As shown in the figure, given that the vertical bisector of the sides of the isosceles triangle ABC and ab intersects AC at point D and ab at e, ab = AC = 8, BC = 6, calculate the circumference of △ BDC

As shown in the figure, given that the vertical bisector of the sides of the isosceles triangle ABC and ab intersects AC at point D and ab at e, ab = AC = 8, BC = 6, calculate the circumference of △ BDC

∵ De is the vertical bisector of ab,
∴AD=BD,
The circumference of △ BDC is BC + CD + BD = BC + CD + ad = BC + AC,
∵BC=6,AC=8,
The circumference of △ BDC is 6 + 8 = 14

In the isosceles triangle ABC, the vertical bisector of one waist AB intersects the other waist AC at g. if AB = 10 and the circumference of △ GBC is 17, then the base BC is () A. 5 B. 7 C. 10 D. 9

Let the midpoint of AB be d,
∵ DG is the vertical bisector of ab
ν GA = GB (the distance from one point on the vertical bisector to the two ends of the line segment is equal),
The circumference of triangle GBC = GB + BC + GC = GA + GC + BC = AC + BC = 17,
And ∵ the triangle ABC is an isosceles triangle, and ab = AC,
∴AB+BC=17,
∴BC=17-AB=17-10=7.
Therefore, B

As shown in the figure, BD is the bisector of the base angle of the isosceles △ ABC. If the base angle ∠ ABC = 72 ° and the waist AB length is 4cm, then the bottom BC length is______ cm.

∵ BD is the bisector of the base angle of ᙽ ABC, if ᙽ ABC = 72 °, a = 36 °, C = 72 °, abd = CBD = 36 °, BDC = 72 °, BD = BC = ad, let BC = x, then ad = BD = x, DC = 4-x, ∵ BCD ∵ ABC,

If the circumference of the isosceles △ ABC is 16 and the height on the base BC is 4, then the area of △ ABC is______ .

Let the base length be 2x
The waist length is 16 − 2x
2=8-x.
Using Pythagorean theorem: (8-x) 2 = x2 + 42,
∴x=3,
The area of △ ABC is 1
2×6×4=12.
So the answer is: 12

If the circumference of the isosceles △ ABC is 16 and the height on the base BC is 4, then the area of △ ABC is______ .

Let the base length be 2x
The waist length is 16 − 2x
2=8-x.
Using Pythagorean theorem: (8-x) 2 = x2 + 42,
∴x=3,
The area of △ ABC is 1
2×6×4=12.
So the answer is: 12

If the circumference of the isosceles △ ABC is 16 and the height on the base BC is 4, then the area of △ ABC is______ .

Let the base length be 2x
The waist length is 16 − 2x
2=8-x.
Using Pythagorean theorem: (8-x) 2 = x2 + 42,
∴x=3,
The area of △ ABC is 1
2×6×4=12.
So the answer is: 12

The base edge BC of the isosceles triangle ABC is 10 and its circumference is 36. The area of () of triangle ABC So how to find the height of the triangle ABC? Who said clearly will give who points!

Waist length = (36-10) × 2 = 13
The height of the triangle = √ (13? - 5?) = 12
Area of triangle = 10 × 12 △ 2 = 60

The waist length of the isosceles triangle ABC is 5 / 6 of the base, and the area of the triangle ABC is 108

Let the bottom edge length a
The waist length of the isosceles triangle ABC is 5 / 6 of the base
The height ad on the bottom edge of ABC = √ [(5 / 6A) 2 - (1 / 2a) 2] = 2 / 3A
∴½×2/3a×a=108
a²=324
∵a>0
∴a=18
The height ad on the bottom edge of the triangle ABC = 18 × 2 / 3 = 12

As shown in the figure, ⊙ o takes one waist ab of the isosceles ⊙ ABC as its diameter, which intersects the other waist AC to e and BC to d Verification: BC = 2DE

Proof: connect ad,
∵ AB is the diameter of ⊙ o,
∴∠ADB=90°,
And ∵ AB = AC,
Ψ B = ∠ C, BD = DC, that is BC = 2dc,
∵ quadrilateral ABDE is a quadrilateral inscribed in a circle,
∴∠CED=∠B,
And ∠ B = ∠ C,
∴∠CED=∠C,
∴DE=DC,
∴BC=2DE.

As shown in the figure, it is known that the base edge BC of the isosceles triangle ABC is 20, D is a point on waist AB, and CD = 16, BD = 12? Look at the analysis carefully and do it again

Let AB = x, ad = x-12ac = x, CD = 16 △ ADC, in which x = (X-12) 2 + 16? X = 50 / 3, perimeter = 50 / 3 × 2 + 20 = 160 / 3 (2) ad