plzzzz answer my question with full method to solve it... what will be the area of an isosceles triangle with base 30 cm?Its a question from 9 class rd book..............plzzz solve it....... m...
plzzzz answer my question with full method to solve it...
what will be the area of an isosceles triangle with base 30 cm?
Its a question from 9 class rd book..............plzzz solve it....... m so confused............hope I get my answer.....
Let ABC be an isiscele triangle:
Let BC be the base:
BC = 30
AB = AC = X
D is midpoint BC such that AD is perpendicular to BC
==> AD^2 = AC^2 + (CD)^2 = x^2 - 15^2
==> AD = sqrt(x^2 - 225)
Then Area of the triangle is:
A = (1/2)*height* base
= (1/2)* AD* BC
= (1/2)*sqrt(x^2 - 225)* 30
= 15sqrt(x^2 - 225)
Then area of the triangle is:
a = 15*sqrt(x^2 - 225) where x is the lenght of the triangle sides.
An isosceles triangle has two equal sides say l each and the other side or base b.
Let ABC be the isosceles triangle with BC = BA = l and AC = b
The area of the isosceles triangle ABC is = (1/2) base*height. = (1/2)bh.........(1)
Let D be the mid point of AC. Then BD = h = sqrt(C^2-CD^2) = sqrt(l^2- (b/2)^2 ) = sqrt(l^2-b^2/4) = (1/2)sqrt(4l^2-b^2). Substitute this value of h in (1) to get the area of the isosceles triangle.
So the area of the isosceles triangle ABC with two equal sides l and base b = (1/2)b*(1/2) sqrt(4l^2 -b^2) = (1/4)b sqrt(4l^2-b^2)
In the given case base b is 30 cm. The other two equal sides are not given.So the other two sides are assumed to be l cm each.
So the area of the given isosceles triangle = (1/4)30sqrt(4l^2-30^2)
Area of isosceles triangle with base, b= 30 and equal sides l is 7.5 sqrt(4l^2-900) sq cm........................(2)
How you use this:
If the two equal sides are 25 cm each, then l =25. Then put l = 25 in the formula at (2) and get the area = 7.5 sqrt(4*25^2-900) = 300 sq cm