Two geometry theorems get a workout here:
1) The sum of the angles of any triangle is 180 degrees,
2) If two angles are supplementary, their sum is 180 degrees.
From this we know that ∠RST + ∠STR + ∠TRS = 180. in ΔRST.
Next, look at line QS with points Q, R, and S. A straight line measures 180 degrees. and any two angles created by the line are supplements.
So, ∠QRT + ∠TRS = 180
Since we have two things equal to 180, we can set them equal to one another through transitivity (if a = b and b = c, then a = c).
∠RST + ∠STR + ∠TRS = ∠QRT + ∠TRS
Now we put in values we know for x.
(9 + x) + 5x + ∠TRS = 9x + ∠TRS
∠TRS was not filled in, but that's okay. If we subtract it from both sides, it won't be there regardless of its measure. The rest of this problem plays out like algebra class.
9 + x + 5x = 9x
9 + 6x = 9x
9 = 3x
x = 3
So x = 3, and we can find our angle measures for all angles in the problem.
Answer:
x=6
Step-by-step explanation:
Line ZW and YX are parallel, which means they have the same measurements.
Therefore:
3x+2=20
subtract two from both sides
3x=18
divide by 3
3x/3=18/3
x=6
hope this helps =D
Answer:
(x+6) (x+8)
Step-by-step explanation:
x² + 14x + 48 = 0
ac = 48
b = 14
(x + 6) (x + 8)
Thus, x= -6, x = -8
Answer:
24 tiles^2
Step-by-step explanation:
Length = 6
width = 4
Area = Length x width
= 6 x 4
= 24 tiles^2
C. â–łADE and â–łEBA
Let's look at the available options and see what will fit SAS.
A. â–łABX and â–łEDX
* It's true that the above 2 triangles are congruent. But let's see if we can somehow make SAS fit. We know that AB and DE are congruent, but demonstrating that either angles ABX and EDX being congruent, or angles BAX and DEX being congruent is rather difficult with the information given. So let's hold off on this option and see if something easier to demonstrate occurs later.
B. â–łACD and â–łADE
* These 2 triangles are not congruent, so let's not even bother.
C. â–łADE and â–łEBA
* These 2 triangles are congruent and we already know that AB and DE are congruent. Also AE is congruent to EA, so let's look at the angles between the 2 pairs of congruent sides which would be DEA and BAE. Those two angles are also congruent since we know that the triangle ACE is an Isosceles triangle since sides CA and CE are congruent. So for triangles â–łADE and â–łEBA, we have AE self congruent to AE, Angles DAE and BEA congruent to each other, and finally, sides AB and DE congruent to each other. And that's exactly what we need to claim that triangles ADE and EBA to be congruent via the SAS postulate.