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If the edge of the regular triangle ABC is 15 and the distance from a point p outside the plane ABC to a, B and C is 20, find the distance from point P to plane ABC
Edge of regular triangle ABC = 15,
therefore
Height = 15 / 2 * √ 3 = 15 √ 3 / 2
therefore
Distance from center of gravity to vertex = 15 √ 3 / 2 × 2 / 3 = 5 √ 3
therefore
Distance from point P to plane ABC = √ (20) &# 178; - (5 √ 3) &# 178; = 5 √ 13
As shown in the figure, in the isosceles triangle ABC, ab = AC, AF is the middle line of BC, D is a point on AF, and the vertical bisector of BD passes through point C and intersects BD with E, proving that BCD is equilateral
∵ CE bisects BD vertically, that is ∠ BEC = ∠ Dec = 90 ° be = De, and CE is the common edge
The value of Δ BEC is equal to Δ Dec
∴BC=CD
In ∵ △ ABC, ab = AC, AF is the middle line of BC, that is, BF = CF, and AF is the common edge
The value of ABF is equal to ACF
∴∠AFB=∠AFC=90°
∵ D is a point on AF, ∠ DFB = ∠ DFC = 90 ° BF = CF, and DF is the common edge
The ∧ DBF is equal to the ∧ DCF
∴BD=CD
And ∵ BC = CD
The results show that BC = CD = BD and △ BCD are equilateral triangles
As shown in the figure, in isosceles △ ABC, ab = AC, the vertical bisector of AB side intersects AC at D. if AB = 8, BC = 6, find the perimeter of △ BCD
The vertical bisector of ∵ AB intersects at D
∴DA=DB
The perimeter of triangle BDC
=BD+CD+BC
=AD+DC+BC
=AC+BC
=8+6=14
As shown in the figure, in △ ABC, ab = AC = 5, BC = 4, the vertical bisector De of AB intersects AB and ab at points E and D respectively, and the perimeter of △ BCD is calculated
De vertical bisection ab
So, ad = BD
The perimeter of △ BCD = BD + DC + BC
=(AD+DC)+BC
=AC+BC
=5+4=9
If we know that the new distance of the cross section of a, B and C on the sphere is equal to half of the radius, ab = AC = BC = 2, we can find the sphere area
It's the area of the ball cap with a quarter of the diameter of the high ball
Make three vertical lines of triangle ABC, find the vertical center P, and connect POA to form a triangle (o is the center of the circle)
From ab = AC = BC = 2, the distance between the perpendicular center P of ABC triangle and vertex angle A
PA = 2 / 3 × (root 3 of triangle height) = 0.667 * root 3
For solving triangle OPA, if R * r = 0.25r * r + PA * PA (i.e. 4 / 3), then 0.75r * r = 4 / 3
R=4/3
Area s = 2 π RH = π * r * r = (16 / 9) * π = 5.58505351
If the length of two right sides of a right triangle is 3 and 4, and the distance from a point in the triangle to each side is equal, then the distance is______ ．
Connect OA, ob, OC, then the distance from point O to the three sides is the high line of △ AOC, △ BOC, and △ AOB. If the distance to the three sides is x, then the sum of the areas of the three triangles is: 12ac · x + 12bc · x + 12ab · x = 12ac · BC, then x = 1
If the length of two right sides of a right triangle is 3 and 4, and the distance from a point in the triangle to each side is equal, then the distance is______ ．
Connect OA, ob, OC, then the distance from point O to the three sides is the high line of △ AOC, △ BOC, and △ AOB. If the distance to the three sides is x, then the sum of the areas of the three triangles is: 12ac · x + 12bc · x + 12ab · x = 12ac · BC, then x = 1
If the length of two right sides of a right triangle is 3 and 4, and the distance from a point in the triangle to each side is equal, then the distance is______ ．
Connect OA, ob, OC, then the distance from point O to the three sides is the high line of △ AOC, △ BOC, and △ AOB. If the distance to the three sides is x, then the sum of the areas of the three triangles is: 12ac · x + 12bc · x + 12ab · x = 12ac · BC, then x = 1
If the length of two right sides of a right triangle is 3 and 4, and the distance from a point in the triangle to each side is equal, then the distance is______ ．
Connect OA, ob, OC, then the distance from point O to the three sides is the high line of △ AOC, △ BOC, and △ AOB. If the distance to the three sides is x, then the sum of the areas of the three triangles is: 12ac · x + 12bc · x + 12ab · x = 12ac · BC, then x = 1
If the length of two right sides of a right triangle is 3 and 4, and the distance from a point in the triangle to each side is equal, then the distance is______ ．
Connect OA, ob, OC, then the distance from point O to the three sides is the high line of △ AOC, △ BOC, and △ AOB. If the distance to the three sides is x, then the sum of the areas of the three triangles is: 12ac · x + 12bc · x + 12ab · x = 12ac · BC, then x = 1
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