Question

Here is a diagram showing triangle ABC and some transformations of triangle ABC .On the left side of the diagram, triangle ABC has been reflected across line AC to form quadrilateral ABCD . On the right side of the diagram, triangle ABC has been rotated 180 degrees using midpoint M as a center to form quadrilateral ABCE. Using what you know about rigid transformations, side lengths and angle measures, label as many side lengths and angle measures as you can in quadrilaterals ABCD and ABCE. *

Answer 1

1. For quadrilateral ABCD:

AB = AD = 2.7, BC = DC = 3.2, and AC = 3.168

<BAC = <DAC
`65.5^(o)`, <ABC = <ADC =
`64.3^(o)`, and <ACB = <ACD =
`50.2^(o)`

2. For quadrilateral ABCE:

AB = EC = 2.7, BC = AE = 3.2, and AC = 3.168

<BAC = <ACE
`65.5^(o)`, <ABC = <AEC =
`64.3^(o)`, and <ACB = <EAC =
`50.2^(o)`

Explanation:

Applying the Cosine rule to triangle ABC,

`/AC/^(2)` =
`/AB/^(2)` +
`/BC/^(2)` - 2/AB/ x /BC/ Cos B

=
`2.7^(2)` +
`3.2^(2)` - 2 x 2.7 x 3.2 Cos 64.3

= 7.29 + 10.24 - 17.28 x 0.4337

= 17.53 - 7.49434

= 10.03566

AC =
`√(10.03566)`

= 3.168

Applying the Sine rule,

`(a)/(Sin A)` =
`(b)/(Sin B)` =
`(c)/(Sin C)`

So that:

`(3.2)/(Sin A)` =
`(3.168)/(Sin 643.^(o) )`

`(3.2)/(Sin A)` =
`(3.168)/(0.9011)`

Sin A =
`(3.2 * 0.9011)/(3.168)`

=
`(2.88352)/(3.168)`

= 0.9102

⇒ A =
`Sin^(-1)` 0.9102

=
`65.5^(o)`

But sum of angles in a triangle is
`180^(o)`, so that;

A + B + C =
`180^(o)`

65.5 + 64.3 + C =
`180^(o)`

129.8 + C =
`180^(o)`

C =
`180^(o)` - 129.8

C =
`50.2^(o)`

1. For quadrilateral ABCD:

AB = AD = 2.7, BC = DC = 3.2, and AC = 3.168

<BAC = <DAC
`65.5^(o)`, <ABC = <ADC =
`64.3^(o)`, and <ACB = <ACD =
`50.2^(o)`

2. For quadrilateral ABCE:

AB = EC = 2.7, BC = AE = 3.2, and AC = 3.168

<BAC = <ACE
`65.5^(o)`, <ABC = <AEC =
`64.3^(o)`, and <ACB = <EAC =
`50.2^(o)`