Answer:
Step-by-step explanation:
![a^{2} + b^{2} = c^{2}](https://tex.z-dn.net/?f=a%5E%7B2%7D%20%2B%20b%5E%7B2%7D%20%3D%20c%5E%7B2%7D)
In this situation a = 1, b = 2
![1^{2} + 2^{2} = c^{2}](https://tex.z-dn.net/?f=1%5E%7B2%7D%20%2B%202%5E%7B2%7D%20%20%3D%20c%5E%7B2%7D)
1 + 4 = ![c^{2}](https://tex.z-dn.net/?f=c%5E%7B2%7D)
5 = ![c^{2}](https://tex.z-dn.net/?f=c%5E%7B2%7D)
c = ![\sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B5%7D)
This is an important mathematical theorem to know. It's called the Pythagorean Theorem.
Answer:
Yes, we can conclude that Triangle ABC is similar to triangle DEF because the measures of the 3 angles of both triangles are congruent.
Step-by-step explanation:
We have the measure of 2 angles from both triangles, and we know that triangles have 180°, so we can solve for the measure of the third angle for both triangles.
Triangle ABC:
Measure of angle A= 60°
Measure of angle C= 40°
Measure of angle B = 180°- (measure of angle A + measure of angle C) = 180° - (60° + 40°) = 80°
Triangle DEF
Measure of angle E= 80°
Measure of angle F= 40°
Measure of angle D= 180° - (measure of angle E + measure of angle F) = 180° - (80° + 40°) = 60°
The measures of the angles in Triangle ABC are: 60°, 40°, and 80°.
The measures of the angles in Triangle DEF are: 60°, 40°, and 80°.
Since the measure of 3 angles of the two triangles are the same, we know that the two triangles are similar.
Answer:
13y-10
Step-by-step explanation:
Expand: -5(-y+2): 5y-10
Add similiar elements: 8y+5y=13y
=13y-10
1. Check the drawing of the rhombus ABCD in the picture attached.
2. m(CDA)=60°, and AC and BD be the diagonals and let their intersection point be O.
3. The diagonals:
i) bisect the angles so m(ODC)=60°/2=30°
ii) are perpendicular to each other, so m(DOC)=90°
4. In a right triangle, the length of the side opposite to the 30° angle is half of the hypothenuse, so OC=3 in.
5. By the pythagorean theorem,
![=\sqrt{9}* \sqrt{3} =3\sqrt{3} (in)](https://tex.z-dn.net/?f=%3D%5Csqrt%7B9%7D%2A%20%5Csqrt%7B3%7D%20%3D3%5Csqrt%7B3%7D%20%20%28in%29)
6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*
![\frac{DO*OC}{2}=4 \frac{3 \sqrt{3} *3}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7BDO%2AOC%7D%7B2%7D%3D4%20%5Cfrac%7B3%20%5Csqrt%7B3%7D%20%2A3%7D%7B2%7D%20)
=
![=2*9 \sqrt{3}=18 \sqrt{3}](https://tex.z-dn.net/?f=%3D2%2A9%20%5Csqrt%7B3%7D%3D18%20%5Csqrt%7B3%7D%20%20)
(
![in^{2}](https://tex.z-dn.net/?f=%20in%5E%7B2%7D%20)
)
FA = FD = 12/2 = 6
FC is given as 8. FA+FC = 14 is half the perimeter.
So the perimeter is 2(6+8) = 28 . . . . . . matches selection A