RS Aggarwal Class 9 Solutions Chapter 7 - Areas

RS Aggarwal Class 9 Chapter 7 - Areas Solutions Free PDF

The Egyptian Mathematician Heron is credited for his work on Mathematics, particularly in the field of mensuration. He derived the ‘Heron’s Formula’ to simplify the formula of finding the area of the triangle easily. This chapter teaches students about the calculation of the area and perimeters of the different types of triangles as well, such as:

• Equilateral triangles
• Isosceles triangles
• Right angled triangle

Some of the major formulas for areas mentioned in this chapter are:

1. Area of a square
2. Area of a rectangle
3. Area of a trapezium
4. Area of polygons
5. Area of circles
6. Area of surfaces of cylinders
7. Area of cones and sphere

RS Aggarwal Class 9 Chapter 7

Q1: Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm

Sol:

Here, b = 24 cm and h = 14.5 cm

Area of triangle= (12×base×height)$(\frac{1}{2}\times base\times height)$ sq units = (12×24×14.5)cm2$\left ( \frac{1}{2}\times 24\times 14.5 \right ) cm^{2}$ = 174 cm2

Q2: The base of a triangular field is 3 times its altitude. If the cost of sowing the field at Rs.58 per hectare is Rs.783 find its base and height.

Sol:

Here, height = x units and base = 3x units

Area of triangle = (12×base×height)$(\frac{1}{2}\times base\times height)$ sq units

= (12×x×3x)squnits$\left ( \frac{1}{2}\times x \times 3x \right ) \; sq \; units$ = 32x2$\frac{3}{2}x^{2}$

We know that, 1 hectare =1000 sq meters

Rate of sowing the triangular field = Rs. 58

Total cost of sowing the triangular field = Rs. 783

$\Rightarrow$ Total cost = Area of the triangular filed ×$\times$ Rs. 58

$\Rightarrow$ 32x2×5810000=783$\frac{3}{2}x^{2}\times \frac{58}{10000}=783$

x2=78358×23×10000sqmetre$\Rightarrow x^{2} =\frac{783}{58}\times \frac{2}{3}\times 10000 \; sq \; metre$
x2=90000sqmetre$\Rightarrow x^{2} =90000 \; sq \; metre$
x=300m$\Rightarrow x =300 \; m$

Hence, the height comes out to be 300 m and base is 3×300m=900m$3\times 300 \; m=900 \; m$

Q3 find the area of the triangle whose sides are 42cm, 34 cm and 20cm in length. Also, find the length of the altitude corresponding to the smallest side.

Sol:

Here , a = 42 cm ,b = 34 cm and c = 20 cm

Therefore, S = 42+34+202$\frac{42+34+20}{2}$ = 48

Area = S(Sa)(Sb)(Sc)$\sqrt{S(S-a)(S-b)(S-c)}$

= 48(4842)(4834)(4820)$\sqrt{48(48-42)(48-34)(48-20)}$

= 48×6×14×28$\sqrt{48 \times 6\times 14 \times 28 }$

= 4×4×3×3×2×14×14×2$\sqrt{4 \times 4 \times 3\times 3 \times 2 \times 14 \times 14 \times 2 }$

= 4×3×2×14$4 \times 3 \times 2 \times 14$ = 336 cm2

Longest side = 42 cm

b=42cm$\Rightarrow b= 42 cm$

Let, h be the height corresponding to the longest side.

Area of triangle = (12×base×height)$(\frac{1}{2}\times base\times height)$ sq units

12×base×height=336cm2$\Rightarrow \frac{1}{2} \times base \times height =336 \; cm^{2}$
42×h=336×2$\Rightarrow 42 \times h =336 \times 2$

Therefore,  h = 16 cm

Q4. Calculate the area of the triangle whose sides are 18 cm 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.

Sol:

Here, a = 18 cm, b = 24 cm , and c = 30 cm

Therefore, S=18+24+302=36$S=\frac{18+24+30}{2}=36$

Area = S(Sa)(Sb)(Sc)$\sqrt{S(S-a)(S-b)(S-c)}$

= 36(3618)(3624)(3630)$\sqrt{36(36-18)(36-24)(36-30)}$

= 36×18×12×6$\sqrt{36 \times 18 \times 12 \times 6 }$

= 6×6×6×3×3×4×6$\sqrt{6 \times 6 \times 6 \times 3 \times 3 \times 4 \times 6 }$

= 6×6×3×2$6 \times 6 \times 3 \times 2$ = 216 cm2

Smallest side = 18 cm

Let, h be the height corresponding to the smallest side.

Area of the triangle =  (12×base×height)$(\frac{1}{2}\times base\times height)$ sq units

12×base×height=216cm2$\Rightarrow \frac{1}{2} \times base \times height = 216 \; cm^{2}$
18×h=216×2$\Rightarrow 18 \times h = 216 \times 2$

Therefore, h = 24 cm

Q5: Find the area of a triangular field whose sides are 91m, 98m and 105m in length. Find the area of the triangle

Sol:

Here, a = 91 m, b = 98 m and c =105 m

Therefore, S=91+98+1052=147$S=\frac{91+98+105}{2}=147$

Area = S(Sa)(Sb)(Sc)$\sqrt{S(S-a)(S-b)(S-c)}$

= 147(14791)(14798)(147105)$\sqrt{147(147-91)(147-98)(147-105)}$

= 147×56×49×42$\sqrt{147 \times 56 \times 49 \times 42 }$

= 49×3×7×2×2×2×49×7×3×2$\sqrt{49 \times 3 \times 7 \times 2 \times 2 \times 2 \times 49 \times 7 \times 3 \times 2}$

= 49×3×2×2×7$49 \times 3 \times 2 \times 2 \times 7$ = 4116 m2

Longest side = 105 m and base = 105 m

Let, h be the height corresponding to the longest side.

Area of the triangle = (12×base×height)$(\frac{1}{2}\times base \times height)$ sq units

12×base×height=4116m2$\Rightarrow \frac{1}{2} \times base \times height = 4116 \; m^{2}$
105×height=2×4116$\Rightarrow 105 \times height = 2 \times 4116$

Therefore, height (h)  = 78.4 m

Q6: The sides of a triangle are in the ratio 5:12:13 and its perimeter is 150m. Find the area of the triangle

Sol:

Let, the sides of the triangle be 5x, 12x and 13x.

Its perimeter = (5x + 12x + 13x) = 30x

Therefore, 30 x = 150 m   [ given ]

x=15030=5m$\Rightarrow x=\frac{150}{30}= 5 \; m$

Thus, the sides of the triangle are:

5x=5×5=25m$5x=5 \times 5=25 \;m$
12x=12×5=60m$12x=12 \times 5= 60 \;m$
13x=13×5=65m$13x=13 \times 5= 65 \;m$

Now, S = 12(a+b+c)$\frac{1}{2}(a+b+c)$

S = 12(25+60+65)=1502$\frac{1}{2}(25+60+65) \; =\frac{150}{2}$ = 75 m

Therefore, area of triangle = S(Sa)(Sb)(Sc)$\sqrt{S(S-a)(S-b)(S-c)}$

= 75(7525)(7560)(7565)$\sqrt{75(75-25)(75-60)(75-65)}$

= 75×50×15×10$\sqrt{75\times 50\times 15 \times 10}$

= 25×3×25×2×5×3×5×2$\sqrt{25 \times 3\times 25 \times 2 \times 5 \times 3 \times 5 \times 2 }$

= 25×5×3×2=750$25 \times 5 \times 3 \times 2 = 750$ sq m.

Therefore, the area of triangle = 750 sq m.

Q.7: The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle and also find the cost of plugging the field at rupees 18.80 per 10 m2.

Sol:

Let, the sides of the triangle be 25x, 17x and 12x.

Its perimeter = (25x +17x + 12x) = 54x

Therefore, 54x = 540 m [given]

x=54054=10m$\Rightarrow x=\frac{540}{54}= 10 \; m$

Thus, the sides of the triangle are:

25x=25×10=250m$25x=25 \times 10=250 \;m$
17x=17×10=170m$17x=17 \times 10= 170 \;m$
12x=12×10=120m$12x=12 \times 10= 120 \;m$

Now, S = 12(a+b+c)$\frac{1}{2}(a+b+c)$

S = 12(25+60+65)=5402=270m$\frac{1}{2}(25+60+65) \; =\frac{540}{2}= 270 \; m$

Therefore, the area of triangle =S(Sa)(Sb)(Sc)$\sqrt{S(S-a)(S-b)(S-c)}$

= 270(270250)(270170)(270120)$\sqrt{270(270-250)(270-170)(270-120)}$

= 270×20×100×150$\sqrt{270 \times 20 \times 100 \times 150 }$

= 3×3×3×10×10×2×10×10×10×5×3$\sqrt{3 \times 3 \times 3 \times 10 \times 10 \times 2 \times 10 \times 10 \times 10 \times 5 \times 3 }$

= 3×3×10×10×10=9000$3 \times 3 \times 10 \times 10 \times 10 = 9000$ sq m.

Therefore, Cost of plugging the field at the rate of Rs. 18.80 per 10 m2$m^{2}$

= 18.8010×9000=Rs.16290$\frac{18.80}{10}\times 9000= Rs. \; 16290$

Therefore, cost of plugging the field = Rs. 16290

Q8: Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find

(i). The area of the field

(ii). The length of the perpendicular from the opposite vertex on the side measuring 154 m.

Sol:

One side of triangular field = 85 m

Second side of triangular field = 154 m

Let the third side of a triangular field be x m.

Its perimeter = 324 m (Given)

Therefore, 85 + 154 + x = 324

x = 324 – 239 = 85 m

Thus, the third sides of the triangle is 85 m,

Let, a = 85 m, b = 154 m and c = 85 m

Now,

S = 12(a+b+c)$\frac{1}{2}(a+b+c)$

S = 12(85+154+85)=3242=162m$\frac{1}{2}(85+154+85) \; =\frac{324}{2}= 162 \; m$

Therefore, area of triangle =S(Sa)(Sb)(Sc)$\sqrt{S(S-a)(S-b)(S-c)}$

= 162(16285)(162154)(16285)$\sqrt{162(162-85)(162-154)(162-85)}$

= 162×77×8×77$\sqrt{162 \times 77 \times 8 \times 77 }$

= 2×9×9×7×11×2×2×2×7×11$\sqrt{2 \times 9 \times 9 \times 7 \times 11 \times 2 \times 2 \times 2 \times 7 \times 11 }$

= 11×11×9×9×7×7×2×2×2×2$\sqrt{11 \times 11 \times 9 \times 9 \times 7 \times 7 \times 2 \times 2 \times 2 \times 2 }$

= 11×9×7×2×2=2772$11 \times 9 \times 7 \times 2 \times 2 = 2772$ sq m.

Therefore, area of triangle = 2772 sq m.

Also area of triangle =  (12×base×height)$(\frac{1}{2}\times base \times height)$

2772 = (12×154×height)$(\frac{1}{2}\times 154 \times height)$

77 × height = 2772

Therefore, h= 36 m

Therefore, the length of the perpendicular from the opposite vertex on the side measuring 154m = 36 m

Q.9: Find the area of an isosceles triangle each of whose equal sides measure 13 cm and whose base measure 20 cm.

Sol:

Let, a = 13 cm, b = 13cm and c = 20 cm

Now,

S = 12(a+b+c)$\frac{1}{2}(a+b+c)$

S = 12(13+13+20)=462=23cm$\frac{1}{2}(13+13+20) \; =\frac{46}{2}= 23 \; cm$

Therefore, area of triangle =S(Sa)(Sb)(Sc)$\sqrt{S(S-a)(S-b)(S-c)}$

= 23(2313)(2313)(2320)$\sqrt{23(23-13)(23-13)(23-20)}$

= 23×10×10×3$\sqrt{23 \times 10 \times 10 \times 3 }$

= 23×10×10×3$\sqrt{23 \times 10 \times 10 \times 3 }$

= 1069$10 \sqrt{69 }$

= 10×8.306=83.06cm2$10 \times 8.306 = 83.06 \; cm^{2}$

Q10: The base of an isosceles triangle measures 80 cm and its area is 360 cm2. Find the perimeter of the triangle.

Sol:

Let, $\bigtriangleup$ ABC be an isosceles triangle and let ALBC$AL \parallel BC$.

Given that BC= 80 cm and Area of ABC=360cm2$\bigtriangleup ABC =360 \; cm^{2}$

Therefore, 12×BC×AL=360cm2$\frac{1}{2}\times BC\times AL= 360\; cm^{2}$

Therefore, 12×80×h=360cm2$\frac{1}{2}\times 80 \times h= 360\; cm^{2}$

Therefore, 40×h=360cm2$40\times h= 360\; cm^{2}$

Therefore, h=36040=9cm$h= \frac{360}{40}=9\; cm$

Now, BL=12BC$BL= \frac{1}{2}BC$

BL=12(80)=40$BL= \frac{1}{2}(80)=40$ cm and AL=9cm$AL =9 \; cm$

a = BL2+AL2$\sqrt{BL^{2}+AL^{2}}$

a = 402+92$\sqrt{40^{2}+9^{2}}$

a=1600+81$\Rightarrow a=\sqrt{1600+81}$
a=1681=41cm$\Rightarrow a=\sqrt{1681}=41 \; cm$

Perimeter = (41+41+80) = 162 cm

Perimeter of the triangle = 162 cm

Q11: The perimeter of an isosceles triangle is 42 cm and its base is 112$1\frac{1}{2}$ times each of the equal sides. Find

(i). The length of each side of a triangle.

(ii). The area of the triangle.

(iii). The height of triangle

Sol:

In an isosceles triangle, the lateral sides are of equal length.

Let the length of lateral side be x cm.

Then, base =32×x$\frac{3}{2}\times x$ cm.

(i). Length of each side of the triangle:

Perimeter of an isosceles triangle = 42 cm

x+x+32x=42cm$x+x+\frac{3}{2}x= 42 \; cm$
2x+2x+3x=84cm$\Rightarrow 2x+2x+3x= 84 \; cm$
7x=84cm$\Rightarrow 7x= 84 \; cm$
x=12cm$\Rightarrow x=12 \; cm$

Therefore, length of lateral side= 12 cm

And base = 32x=32×12=18cm$\frac{3}{2}x=\frac{3}{2}\times 12=18 \; cm$

Therefore, the length of each side of a triangle= 12cm , 12 cm and 18 cm.

(ii) Area of the triangle:

Let, a = 12cm, b = 12cm and c = 18 cm

Now,

S = 12(a+b+c)$\frac{1}{2}(a+b+c)$

S = 12(12+12+18)=422=21cm$\frac{1}{2}(12+12+18) \; =\frac{42}{2}= 21 \; cm$

Therefore, area of triangle =S(Sa)(Sb)(Sc)$\sqrt{S(S-a)(S-b)(S-c)}$

= 21(2112)(2112)(2118)$\sqrt{21(21-12)(21-12)(21-18)}$ = 21×9×9×3$\sqrt{21 \times 9 \times 9 \times 3 }$

= 3×7×9×9×3$\sqrt{3 \times 7 \times 9 \times 9 \times 3 }$ = 277$27 \sqrt{7 }$ = 71.42 cm2

Therefore, the area of triangle = 71.42 cm2

(iii) Height of the triangle:

Area of a triangle = 12×base×height$\frac{1}{2}\times base \times height$

71.42=12×18×h$71.42 =\frac{1}{2}\times 18\times h$
71.42=9×h$71.42 =9 \times h$

i.e. h = 7.94 cm

Therefore, the height of the triangle = 7.94 cm.

Q.12: If the area of an equilateral triangle is 363cm2$36\sqrt{3} \;cm^{2}$, find its perimeter.

Sol:

Let, ‘a’ be the length of a side of an equilateral triangle.

Therefore, the Area of an equilateral triangle = 3×a24$\frac{\sqrt{3}\times a^{2}}{4}$ sq units

Area of the equilateral triangle = 363cm2$36\sqrt{3} \;cm^{2}$         (given)

3×a24=36×3$\Rightarrow \frac{\sqrt{3}\times a^{2}}{4}=36\times \sqrt{3}$
a2=36×3×43$\Rightarrow a^{2}=\frac{36\times \sqrt{3}\times 4}{\sqrt{3}}$

i.e. a = 12 cm

Perimeter of an equilateral triangle = 3×a$3\times a$

Since, a = 12 cm

Perimeter = 3×12=36cm$3\times 12 \; =36 \; cm$

Q.13: In the area of an equilateral triangle is 813cm2$81\sqrt{3} \;cm^{2}$, find its height.

Sol:

Let, a be the length of the side of an equilateral triangle.

Therefore, the area of equilateral triangle = 34a2$\frac{\sqrt{3}}{4}a^{2}$ sq units

Area of an equilateral triangle = 813cm2$81\sqrt{3} \; cm^{2}$      (given)

813cm2=34a2$\Rightarrow 81\sqrt{3}\; cm^{2}=\frac{\sqrt{3}}{4}a^{2}$
a2=813×43=324$\Rightarrow a^{2}=\frac{81\sqrt{3}\times 4}{\sqrt{3}}=324$
a=324=18cm$\Rightarrow a= \sqrt{324}=18 \;cm$

Height of an equilateral triangle = 32a$\frac{\sqrt{3}}{2}a$

Since a = 18 cm

Therefore, Height of the equilateral triangle= 32×18=93$\frac{\sqrt{3}}{2}\times 18=9\sqrt{3}$ cm

Q.14: The base of a right angle triangle measures 48 cm and its hypotenuse measures 50 cm. find the area of the triangle.

Sol:

Base of the right triangle is BC = 48 cm

Hypotenuse of the right triangle is AC = 50 cm

Let, AB = x cm

By Pythagoras theorem, we have:

AC2=AB2+BC2$AC^{2}=AB^{2}+BC^{2}$

i.e. 502=x2+482$50^{2}=x^{2}+48^{2}$

x2=502482$\Rightarrow x^{2}= 50^{2}-48^{2}$
x2=196$\Rightarrow x^{2}=196$

x=14$\Rightarrow x=14$ cm

Therefore, the area of the right triangle =12×base×height$=\frac{1}{2}\times base\times height$

= 12×48×14$\frac{1}{2}\times 48 \times 14$

= 24×14cm2$24 \times 14 cm^{2}$ = 336 cm2

Therefore, the area of the triangle = 336 cm2

Q.15: Each side of an equilateral triangle measures 8 cm. Find

(i). The area of the triangle, correct to 2 decimal places

(ii). The height of the triangle, correct to 2 places of decimal. ( Take root 3 is 3=1.732$\sqrt{3}=1.732$ )

Sol:

(i) Area of an equilateral triangle =34×a2$\frac{\sqrt{3}}{4}\times a^{2}$ [a is the side of the equilateral triangle]

Area = 34×82$\frac{\sqrt{3}}{4}\times 8^{2}$ = 34×64$\frac{\sqrt{3}}{4}\times 64$

3×16$\Rightarrow \sqrt{3}\times {16}$ = 1.732×16$1.732\times {16}$ = 27.71 cm2

(ii) Height of an equilateral triangle = 32a$\frac{\sqrt{3}}{2}a$

32×8$\frac{\sqrt{3}}{2}\times 8$

3×4$\sqrt{3}\times 4$ = 1.732×4$1.732 \times 4$ =6.928 cm

Practise This Question

The maximum number of electrons in the nth orbit can be calculated by the formula: