9th Maths 12.1

NCERT Class 9th solution of Exercise 12.1

NCERT Class 9th Maths Projects

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. :

Let side

$a=18cm, b=10cm$

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. :

Let side

$a=12x, b=17x$ and

$c=25x$

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. :

Let side

$a=12$ , b=12`

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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9th Maths 12.1

NCERT Class 9th solution of Exercise 12.1

NCERT Class 9th Maths Projects

Exercise 12.1

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