## 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) sq units = (12×24×14.5)cm2 = 174 cm^{2}

**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(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√

= 48(48−42)(48−34)(48−20)−−−−−−−−−−−−−−−−−−−−−−−√

= 48×6×14×28−−−−−−−−−−−−−√

= 4×4×3×3×2×14×14×2−−−−−−−−−−−−−−−−−−−−−−−−√

= 4×3×2×14 = 336 cm^{2}

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(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√

= 36(36−18)(36−24)(36−30)−−−−−−−−−−−−−−−−−−−−−−−√

= 36×18×12×6−−−−−−−−−−−−−√

= 6×6×6×3×3×4×6−−−−−−−−−−−−−−−−−−−√

= 6×6×3×2 = 216 cm^{2}

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(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√

= 147(147−91)(147−98)(147−105)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

= 147×56×49×42−−−−−−−−−−−−−−−√

= 49×3×7×2×2×2×49×7×3×2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

= 49×3×2×2×7 = 4116 m^{2}

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(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√

= 75(75−25)(75−60)(75−65)−−−−−−−−−−−−−−−−−−−−−−−√

= 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 m ^{2}.**

**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(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√

= 270(270−250)(270−170)(270−120)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

= 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(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√

= 162(162−85)(162−154)(162−85)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

= 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(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√

= 23(23−13)(23−13)(23−20)−−−−−−−−−−−−−−−−−−−−−−−√

= 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 cm ^{2}. Find the perimeter of the triangle.**

Sol:

Let, △ ABC be an isosceles triangle and let AL∥BC.

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

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(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√

= 21(21−12)(21−12)(21−18)−−−−−−−−−−−−−−−−−−−−−−−√ = 21×9×9×3−−−−−−−−−−−√

= 3×7×9×9×3−−−−−−−−−−−−−√ = 277–√ = 71.42 cm^{2}

Therefore, the area of triangle = 71.42 cm^{2}

**(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 363–√cm2, 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 = 363–√cm2 (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 813–√cm2, find its height.**

**Sol:**

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

Therefore, the area of equilateral triangle = 3√4a2 sq units

Area of an equilateral triangle = 813–√cm2 (given)

⇒813–√cm2=3√4a2

⇒a2=813√×43√=324

⇒a=324−−−√=18cm

Height of an equilateral triangle = 3√2a

Since a = 18 cm

Therefore, Height of the equilateral triangle= 3√2×18=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

By Pythagoras theorem, we have:

AC2=AB2+BC2

i.e. 502=x2+482

⇒x2=502−482

⇒x2=196

⇒x=14 cm

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

= 12×48×14

= 24×14cm2 = 336 cm^{2}

Therefore, the area of the triangle = 336 cm^{2}

**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 =3√4×a2 [a is the side of the equilateral triangle]

Area = 3√4×82 = 3√4×64

⇒3–√×16 = 1.732×16 = 27.71 cm^{2}

**(ii)** Height of an equilateral triangle = 3√2a

3√2×8

3–√×4 = 1.732×4 =**6.928 cm**