The answer is D. How do I get that?

Being an Isoscles Triangle (two equal sides) , dropping a perpendicular to the base of '2' will bisect the base line, into 1 + 1 = 2 and intersect at right angles.  So we have two right angled triangles 'inside' the Isosceles triangle.

Next apply Pythagoras, to find the  perpendicular height . 

6^2 = 1^2 + h^2 

h^2 = 6^2 - 1^2 

h^2 = (6 - 1)(6 + 1) 

h^2 = 5(7)

h^2 = 35 

h = sqrt(35) 

Next remember the Area = 0.5 X base X perpendicular height

A = 0.5 X 2 X sqrt(35) 

A = sqrt(35)  

NB 0.5 X 2 = 1 

Answer 'D'. 

h = √6^1-(2/2)^2 = √35

area = b*h/2 = 2√35 / 2 = √35

or by applying Heron's formula 

semi-perimeter sp = (6+6+2)/2 = 7 

area = √sp*(sp-a)*(sp-b)*(sp-c) = √7*(7-6)*(7-6)*(7-2) = √7*1*1*5 = √ 35 

s=(2*6+2)/2=7

The area of the triangle=

sqr[s(s-2)(s-6)^2]

=

(7-6)sqr[7*5]

=

sqr(35) sq.yds

(D)

1st you need to calculate the height of the triangle using Pythagorean theorem.

we 1st get the half of the base to enables us to solve the height of the right triangle.

b = 2/2 =1

applying Pythagorean theorem

h =  √(6^2 - 1^2)

h = √(35)

Now, let us solve for the area of the triangle.

A = 1/2bh

A = 1/2(2)(√(35))

A = √(35) yd^2  Answer//

Random stab I would guess. 

Or you can use Herron's Formula to get the area of the triangle given the length of the three sides:

A = √[s(s - a)(s - b)(s - c)] where s = (a + b + c) / 2

So we can find s, then find A:

s = (a + b + c) / 2

s = (6 + 6 + 2) / 2

s = 14 / 2

s = 7

A = √[s(s - a)(s - b)(s - c)]

A = √[7(7 - 6)(7 - 6)(7 - 2)]

A = √[7(1)(1)(5)]

A = √35 yd²

Cut in half. flip around. Now you have a rectangle.

width = 1

height = sqrt(6^2 - 1^2) = sqrt(35)

Area = width * height = 1 * sqrt(35)

Draw a line down the center, splitting the triangle into two smaller right triangles.

Each* right triangle has a base of length 1 and a hypotenuse of 6. You can thus apply the Pythagorean Theorem to find the height.

Finally, apply the formula for area of a triangle.

* For proof that the perpendicular line does bisect the base: 

Whatever the height is, both right triangles have the same height and hypotenuse; therefore by Pythagoras their bases are identical.

There are 2 ways to find the area

1) Using Pythagorean theorem to find height

Drop perpendicular from top vertex to the base. 

Since this is an isosceles triangle, the perpendicular will divide the base in equal length of 1 yd

height, h = √(6² - 1²) = √35 yd

Area = ½bh = ½(2)(√35) = √35 yd²

2) Using Heron's formula

A = √[s(s-a)(s-b)(s-c)]

where a, b, c are the sides and s is half Perimeter

s = perimeter/2 = (6+6+2)/2 = 7

A = √[7(7-6)(7-6)(7-2)]

A = √[7(1)(1)(5)

A = √35 yd²

Cut the triangle down the middle.  You then have two right triangles.  In either of these right triangles, the short side is 1 yard and the hypotenuse is 6 yards, so the Pythagorean Theorem says the tall side is sqrt(6*6 - 1*1) = sqrt(35).  

Then the area of the original isosceles triangle is

(1/2)*(2 yards)*sqrt(35) yards 

= [sqrt(35)] yd^2.

Answer D.