## 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= ^{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 =

=

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

Hence, the height comes out to be 300 m and base is

** 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 =

Area =

=

=

=

= ^{2}

Longest side = 42 cm

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

Area of triangle =

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,

Area =

=

=

=

= ^{2}

Smallest side = 18 cm

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

Area of the triangle =

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,

Area =

=

=

=

= ^{2}

Longest side = 105 m and base = 105 m

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

Area of the triangle =

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 ]

Thus, the sides of the triangle are:

Now, S =

S =

Therefore, area of triangle =

=

=

=

=

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]

Thus, the sides of the triangle are:

Now, S =

S =

Therefore, the area of triangle =

=

=

=

=

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

=

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 =

S =

Therefore, area of triangle =

=

=

=

=

=

Therefore, area of triangle = 2772 sq m.

Also area of triangle =

2772 =

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 =

S =

Therefore, area of triangle =

=

=

=

=

=

**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,

Given that BC= 80 cm and Area of

Therefore,

Therefore,

Therefore,

Therefore,

Now,

a =

a =

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 =

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

Perimeter of an isosceles triangle = 42 cm

Therefore, length of lateral side= 12 cm

And base =

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 =

S =

Therefore, area of triangle =

=

= ^{2}

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

**(iii) Height of the triangle:**

Area of a triangle =

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 =

Area of the equilateral triangle =

i.e. a = 12 cm

Perimeter of an equilateral triangle =

Since, a = 12 cm

Perimeter =

**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 =

Area of an equilateral triangle =

Height of an equilateral triangle =

Since a = 18 cm

Therefore, Height of the equilateral triangle=

**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:

i.e.

Therefore, the area of the right triangle

=

= ^{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 =

Area =

^{2}

**(ii)** Height of an equilateral triangle =

**6.928 cm**

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