Please solve q 12 ?

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1 Answer
CW
May 13, 2018

anglePQR=112.5^@∠PQR=112.5∘

Explanation:

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Sum of interior angles of a pentagon =3xx180=540^@=3×180=540∘,
given that angleA=angleB=angleD=90^@, and angleC=angleE∠A=∠B=∠D=90∘,and∠C=∠E,
=> angleC=angleE=(540-3xx90)/2=135^@⇒∠C=∠E=540−3×902=135∘,
AsangleABC=90^@, => angleKBC=180-90=90^@∠ABC=90∘,⇒∠KBC=180−90=90∘,
given PQPQ bisects angleKBC, => angleQBC=(angleKBC)/2=90/2=45^@∠KBC,⇒∠QBC=∠KBC2=902=45∘,
as angleBCD=135^@,=> angleLCD=180-135=45^@∠BCD=135∘,⇒∠LCD=180−135=45∘
given QRQR bisects angleLCD∠LCD,
=> angleRCD=angleRCL=(angleLCD)/2=45/2=22.5^@⇒∠RCD=∠RCL=∠LCD2=452=22.5∘,
In DeltaBQC, angleBQC=180-45-22.5=112.5^@

Hence, anglePQR=angleBQC=112.5^@

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