Question

Which statement correctly names the congruent triangles and justifies the reason for congruence?

a coordinate plane with triangles ABC and EFD at A 3 comma 2, B 2 comma 4, C 7 comma 4, D 9 comma 5, E 7 comma 1, and F 9 comma 0, segment AC measures 4 and 5 tenths units, angle A is a right angle, segment ED measures 4 and 5 tenths units, angle E is a right angle

ΔABC ≅ ΔEFD by HL
ΔABC ≅ ΔDEF by HL
ΔABC ≅ ΔEFD by AAS
ΔBAC ≅ ΔDEF by AAS

Answer 1

Answer: Answer is A - ΔABC ≅ ΔEFD by HL

The answer up above is wrong it’s not B it’s A

Answer 2

The answer is

ΔABC ≅ ΔEFD by HL

According to the given condition  

Two triangles ΔABC and ΔEDF are given

The coordinates of different points are given as follows

A (3,2)

B (2,4)

C (7,4)

D (9,5)

E  (7,1)

F (9,0)

∠ A is a right angle

∠E is a right angle

AC= ED

So considering given situation from distance formula we can write that

$\rm For \; points \; (x_1,y_2) \; and \; (x_2,y_2) \; the \; distance\; formula\; can\; be\; written\; as \\\\\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}........(1)$

The Side opposite to ∠ A and ∠E  are the hypotenuses

BC represents the  hypotenuse of ΔABC

DF  represents the  hypotenuse of ΔEDF

So from equation (1)  and (2) we can write

$\rm BC = \sqrt{(7-2)^2+(4-4)^2} = 5 \\Similarly\\DF = \sqrt{(9-9)^2+(5-0)^2} = 5$

Hence  we can conclude that hypotenuse for both the triangles are equal.

According to the Hypotenuse Leg (HL) postulate two right  triangles are congruent when they have equal hypotenuse.

hence

ΔABC ≅ ΔEFD by HL

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