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In the triangle ABC, AC = 6cm, BC = 8cm, ab = 10cm, D, e, f are the midpoint of AB, BC and Ca, respectively. Calculate the area of triangle def
I don't know what grade you are in
If you have studied the median line, you can know that D and E are the midpoint of AB and BC respectively, so De is the median line of triangle
Similarly, EF = AB / 2, DF = BC / 2
Because de: AC = EF: ab = DF: BC = 1:2
Therefore, the similarity ratio is 1:2
Because the area ratio of similar triangles is the square of similarity ratio, the area ratio of two triangles is 1:4
In △ ABC, BC? 2 + AC? 2 = AB? Is a right triangle. The two right sides are AC and BC
So the area of △ ABC is 1 / 2 × 6 × 8 = 24 square centimeter
The area of △ DEF is 6 square centimeters
If you have not learned the knowledge of median line: connecting De,
D is the midpoint of AB, so BD: ab = 1:2; E is the midpoint of BC, so be: BC = 1:2. BD: ab = be: BC
And ∠ B = ∠ B
So △ ABC ∽ BDE, de: AC = BD: ab = 1:2
Similarly, EF = AB / 2, DF = BC / 2
In the known triangle ABC, the angle c is equal to 90 degrees, ab = 10cm, BC = 6cm, AC = 8cm. Starting from C, the moving point P moves along Ca and ab at a speed of 1cm per second to B 1. Suppose that the distance of point P from point C is x cm, and the area of triangle BPC is y square cm, and Y is expressed as a function of X 2. When writing from point C for a few seconds, the triangle BPC = 0.25 triangle ABC Second year students can understand,
I'm glad to help you. First of all, this topic needs to be discussed in categories,
1、 (1) when P-point is on edge Ca, y = 3x;
(2) When p is on the edge AB, so BP = 18-x. if the height of the triangle BPC is h, we can get H / AC = BP / AB (this is understandable), that is, (18-x) / 10 = H / 8, and then H = 4 / 5 * (18-x) can be obtained by simplification, so y = 12 / 5 * (18-x),
(3) When p is on BC, y = 0;
(so I'd like to wrap the three cases in curly brackets at the end of the paper.)
2、 From the first question, if y = 0.25, then x = 1 / 12 in (1), so the time is 1 / 12, (2) x = 859 / 48, time is 859 / 48,
So to sum up, the time is 1 / 12 or 859 / 48
Landlord, very attentive to help you solve the problem, hope to adopt
In the right triangle ABC, the angle c = 90 degrees AB = 10cm BC = 6cm point e starts from point C and moves along CA direction at the speed of 1cm per second and stops at a; point F starts from point a and moves along AB direction at a speed of 2cm per second, and stops moving until B. if the point E.F starts moving at the same time, the triangle AEF is a right triangle? 2 if point E.F moves at the same time, So whether there is a certain moment that the triangle AEF is exactly an isosceles triangle, please write it directly (there are 5 answers to two questions.)
According to the triangle Pythagorean theorem, AC = 8cm
（1） Suppose AEF is a right triangle,
Then (1) when ∠ e is a right angle, AE / AC = AF / ab
That is (8-x) / 8 = 2x / 10
The solution is x = 40 / 13
(2) When ∠ AFE is a right angle, AF / AC = AE / ab
That is, 2x / 8 = (8-x) / 10
X = 16 / 7
（2） Let AEF be an isosceles triangle when moving for M seconds
(1) When ∠ A is the vertex angle, AE = AF
8-m = 2m
M = 8 / 3
(2) When ∠ A is the base angle and the vertex is f,
[(8-M)/2]/8=2M/10
M = 20 / 13
(3) When ∠ A is the base angle and the vertex is e,
[2M/2]/8=(8-M)/10
M = 32 / 19
(knowledge points: triangle similarity)
In RT △ ABC, ab = 10 cm, BC = 6 m, angle c = 90 degrees, AC = 8 cm. Starting from C, point P moves along Ca and ab at a speed of 1 cm / s to point B (1), starting from C In RT △ ABC, ab = 10cm, BC = 6m, angle c = 90 degrees, AC = 8cm. Starting from C, point P moves along Ca and ab at a speed of 1cm / s to point B (1) Let P move from C to xcm, and the area of triangle BCP is YCM ^ 2, and Y is expressed as a function of X (2) After a few seconds from point C, the area of the triangle BCP = the area of the quarter triangle ABC
When 0 ≤ x ≤ 8, y = 3x; when 8 ≤
When x ≤ 18
Y = - 12 / 5x + 216 / 5 Analysis: at this time, P is on AB, and the vertical line of BC is made through P, and the height of △ BCP is obtained by similarity
2. Let the area be 6 after T seconds
T1 = 2, T2 = 15.5
In RT △ ABC, ∠ C = 90 °, BC = 6cm, CA = 8cm. The moving point P starts from C and moves along Ca and ab at the speed of 2cm per second to point B, then starts from C______ S △ BCP = 1 2S△ABC．
If s △ BCP = 1
2S △ ABC, then point P is at the midpoint of AC or ab,
In RT △ ABC, ab = 10cm can be obtained by Pythagorean theorem,
Therefore, the distance of point P movement is 4cm or 13cm,
Therefore, the time of exercise is: 2 seconds or 6.5 seconds
In RT △ ABC, ∠ C = 90 °, BC = 6cm, CA = 8cm. The moving point P starts from C and moves along Ca and ab at the speed of 2cm per second to point B, then starts from C______ S △ BCP = 1 2S△ABC．
If s △ BCP = 1
2S △ ABC, then point P is at the midpoint of AC or ab,
In RT △ ABC, ab = 10cm can be obtained by Pythagorean theorem,
Therefore, the distance of point P movement is 4cm or 13cm,
Therefore, the time of exercise is: 2 seconds or 6.5 seconds
In RT △ ABC, ∠ C = 90 °, BC = 6cm, CA = 8cm. The moving point P starts from C and moves along Ca and ab at the speed of 2cm per second to point B, then starts from C______ S △ BCP = 1 2S△ABC．
If s △ BCP = 1
2S △ ABC, then point P is at the midpoint of AC or ab,
In RT △ ABC, ab = 10cm can be obtained by Pythagorean theorem,
Therefore, the distance of point P movement is 4cm or 13cm,
Therefore, the time of exercise is: 2 seconds or 6.5 seconds
In △ ABC, ∠ C = 90 °, BC = 8cm, ab = 10cm. Starting from B, the moving point P moves to C at the speed of 2cm / s, and Q moves from C to a at the speed of 1cm / s?
If two triangles are similar after T seconds, then
CP / CQ = BC / AC, or CP / CQ = AC / BC
That is (8-2t) / T = 8 / 6 or (8-2t) / T = 6 / 8, where AC = 6
t=12/5 t=32/11
It is known that: as shown in the figure, in the triangle ABC, ∠ B = 90 degrees, ab = 5cm, BC = 7cm. Point P starts from point a and moves along the edge of AB at a speed of 1cm / s, and point Q Starting from point B, move along the edge of BC to C at a speed of 2 cm / s (1) After a few seconds, the area of the △ PBQ is 4m? (2) After a few seconds, the length of PQ is equal to two pieces of 10cm? (3) In (1), can the area of triangle pqb equal to 7cm? Explain the reason Don't copy from Baidu. I don't understand
After x seconds
(1) It can be seen from the meaning of the title
PB=AB-AP=5-X
BQ=2X
S=1/2ah=(5-X)X=5X-X^2
When s = 4
5X-X^2=4
The solution is X1 = 1, X2 = 4
Because after 3.5 seconds, it stops
So let's leave off x2. After one second, the area is four
(2) PQ is 2 pieces of 10 pieces
So (5-x) ^ 2 + (2x) ^ 2 = 4 * 10
X1 = 3, X2 = - 1 (omitted)
After 3 seconds, PQ is 2 pieces and 10 pieces
(3) No
Let s = 7, then 5x-x ^ 2 = 7
X has no solution
So we can't
In △ ABC, AB: AC: BC = 3:4:5, with a circumference of 36cm, point P moves at a speed of 1cm per second from a to B, and point Q moves at a speed of 2cm per second from B to C. at the same time, after starting for 3 seconds, the area of △ PBQ is calculated?
According to the title, ab = 3 * 12 / 36 = 9cm, AC = 12cm, BC = 15cm, and △ ABC is RT △
Taking BP as the base of △ bpq, BP = 9-3 * 1 = 6
If the vertical line QH of AB is drawn through the point Q, from the similarity of triangles, △ BQH ≓ △ BCA, then QH = 12 * 2 * 3 / 15 = 4.8
/ / s △ BQP = bottom * height / 2 = 14.4cm
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