(i) The area of an equilateral triangle is 100m². Find the perimeter of the triangle.
(ii)  Find the base of an isosceles triangle whose area is 60 sq.cm and length of equal sides is 13 cm.
(iii) The area of an isosceles right triangle is 200 m². Find its hypotenuse.
(iv) The ratio of the base to the altitude of a triangular field is 3 : 1. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10. Find its base and height.

(i) Area of the equilateral triangle = (side)².
                                 100 = (a)²
                                                                Þ a = 20 m 
∴ Perimeter of the equilateral triangle = 3a = 3 ´ 20 = 60 m
(ii) We are given an isosceles ΔABC in which    

AB = AC = 13 cm.
Let AD ^ BC and BC = 2x
⇒ BD = DC = x cm  
AD =
Area of ΔABC =   ´ base ´ height
⇒ 60 =
⇒ x = 12 or 5   
.^.. base = 24 cm or 10 cm
(iii)

Let AB=BC=x cm 
   AC² = AB² + BC²     
        = x² + x²
 ⇒ AC = 𢆢x
Area of the triangle =  ´ base ´ height
                Or 200 =  × x × x ⇒ x = 20
.^.. Hypotenuse 𢆢x = 𢆢 휠 = 28.28 m
(iv) Let the altitude be = x m
Then the base of the triangle = 3x m
Area of the field =
                     = 13.5 hectare
                     = 135000 m²
Þ  ´ base × height = 135000
Þ  ´ 3x × x = 135000
⇒ x = 300
∴ Base(3x) = 900 m
Altitude (x) = 300 m