A cone-shaped grain pile has a bottom radius of 3 m and a height of 2 M. how high can the pile be placed in a cylindrical grain bin with a bottom diameter of 4 m?

2012_ End_ Night,
Grain pile volume: 3.14 × 3 × 3 × 2 × 1 / 3 = 18.84 (M3)
Grain storage area: 3.14 × (4 △ 2) ^ 2 = 12.56 (M2)
Stacking height: 18.84 △ 12.56 = 1.5 (m)

It is proved that the heights of the two waists of an isosceles triangle are equal

ZM: as shown in the figure, in △ BDC and △ CEB, ∵ ∠ DBC = ∠ ECB, ∵ BDC = ∠ CEB = 90 °, BC = BC, ≌ BDC ≌ CEB, CD = be

It is proved that the midlines on the two waists of an isosceles triangle are equal

It is known that ab = AC, ad = DC, AE = EB in isosceles △ ABC, and BD = CE is proved. It is proved that ∵ AB = AC, ad = DC, AE = EB, ∵ DC = be, ∵ DCB = ∵ EBC. ∵ BC = CB, ≌ BDC ≌ CEB (SAS) ≌ BD = CE. That is to say, the midlines on the two waists of isosceles triangle are equal

It is proved that the heights of the two waists of an isosceles triangle are equal

ZM: as shown in the figure, in △ BDC and △ CEB, ∵ ∠ DBC = ∠ ECB, ∵ BDC = ∠ CEB = 90 °, BC = BC, ≌ BDC ≌ CEB, CD = be

It is proved that the heights of the two waists of an isosceles triangle are equal

ZM: as shown in the figure, in △ BDC and △ CEB, ∵ ∠ DBC = ∠ ECB, ∵ BDC = ∠ CEB = 90 °, BC = BC, ≌ BDC ≌ CEB, CD = be

Is the bisector of the two base angles of an isosceles triangle equal? What about the middle line on the two waists? What about the height on the two waists? Prove one of the conclusions
It's best to draw

It must be equal
Let's take the middle line of the two waists as an example. Triangle ABC, ab = AC, D is the midpoint of AB, e is the midpoint of AC, connecting CD and be. Obviously, triangle DBC is similar to triangle ECB, then CD = be

Is the bisector of the two base angles of an isosceles triangle equal? What about the middle line on the two waists? What about the height on the two waists? Prove one of the knots

Prove: the bisectors of two base angles of isosceles triangle are equal proposition: triangle ABC is isosceles triangle, BD and CE are bisectors of angle B and angle C respectively prove BD = CE prove: because triangle ABC is isosceles triangle, so AB = AC, angle B = angle c, and because BD bisectors angle B, CE bisectors angle c, so angle abd = angle ace in three

The distance from the vertex of an isosceles triangle to the midline of the two waists is equal

As shown in the figure:
In the triangle ABC, D and E are the middle points of AB and AC respectively, then CD and be are the middle lines of the triangle ABC. Suppose CD and be intersect at point O, AF and Ag are the vertical lines from point a to CD and be respectively,
It is proved that AF = AG
Connect Ao
Because angle BAE = angle CAD, ab = AC, ad = 1 / 2Ab = 1 / 2Ac = AE, triangle Abe and triangle ACD are congruent,
So angle Abe = angle ACD, and because angle BOD = angle Coe,
So angle ADF = angle Abe + angle BOD = angle ACD + angle COE = angle AEG
Because angle AFD = angle age = 90 ° ad = AE
So the triangle ADF is congruent with the triangle AEG, so AF = AG, that is, the distance from the vertex of an isosceles triangle to the midline of the two waists is equal

Proof: the distance from the vertex of the top angle of an isosceles triangle to the bisector of the two bottom angles is equal
Verification with graph

In the triangle ABC, ab = AC, CD is the bisector of angle ACB, be is the bisector of angle ABC, AF is vertical to CD, Ag is vertical to be. The proof is as follows: because AB = AC, so angle ABC = angle ACB; because CD and be are the bisectors of two low angles, so angle ACF = angle ABG, angle AFC = AGB = 90 degrees, AC = AB, so triangle ag

Verification: the distance between the two vertices of a triangle and the middle line of the third side is equal
As above, good additional points

Let's just draw a picture. It is known that the middle line on the sides of ABC and BC is ad. prove that the distances from B and C to AD are the same. Prove that through B and C, be and CF are perpendicular to e and f respectively. Then: in triangle bed and triangle CEF, BD = BD, angle bed = angle CEF = 90 degrees, angle EBD = angle CFD (be / / CF), so triangle bed and triangle C

A cone-shaped grain pile has a bottom radius of 3 m and a height of 2 M. how high can the pile be placed in a cylindrical grain bin with a bottom diameter of 4 m?