This is the isosceles triangle. Therefore, the angles at the base are congruent.
We know that the sum of the measures of the angles in a triangle is 180°.
Therefore we have the equation:
![38^o+x^o+x^o=180^o](https://tex.z-dn.net/?f=38%5Eo%2Bx%5Eo%2Bx%5Eo%3D180%5Eo)
<em>subtract 38 from both sides</em>
<em>divide both sides by 2</em>
![x^2=71^o](https://tex.z-dn.net/?f=x%5E2%3D71%5Eo)
<h3>Answer: x = 71.</h3>
Answer:10%
Step-by-step explanation:
Answer:
slope = - ![\frac{7}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B5%7D)
Step-by-step explanation:
Calculate the slope m using the slope formula
m = ![\frac{y_{2}-y_{1} }{x_{2}-x_{1} }](https://tex.z-dn.net/?f=%5Cfrac%7By_%7B2%7D-y_%7B1%7D%20%20%7D%7Bx_%7B2%7D-x_%7B1%7D%20%20%7D)
with (x₁, y₁ ) = (3, 4 ) and (x₂, y₂ ) = (8, - 3 )
m =
=
= - ![\frac{7}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B7%7D%7B5%7D)
The an answer is 2x
explanation :
12+3x-x-12
= 3x-x
= 2x
Answer:
(A) AA Similarity Theorem
Step-by-step explanation:
Given: AB ∥ DE
To Prove: ![\triangle ACB \sim \triangle DCE](https://tex.z-dn.net/?f=%5Ctriangle%20ACB%20%5Csim%20%5Ctriangle%20DCE)
Given Triangle ABC with Line DE drawn inside of the triangle and parallel to side AB. The line DE forms a new triangle DCE.
Because AB∥DE and segment CB crosses both lines, we can consider segment CB a transversal of the parallel lines.
Angles CED and CBA are corresponding angles of transversal CB and are therefore congruent, so ∠CED ≅ ∠CBA.
We can state ∠C ≅ ∠C using the reflexive property.
Therefore,
by the AA similarity theorem.
Remark: In the diagram, we can see that the two triangles share Angle C and have two equal angles at E and B. Therefore, they are similar by the Angle-Angle Similarity Theorem.