It is known that AB is the diameter of the circle O, C is the point on the circumference different from a and B, PA is perpendicular to the plane where the circle O is located, AE ⊥ PC is in E, find the plane Abe ⊥ plane PBC

It is known that AB is the diameter of the circle O, C is the point on the circumference different from a and B, PA is perpendicular to the plane where the circle O is located, AE ⊥ PC is in E, find the plane Abe ⊥ plane PBC

PA is perpendicular to the plane of the circle
So, PA is perpendicular to AB, BC
AB is the diameter, C is different from a, B
So, BC perpendicular to AC
So BC is perpendicular to the plane PAC
So BC is perpendicular to the line AE
AE perpendicular to PC
So the two planes are perpendicular to each other

As shown in the figure, BD is the bisector of angle ABC, De is perpendicular to AB and E, area of triangle ABC = 36, ab = 18, BC = 12, then de=____ .

Let d be DF ⊥ BC to F, ∵ BD be the bisector, ≁ DF = De, s (△ ABD) + s (△ CBD) = s (△ ABC) = 36, ≁ 1 / 2Ab * de + 1 / 2BC * DF = 36,1 / 2 * 18 * de + 1 / 2 * 12 * DF = 36, ≁ de = 12 / 5, please click [adopt as satisfactory answer]; if you are not satisfied, please point out that I will

As shown in the figure, BD is the bisector of angle ABC, and De is perpendicular to AB and E. if the area of triangle ABC is 36 and ab is 12, find the length of de

Is there a condition AB = 18?

As shown in the figure, BD is the angular bisector of ∠ ABC, and de ⊥ AB is calculated when e AB = 36cm BC = 24cms △ ABC = 144 square cm

Because BD is the bisector de ⊥ AB DF ⊥ BC of ∠ ABC
So de = DF
Because de ⊥ ab
So s △ abd = 1 / 2 * AB * De
Because DF ⊥ BC
So s △ BCD = 1 / 2 * BC * DF
Since s △ ABC = s △ BCD + s △ abd
So s △ BCD + s △ abd = 36
Then 1 / 2 * AB * de + 1 / 2 * BC * DF = 36
According to de = De
DE=2.4cm

In the triangle ABC, angle c = 90, BD is the bisector of angle ABC, De is perpendicular to AB, e, ab = 36, BC = 24, AE =?

∵ CD=DE BD=BD ∠C=∠BED=90∴ △BCD ≌ △BED BE = BC = 24AE = AB - BE = 36-24 =12

In △ ABC, BD bisects ∠ ABC, de ⊥ AB at point E, ab = 6 cm, BC = 24 cm, the area of △ ABC is 144 square cm, and the length of De is calculated
ABC is an acute triangle

Make DF perpendicular to BC and F
∵ BD bisection ∠ ABC de ⊥ AB at point e
∴DE=DF
（DE*AB）/2 +(DF*BC)/2 =144
∴DE=9.6cm

As shown in the figure, BD is the bisector of ∠ ABC, de ⊥ AB is equal to e, s △ ABC = 36cm2, ab = 18cm, BC = 12cm, then the length of De is ()
A. 2cmB. 4cmC. 1.2cmD. 2.4cm

As shown in the figure, through the point D, make DF ⊥ BC in F, ∵ BD is the bisector of ∠ ABC, de ⊥ AB, ∵ de = DF, ∵ AB = 18cm, BC = 12cm, ∵ s △ ABC = 12 × 18 · de + 12 × 12 · de = 36, the solution is de = 2.4cm

Given that in the right triangle ABC, ∠ C = 90 °, AC = 3cm, BC = 4cm, the midpoint of BC is O, Po ⊥ plane ABC, and Po = 5cm, find the distance from point P to ab
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In the triangle, make od perpendicular to point D, and the line PD is the distance from point P to ab
From the plane geometry knowledge, it is easy to know: ab = 5, OB = 2, OD = 6 / 5
PD^2=PO^2+OD^2=25+36/25=661/25
PD = radical (661) / 5
The main question I want to ask is how to calculate OD

Because ∠ cab = ∠ DOB,
COS∠CAB=3/5
COS∠DOB=OD/OB
Known ob = 2
Then od / 0b = 3 / 5
OD=3/5*2
=6/5

Given the length of three sides of right triangle ABC, AC = 5cm, ab = 3cm, BC = 4cm, the height is BD, AB and BC are two right sides, the height of AC side is obtained

BD=BC*AB/AC=2.4

In RT △ ABC, ∠ C = 90 °, AC = 2.5cm, BC = 6cm, find the length of ab
The book gives the answer, that is, it changes the decimal into the fraction (2 / 5) &# 178;, and then the result is 4 / 169 under the root sign. Let me ask you how 169 is calculated,
I have the wrong number. It should be 169 / 4

It's impossible. 169 / 4 equals 13 / 2 under the root sign. The answer is not even simplified?
Decimals are 5 / 2, not 2 / 5
The problem is very simple, 2,5 and 178; = (5 / 2) & 178; = 25 / 4
6² =36=144/4
So 2,5 and 178; + 6 and 178; = 169 / 4
AB is 169 / 4 = 13 / 2 under the root sign