lenpol7
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'.
oubaas
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
llaffer
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
Daniel H
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)
Amy
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.