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As shown in the figure, the triangle ABC is a right triangle. The area of shadow 1 is 23cm square. Find the length of BC
Let's give you a picture
The area of the shadow of the right angle of ABC is less than that of the square part of the triangle______ Centimeter
The area of the semicircle is 3.14 × (40
2)2×1
2 = 628 (square centimeter),
The area of triangle ABC is 628 + 28 = 656 (square centimeter)
The length of BC is 656 × 2 △ 40 = 32.8 (CM)
So the answer is: 32.8
In the triangle ABC, the points D, e and F are the midpoint of BC, ad and CE respectively, and s △ ABC = 4cm square. Calculate the BEF area of shadow part Really no one will?
If s △ abd = 1 / 2S △ ABC, then s △ AEB = 1 / 2S △ abd = 1 / 4S △ ABC. S △ BFC = 1 / 4abc. From s △ ADC = 1 / 2S △ ABC, we can get s △ AEC = 1 / 2S △ ADC = 1 / 4S △ ABC
So s △ bef = s △ abc-s △ aeb-s △ bfc-s △ AEC = 4 - (1 / 4 + 1 / 4 + 1 / 4) 4 = 1
In the triangle ABC, D is the midpoint of BC, e is the midpoint of AD, and the area of triangle ABC is 72 square centimeters. What is the area of shadow part
If D is the midpoint of BC, then BD = CD and the heights of DB and CD are equal
So s Δ abd = s Δ ACD = s Δ ABC / 2
In addition, e is the midpoint of AD
SΔabe=SΔade=SΔabd/2=SΔacd/2=SΔabc/4
You should be able to figure out which shadow is
As shown in the figure ad is the height of △ ABC, the points g and H are on the edge of BC, the point E is on the edge of AB, the point F is on the edge of AC, BC = 10cm, ad = 8cm, and the quadrilateral efhg is a rectangle with an area of 15cm2. Find the length and width of this rectangle
Let the length of the rectangle efhg be xcm,
∵ the quadrilateral efhg is a rectangle with an area of 15cm2,
The width of the rectangular efhg is: 15
xcm，
That is, EF = GH = xcm, eg = FH = 15
xcm，
∵ ad is the height of ∵ ABC, and the quadrilateral efhg is a rectangle,
∴EF∥BC，KD=EG=15
xcm，
∴AD⊥EF，AK=AD-KD=（8-15
x）cm，
∴△AEF∽△ABC，
∴AK
AD＝EF
BC，
∴8−15
X
8＝x
10，
That is, 4x2-40x + 75 = 0,
∴（2x-15）（2x-5）=0，
The solution is: x = 15
2 or x = 5
2，
When x = 15
2:00, 15:00
x=2；
When x = 5
2:00, 15:00
x=6．
The length and width of the rectangle are: 15
2, 2 or 6, 5
2．
As shown in the figure, the ratio of the length to width of the rectangle is 3:2, and the area of the triangle ABC is 6 square centimetres. It is also known that point C divides be into 1:2. D is the midpoint of EF
(1) Because point C divides be into 1:2, the area ratio of triangle ABC and triangle ace is 1:2, so the area of triangle ace = 6 × 2 = 12 (square centimeter) (2) because D is the midpoint of EF, the area of triangle ade is half of the area of triangle AFE, so the area product of triangle ade = (6 + 12
As shown in the figure ad is the height of △ ABC, the points g and H are on the edge of BC, the point E is on the edge of AB, the point F is on the edge of AC, BC = 10cm, ad = 8cm, and the quadrilateral efhg is a rectangle with an area of 15cm2. Find the length and width of this rectangle
Let the length of the rectangle efhg be xcm, ∵ the quadrilateral efhg is the rectangle with an area of 15cm2, ᙽ the width of the rectangle efhg is 15xcm, that is, EF = GH = xcm, eg = FH = 15xcm, ∵ ad is the height of △ ABC, the quadrilateral efhg is a rectangle, ∵ EF ∵ BC, Kd = eg = 15xcm,
As shown in the figure ad is the height of △ ABC, the points g and H are on the edge of BC, the point E is on the edge of AB, the point F is on the edge of AC, BC = 10cm, ad = 8cm, and the quadrilateral efhg is a rectangle with an area of 15cm2. Find the length and width of this rectangle
Let the length of the rectangle efhg be xcm,
∵ the quadrilateral efhg is a rectangle with an area of 15cm2,
The width of the rectangular efhg is: 15
xcm，
That is, EF = GH = xcm, eg = FH = 15
xcm，
∵ ad is the height of ∵ ABC, and the quadrilateral efhg is a rectangle,
∴EF∥BC，KD=EG=15
xcm，
∴AD⊥EF，AK=AD-KD=（8-15
x）cm，
∴△AEF∽△ABC，
∴AK
AD＝EF
BC，
∴8−15
X
8＝x
10，
That is, 4x2-40x + 75 = 0,
∴（2x-15）（2x-5）=0，
The solution is: x = 15
2 or x = 5
2，
When x = 15
2:00, 15:00
x=2；
When x = 5
2:00, 15:00
x=6．
The length and width of the rectangle are: 15
2, 2 or 6, 5
2．
As shown in the figure ad is the height of △ ABC, the points g and H are on the edge of BC, the point E is on the edge of AB, the point F is on the edge of AC, BC = 10cm, ad = 8cm, and the quadrilateral efhg is a rectangle with an area of 15cm2. Find the length and width of this rectangle
Let the length of the rectangle efhg be xcm,
∵ the quadrilateral efhg is a rectangle with an area of 15cm2,
The width of the rectangular efhg is: 15
xcm，
That is, EF = GH = xcm, eg = FH = 15
xcm，
∵ ad is the height of ∵ ABC, and the quadrilateral efhg is a rectangle,
∴EF∥BC，KD=EG=15
xcm，
∴AD⊥EF，AK=AD-KD=（8-15
x）cm，
∴△AEF∽△ABC，
∴AK
AD＝EF
BC，
∴8−15
X
8＝x
10，
That is, 4x2-40x + 75 = 0,
∴（2x-15）（2x-5）=0，
The solution is: x = 15
2 or x = 5
2，
When x = 15
2:00, 15:00
x=2；
When x = 5
2:00, 15:00
x=6．
The length and width of the rectangle are: 15
2, 2 or 6, 5
2．
As shown in the figure, we know that in △ ABC, ab = AC = 10cm, BC = 8cm, and point D is the midpoint of ab. (1) if point P is on the line BC at a speed of 3 cm / s from point B As shown in the figure, in △ ABC, ab = AC = 10cm, BC = 8cm, and point D is the midpoint of ab (1) If point P moves from point B to point C at a speed of 3 cm / s on line BC, and point Q moves from point C to point a on line ca ① If the motion speed of point q is equal to that of point P, after 1 second, try to compare the size relationship between line DP and PQ. Please explain the reason; ② If the velocity of point q is not equal to that of point P, what is the velocity of point Q that can make △ BPD and △ CQP congruent? (2) If point Q starts from point C at the speed of (2) and point P starts from point B at the same time at the original speed, they all move anticlockwise along the three sides of △ ABC. How long does it take for point P and point Q to meet on which side of △ ABC for the first time?
(1) According to the formula of distance = velocity × time, the time of point P's movement is obtained, and then the velocity of point q is obtained
RELATED INFORMATIONS
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