In △ ABC, ab = 4 3，AC=2 3, ad is the center line of BC side, and ∠ bad = 30 °, then BC=______ .

In △ ABC, ab = 4 3，AC=2 3, ad is the center line of BC side, and ∠ bad = 30 °, then BC=______ .

Take the midpoint e of AB, get be = AE = 12ab = 23, connect De, we can get the median line of △ ABC,  de ∵ AC,  de = 12ac = 3, that is, de = 12ae, ? bad = 30 ° and ∠ EDA = 90 ° according to the Pythagorean theorem: ad = AE2 − ED2 = 3, ∵ ed ∥ AC, ∵ DAC = ∠ ade = 90 ° according to the Pythagorean theorem

In △ ABC, ad is the height of BC edge, ∠ C = 30 °, BC = 2 + radical 3, tanb = 1 / 2. Find the length of ab

Ad = 1 / 2 * AC is obtained from ∠ C = 30 °;
From tanb = 1 / 2, AD / BD = 1 / 2, ad = 1 / 2 * BD;
AC = BD
In the right triangle ADC, according to Pythagorean theorem, BD ^ 2 = (bc-bd) ^ 2 + (1 / 2bd) ^ 2,
After substituting BC, BD = 2
Ad = 1
In the right triangle ADB, ab = √ 5 is obtained by Pythagorean theorem

As shown in the figure, in △ ABC, ad is the height on BC side, ∠ C = 30 ° and BC = 2+ 3，tanB＝1 2. So what is the length of ad______ .

It is known that ad is the height on BC, ∠ C = 30 ° and BC = 2+
3，tanB＝1
2，
Let ad = X,
Then BD = 2x, X
BD=tanB，
Result: CD=
3x，
∴2x+
3x=2+
3，
∴x=1．
Ad = 1,
So the answer is: 1

In the isosceles triangle ABC, ab = AC, height ad = 4, the circumference of △ ABC is 16 On Pythagorean theorem

If the circumference of △ abd is 16, then the circumference of △ ABC is 24,
∵ ad = 4 (known)
∴AB+BD=16-4=12
∵ △ ABC is an isosceles triangle (known)
/ / BD = CD (isosceles triangle with three lines in one)
The circumference of △ ABC is 24

In isosceles △ ABC, we know AB = AC = 5, BC = 6. If we fold △ ABC along the broken line BD so that the point C falls on C1 on the straight line AC, then AC1=______ .

As shown in the figure, pass point a as AE ⊥ BC at E,
∵AB=AC，BC=6，
∴CE=1
2BC=1
2×6=3，
∴cos∠C=CD
BC=CE
AC，
CD
6=3
5，
CD = 18
5，
∵ △ ABC along the broken line BD, the turning point C falls at C1 on the straight line AC,
∴C1D=CD=18
5，
∴AC1=CC1-AC=18
5×2-5=11
5．
So the answer is: 11
5．

If the height of the base edge of the isosceles triangle ABC is ad = half BC, ab = radical 2, then the area of △ ABC is obtained A. Radical 2 B.1 C.2 D

The answer is B
Since the triangle is an isosceles triangle, point D is also the midpoint of the bottom BC. Ad = half BC, so ad= BD.AB So ad = BD = 1, that is BC = 2. Area = BC * ad / 2 = 1

As shown in the figure, in the isosceles triangle ABC, the base edge BC = 24cm, and the area of △ ABC is equal to 60cm2. Please calculate the length of waist ab

Ad ⊥ BC in D
Then s △ ABC = 1
2AD•BC=60，
∵BC=24，
∴AD=5，
In RT △ abd, ab=
AD2+BD2＝
52+122＝13cm．

As shown in the figure, in the isosceles triangle ABC, the base edge BC = 24cm, and the area of △ ABC is equal to 60cm2. Please calculate the length of waist ab

Ad ⊥ BC in D
Then s △ ABC = 1
2AD•BC=60，
∵BC=24，
∴AD=5，
In RT △ abd, ab=
AD2+BD2＝
52+122＝13cm．

In the isosceles triangle ABC, the base edge BC = 24cm, and the area of △ ABC is equal to 60cm2. Calculate the length of waist ab

If the height ad is made through BC, then 60 = 1 / 2 * 24 * ad is obtained from the area formula, so ad = 5 and the triangle is isosceles, so the high lines and the central lines coincide, so BD = 1 / 2BC = 12 can be obtained from the Pythagorean theorem: ab = 13

In the isosceles triangle ABC, the base edge BC = 2cm, and the area of triangle ABC is equal to 60 square centimeters. Please calculate the waist ab

3601 square root cm