) Area = p(p - x) (p - y) (p - z)
    (b) Area = p(p+x)(p+y)(p+z)\sqrt{p(p + x)(p + y)(p + z)}
    (c) Area = (p−x)(p−y)(p−z)\sqrt{(p - x)(p - y)(p - z)}
    (d) Area = p(p−x)(p−y)(p−z)\sqrt{p(p - x)(p - y)(p - z)}

Mensuration

1 Like

Answer

1. Let a = 122, b = 120, c = 22.

s = a+b+c2=122+120+222\dfrac{a + b + c}{2} = \dfrac{122 + 120 + 22}{2}

= 2642\dfrac{264}{2} = 132 m.

Hence, option (b) is the correct option.

2. By formula,

Area of triangle=s(s−a)(s−b)(s−c)=132(132−122)(132−120)(132−22)=132(10)(12)(110)=1742400=1320 m2.\text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{132(132 - 122)(132 - 120)(132 - 22)} \\[1em] = \sqrt{132(10)(12)(110)} \\[1em] = \sqrt{1742400} \\[1em] = 1320 \text{ m}^2.

Hence, option (d) is the correct option.

3. Rate per year = ₹ 1,000 per m2

Cost = Area × Rate per m2

= 1320 × 1000 = ₹ 13,20,000 per year.

For 4 months, cost will be :

⇒ 13,20,000 × 412\dfrac{4}{12}

⇒ 13,20,000 × 13\dfrac{1}{3}

⇒ ₹4,40,000.

Hence, option (c) is the correct option.

4. Yearly rent = ₹13,20,000

Monthly rent = 13,20,00012\dfrac{13,20,000}{12} = ₹1,10,000

∴ No. of months = Amount paidCharge per month=5,50,0001,10,000\dfrac{\text{Amount paid}}{\text{Charge per month}} = \dfrac{5,50,000}{1,10,000} = 5.

Hence, option (a) is the correct option.

5. By formula,

For a triangle with sides x, y, z and semi perimeter p.

Area = p(p−x)(p−y)(p−z)\sqrt{p(p - x)(p - y)(p - z)}

Hence, option (d) is the correct option.

Answered By

3 Likes

Related Questions