The area of an isosceles triangle is 60 cm^2 and the length of each one of its equal sides is 13 cm.

Area of triangle = sqrt [ s(s-a)(s-b)(s-c) ]
a = c = 13 , s = (a+b+c)/2 = (26 + b)/2
(s - a) = (s - c) = (b/2) , (s - b) = (26-b)/2

=> 60^2 = (26 + b)(26 - b)b^2 / 16
=> 16*60^2 = b^2*(26^2 - b^2)
=> b = 10,24

Having drawn your isosceles triangle with the length of the equal sides equal to 13 cm, the width of the base and the altitude equal to 2.x and h respectively, then we may write for the area A of the triangle

A = h.x = 60 cm²

so that h = 60/x

and from Pythagoras applied to either of the right-angled triangles

x² + h² = 13²

Substituting for h from above gives

x² + (60/x)² - 13² = 0

Multiplying through by x² and rearranging

x^4 - 169.x² + 3600 = 0

which is a quadratic in z = x² with solutions

z = [169 ±√(169² - 4*3600)]/2

giving z = [169 ± 119]/2 = 25 and 144

so that the solutions for x are 5 and 12, both of which are valid,

The base of the triangle may therefore be either 2.x = 10 or 24 cm in length.

Divide the isosceles triangle in two halves by dividing the base side from middle .
Now we have 2 right angle triangles and the area of each of two resultant triangles is 30cm^2.
Using prognathous theorem where hypotenuse is fixed as 13 and also 1/2*base*height= 30
We get two equations:
b^2 + h^2 = 13^2
1/2 *b*h= 30
=> b*h = 60

Solving we get 5, 12. As we have divided the base in two equal parts we need to multiply by 2
Hence answer is 5*2, 12*2 = 24, 10

Let x represent the measure of segement AD. Let y represent segment BD. Since we know from a common geometric proof that the perpendicular from the apex to the base of an isosceles triangle bisects the base, the apex angle, and the triangle itself. Therefore the area of triangle ABD is equal to the area of triangle BCD and the area of each is one-half of the area of the entire triangle ABC.
2x=24,10
Area =xy
x^2+y^2=13
x^2+3600/x^2=169
we get x=12,5
so,
segment BD bisects segment AC, and by construction, AD plus DC is equal to AC, we can deduce that AC is equal to two times AD. Therefore the base measurement that we seek as an answer to this problem is two times the value of
or
2x=24,10

Let x represent the measure of segement AD. Let y represent segment BD. Since we know from a common geometric proof that the perpendicular from the apex to the base of an isosceles triangle bisects the base, the apex angle, and the triangle itself. Therefore the area of triangle ABD is equal to the area of triangle BCD and the area of each is one-half of the area of the entire triangle ABC.
2x=24,10
Area =xy
x^2+y^2=13
x^2+3600/x^2=169
we get x=12,5
so,
segment BD bisects segment AC, and by construction, AD plus DC is equal to AC, we can deduce that AC is equal to two times AD. Therefore the base measurement that we seek as an answer to this problem is two times the value of
or
2x=24,10

Area of an Isosceles triangle=(b/4)sqrt(4a^2-b^2)
where a=Each one of equal sides
b=Base
Equation will be b^4-676b^2+240^2=0
so b^2=576,100
Thus b=24,10.

its equal sides are 13 cm..
by Pythagoras theorem, in right triangle with median and base, the median length is of 12 or 5 cm . similarly the half of base is 5 or 12 cm. so total of its base is 12*2 = 24; or 5*2 = 10;
so option 2 & 7 both are correct.

becoz hight is x and base is y then 1/2xy=60
and in right angle tringle (x/2)^2+(y/2)^2=13^2 so by eq. 1 and 2 base may be 24 or 10

Let x be the base and y be the height.
So 1/2(xy)=60.....(1)
and by Pythagoras (x/2)^2 + y^2= 13^2 = 169 (2)
Solving (1)&(2)
We get x= 24 or x=10.
Hence option no. 2.