DEF is a right triangle. A. True B. False

Mathematics

2 answers:

Vika [28.1K]3 years ago

8 0

------------ False-----------

~ APEX

~ took the quiz

----------------- <3---------------------

Sonbull [250]3 years ago

5 0

Given triangle DEF with the lengths of all sides. If you want to check whether this triangle is right, use the Pythagorean theorem. The greatest side is EF, then only this side could be the hypotenuse.

Check, whether

$EF^2=ED^2+DF^2.$

Substitute ED=25, EF=48 and DF=27:

$DF^2+DE^2=27^2+25^2=729+625=1354,\\ \\EF^2=48^2=2304.$

Since $1354\neq 2304,$ the triangle DEF is not right triangle.

Answer: correct choice is B

You might be interested in

Find value of P and Q

ch4aika [34]

Answer:

$P= 6 \frac{1}{7}\\\\Q = 4 \frac{2}{7}$

Step-by-step Explanation

$\frac{5 + 3 \sqrt{2} }{5 - 3 \sqrt{2} } = P+Q\sqrt 2 \\ \\ \frac{(5 + 3 \sqrt{2}) }{(5 - 3 \sqrt{2}) } \times \frac{(5 + 3 \sqrt{2}) }{(5 + 3 \sqrt{2} )} = P+Q\sqrt 2 \\ \\ \frac{ {(5 + 3 \sqrt{2} )}^{2} }{ {(5)}^{2} - (3 \sqrt{2})^{2} } = P+Q\sqrt 2 \\ \\ \frac{ {5}^{2} + {(3 \sqrt{2} )^{2} + 2.5.3 \sqrt{2} } }{25 - 9 \times 2} = P+Q\sqrt 2 \\ \\\frac{ 25 + 18 + 30 \sqrt{2} }{25 - 18} = P+Q\sqrt 2 \\ \\\frac{ 43 + 30 \sqrt{2} }{7} = P+Q\sqrt 2 \\ \\ \frac{43}{7} + \frac{30 }{7} \sqrt{2} = P+Q\sqrt 2 \\ \\ equating \: like \: terms \: on \: both \: sides \\ \\ P = \frac{43}{7} \\ \\ \huge \red { \boxed{ P= 6 \frac{1}{7}}} \\ \\ Q = \frac{30}{7} \\ \\ \huge \orange { \boxed{ Q = 4 \frac{2}{7} }}$