Question

Given: ΔABC ≅ ΔEDF

What is the length of rounded to the nearest tenth?



a 1.1
b 3
c 3.2
d 4

Answer 1

I think the figure is that of a right triangle.

Its short leg = 2 - 1 = 1 unit This is the value of y.
Its long leg = 3 - 0 = 3 units This is the value of x.

Use the Pythagorean theorem to solve for the hypotenuse.

a² + b² = c²
1² + 3² = c²
1 + 9 = c²
10 = c²
√10 = √c²
3.16 = c

The length of the hypotenuse is C.) 3.2 rounded to the nearest tenth.

Answer 2

we know that

the triangle ABC and triangle EDF are congruent triangles-----> given problem

therefore

`AC=EF\\AB=ED\\BC=DF`

we know that

the distance between two points is equal to

`d=\sqrt{(y2-y1)^(2)+(x2-x1)^(2)}`

Step
`1`

Find the distance AB

`A(0,1)\\B(0,2)`

substitute in the formula of the distance

`d=\sqrt{(2-1)^(2)+(0-0)^(2)}`

`d=\sqrt{(1)^(2)+(0)^(2)}`

`dAB=1\ unit`

Step
`2`

Find the distance BC

`B(0,2)\\C(3,2)`

substitute in the formula of the distance

`d=\sqrt{(2-2)^(2)+(3-0)^(2)}`

`d=\sqrt{(0)^(2)+(3)^(2)}`

`dBC=3\ units`

Step
`3`

Find the distance AC

we know that

the triangle ABC is a right triangle

so

Applying the Pythagorean Theorem

`AC^(2)=AB^(2) +BC^(2)`

substitute the values in the formula

`AC^(2)=1^(2) +3^(2)`

`AC=√(10)\ units=3.16\ units`

round to the nearest tenth

`AC=3.2\ units`

therefore

`EF=3.2\ units`

`ED=1\ unit`

`DF=3\ units`

the answer is

The length of the hypotenuse is equal to
`3.2\ units`