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As shown in the figure, in the triangle ABC, ad is the height on the bottom BC, the area of triangle ABC = 6 square centimeters, ad = 2 centimeters, and the distance from point C to ad is 2 Cm, find the distance from point B to AD
First, BC = 6cm, CD = 2cm;
There are three cases
1. D on BC, BD = bc-cd = 4cm;
2. D is on the BC extension line, BD = BC + CD = 8 cm;
3. D on CB extension line, BD = cd-bc = - 4 is impossible
The area of triangle ABC is 18 square centimeters, BC = 10 centimeters. What is the distance from point a to point BC
Let the distance from a to point BC be x, then
10*x/2＝18
The solution is x = 3.6
It is known that in the triangle ABC, ab = 2, AC = 1, and the angle bisector ad = 1 of angle a, we can find the area of the triangle
cos∠BAD=(2²+1²-BD²)/(2*2*1)=(5-BD²)/4
cos∠DAC=(1²+1²-DC²)/(2*1*1)=(2-DC²)/2
Because ∠ bad = ∠ DAC,
So (5-bd & sup2;) / 4 = (2-DC & sup2;) / 2, BD & sup2; - 2dc & sup2; = 1
And AB / AC = BD / DC, 2 / 1 = BD / DC, BD = 2dc
(2DC)²-2DC²=1,DC=√2/2,BD=2*√2/2=√2,BC=√2/2+√2=3√2/2≈2.12.
(2+1+2.12)/2=2.56
S△ABC=√[2.56*(2.56-2)(2.56-1)(2.56-2.1)]≈1.01
Help solve a problem. In the triangle ABC, angle c = 90 degrees. AC = 2.1 cm.BC=2 8 cm. [1] find the length of the high Cd on the hypotenuse ab of this triangle!
[2] Ask hypotenuse to be divided into two parts AD and ad long.. urgent!
The answer is 2cm
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