1. You can make use of the Pythagorean theorem considering "a" to be the longer leg of the largest right triangle. Then the value of "a" can be calculated from
(16+9)² = 15² +a²
625 -225 = a²
√400 = a = 20
2. You can find the altitude of the largest triangle (the length of the unmarked vertical line), then find "a" as the hypotenuse of the medium-sized right triangle.
That altitude is ...
√(15² -9²) = √144 = 12
so the length "a" is ...
a² = 12² +16² = 144 +256 = 400
a = √400 = 20
3. Based on the 15 and 9 dimensions of the smallest right triangle, you can realize that these triangles are multiples of the 3-4-5 right triangle. Then you can use relevant ratios of the side lengths of any of the triangles of which "a" is a part.
a = 4/3×15 = 4/5×(9+16) = 5/4×16 = 20
(23•34y17) would be the answer.
Xy/w
-3(-4)/-6
=12/-6
=-2
The answer is B. Negative 2
Answer:
Option B) ∠DCE is congruent to ∠BCA by the Vertical Angles Theorem and 15 over 5 equals 12 over 4 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SAS Similarity Postulate
Step-by-step explanation:
we know that
The SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar
we have that
m∠DCE≅m∠BCA -----> by vertical angles
substitute the values
----> is true
so
Two sides in triangle ABC are are proportional to two sides in triangle EDC and the included angle in both triangles are congruent
therefore
ΔABC and ΔEDC are similar by SAS Similarity Postulate