Question

19) In the given figure AB//CD
Find the value of x.
​

Answer 1

$\underline{\textsf{\textbf{\purple{ \mapsto Given:}}}}$

• In the figure AB || CD .
• Measure of ∠FEP is 85°.
• Measure of ∠CHQ is 115°.

$\underline{\textsf{\textbf{\purple{ \mapsto To\:Find:}}}}$

• The value of x .

$\underline{\textsf{\textbf{\purple{ \mapsto Concept\:Used:}}}}$

• We know that when parallel lines are intersected by a transversal then the corresponding angles so formed are equal .
• Also we know that the measure of a straight line is 180° .

$\underline{\textsf{\textbf{\purple{ \mapsto Answer:}}}}$

$\underline{\sf{\red{\hookrightarrow Figure :-}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,8)\put(0,5){\line(1,0){6}}\put(6,5){\line(-1,-1){4.5}}\put(2,6){\line(0,-1){5.5}}\put(5,6){\line(0,-1){5.5}}\put(2,0.2){ \sf A }\put(1.2,0.2){ \sf R }\put(5,0.2){ \sf C }\put(2,6.2){ \sf B }\put(5,6.2){ \sf D }\put(6.2,5){ \sf Q }\put(1.7,5.2){ \sf E }\put(1.7,1){ \sf F }\put(4.7,4.2){ \sf H }\put(4.6,5.2){ \sf G }</p><p>\qbezier(2,4.5)(1.5,4.5)(1.6,5)\put(1,4.5){ \sf 85^{\circ} }\qbezier(5,3.8)(5.5,4)(5.4,4.4)\put(5.5,3.6){ \sf 115^{\circ} }\put(2,5){\vector(1,0){1.5}}\end{picture}$

$\rule{200}3$

Now from the figure we can see that ∠FEP and ∠HGE are corresponding angles ( since AB || CD ) . Hence ∠FEP = ∠HGE = 85° .

Now , again we can see that ∠HGE and ∠HGQ are angles of linear pair . So ,

➳ ∠HGE + ∠HGQ = 180°.

➳ 85° + ∠HGQ = 180° .

➳ ∠HGQ = 180° - 85°.

➳ ∠ HGQ = 95° .

$\rule{200}3$

Similarly ∠CHQ and ∠QHG are angles of linear pair . So their sum will be 180° ,

➳ ∠CHQ + ∠QHG = 180° .

➳ 115° + ∠QHG = 180° .

➳ ∠QHG = 180° - 115°.

➳ ∠QHG = 65°.

$\rule{200}3$

$\underline{\sf{\red{\hookrightarrow Now\:in\:\triangle QHG :-}}}$

We know that the angle sum property of a triangle is 180° .

➳ ∠QHD + ∠QGG + ∠HQD = 180°.

➳ ∠HQD + 65° + 85° = 180°

➳ ∠HQD + 150° = 180°

➳ ∠HQD = 180° - 150°

➳ x = 30° .

$\boxed{\green{\bf\pink{\dag}\:Hence\:value\:of\:x\:is\:30^{\circ}}}$