The side length of equilateral triangle ABC is a, and the square defg is inscribed with △ ABC, D, E on AB, AC, G, f on BC respectively. Find the side length of square defg

The side length of equilateral triangle ABC is a, and the square defg is inscribed with △ ABC, D, E on AB, AC, G, f on BC respectively. Find the side length of square defg

Solution: ah through a and BC at h, then
AH=AB*sin60°=√3a/2
Let the side length of square defg = x, then
Δ ade ∽ ABC, ratio of bottom edge = ratio of high edge
x/a=(AH-x)/AH=(√3a/2-x)/(√3a/2)
x=√3a/(2+√3)=(2√3-3)a
Answer: side length of square defg = (2 √ 3-3) a

If the perimeter of △ ABC is 30 and its area is equal to 30, the trilateral length can be obtained
It's wrong to write. It's three trigonometric functions for finding vertex angle a in known isosceles triangle ABC, ab = AC = 5, BC = 3

Let two right angle sides be a and B respectively, then the length of oblique side is 30-a-b
a²+b²=(30-a-b)²
ab/2=30
The results of solving equations
A = 5, B = 12 or a = 12, B = 5
The lengths of the three sides are 5, 12 and 13 respectively
AB=AC=5,BC=3,
sin(A/2)=3/10
cos(A/2)=√91/10
Three trigonometric functions of vertex angle a
sinA=3√91/50
cosA=41/50
tanA=3√91/41

One side ab of triangle ABC is extended by one time to D, the other side AC is extended by two times to e, and a larger triangle ade is obtained. The area of triangle ABC is triangle ade

Let f be the midpoint of CE and DF be connected. It is known that BC is parallel to DF and is the median line of triangle ADF. Let G be the midpoint of DF and H be the perpendicular line of DF through e. the area of triangle BDG is half of that of triangle DFE, so the area of triangle ABC is one sixth of that of triangle ade

In triangle ABC, angle c = 90 degrees, D is the midpoint of AB, ED is perpendicular to AB, ab = 20, AC = 12, find the area of quadrilateral Adec
Such as the title

From ab = 20, AC = 12, BC = 16
D is the midpoint of AB, BD = 10
Make CF perpendicular to AB, intersect AB with F, and calculate CF = 48 / 5,
In right triangle BFC, BF = 64 / 5,
Ed / / CF so de / CF = BD / BF, de = 30 / 4
BDE area = 1 / 2 x BD x de = 75 / 2
Area of Adec = abc-bde = 53 / 2