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Question
In an isosceles triangle PQR, the length of equal sides PQ and PR is 13 cm and base QR is 10 cm. Find the length of perpendicular bisector drawn from vertex P to side QR.
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Solution
Given: In an isosceles triangle PQR, PQ = PR = 13 cm and base QR = 10 cm.
Since, PS is the perpendicular bisector,
∴ QS = SR = `(QR)/2 = 10/2` = 5
Now, using Pythagoras' theorem, in a right-angled triangle PSQ,
PS2 + SQ2 = PQ2
Substituting the values of SQ and PQ in equation (i),
PS2 + 52 = 132
⇒ PS2 = 132 – 52
⇒ PS2 = 169 – 25
⇒ PS2 = 144
∴ PS = `sqrt(144)` = 12
As a result, the length of the perpendicular bisector PS is 12 cm.
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