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If A,B,C and D, the measures of the four angles of a quadrilateral, are in the ratio 3:5:4:6, then the value of 3A+2B is:
340°
360°
380°
330°
380°
Sum of ratios is 3+5+4+6=18. Each part is 360/18=20°. Thus, A=3(20)=60°, B=5(20)=100°. Then 3A+2B=3(60)+2(100)=180+200=380°.
The four angles A, B, C, and D of a quadrilateral are in the ratio 3 : 5 : 4 : 6.
A+B+C+D=360°
Sum of ratios is 3+5+4+6=18. Each part is 360/18=20°. Thus, A=3(20)=60°, B=5(20)=100°. Then 3A+2B=3(60)+2(100)=180+200=380°.
Many students forget that the sum of the interior angles of a quadrilateral is 360° and mistakenly use 180°.
Define angles in terms of x
Since the ratio of angles A, B, C, and D is 3: 5: 4: 6, let the angles be 3x,5x,4x,6x.
A=3x,B=5x,C=4x,D=6x
Calculate sum of angles
The sum of interior angles of a quadrilateral is 360°. So, 3x+5x+4x+6x=360°.
18x=360°⟹x=20°
Find individual angles
Substitute x=20° back into the expressions for A and B.
A=3×20°=60°,B=5×20°=100°
Evaluate the expression
Calculate the value of 3A+2B using the derived angles.
3(60°)+2(100°)=180°+200°=380°
C is correct because the angles are 60° and 100° respectively, leading to 3(60°)+2(100°)=380°.
This ratio method is universal for geometry problems involving interior angles of any polygon with sum (n−2)×180°.
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