Question

In the given triangle PQR, QPR = 90 0 , PQ= 24 cm Fig 1 and QR = 26 cm and in ∆PKR , PKR = 90 0 and KR = 8 cm, find PK.

Answer 1

in ΔPQR,PQ=24cm
                QR=26cm
                ∠QPR=90°
By using pythagoras theorem,
    ⇒QR²=QP²+PR²
    ⇒26²=24²+PR²
   ⇒676=576+PR²
   ⇒PR²=100
   ⇒PR=10cm---------------------------eq1
in ΔPKR,KR=8cm
               ∠PKR=90°
By using pythagoras theorem,
⇒PR²=PK²+KR²
⇒10²=PK²+8²
⇒100=PK²+64
⇒PK²=36
⇒PK=6cm

Answer 2

Answer:

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