Find the area of the blue triangle

Find the area of the blue triangle.

[@minxythemerciless]

Easier is to create a rectangle ABCF where F is 6 units above A. Then note that the blue area is (6x18)/2 minus the area of triangle AFD = 6 * (18 - x) / 2 => Area 6x/2 = 30

[@hongningsuen1348]

There are 3 different methods to find the top side of trapezium x = 10:
1. In triangle ADE, by Pythagoras theorem, x^2 = 6^2 + (18-x)^2
2. Triangle ABC similar to triangle CHD, by corresponding sides proportionality equation
x/6sqrt10 = 3sqrt10/18 (DH perpendicular bisector of isosceles triangle ACD)
3. In triangle CHD, x = 3sqrt10/cosB (In triangle ABC, cos B = 18/6sqr10)

[@MathEducation100M]

Nice solution

[@davidellis1929]

Once you find that the legs of the shaded triangle have length 10, there's a much simpler way to finish the solution. The shaded area is the trapezoid area less the right triangle at its bottom. This works out to (1/2)(6)(10+18)-(1/2)(6)(18), which is (1/2)(6)(10) = 30.

[@mathu6514]

very nice, thank you for watching.

[@Antony_V]

There's different ways to find DC. One of these, for example, is that CH:HD=3:1. Once we have CD=10 shaded area = 1/2*CD*BC = 30 squ

[@plamenpenchev262]

Upper line segment, x, is 10
(18^2 + 6^2 = x^2)
The oblique segment, y, is 6×sqrt(10)
(y^2 = 6^2 + 18^2)
Sin(t) = 6/(6×sqrt(10)), then
2S = 10×6×sqrt(10)×1/(sqrt(10))

[@user-sw9lb2zs6e]

Draw a perpendicular from D to AB. Then Pythagoras (18-a)^2+36=a^2 , Hence a=10. Then subtract triangle from trapezoid

[@carloshenriquedesouzacoelho]

Isosceles triangle

[@youssefelyousfi4929]

(30)u^2