RS Aggarwal Class 9 Areas

RS Aggarwal Class 9 Chapter 7

Areas Exercise 7.1

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) sq units = (12×24×14.5)cm2 = 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) sq units

= (12×x×3x)squnits = 32x2

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

Total cost = Area of the triangular filed × Rs. 58

32x2×5810000=783

x2=78358×23×10000sqmetre x2=90000sqmetre x=300m

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

 

 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 = 48

Area = S(Sa)(Sb)(Sc)

= 48(4842)(4834)(4820)

= 48×6×14×28

= 4×4×3×3×2×14×14×2

= 4×3×2×14 = 336 cm2

Longest side = 42 cm

b=42cm

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

Area of triangle = (12×base×height) sq units

12×base×height=336cm2 42×h=336×2

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

Area = S(Sa)(Sb)(Sc)

= 36(3618)(3624)(3630)

= 36×18×12×6

= 6×6×6×3×3×4×6

= 6×6×3×2 = 216 cm2

Smallest side = 18 cm

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

Area of the triangle =  (12×base×height) sq units

12×base×height=216cm2 18×h=216×2

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

Area = S(Sa)(Sb)(Sc)

= 147(14791)(14798)(147105)

= 147×56×49×42

= 49×3×7×2×2×2×49×7×3×2

= 49×3×2×2×7 = 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) sq units

12×base×height=4116m2 105×height=2×4116

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

Thus, the sides of the triangle are:

5x=5×5=25m 12x=12×5=60m 13x=13×5=65m

Now, S = 12(a+b+c)

S = 12(25+60+65)=1502 = 75 m

Therefore, area of triangle = S(Sa)(Sb)(Sc)

= 75(7525)(7560)(7565)

= 75×50×15×10

= 25×3×25×2×5×3×5×2

= 25×5×3×2=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 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

Thus, the sides of the triangle are:

25x=25×10=250m 17x=17×10=170m 12x=12×10=120m

Now, S = 12(a+b+c)

S = 12(25+60+65)=5402=270m

Therefore, the area of triangle =S(Sa)(Sb)(Sc)

= 270(270250)(270170)(270120)

= 270×20×100×150

= 3×3×3×10×10×2×10×10×10×5×3

= 3×3×10×10×10=9000 sq m.

Therefore, Cost of plugging the field at the rate of Rs. 18.80 per 10 m2

= 18.8010×9000=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)

S = 12(85+154+85)=3242=162m

Therefore, area of triangle =S(Sa)(Sb)(Sc)

= 162(16285)(162154)(16285)

= 162×77×8×77

= 2×9×9×7×11×2×2×2×7×11

= 11×11×9×9×7×7×2×2×2×2

= 11×9×7×2×2=2772 sq m.

Therefore, area of triangle = 2772 sq m.

Also area of triangle =  (12×base×height)

2772 = (12×154×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)

S = 12(13+13+20)=462=23cm

Therefore, area of triangle =S(Sa)(Sb)(Sc)

= 23(2313)(2313)(2320)

= 23×10×10×3

= 23×10×10×3

= 1069

= 10×8.306=83.06cm2

 

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

Sol:

Let, ABC be an isosceles triangle and let ALBC.

Given that BC= 80 cm and Area of ABC=360cm2

Therefore, 12×BC×AL=360cm2

Therefore, 12×80×h=360cm2

Therefore, 40×h=360cm2

Therefore, h=36040=9cm

 

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Now, BL=12BC

BL=12(80)=40 cm and AL=9cm

a = BL2+AL2

a = 402+92

a=1600+81 a=1681=41cm

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 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 cm.

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

Perimeter of an isosceles triangle = 42 cm

x+x+32x=42cm 2x+2x+3x=84cm 7x=84cm x=12cm

Therefore, length of lateral side= 12 cm

And base = 32x=32×12=18cm

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)

S = 12(12+12+18)=422=21cm

Therefore, area of triangle =S(Sa)(Sb)(Sc)

= 21(2112)(2112)(2118) = 21×9×9×3

= 3×7×9×9×3 = 277 = 71.42 cm2

Therefore, the area of triangle = 71.42 cm2

(iii) Height of the triangle:

Area of a triangle = 12×base×height

71.42=12×18×h 71.42=9×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, 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 sq units

Area of the equilateral triangle = 363cm2         (given)

3×a24=36×3 a2=36×3×43

i.e. a = 12 cm

Perimeter of an equilateral triangle = 3×a

Since, a = 12 cm

Perimeter = 3×12=36cm

 

Q.13: In the area of an equilateral triangle is 813cm2, find its height.

Sol:

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

Therefore, the area of equilateral triangle = 34a2 sq units

Area of an equilateral triangle = 813cm2      (given)

813cm2=34a2 a2=813×43=324 a=324=18cm

Height of an equilateral triangle = 32a

Since a = 18 cm

Therefore, Height of the equilateral triangle= 32×18=93 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

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By Pythagoras theorem, we have:

AC2=AB2+BC2

i.e. 502=x2+482

x2=502482 x2=196

x=14 cm

Therefore, the area of the right triangle =12×base×height

= 12×48×14

= 24×14cm2 = 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 )

Sol:

(i) Area of an equilateral triangle =34×a2 [a is the side of the equilateral triangle]

Area = 34×82 = 34×64

3×16 = 1.732×16 = 27.71 cm2

(ii) Height of an equilateral triangle = 32a

32×8

3×4 = 1.732×4 =6.928 cm

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