Q.1 Find the area of a triangle, two sides of which are 8 cm and 11 cm and

the perimeter is 32 cm ?

EXPLANATION : Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm.

Third side c = 32 cm – (8 + 11) cm = 13 cm

So, 2s = 32, i.e., s = 16 cm,

s – a = (16 – 8) cm = 8 cm,

s – b = (16 – 11) cm = 5 cm,

s – c = (16 – 13) cm = 3 cm.

Therefore, area of the triangle = s(s ?a) ( s ??b) (s ??c)

= 16??8??5?3 cm2 ??8 30 cm2

Q.2 A triangular park ABC has sides 120m, 80m and 50m. A

gardener Dhania has to put a fence all around it and also plant grass inside. How

much area does she need to plant? Find the cost of fencing it with barbed wire at the

rate of Rs 20 per metre leaving a space 3m wide for a gate on one side.

Solution : For finding area of the park, we have

2s = 50 m + 80 m + 120 m = 250 m.

i.e., s = 125 m

Now, s – a = (125 – 120) m = 5 m,

s – b = (125 – 80) m = 45 m,

s – c = (125 – 50) m = 75 m.

Therefore, area of the park = s(s ?a) (s ??b) (s ??c)

= 125 ??5 ??45 ??75 m2

= 375 15m2

Also, perimeter of the park = AB + BC + CA = 250 m

Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate)

= 247 m

Q.3 The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter is 300 m. Find its area.

Suppose that the sides, in metres, are 3x, 5x and 7x .

Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle)

Therefore, 15x = 300, which gives x = 20.

So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 m

i.e., 60 m, 100 m and 140 m.

Can you now find the area [Using Heron’s formula]?

We have s =

60 100 140

2

???

m = 150 m,

and area will be 150(150??60) (150 ??100) (150 ?140) m2

= 150 ??90 ?50 ??10 m2

= 1500 3m2

Q.4 Kamla has a triangular field with sides 240 m, 200 m, 360 m, where she grew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to the previous field, she wanted to grow potatoes and onions (see Fig. 12.11). She divided the field in two parts by joining the mid-point of the longest side to the opposite vertex and grew patatoes in one part and onions in the other part. How much area (in hectares) has been used for wheat, potatoes and onions? (1 hectare = 10000 m2)

Solution : Let ABC be the field where wheat is grown. Also let ACD be the field

which has been divided in two parts by joining C to the mid-point E of AD. For the

area of triangle ABC, we have

a = 200 m, b = 240 m, c = 360 m

Therefore, s =

200 +240+ 360 / 2

m = 400 m.

So, area for growing wheat

= 400(400 ??200) (400 – 240) (400 – 360) m2

= 400 ??200 ??160 ??40 m2

= 16000 2m2 = 1.6 × 2 hectares

= 2.26 hectares (nearly)

Let us now calculate the area of triangle ACD.

Here, we have s =

240+ 320 +400/2

m = 480 m.

So, area of ??ACD = 480(480 ??240) (480 ??320) (480 ??400) m2

= 480 ??240 ??160 ??80 m2 = 38400 m2 = 3.84 hectares

We notice that the line segment joining the mid-point E of AD to C divides the

triangle ACD in two parts equal in area. Can you give the reason for this? In fact, they

have the bases AE and ED equal and, of course, they have the same height.

Therefore, area for growing potatoes = area for growing onions

= (3.84 ??2) hectares = 1.92 hectares.

Q.5 Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA (see Fig. 12.12). Then they cleaned

the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m

and ??B = 90º, which group cleaned more area and by how much? Find the total area

cleaned by the students (neglecting the width of the lanes).

Solution : Since AB = 9 m and BC = 40 m, ??B = 90°, we have:

AC = 92 ??402 m

= 81 ??1600 m

= 1681 m = 41m

Therefore, the first group has to clean the area of triangle ABC, which is right angled.

Area of ??ABC =

1

2 × base × height

=

1

2 × 40 × 9 m2 = 180 m2

The second group has to clean the area of triangle ACD, which is scalene having sides

41 m, 15 m and 28 m.

Here, s =

41+ 15 +28 /2

m = 42 m

Therefore, area of ??ACD = s(s – a) ( s – b) (s – c )

= 42(42 – 41) (42 – 15) (42 – 28) m2

= 42 ?1 ??27 ??14 m2 = 126 m2

So first group cleaned 180 m2 which is (180 – 126) m2, i.e., 54 m2 more than the area

cleaned by the second group.

Total area cleaned by all the students = (180 + 126) m2 = 306 m2.