Idk if someone wants to help to solve this but- it’s worth a try I think it’s 15 points or so and I’ll mark brainliest to first
person who answers !
1 answer:
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The expansion of a perfect square is
In words, the square of a sum of two terms is the sum of the squares of the two terms ( and
), plus twice the product of the two terms (
)
So, when determining if you have a perfect square trinomial, you should have two perfect squares. Note that they don't have to be the first and third term, since you can rearrange terms as you prefer.
I draw the two triangles, see the picture attached.
As you can see, angle 1 and 2 are vertically opposite angles because they are formed by the same two crossing lines and they face each other.
Angles <span>ABQ and QPR, as well as angles BAQ and QRP, are alternate interior angles because they are formed by </span><span>two parallel lines crossed by a transversal, and they are inside the two lines on opposite sides of the transversal.</span>
Hence, Allison's correct claims are:
1 = 2 because they are vertically opposite angles.
BAQ = QRP because they are alternate interior angles.
Therefore Allison, in order to prove her claim, can use the AA similarity theorem: if two angles of a triangle are congruent to two angles of the other triangle, then the two triangles are similar.
As you can see, angle 1 and 2 are vertically opposite angles because they are formed by the same two crossing lines and they face each other.
Angles <span>ABQ and QPR, as well as angles BAQ and QRP, are alternate interior angles because they are formed by </span><span>two parallel lines crossed by a transversal, and they are inside the two lines on opposite sides of the transversal.</span>
Hence, Allison's correct claims are:
1 = 2 because they are vertically opposite angles.
BAQ = QRP because they are alternate interior angles.
Therefore Allison, in order to prove her claim, can use the AA similarity theorem: if two angles of a triangle are congruent to two angles of the other triangle, then the two triangles are similar.