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Sindrei [870]
3 years ago
12

Based only on the information given in the diagram, which congruence theorems or postulates could be given as reasons why DEF=JK

L?

Mathematics
2 answers:
NISA [10]3 years ago
7 0

Answer:

A) AAS; B) LA; C) ASA

Step-by-step explanation:

AAS is the Angle-Angle-Side congruence statement.  It says that if two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of a second triangle, then the triangles are congruent.  In these triangles, ∠E≅∠K, ∠F≅∠L, and DE≅JK.  These are two angles and a non-included side; this is AAS.

LA is the leg-acute theorem.  It states that if a leg and acute angle of one triangle is congruent to the corresponding leg and acute angle of another triangle, then the triangles are congruent.

The leg we have congruent from each triangle is DE and JK.  We also have ∠E≅∠K and ∠F≅∠L, both pairs of which are acute.  This is the LA theorem.

ASA is the Angle-Side-Angle congruence statement.  It says that if two angles and an included side of one triangle are congruent to the corresponding two angles and included side of another triangle, then the triangles are congruent.  

We have that ∠D≅∠J, DE≅JK and ∠E≅∠K.  This gives us two angles and an included side, or ASA.

ASHA 777 [7]3 years ago
6 0

Answer: AAS

LA

ASA

Step-by-step explanation:

In the given picture, we have two right triangles in which one side and one angle is equal to the corresponding side and one angle of the another right triangle,

So by LA theorem

ΔDEF≅ΔJKL

Also, two angles of ΔDEF is equal to the corresponding two angles of ΔJKL and one non included side, So by AAS theorem they are congruent.

If we consider right angle also, then two angles of ΔDEF is equal to the corresponding two angles of ΔJKL and one included side, So by ASA postulate they are congruent.

In ΔDEF and ΔJKL no two sides are given congruent thus we cannot apply SAS or LL or HL (hypotenuse are not given to be equal).

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Then you will find the area of the circle using the given radius 20.5.
Lets work it out.
84×92=7728
7728÷2=3864 ft²
3864 ft² is the area of the backyard including the pool.
A=pi(20.5)²
A=1320.25 ft²
1320.25 ft² is the area of the pool.
Now we have to subtract to get the area of the backyard without the pool.
3864-1320.25=2543.75 ft²
2543.75 ft² is your answer.
Hope This Helps and God Bless!
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