Construct APQR, in which QR = 7.2 cm PQ-PR=3.2 cm and /PQR = 45'
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Answer: just simple changes but this is solution
Step-by-step explanation:
Step 1: Draw line segment QR=3.2cm.
Step 2: Draw RX¯¯¯¯¯¯, such that ∠QRX=(90∘+∠R−∠Q2)=90∘+30∘=120∘.
Step 3: Taking Q as the center and a radius equal to 5.9cm (PQ + PR), draw an arc intersecting RX¯¯¯¯¯¯ at S.
Step 4: Draw RY, such that
∠QRY=∠R−∠Q2=30∘, intersecting QS at T.
Step 6: Draw the perpendicular bisector of RT¯¯¯¯¯ intersecting QS¯¯¯¯¯ at P.
Step 7: Join P and R. PQR is the required triangle.
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