View Full Version : Geometry Question

JOSHIE63

05-27-2016, 11:21 PM

I have a quick question for anyone that has knowledge of geometry: can you walk me through this one?

125474

I am trying to find the area. Whether I use Heron's formula or find the area of the two smaller triangles and add does not matter.

Given the 15 as the altitude and the 18 as a hypotenuse, I concluded that the leg in the bottommost triangle is 3√11. The larger triangle that encompasses both smaller triangles is in fact isosceles; however, without any given angle measures, no trigonometric ratios can be set up, much less using the isosceles triangle theorem.

I figured I would ask here as this is pretty specific, and there has to be at least a few forumers that take geometry.

xxgrowzz

05-27-2016, 11:35 PM

Im not sure, but try to use sin cos tan to find the angle.

It turns out the smaller bottom triangle is a phytagoras possible triangle (which means it has 90 degree on one of the side)

Hope it helps

So basically I'm almost failing geometry myself, so you'll have to check my work.

But anyways we just started learning trig ourselves, and my friend who is a class ahead of me just gave me a great lesson on it this morning.

Let's start with the bottom triangle. First we need to find all of the angles. we already know one of them is 90. To find the other one (let's say the one on the left), you'll need to identify the opposite and adjacent angles. The opposite would be 3 sqrt 11, or ~9.949. The adjacent would be 15. Using SOHCAHTOA we would find out that we need to use tan. Opposite over adjacent = 9.949 / 15. That would give us 0.663 we would then use tan-1(0.663) giving us an angle of 33.544 or rounding, about 34 degrees.

There are 180 degrees in a triangle. Take 180 - (34+90) and you get the final angle to equal 56.

Using AAS we can conclude the area of the lower triangle is about 752.

To be honest I have no idea if I got any of that right, and I have not a clue on how one would solve the top triangle. I'm mostly doing this to practice it myself.

EDIT: might have figured it out! Both angles on the bottom would have to be 56 degrees since they are congruent. We would do 180 - (56*2) to get 68 for the top angle.Using AAS again, we would get the total area of both triangles combined to be 120.

Again though, I'm most probably wrong.

JOSHIE63

05-28-2016, 02:38 AM

So basically I'm almost failing geometry myself, so you'll have to check my work.

But anyways we just started learning trig ourselves, and my friend who is a class ahead of me just gave me a great lesson on it this morning.

Let's start with the bottom triangle. First we need to find all of the angles. we already know one of them is 90. To find the other one (let's say the one on the left), you'll need to identify the opposite and adjacent angles. The opposite would be 3 sqrt 11, or ~9.949. The adjacent would be 15. Using SOHCAHTOA we would find out that we need to use tan. Opposite over adjacent = 9.949 / 15. That would give us 0.663 we would then use tan-1(0.663) giving us an angle of 33.544 or rounding, about 34 degrees.

There are 180 degrees in a triangle. Take 180 - (34+90) and you get the final angle to equal 56.

Using AAS we can conclude the area of the lower triangle is about 752.

To be honest I have no idea if I got any of that right, and I have not a clue on how one would solve the top triangle. I'm mostly doing this to practice it myself.

EDIT: might have figured it out! Both angles on the bottom would have to be 56 degrees since they are congruent. We would do 180 - (56*2) to get 68 for the top angle.Using AAS again, we would get the total area of both triangles combined to be 120.

Again though, I'm most probably wrong.

The inverse of the sine ratio — why could I not think of that. Of course I can find the angles now and use Heron's formula to calculate the area with the last full side.

Thanks a ton! I will be receiving three extra credit points now.

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