9th Maths 12.1
NCERT Class 9th solution of Exercise 12.1
Exercise 12.1
Q1. A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side 'a'. Find the area of the signal board, using Heron's formula. If its perimeter is `180cm`, what will be the area of the signal board?
Sol. :
Semi persmeter `S=(a+a+a)/2=(3a)/2`
By Heron's formula:
Area of `triangle=sqrt(S(S-a)(S-b)(S-c))`
Area of `triangle=sqrt((3a)/2((3a)/2-a)((3a)/2-a)((3a)/2-a))`
Area of `triangle=sqrt((3a)/2(a/2)(a/2)(a/2))`
Area of `triangle=sqrt(3)/4a^2` square unit.
The Perimeter of equilateral `triangle=3a`
`180=3a`
`a=180/3`
`a=60`
Area of `triangle=sqrt3/4a^2`
Area of `triangle=(60)^2/4sqrt3cm^2`
Area of `triangle=(60times60)/4sqrt3cm^2`
Area of `triangle=900sqrt3cm^2`
Answer :
Area of the signal board `triangle=900sqrt3cm^2.`
Q2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are `122m, 22m` and `120m`[see figure]. The advertisements yield an earning of `₹5000` per `m^2`per year. A company hired one of its walls for `3` months. How much rent did it pay?
Sol. :
Let side `a=122m, b=120m` and `c=22m``S=(a+b+c)/2`
`S=(122+120+22)/2`
`S=264/2`
`S=132`
By Heron's formula:
Area of `triangle=sqrt(S(S-a)(S-b)(S-c))`
Area of `triangle=sqrt(132(132-122)(132-120)(132-22))`
Area of `triangle=sqrt(132(10)(12)(110))`
Area of `triangle=sqrt(11times12times10times12times11times10)`
Area of `triangle=10times11times12`
Area of `triangle=1320m^2`
Rent for `3` months = Area`times`Rate
`=₹1320times(5000times3)/12`
`=₹110times15000`
`=₹165000`
Answer:
Rent for `3` months `₹165000`.
Q3. 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 figure]. If the sides of the wall are `15m`, `11m`, and `6m`. find the area painted in colour.
Sol. :
Let side `a=15m, b=11m` and `c=6m`
`S=(a+b+c)/2`
`S=(15+11+6)/2`
`S=32/2`
`S=16m`
By Heron's formula:
Area of `triangle=sqrt(S(S-a)(S-b)(S-c))`
Area of `triangle=sqrt(16(16-15)(16-11)(16-6))`
Area of `triangle=sqrt(16times1times5times10)`
Area of `triangle=sqrt(4times4times5times5times2)`
Area of `triangle=4times5sqrt2m^2`
Area of `triangle=20sqrt2m^2`
Answer:
The area of painted in colour `20sqrt2m^2`.
Q4. Find the area of a triangle two sides of which are `18cm` and `10cm` the perimeter is `42cm`.
Sol. :
Perimeter of `triangle=a+b+c`
`42=18+10+c`
`c=42-28`
`c=14`
`S=42/2`
`S=21`
By Heron's formula:
Area of `triangle=sqrt(S(S-a)(S-b)(S-c))`
Area of `triangle=sqrt(21(21-18)(21-10)(21-14))`
Area of `triangle=sqrt(21times3times11times7)`
Area of `triangle=21sqrt11cm^2`
Answer:
Area of `triangle=21sqrt11cm^2`.
Q5. Sides of a triangle are in the ratio of `12:17:25` and its perimeter is `540cm`. Find its area.
Sol. :
Perimeter of `triangle=a+b+c`
`540=12x+17x+25x`
`540=54x`
`x=540/54`
`x=10`
`a=12times10=120`
`b=17times10=170`
`c=25times10=250`
`S=540/2=270`
Area of `triangle=sqrt(S(S-a)(S-b)(S-c))`
Area of `triangle=sqrt(270(270-120)(270-170)(270-250))`
Area of `triangle=sqrt(270times150times100times20)`
Area of `triangle=9000cm^2`
Answer:
Area of `triangle=9000cm^2`.
Q6. An isosceles triangle has perimeter `30cm` and each of the equal sides is `12cm`. Find the area of the triangle.
Sol. :
Perimeter of `triangle=a+b+c`
`30=12+12+c`
`c=30-24`
`c=6`
`S=30/2=15`
By Heron's formula:
Area of `triangle=sqrt(S(S-a)(S-b)(S-c))`
Area of `triangle=sqrt(15(15-12)(15-12)(15-6))`
Area of `triangle=sqrt(15times3times3times9)`
Area of `triangle=9sqrt15`
Answer:
Area of `triangle=9sqrt15cm^2`.
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