Question

10. In the given figure AB parallel to CD. Find the value of x
A
+
B
+
140
C
35
+
D
C​

Answer 1

Step-by-step explanation:

Draw a line EF perpendicular to AB and CD passing through M

In △BME

∠BME+∠MBE+∠BEM=180

∘

(Angle sum proprerty)

∠BME+35

∘

+90

∘

=180

∘

∠BME=180

∘

−125

∘

∠BME=45

∘

In △DMF

∠DMF+∠MDF+∠MFD=180

∘

(Angle sum property )

∠DMF+75

∘

+90

∘

=180

∘

∠DMF=180

∘

−165

∘

∠DMF=15

∘

x=∠DMF+∠BME+∠EMF

x=15

∘

+45

∘

+180

∘

x=240

Answer 2

$\LARGE{ \underline{ \orange{ \sf{Required \: answer:}}}}$

Draw line EF perpendicular to AB and CD passing through M.

In ∆BME

angle BME + angle MBE + angle BEM = 180°

(Angle sum property )

BME + 35° + 90° = 180°

BME= 180°- 125°

BME = 45°

In ∆DMF

angle DMF +angle MDF + angle MFD =180°

(Angle sum property)

DMF + 75° +90°= 180°

DMF = 180 ° - 165°

DMF = 15°

x = DMF + BME + EMF

x = 15 degree + 45 degree + 180 degree

x=240°

hope it helps........✨