Thread: Geometry Help

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.

Ok thanks, I am going to flunk the test. Thanks a lot.

For question 3, you need exactly 3 points in order to "determine" a plane. What you need to figure out is how many trios of the 5 vertices can you form without repetition. i.e., ABC = BCA = ACB = etc.

For question 1, what you have is x, 3x/2, and 3x/2+4. So x+3x/2+3x/2+4=180. x being the smallest of all angles. Solve for x, use it to find the largest of angles, and then subtract it from 360.

For question 2, I don't understand your drawing...

Originally Posted by AirForceOne

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!)
2.

2.a If AD is perpendicular to m then angle ADC= angle ADE = 90 degrees.
But angle ADC=x+88, and angle ADE=74-8x, so both cannot be 90 degrees.
Therefore AD is not perpendicular to m.

To do the remaining parts there is insufficient information, but lets assume
you intend that D lies on segment AB.

Then angles ADE and BDE are supplementary - that is they sum to
180 degrees. So:

74 - 8x + 2x +94 =180,

or:

x = -2.

2.b No segments AD and CD are not perpendicular, as angle ADC=86 degrees.

2.c Yes segments AD and DE are perpendicular as angle ADE=90 degrees.

RonL

Thanks, read the replies yesterday night