##60## ##”square units”##

I have two solutions for you, using the formulas,

##A=1/2″*base*height=1/2bh##

-and using –

##A=sqrt(s(s-a)(s-b)(s-c))##

where: ##a, b and c =>## sides of the triangles ##s=>## semi-perimeter ##=P/2=(a+b+c)/2##

Since it is an isosceles triangle, the measure of the sides given its periemter would be ##13##, ##13## and ##10##. **Because two sides of an isosceles triangle is congruent.

Using ##A=1/2bh## :

First is to find the height.

Using this figure:

Know that the height of the triangle bisects the side opposite the vertex.

We can use pythagorean formula to find the height:

##h=sqrt((13^2)-(5^2))##

##h=12##

then, we solve for the area:

##A=1/2bh=1/2(10)(12)##

##A=60## square units

Using Heron’s Formula:

Let the sides of the triangle be: ##a=13## ##b=13## ##c=10##

##s=(a+b+c)/2=(13+13+10)/2=18##

##A=sqrt(s(s-a)(s-b)(s-c))##

##A=sqrt(18(18-13)(18-13)(18-10))##

##A=60## square units

##60## ##”square units”##

I have two solutions for you, using the formulas,

##A=1/2″*base*height=1/2bh##

-and using –

##A=sqrt(s(s-a)(s-b)(s-c))##

where: ##a, b and c =>## sides of the triangles ##s=>## semi-perimeter ##=P/2=(a+b+c)/2##

Since it is an isosceles triangle, the measure of the sides given its periemter would be ##13##, ##13## and ##10##. **Because two sides of an isosceles triangle is congruent.

Using ##A=1/2bh## :

First is to find the height.

Using this figure:

Know that the height of the triangle bisects the side opposite the vertex.

We can use pythagorean formula to find the height:

##h=sqrt((13^2)-(5^2))##

##h=12##

then, we solve for the area:

##A=1/2bh=1/2(10)(12)##

##A=60## square units

Using Heron’s Formula:

Let the sides of the triangle be: ##a=13## ##b=13## ##c=10##

##s=(a+b+c)/2=(13+13+10)/2=18##

##A=sqrt(s(s-a)(s-b)(s-c))##

##A=sqrt(18(18-13)(18-13)(18-10))##

##A=60## square units