As shown in the figure, in the triangle ABC, ab = AC = a, take BC as the edge, make an equilateral triangle outward, and find the maximum value of AD

As shown in the figure, in the triangle ABC, ab = AC = a, take BC as the edge, make an equilateral triangle outward, and find the maximum value of AD

2a
Ad, BC intersect with E, angle BAP = α
AE=asinα,ED=√3 * a cos α
AD=a sin α + √3 a cos α = 2 * a * sin (α + π/6) ≦ 2a

In the triangle ABC, B = 60 °, D is a point on the edge BC, ad = AB, AC = 4 times root 3, find the maximum DC perimeter of triangle a?

Guess: the topic requires the maximum perimeter of the triangle ADC
Let AB = x, DC = y, ∠ CAD = α
From the sine theorem, we can get that:
x/sin(60-α)=y/sinα=4√3/sin120°=8
x=8sin(60°-α)
y=8sinα
The perimeter of triangle ADC is C = x + y + 4 √ 3 = 8sin (60 ° - α) + 8sin α + 4 √ 3
=8(sin60cosα-cos60sinα+sinα)+4√3
=8sin(α+60°)+4√3
∵0

Let △ ABC be the inscribed triangle of a circle with radius r, and ab = AC, ad ⊥ BC in D, then find the maximum value of AD + BC
Such as the title

Let BC = 2x, ad = Ao + od = R + √ (R ^ 2-x ^ 2),
Let y = AD + BC = R + √ (R ^ 2-x ^ 2) + 2x,
If you want the maximum value, you need to find the derivative of X and find the stationary point,
y'=2-x(R^2-x^2)^(-1/2),
Let y '= 0,
x=2√5R/5,
Because the sign of the first derivative of the left and right adjacent points at x0 = 2 √ 5R / 5 is different, from positive to negative, there is a maximum,
y(max)=(√5+1)R.
The maximum value of AD + BC is (√ 5 + 1) r

If the area of a parallelogram is 24 square decimeters, the area of a parallelogram is larger than that of a triangle ()

Triangle area formula: low * high / 2
Parallelogram area: low * high
The parallelogram area given in the title is 24, so the triangle area is his half 12
The area of quadrilateral is larger than that of triangle: 24-12 = 12 (DM & # 178;)

The area of a triangle is 7.5 square decimeters, which is equal to the area of a parallelogram whose height is 2.5 decimeters?

7.5 * 2 / 2.5 = 6dm

The base and height of a triangle and a parallelogram are equal. It is known that the area of a triangle is 8 square decimeters, and the area of a parallelogram is ()

The base and height of a triangle and a parallelogram are equal. It is known that the area of a triangle is 8 square decimeters and that of a parallelogram is (16 square decimeters)

The area of a triangle is 45 square decimeters less than that of a rectangle with the same base and height. The area of this triangle is ()
What is it? It's urgent!

Triangle area
=45÷（2-1）
=45÷1
=45 square decimeters
 

A rectangle and a triangle have the same base and height. It is known that the area of a triangle is 24 square centimeters. How many square decimeters and how many square meters is the area of a rectangle?
A rectangular vegetable field covers an area of 3.6 hectares. The width of the vegetable field is 180 meters. How long is the vegetable field?

A rectangle and a triangle have the same base and height
So the area of the cuboid is twice the area of the triangle = 2 * 24 = 48CM ^ 2 = 0.048 square decimeter = 0.000048 square meter
1 ha = 10 000 m2
Length = area / width = 3.6 * 10000 / 180 = 200m

If the area of a triangle is 45 square decimeters less than that of a rectangle with the same base and height, then the area of the triangle is ()

22.4

The top of trapezoid is 12cm, the bottom is 18cm, and the height is 8cm. Cut the largest triangle from it. What is the area of the remaining figure in square centimeter?
If you want to have a test, you can't use an equation!

The longest edge is the bottom edge. Use the bottom edge to make the bottom edge of the triangle. Because the upper and lower bottom edges are parallel and the distance between parallel lines is equal, no matter which point connects the upper and bottom edges, the area of the triangle is equal, that is, 18 / 30 of the trapezoid area. The rest is 12 / 30
That is (12 / 30) * (12 + 18) * 8 / 2 = 48