Please Hurry. 3 Geometry Problems. I have Test Tomorrow. Hur

[Guest]

Hi I am in a hurry. I am in 9th Grade Geometry Honors and need help with 3 problems. Please hurry. I have a test tomorrow. I have quickly drew the probs 2 and 3 in paint, (sorry for lack of qualilty, i'm trying to study!)

1. The Measures of two angles of a triangle are in the ratio 2:3. If the third angle is 4 degrees larger than the larger of the other two angles, find the measures of an exterior angle at the third vertex.

2.

3.

Thanks.

 

[Denis]

I'll do #1:
2k + 3k + 3k+4 = 180
8k = 176
k = 22

2k: 44
3k: 66
3k+4: 70

 

[soroban]

Re: Please Hurry. 3 Geometry Problems. I have Test Tomorrow.

Hello, AirForceOne!

Here's my explanation for #3 . . .

3.

Three points will determine a plane.

We have five points to choose from: \(\displaystyle \{A,B,C,D,P\}\)
. . So there are \(\displaystyle C(5,3)\,=\,\frac{5!}{3!\,2!}\,=\,10\) ways.

But four of them determine the same plane:
. . \(\displaystyle ABC,\,ABD,\,ACD,\,BCD\) all determine the base-plane.

So there are seven planes: \(\displaystyle ABP,\,BCP,\,CDP,\,DAP,\,ACP,\,BDP\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Another approach . . .

We can see five planes already . . .
. . those determined by the base and the four triangular faces.

And there are two more . . . determined by triangles PAC and PBD.

 

[soroban]

Re: Please Hurry. 3 Geometry Problems. I have Test Tomorrow.

Hello again, AirForceOne!

If I read the diagram correctly, here is #2 . . .

2.

I assume (always dangerous!) that \(\displaystyle ADB\) is a straight line.

We're told that: \(\displaystyle \,\angle ADE\:=:,74\,-\,8x\) degrees.
and we're told: \(\displaystyle \,\angle BDE\:=\:2x\,+\,94\) degrees.

Since \(\displaystyle \angle ADE\,+\,\angle BDE\:=\:180^o\), we have: \(\displaystyle \74 - 8x)\,+\,(2x + 94)\:=\:180\;\;\Rightarrow\;\;x\,=\,-2\)

Hence: \(\displaystyle \,\angle ADE\:=\:74\,-\,8(-2)\:=\:90^o\)
and: \(\displaystyle \,\angle BDE\:=\:2(-2)\,+\,94\:=\:90^o\)

c. Are segments \(\displaystyle \overline{AD}\) and \(\displaystyle \overline{DE}\) perpendicular? . Yes, angle ADE = 90°


Then \(\displaystyle \angle ADC\:=\-2)\,+\,88\:=\:86^o\)

b. Are segments \(\displaystyle \overline{AD}\) and \(\displaystyle \overline{CD}\) perpendicular? . No, angle ADC ≠ 90°


a. Are segment \(\displaystyle \overline{AD}\) and plane \(\displaystyle m\) perpendicular? . No

If \(\displaystyle \overline{AD}\) is perpendicular to the plane, then it is perpendicular to every line in the plane.
. . And we've seen that \(\displaystyle \overline{AD}\) is <u>not</u> perpendicular to \(\displaystyle \overline{CD}\).

 

[Guest]

Re: Please Hurry. 3 Geometry Problems. I have Test Tomorrow.

So there are seven planes: \(\displaystyle ABP,\,BCP,\,CDP,\,DAP,\,ACP,\,BDP\)

Thanks
But you put only 6 planes? What about the base ones? (or one)?
Is the base made of 2 planes or can 1 of the planes of the base represent the whole base as a plane? Like, since one of the answers is ABC (in the base), they did not put ADC also (also in base)

Thanks to all who have help me!