-
Passing the point D is made of DG, and the vertical BC is crossed with G. Then the triangle BEF is similar to the triangle chain ascending GDC and gc=4-1=3=dg=ab=2eb. So EB = 3/2. Then from the Pythagorean shed scattered old slag digging theory, EF = 3/2 of the number 2
-
Pythagorean theorem.
1.If the leak is ab on the side, the height will always be bc
Then the triangle area is 10x10 2=10
2.Take BC as one side. Then be=15, bc=10, using the Pythagorean theorem to obtain ce=root number 125Only socks, i.e., 5 roots, 5 three-finger liquids, with an area of 10x5, 5 roots, 5 2 = 25, 5 roots, 5 <>
-
Because ABC is an isosceles ladder sedan deformed, angle b = angle trembling c, as df ab, then the closed angle c = angle dfc, ab = df = cd = 6, ad = bf = 5, cf = bc-bf = bc-ad = 3
So the circumference of the triangle is 6+6+3=15
-
As AE DC, the angle c=70°, so the angle AEB=70°
In AEB, the angle b=55°, so the angle EAB=55°, equiangular to the equiside, AE=BE=5
So dc=ae=5
-
In the isosceles coarse reed trapezoidal ABCD, bending delay a= d, c = bad bc
Rock belt c + d = 180
then c+ a=180
-
Because AD is parallel to BC angle ACB = angle DAC
And the angle b = the angle AED, so the triangle ABC is similar to the ADE.
And because of the pretense ab=3 bc=4, so ac=5
By the source of the lack of Chang in the similar AD than AC and hail de than AB equal. de=
-
∠b=90º
ac= (ab +bc )=3 +4 wreckage) = 5 abc area = ab bc 2=3 4 2=6 trapezoidal area = (ad+bc) ab 2=(2+4) 3 2=9 adc area = trapezoidal area - abc area = 9-6=3de acde ac 2=3
i.e.: de 5 2 = 3
de=6/5
ac*de=ad*ab means that the area of the triangle with the same base is equal.
-
The area of the ADC can be based on AC, and DE can be the height on the edge of AC; You can also use AD as the bottom, and AB is the height on the side of AD.
-
1 2ac*de=1 2ad*ab are all areas of a triangular acd, so ac*de=ad*ab
-
<> do and bury de bc to hand over bc to e
de=ab=ad=be=1
ec = cd -de = 3 -1 = 2 2 trapezoidal ABCD full area.
1+1+2 2 *1 2=1+ first 2
Because ab=ac=bc
So, angular BAC = angular ACD = 60 degrees. >>>More
Connect OA, OE, of AE, OA, OE is the bisector of Bao and FEC. >>>More
n,k n, then the coordinates of point e are: 0, and the coordinates of k m f point are: n,0 , and the equation of the straight line can be found from two points: >>>More
Proof: 1. From the known: afg= afe= b=90°ab=af=6,ag=ag
abg≌△afg (1) >>>More