An isoceles field whose base length is b and equal length a then it's area?

Question

High school level

Answer 1

Please rephrase your question.

Answer 2

Willie is correct except that the area of an isosceles triangle is = to 1/2 b*height Expl: Imagine the triangle cut into 2 pieces down the middle(apex to 1/2b) then the side a put together with the other side a makes a square of 1/2b*H.

Answer 3

the area of a triangle is 1/2 the area of a related square

Answer 4

I don't entirely get the question, but for an isosceles triangle, the area=1/2ab. The area is one half the product of the two congruent sides.

Answer 5

You have to first calculate the heighth (h) which is the line from the middle of the base to the apex of the triangle.

a^2 = h^2 + (1/2b)^2
h^2 = a^2 - (1/2b)^2
h = sq. root quantity (a^2 - (1/2b)^2
area = 2 x (1/2b x h)

Answer 6

what I understand is......two sides have length = a

and one side has length b.......

we know that area of a triangle with a, b, c as sides is

sqrt [ s(s-a)(s-b)(s-c) ] ......and s = (a+b+c)/2

here b and c are equal......say b......s = (a+2b)/2 = b +a/2

so,

area =

sqrt [ (b +a/2)(b+a/2-a)(b+a/2-b)
(b+a/2-b) ]

= sqrt [(b+a/2)(b-a/2)(a/2)(a/2)]

= a/2 x sqrt [b^2- (a/2)^2]

= a/4 x sqrt [4b^2-a^2]

so, area of the triangle = a/4 x sqrt [4b^2-a^2]

Answer 7

Find it's height First. Then 1/2 * b * h.

Answer 8

assuming you're talking about a triangle, a sq. /2 or b sq. /2 since a and b are equal lengths.

Answer 9

(b/2)[(a^2-(b/2)^2)^(1/2)]

Answer 10

I hope you are asking for its area?

.5*sqrt(a^2-(.5b)^2)*b

Answer 11

1/2a2*b ... jst try it out!!