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Question 4 : On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is [TITA]
Let ABC be the triangle on which a circle of diameter BC is drawn, intersecting AB and AC at points P and Q respectively.The lengths of AB ,AC and CP are 30cm , 25cm and 20 cm respectively we have to find the length of BQ in cm.
Key thing here is ,this is semicircle so this angle P and Q should be 90° and now we are looking to do Pythagoras theorem
So think about triangle PAC, by Pythagoras theorem
AC2 = AP2 + PC2 we can find that AP = 15
Since AP = 15 we can find BP by
AB = AP + PB
30 = 15 + PB
PB = 15
Now we can look at the triangle BPC, by Pythagoras theorem we can find that BC = 25
Now we have to find BQ .So we can take the area of the triangle formula which is equal to \\frac{1}{2}) × base × height
area of the triangle formula = \\frac{1}{2}) × base × height
\\frac{1}{2}) × AB × PC = \\frac{1}{2}) × AC × BQ
\\frac{1}{2}) × 30 × 20 = \\frac{1}{2}) × 25 × BQ
BQ = 24 cm
The question is "On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is [TITA] "
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