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As shown in the figure, in the right triangle ABC, ∠ ACB = 90 °, CD is the height on AB side, ab = 13cm, BC = 12cm, AC = 5cm (1) The area of ABC; (2) The length of CD; (3) Make the center line be on the edge AC of △ ABC and calculate the area of △ Abe; (4) When BD = 11cm, try to find the length of DF
（1）∵∠ACB=90°，BC=12cm，AC=5cm，∴S△ABC=12BC×AC=30cm2，（2）∵S△ABC=12AB×CD=30cm2，∴CD=30÷12AB=6013cm，（3）S△ABE=12S△ABC=12×30=15cm2，（4）∵S△BCD=12BD×CD=12BC•DF，∴BD•CD=BC•DF，∴...
In △ ABC, ed intersects AB at point E, AC at point D, ad = AC, AE = 3 / 5, and the difference between the circumference of △ ABC and △ ade is 16cm. Calculate the perimeter of △ ABC and △ ade
∵AD/AB=AE/AC ∠A=∠A
The similarity ratio of △ ad e ∽ △ ABC = 3 / 5
The circumference of △ ad E / △ ABC = 3 / 5
∵ △ perimeter of ABC - △ ade = 16
The circumference of △ ABC = 40
Perimeter of △ ade = 24
If in △ ABC and △ ade, and AB / ad = AC / AE, and the circumference of △ ABC is 36, calculate the circumference of △ ade
Your question is not good condition, AB / ad = AC / AE =? The question is to investigate the relationship between the similarity ratio of two similar triangles and the perimeter ratio of two triangles. What is the ratio of similarity ratio? This is the key to solve the problem
If AB: ad = AC: ae-bc: de = 6:5, and the circumference difference between △ ABC and △ ade is 4, calculate the circumference of △ ade
Twenty
It is known that: as shown in the figure, De is the median line of △ ABC. If ad = 4, AE = 5, BC = 12, the circumference of △ ade is () A. 7.5 B. 15 C. 30 D. 24
∵ De is the median line of ∵ ABC,
∴DE=1
2BC=1
2×12=6，
The circumference of △ ade is 4 + 5 + 6 = 15
Therefore, B
In the parallelogram ABCD, BC = 10cm, AC = 8cm, BD = 10cm, what is the AOD perimeter of triangle?
AO+OD=1/2(AC+BD)=9
AD=BC=10
The perimeter comes out 19
If the circumference of the parallelogram ABCD is 40 cm and that of the triangle ABC is 27 cm, what is the length of AC? A13cm B3cm C7cm D11.5cm
C7cm
Circumference of parallelogram ABCD =2 (ab+bc) =40cm
ab+bc=20cm
The circumference of triangle ABC = AB + BC + AC = 27cm
Because AB + BC = 20cm, so
ac=27-ab-bc=27=20=7cm
If the circumference of the parallelogram ABCD is 10 cm and the circumference of the triangle ABC is 8 cm, then AC=
AB+BC=10÷2=5
∴AC=8-5=3
As shown in the figure: ▱ the circumference of ABCD is 28cm, the circumference of △ ABC is 22cm, then the length of AC is () A. 6 cm B. 12 cm C. 4 cm D. 8 cm
The circumference of ABCD is 28cm,
∴AB+AD=14cm，
The circumference of ABC is 22cm,
∴AB+BC+AC=22cm，
∴AC=（AB+BC+AC）-（AB+AC）=22-14=8cm．
Therefore, D
As shown in the figure, in the parallelogram ABCD, AC and BD intersect at point O. it is known that ab = 8, BC = 6, and the circumference of △ AOB is 18 The circumference of △ AOD is______ .
∵ quadrilateral ABCD is a parallelogram,
∴AO=CO，DO=BO，AD=BC=6，
The circumference of AOB is 18, ab = 8,
∴AO+BO=AO+DO=18-8=10，
The circumference of △ AOD is Ao + do + ad = 10 + 6 = 16,
So the answer is 16
RELATED INFORMATIONS
- 1. As shown in the figure, in △ ABC, ab = AC, ad is the angular bisector of ∠ BAC, ad = 8cm, BC = 6cm, and points E and F are two points on ad, then the area of shadow part in the figure is () A. 48 B. 24 C. 12 D. 6
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- 8. As shown in the figure, in △ ABC, point E is on AB, point D is on BC, BD = be, ∠ bad = ∠ BCE, ad and CE intersect at point F. try to judge the shape of △ AFC and explain the reasons
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