Find the area of triangle PQR in which angle PQR=90 degree, angle PRQ=45 degree and PR=8cm
mary154:
can u just give me the height?
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In triangle PQR, ∠PQR=90° and ∠PRQ=45° and PR=8cm
Then, PQ/PR=sin45° [PQ=perpendicular, PR=hypotenuses]
∴, PQ=PRsin45°=8×1/√2=8/√2=4√2 cm
Similarly, QR/PR=cos45° [QR=base]
∴, QR=PRcos45°=8×1/√2=8/√2=4√2 cm
Area of the triangle
=1/2×base×height
=1/2×4√2×4√2
=(16×2)/2=16 cm²
Then, PQ/PR=sin45° [PQ=perpendicular, PR=hypotenuses]
∴, PQ=PRsin45°=8×1/√2=8/√2=4√2 cm
Similarly, QR/PR=cos45° [QR=base]
∴, QR=PRcos45°=8×1/√2=8/√2=4√2 cm
Area of the triangle
=1/2×base×height
=1/2×4√2×4√2
=(16×2)/2=16 cm²
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