Question

In a right triangle ABC, the legs are AB = 5 and BC = 12, and in the right triangle DEF, the hypotenuse is DE = 10, and EF = 8. Which is DF + AC

Answer 1

\[5^2+12^2=25+144=169\] so hypotenuse of first triangle is
\[13\] you will often see a 5-12-13 right triangle.

Answer 2

second triangle is a 3-4-5 right triangle, only in this case you have 6-8-10

Answer 3

so i guess you add to get 6+13=19

Answer 4

AC is hypotenuse in the triangle ABC
\[AC=\sqrt{AB ^{2}+BC^{2}}\] - Pythagorean theorem
\[AC=\sqrt{5 ^{2}+12^{2}}\]
\[AC=\sqrt{25+144}\]
\[AC=\sqrt{169}\]
\[AC=13\]

In DEF triangle DE=10 is hypotenuse and leg EF =8
\[DE ^{2}=EF ^{2}+DF ^{2}\]
\[DF ^{2}=DE ^{2}-EF ^{2}\]
\[DF=\sqrt{DE ^{2}-EF ^{2}}\]
\[DF=\sqrt{10^{2}-8^{2}}\]
\[DF=\sqrt{100-64}\]
\[DF=\sqrt{36}\]
DF=6

DF+AC=6+13=19

Answer 5

can any of you help me on my next question please