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If the perimeter of a right triangle is 4 + 2, the root is 6, and the length of the central line on the hypotenuse is 2, then the area of the triangle is
Let three sides be a / B / C (C is bevel)
∵ the length of the center line is 2 ∵ C=4;
∴A+B=2*SQRT(6);
A ^ 2 + B ^ 2 = C ^ 2 = 16
∴A*B=[(A+B)^2-A^2+B^2]/2=4
∴S=1/2*A*B=2
Given that the circumference of the right triangle ABC is 4 + 3 and the root is 3, and the length of the central line on the hypotenuse is 2, what is the area of the triangle ABC?
a+b+c=4+3√3
C=4
a+b=3√3
(a+b)^2=27
a^2+b^2+2ab=27
A ^ 2 + B ^ 2 = 16
So AB = 11 / 2
S=1/2*ab=11/4
In the triangle ABC, the angle a = 90 degrees, BC = 2, the circumference of the triangle ABC is 2, the root sign is 6, and the area of the triangle ABC is calculated
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Connect AC
∵AF∥CD，
∴∠ACD=180°-∠CAF，
And ∠ ACB = 180 ° - ∠ B - ∠ BAC,
∴∠BCD=∠ACD+∠ACB=180°-∠CAF+180°-∠B-∠BAC=360°-120°-80°=160°．
Connect BD
∵AB∥DE，
∴∠BDE=180°-∠ABD．
And ? BDC = 180 ° - ∠ BCD - ∠ CBD,
∴∠CDE=∠BDC+∠BDE=180°-∠ABD+180°-∠BCD-∠CBD=360°-80°-160°=120°．
As the right graph, in △ ABC, ∠ BAC = 135 °, ab = radical 2, AC = 1, D is a point on edge BC, DC = 2bd, then vector ad multiplied by vector BC is equal to Novice, everyone help ah, just registered
Vector ad multiplied by vector BC
=(AB+BD)·(BA+AC)
=(AB+1/3BC)·(AC-AB)
=(AB+1/3AC-1/3AB)·(AC-AB)
=(2AB+AC)·(AC-AB)/3
=(AB·AC+AC^2-2×AB^2)/3
=(-√2×√2/2+1-2×2)/3
=-4 / 3 agreed
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2012-1-8 12:06 1609805210 | grade III 54%
(1) Find BC: cos ∠ BAC = 135 ° = [AB * AB + AC * ac-bc * BC] / (2 * AB * AC) to obtain BC = √ 5
D is the trisection point of BC, BD = √ 5 / 3, CD = 2 √ 5
（2） Cos ﹤ B= (2+BD*BD-AD*AD) / (2 ﹤ 2) = (2+5-1) / (2 ﹤ 2* ﹣ 5) =3/ ﹣ 10 the solution is: AD= ﹣ 5/3
cos∠BDA=[AD*AD+BD*BD-BA*BA]/(2*AD*BD)=-4/5
Vector ad multiply vector BC = vector ad multiply vector BD multiply 3
= 3*AD*BD*cos(∠BDA)
=3*√5/3*√5/3 *(-4/5)
=-4/3
In △ ABC, ∠ BAC = 120 degrees, ab = 2, AC = 1, D is a point on edge BC, DC = 2bd, find the value of vector ad * BC (vector) There has to be a process
2 / 3 this problem is solved by cosine theorem. First, ∠ BAC = 120 degrees, ab = 2, AC = 1, BC = radical 7, thus BD = radical 7 / 3DC = 2 times radical 7 / 3. Then, the cosine theorem is applied to the triangle abd and the triangle ADC respectively. Let ∠ ADC = a ∠ ADB = 180-a, let ad = M. the formula 4 = (radical 7
In the triangle ABC, the angle BAC = 120 degrees, ab = 2, AC = 1, D is a point on the edge of BC, and DC = 2bd, then Ad vector click BC vector equals to
2 / 3 this problem is solved by cosine theorem. First, ∠ BAC = 120 degrees, ab = 2, AC = 1, we can calculate BC = radical 7, thus BD = radical 7 / 3, DC = 2 times radical 7 / 3. Then, we apply cosine theorem to triangle abd and triangle ADC respectively, let ∠ ADC = a ∠ ADB = 180-a, let ad = m two cosine theorems listed formula 4 = (...)
In the triangle ABC, the angle BAC = 120 ° AB = 2, AC = 1, D is a point on the edge of BC and DC = 2bd. Find the value of vector ad multiplied by vector BC
In this paper, we use the cosine theorem to solve the problem. First, we can calculate BC = radical 7, so BD = root 7 / 3, x0 DC = 2 times root 7 / 3. Then we apply the cosine theorem to the triangle abd and the triangle ADC respectively. Let ∠ ADC = a ∠ ADB = 180-a, let ad = M. the formula listed in the two cosine theorems ﹣ x0d4 = (root sign
In the triangle ABC, vector AB = C, vector AC = B. if point d satisfies vector BD = 2 times vector DC, then vector ad
D can only be on the side of BC, then
Vector ad = vector AB + vector BD
=Vector C + 2 / 3 vector BC
=Vector C + 2 / 3 (vector b-vector C)
=2 / 3 vector B + 1 / 3 vector C
In the triangle ABC, if angle a = 120 °, vector AB dot multiplication vector AC = - 2, D is the midpoint of BC, then the minimum value of modulus of AD
Vector AB dot multiplication vector AC = | ab| * | ac* cosa = - 2
|AB|*|AC|=4
Make parallelogram with AB and AC as adjacent sides ABEC AC = be | ab | * | be | = 4
Then AE = 2ad
Cosine theorem
AE^2=|AB|^2+|BE|^2-2|AB|*|BE|*cos60°
=|AB|^2+|BE|^2-|AB|*|BE|
>=2|AB|*|BE|-|AB|*|BE|
=|AB|*|BE|
=4
|AE|>=2
|AD|>=1
The minimum value of modulus of ad = 1
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