pqr is a triangle right angled at p with PQ= 6 centimetre and PR = 8 centimetre.
find the area of triangle pqr. also, find the length of perpendicular PS on QR.
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Area of ΔPQR=1/2xbasexheight
=1/2x8x6
=24cm
By Pythagoras theorem
QR²=PQ²+PR²
QR=√PQ²+PR²
=√6²+8²
=√100
=10cm
∴QS=5cm
Now,apply Pythagoras theorem in ΔPRS
PS²=PR²-MSR²
PS=√8²-5²
=√39
Hope its helpful.....
=1/2x8x6
=24cm
By Pythagoras theorem
QR²=PQ²+PR²
QR=√PQ²+PR²
=√6²+8²
=√100
=10cm
∴QS=5cm
Now,apply Pythagoras theorem in ΔPRS
PS²=PR²-MSR²
PS=√8²-5²
=√39
Hope its helpful.....
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