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1. The base of a triangulr field is 880 metres and its height 550 metres. Find the area of the field. Also calculate the charges for supplying water to the field at the rate of Rs 24.25 per sq hectometre.
  A.  Area: 20 sq hectometres; Charges: Rs 448.5
  B.  Area: 22.2 sq hectometres; Charges: Rs 546
  C.  Area: 25 sq hectometres; Charges: Rs 583.85
  D.  Area: 24.2 sq hectometres; Charges: Rs 586.85
     
   
View Answer

Shortcut:
The area of triangle when its base (b) and height (h) are given.
Area =
1 / 2
x b x h
Here, b = 880, h = 550
Using these values in the shortcut, we get:
A =
1 / 2
x 880 x 550 = 242000 m2
Converting the unit into hectometres.
242000 / 100 x 100
= 24.2 sq hectometre
Cost of supplying water to 1 sq hectometre is Rs 24.25
∴ cost of supplying water to 24.20 sq hectometre = 24.20 x 24.25 = Rs 586.85

2. The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.1, find its base and height.
  A.  Height: 300 m; Base: 900 m
  B.  Height: 200 m; Base: 800m
  C.  Height: 350 m; Base: 950m
  D.  None of the above
     
   
View Answer

Shortcut:
The area of triangle when its base (b) and height (h) are given.
Area =
1 / 2
x b x h
Here, Area of the field =
332.10 / 24.60
= 13.5 hectares
or, 13.5 x 10000 = 135000 sq m.
Let, the altitude be y metres. Then, base = 3y metres.
Area =
1 / 2
x y x 3y
135000 =
3(y)2 / 2

y2 = 90000
or, y = √(90000)
or, y = 300
∴ altitute = 300 m and base = 900 m

3. A lawn is in the form of a triangle having its base and height in the ratio 2 : 3. The area of the lawn
is
3 / 12
hectare. Find the base and the height of the lawn.
  A.  Height 55 m; Base: 33
4 / 3
  B.  Height 50 m; Base: 33
1 / 3
  C.  Height 40 m; Base: 31
1 / 3
  D.  Height 60 m; Base: 33
1 / 5
     
   
View Answer

Shortcut:
The area of triangle when its base (b) and height (h) are given.
Area =
1 / 2
x b x h
Here,
Let the base be 2y metres and height 3y metres.
Then,
1 / 2
(2y x 3y) =
1 / 12
x 10000 [Since 1 hectare = 10000 sq m]
or, 2y x 3y =
1 / 6
x 10000
or, y2 =
10000 / 6 x 6

y =
√(
100 x 100 / 6 x 6
)

or, y =
100 / 6
=
50 / 3
= 33
1 / 6

Hence, the base is 33
1 / 6
m
Height =
3 x 50 / 3
= 50 m

4. Find the area of a triangle whose sides are 50 metres, 78 metres, 112 metres respectively and also find the perpendicular from the opposite angle on the side 112 metres.
  A.  Area: 680 sq metres; Perpendicular: 30 metres
  B.  Area: 1680 sq metres; Perpendicular: 35 metres
  C.  Area: 1680 sq metres; Perpendicular: 30 metres
  D.  Area: 1780 sq metres; Perpendicular: 30 metres
     
   
View Answer

Shortcut:
If a, b and c are the lengths of the sides of a triangle and
S =
1 / 2
(a + b + c), then
Area of triangle = √[s(s − a)(s − b)(s − c)]
Here, a = 50, b = 78, c = 112,
s =
1 / 2
(50 + 78 + 112) = 120
Using these values in the shortcut, we get:
Area = √[120(120 − 50)(120 − 78)(120 − 112)]
= √(120 x 70 x 42 x 8) = 1680
Hence, area of triangle is 1680 sq metre.
Perpendicular = 2 x
Area / Base
=
1680 x 2 / 112
= 30
Hence, the perpendicular is 30 m

5. If the sides of a triangle are doubled, its area
  A.  becomes 4 times
  B.  becomes 5 times
  C.  becomes 2 times
  D.  No changes in its area
     
   
View Answer

If a, b and c are the lengths of the sides of a triangle and
S =
1 / 2
(a + b + c), then
Area of triangle = √[s(s − a)(s − b)(s − c)]
Let the original sides be a, b and c then s =
1 / 2
(a + b + c)
Area of triangle = √[s(s − a)(s − b)(s − c)]
For new triangle, the sides are 2a, 2b and 2c & s = 2s
Area of new triangle = √[2s(2s − 2a)(2s − 2b)(2s − 2c)]
= √[16s(s − a)(s − b)(s − c)]
= 4√[s(s − a)(s − b)(s − c)]
= 4 x area of original triangle

6. Two sides of a triangular field are 85 metres and 154 metres respectively and its perimetre is 324 metres. Find the area of the field and cost of levelling the field at the rate of Rs 5 per sq metre.
  A.  None of these
  B.  Area: 2872 sq metres; Cost: Rs 14360
  C.  Area: 2700 sq metres; Cost: Rs 13500
  D.  Area: 2772 sq metres; Cost: Rs 13860
     
   
View Answer

Shortcut:
The third side of the triangle = 324 − (154 + 85) = 85 m
a = 85, b = 154, c = 85,
s =
1 / 2
(85 + 154 + 85) = 162
Area of triangle = √[162(162 − 85)(162 − 154)(162 − 85)]
= √162 x 77 x 8 x 77)
= √(9 x 9 x 2 x 77 x 2 x 2 x 2 x 77)
= 9 x 4 x 77 = 2772
Hence, the area of field is 2772 sq metre
Cost of levelling 2772 sq m = 2772 x 5 = Rs 13,860.

7. Find the area of an equilateral triangle each of whose sides measures 8 cm. Also find perimeter of the equilateral triangle.
  A.  Area 10√3 sq cm; Perimeter: 18 cm
  B.  Area 16√3 sq cm; Perimeter: 24 cm
  C.  Area 12√3 sq cm; Perimeter: 20 cm
  D.  Area 16√5 sq cm; Perimeter: 24.5 cm
     
   
View Answer

Shortcut:
Area of equilateral triangle =
√3 / 4
x (side)2 and
perimeter of an equilateral triangle = 3 x side.
Here, side = 8
Using these values in the shortcut, we get:
Area =
√3 / 4
x (8)2 = 16√3
Perimeter of equilateral triangle = 3 x side = 3 x 8 = 24.
Hence, the area of equilateral triangle is 16√3 sq cm and its perimeter is 24 cm

8. Each side of an equilateral triangle is increased by 1.5%. The percentage increase in its area is:
  A.  2.5%
  B.  5%
  C.  1.5%
  D.  2%
     
   
View Answer

Shortcut:
Let the original length of each side = a
Then, Area of equilateral triangle =
√3 / 4
x (a)2
Now its each side is increased by 1.5%. Then,
New Area =
√3 / 4
x (
101.5 / 100
a)2
=
√3 / 4
x
20.3 / 20
a2 =
20.3 / 20
A
Increase in area =
(
0.3 / 20
A x
I / A
x 100
)
% = 1.5%

9. The perimeter of an isosceles triangle is equal to 14 cm; the lateral side is to the base in the ratio 5 to 4. The area in cm2, of the triangle is:
  A.  2√21
  B.  √22
  C.  
2 / 3
√21
  D.  
1 / 2
√21
     
   
View Answer

Shortcut:

Area of an isosceles triangle =
b / 4
√(4a2 − b2)
(ii) Height of isosceles triangle =
√[
a2 −
(
b / 2
)
2
]

(iii) Perimeter of isosceles triangle = (2a + b)
where a = sides & b = base
Let lateral side = 5y and base = 4y
Then, 5y + 5y + 4y = 14
⇒ y = 1
∴ The sides are 5 cm, 5 cm, 4 cm
Now, h2 = 52 - 22
⇒ h = √21
∴ Area =
1 / 2
x 4 x √21 = 2√21 cm2.

10. A plot of land is in the shape of a right angled isosceles triangle. The lenght of hypotenuse is 50√2 metres. If the rate of fencing is Rs 3 per metre, find the cost of fencing the plot
  A.  Incomplete data
  B.  less than Rs 350
  C.  less than Rs 500
  D.  more than Rs 500
     
   
View Answer

Let each of the equal sides be 'a' metre long.
Then, a2 + a2 = (50√2)2 = 5000
⇒ a2 = 2500
⇒ a = 50
∴ Perimeter of the triangle = ( 50 + 50 + 50√2)
= 100 + 20 x 1.4146 = 170.73 m
∴ Cost of fencing = Rs (170.73 x 3) = Rs 512.19

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