In the triangle ABC, D and E are close to the trisection point of a and B respectively. How many times is the area of triangle ABC?

In the triangle ABC, D and E are close to the trisection point of a and B respectively. How many times is the area of triangle ABC?

The area of triangle ABC is 9 times of triangle ade

It is known that in △ ABC, points D and F are on edge AB, points E and G are on edge AC, and lines de and FG parallel to BC divide the area of △ ABC into three equal parts
BC = 15 cm to find the length of De and FG

S△ADE/S△ABC=1/3
Similarity △ area ratio = square of length ratio
（DE/BC）^2=1/3
DE=5√3
S△AFG/S△ABC=2/3
(FG/BC)^2=2/3
FG=5√6

As shown in the figure, we know that in △ ABC, points D and F are on edge AB, and points E and G are on edge AC. lines de and FG parallel to BC divide the area of △ ABC into three equal parts
BC = 15, find the length of De and FG. Do not copy

Knowledge point: the area ratio of similar triangle is equal to the square of similar ratio
∵SΔADE/SΔABC=1/3=(DE/BC)^2,
∴DE/15=(1/√3),DE=5√3,
∵SΔAFG/SΔABC=2/3=(FG/BC)^2,
∴FG/15=√(2/3)
FG=5√6.

The area of triangle ABC is 16, ab = 4, D is any point of AB, f is the middle point of BD, de ∥ BC, FG ∥ BC, intersecting AC with e ` g respectively, let ad = X
(1) The area S1 of triangle ade is expressed by an algebraic formula containing X
(2) The area S2 of trapezoidal dfge is expressed by an algebraic formula containing X

Triangle ade is similar to triangle ABC. The area ratio is equal to the square of the similarity ratio. The similarity ratio is the ratio of corresponding sides, that is, ad: ab = x: 4, so S1: S = x ^ 2:16, S1 = x ^ 2
S2 = area of triangle AFG - area of triangle ade
The area algorithm of triangle AFG is the same as above. AF = (4-x) / 2 + x = x / 2 + 2
AF:AB=(x/2+2):4
Area of AFG: area of ABC = (4 + x) ^ 2 / 4:16 = (4 + x) ^ 2 / 64
So the area of AFG = (4 + x) ^ 2 / 4
S2=（4+x）^2/4-X^2=（-x^2+8x+16）/2

As shown in the figure, in triangle ABC, FG is parallel, De is parallel to BC, and BD = DF = FA. Prove de + FG = BC

The line EM is parallel to AB, and the point m intersects the edge BC
because
FG / / de / / BC, and BD = DF = FA
therefore
AG=GE=CE=1/3 AC
And because
EM//AB
therefore
Δ CEM congruent △ GAF
BM=DE
therefore
CM=FG
And because
BC=BM+MC
therefore
DE+FG=BC

In triangle ABC, de / / FG / / BC, ad: DF: FB = 1:1:1, s △ ade: s △ quadrilateral defg: s quadrilateral fgcb=______

Analysis, because, de ∥ FG ∥ FB, ad: DF: FB = 1:1:1
So, ad: AF: ab = 1:2:3
So, s △ ade ∽ s ∽ AFG ∽ s ∽ ABC, and s ∽ AFG ∽ s ∽ ABC ADE:S △ AFG:S △ABC=1:2²:3²=1:4:9
Let s △ ade = a, then s △ AFG = 4A, s △ ABC = 9A
S (quadrilateral defg) = s △ afg-s △ ade = 3A
S (quadrilateral fgcb) = s △ abc-s △ AFG = 5A
So, s Delta ADE:S (quadrilateral defg): s (quadrilateral fgcb) = A: 3A: 5A = 1:3:5

In the acute triangle ABC, BC = 6, the area of the triangle ABC is equal to 12, the two moving points m and N slide on the sides AB and AC respectively, and Mn is parallel to BC. Take Mn as the side and make a square mpqn downward. Let the side length be x, and the area of the common part of the square mpqn and the triangle ABC be y (y)

Let the height of BC be ad
1/2BC*AD=12
1/2*6*AD=12
AD=4
When ad intersects Mn with E, AE is the highest value on Mn
AE：MN=AD：BC
AE:x=4:6
AE=2x/3
DE=AD-AE=4-2x/3
y=MN*DE=x*(4-2x/3)
y=4x-2x²/3 (0

In the acute triangle ABC, AB is equal to 3, AC is equal to 4, area is equal to 3 times the root sign 3, find the length of the sides of angle A and BC

You should draw a picture yourself
first,
Making CD ⊥ AB in D
obtain
S=3*CD/2=3√3,CD=2√3
So ad ^ 2 = AC ^ 2-CD ^ 2 = 16-12 = 4
That is, ad = 2
BD=1
BC=√13
Because ad is half of AC, DCA = 30 degrees
Angle a = 60 degrees
Therefore, angle a is 60 degrees and BC side length is √ 13

In the triangle ABC, a is an acute angle, AB is equal to AC plus 6, AB multiplied by AC is equal to 64, and the area of the triangle ABC is equal to 16 times the root sign 3, find the hyperbola with BC as the focus and passing through a

A is an acute angle, ab = AC = 6, AB * AC = 64S = 1 / 2 * AB * ac * Sina = 16 √ 3, Sina = √ 3 / 2A = π / 3, then according to cosine theorem, cosa = (AB * AB + AC * AC BC * BC) / (2Ab * AC) = 1 / 2 (AB AC) ^ 2 = AB ^ 2 + AC ^ 2-2ab * AC = 36, so AB ^ 2 + AC ^ 2 = 36 + 2 * 64 = 164 is substituted to get 164 BC ^ 2 = 64, so BC = 10

Given the triangle ABC, the vector AB = (cos23 ° cos67 ° and the vector BC = (2cos68 ° and 2cos22 ° respectively), we can find the area of the triangle

Replace with
Vector AB = (cos23 ° sin23 °), vector BC = (2cos68 ° 2sin68 °),
Obviously, the vectors AB and BC are at 23 ° and 68 ° angles to the x-axis, and their lengths are 1 and 2, respectively. The angle between them is 68-23 = 45 degrees, so the area is
0.5 * 1 * 2 * sin 45 ° = 0.5 * radical 2