M angles = (x+4) and m angle Q = (5x-4)
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It is given that the measures of the angles of triangle QRS are m∠Q=2x+4, m∠R=4x−12 and m∠S=3x+8 and also we know that the sum of the measures of the angles of a triangle is 180.
Therefore, 2x+4+4x−12+3x+8=180
⇒9x=180
⇒x=20
Substitute the value of x in the measures of the angles that is:
m∠Q=2(20)+4=44
⇒m∠R=4(20)−12=68
⇒m∠S=3(20)+8=68
Therefore, △QRS is isosceles and QR=QS which means
y+9=3y−13
⇒3y−y=13+9
⇒2y=22
⇒y=11
Now consider the sum of the sides that is:
QR+RS+QS=y+9+2y−7+3y−13=6y−11
Hence, the perimeter of triangle QRS is:
6y−11=6(11)−11=66−11=55
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