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An obtuse triangle ABC, ∠ B is an obtuse angle, where AB = 10, BC = 9, AC = 17, find the height on BC side
Make AE perpendicular to BC and the extension of BC at point E
Let be = X
Then 10? X? 2 = 17? (9 + x) 2
The solution is x = 6
So AE? 2 = 10? - 6? 2 = 64
AE=8
The height on the BC side is 8
In △ ABC, ab = 25, BC = 12, AC = 17
Let BD = x, then ad = 25-x. according to the Pythagorean theorem, we can get that: BC 2 - BD 2 = CD 2 = AC 2 - Ad 2, ν12 - x 2 = 17 - 2 - (25 - x) 2. The solution is: x = 48 / 5, CD 2 = 12 2 - (48 / 5) 2
In the obtuse triangle ABC, we know that AB is equal to 5, AC is equal to 3, and the median line ad of BC is 2?
3×4÷2=6
Obtuse angle triangle ABC, ab = 17, BC = 9, AC = 10, angle B is obtuse angle, calculate ABC area
C is an obtuse angle. The big side is opposite to the big angle
Method 1 s = √ [P (P-A) (P-B) (P-C)] and P in the formula is half circumference: P = (a + B + C) / 2
Method 2 lengthen AC through B to make vertical line BD
If DC = x, then
17 * 17 - (x + 10) square = 9 * 9-x * x x = 5.4 high BD = 7.2
So s = 1 / 2 * 10 * 7.2 = 36
As shown in the figure, in the obtuse triangle ABC, ab = 6cm, AC = 12cm, the moving point d starts from point a to point B, and the moving point e starts from point C to point a. the velocity of point D is 1cm / s and that of point E is 2cm / s. if two points move at the same time, then when the triangles with points a, D and E as vertices are similar to △ ABC, the movement time is () A. 3 seconds or 4.8 seconds B. 3 seconds C. 4.5 seconds D. 4.5 seconds or 4.8 seconds
According to the meaning of the question, suppose that the triangle with points a, D, e as its vertices is similar to △ ABC, the time of motion is x seconds,
① If △ ade ∽ ABC, then
AD
AB
=
AE
AC
,
Qi
X
Six
=
12−2x
Twelve
,
The solution is: x = 3;
② If △ ade ∽ ACB, then
AD
AC
=
AE
AB
,
Qi
X
Twelve
=
12−2x
Six
,
The solution is: x = 4.8
When the triangle with points a, D and E as the vertices is similar to △ ABC, the movement time is 3 seconds or 4.8 seconds
Therefore, a
It is known that in the equation AX ^ 2-radical 2bx + C = 0, a, B, C are the edges to which a, B, C of the three inner angles of the obtuse angle triangle ABC respectively, and B is the largest edge (1) (2) let the equation have two unequal heels x, Y. if the triangle ABC is an isosceles triangle, find the value range of X-Y
A is not equal to 0
△=2b^2-4ac=2(b^2-2ac)
Because B is the largest side
So B ^ 2 > A ^ 2 + C ^ 2
Namely △ > (a ^ 2 + C ^ 2-2ac) = (a + C) ^ 2 > 0
So the equation has two unequal roots
Because X1 + x2 = √ 2A / 2B > 0
x1*x2=a/c>0
So these two are positive roots
Question two
(X-Y)^2=(X+Y)^2 -4XY=(2b^2-4a^2)/a^2=2(b/a)^2 -4>0
So (X-Y) ^ 2 > 0
There is something wrong with this question
In △ ABC, if ∠ B = 135 °, AC = 2, then the radius of the circumscribed circle of the triangle is () A. 1 B. Two Two C. Two D. 2
∵∠B=135°，AC=
2，
From the sine theorem AC
SINB = 2R, i.e. R=
Two
2 x
Two
2=1．
So choose a
The circumcircle radius of triangle ABC is 2, ab = 2, radical 3, angle a = π / 6, find AC?
Find the center O, connect ob, OC
Since the angle a = π / 6, the angle BOC = π / 3 can be deduced
Because ob = OC = 2, the triangle BOC is a regular triangle
So BC = 2
From BC = 2, ab = 2, radical 3, angle a = π / 6, the angle ABC = π / 2 can be deduced
Understanding the side length of right triangle AC = 2BC = 4 from Pythagorean definition
[help] if AB = 2, BC = 3, AC = radical 7, then the circumcircle radius r of the triangle ABC is r=
According to the sine theorem, a / Sina = B / SINB = C / sinc = 2R, R is the radius of the circumscribed circle of the triangle,
According to the cosine theorem, CoSb = (BC ^ 2 + AB ^ 2-ac ^ 2) / (2 * BC * AB) = 1 / 2, so B = 60 °,
ν SINB = (root 3) / 2, ν r = (root 21) / 3
If AB = AC = 4 times root sign 2 and height ad = 4, then the circumcircle radius of triangle ABC is It is not said that triangle ABC is a right triangle
Drawing
BD = 4 is obtained by Pythagorean theorem
∠B=45°
Using the sine theorem a / Sina = B / SINB = C / sinc = 2R
R = 4
I'm talking about △ abd
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