plz solve this qustion
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4AP = 24cm. 3PB = 24cm
AP = 6cm. PB = 8cm
In ∆ APB
Angle P = 90° [AB is the diameter]
AP^2 + PB^2 = AB^2
6^2 + 8^2 = AB^2
36 + 64 = AB^2
√100 = AB
AB = 10cm.
Radius = 1/2 Diameter.
= 5 cm.
Hope it helps................
AP = 6cm. PB = 8cm
In ∆ APB
Angle P = 90° [AB is the diameter]
AP^2 + PB^2 = AB^2
6^2 + 8^2 = AB^2
36 + 64 = AB^2
√100 = AB
AB = 10cm.
Radius = 1/2 Diameter.
= 5 cm.
Hope it helps................
Answered by
1
the triangle is inside the circle and one of its side is the diameter of the circle. So according to the property of angles inside a semicircle angle APB= 90 degrees.
applying pythagoras theorem to triAPB.
AP^2+PB^2=AB^2
6^2+8^2=(2OA)^2 (OA=OB radius of the circle)
36+64=4OA^2
OA^2= 100/4
OA=5
applying pythagoras theorem to triAPB.
AP^2+PB^2=AB^2
6^2+8^2=(2OA)^2 (OA=OB radius of the circle)
36+64=4OA^2
OA^2= 100/4
OA=5
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