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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
Correct answer is '24'. Can you explain this answer?
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Verified Answer
On a triangle ABC, a circle with diameter BC is drawn, intersecting AB...
Let us draw the diagram according to the available information.
We can see that triangle BPC and BQC are inscribed inside a semicircle. Hence, we can say that
In triangle ABC,
Area of triangle = (1/2)*Base*Height = (1/2)*AB*CP = (1/2)*AC*BQ
We can see that triangle BPC and BQC are inscribed inside a semicircle. Hence, we can say that
In triangle ABC,
Area of triangle = (1/2)*Base*Height = (1/2)*AB*CP = (1/2)*AC*BQ
Most Upvoted Answer
On a triangle ABC, a circle with diameter BC is drawn, intersecting AB...
Problem: Given 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:
Solution:
Step 1: Draw the diagram
Step 2: Find the radius of the circle
The diameter of the circle is BC. Therefore, the radius of the circle is BC/2. Using the Pythagorean theorem, we can find BC as follows:
BC^2 = AB^2 + AC^2
BC^2 = 30^2 + 25^2 = 1625
BC = sqrt(1625)
Radius = BC/2 = sqrt(1625)/2
Step 3: Find the length of AP and AQ
The line AP is a tangent to the circle. Therefore, the radius is perpendicular to AP. Using the Pythagorean theorem, we can find AP as follows:
AP^2 = AB^2 - (radius)^2
AP^2 = 30^2 - (sqrt(1625)/2)^2 = 675/4
AP = sqrt(675)/2 = 5sqrt(3)
Similarly, using the Pythagorean theorem, we can find AQ as follows:
AQ^2 = AC^2 - (radius)^2
AQ^2 = 25^2 - (sqrt(1625)/2)^2 = 625/4
AQ = sqrt(625)/2 = 5
Step 4: Find the length of PQ
PQ is the difference between AP and AQ:
PQ = AP - AQ = 5sqrt(3) - 5 = 5(sqrt(3) - 1)
Step 5: Find the length of BQ
BQ = BP + PQ
BP = CP - CB = 20 - sqrt(1625)/2
BQ = 20 - sqrt(1625)/2 + 5(sqrt(3) - 1)
BQ = 5sqrt(3) - sqrt(1625)/2 + 15
BQ = 24 (approx.)
Solution:
Step 1: Draw the diagram
Step 2: Find the radius of the circle
The diameter of the circle is BC. Therefore, the radius of the circle is BC/2. Using the Pythagorean theorem, we can find BC as follows:
BC^2 = AB^2 + AC^2
BC^2 = 30^2 + 25^2 = 1625
BC = sqrt(1625)
Radius = BC/2 = sqrt(1625)/2
Step 3: Find the length of AP and AQ
The line AP is a tangent to the circle. Therefore, the radius is perpendicular to AP. Using the Pythagorean theorem, we can find AP as follows:
AP^2 = AB^2 - (radius)^2
AP^2 = 30^2 - (sqrt(1625)/2)^2 = 675/4
AP = sqrt(675)/2 = 5sqrt(3)
Similarly, using the Pythagorean theorem, we can find AQ as follows:
AQ^2 = AC^2 - (radius)^2
AQ^2 = 25^2 - (sqrt(1625)/2)^2 = 625/4
AQ = sqrt(625)/2 = 5
Step 4: Find the length of PQ
PQ is the difference between AP and AQ:
PQ = AP - AQ = 5sqrt(3) - 5 = 5(sqrt(3) - 1)
Step 5: Find the length of BQ
BQ = BP + PQ
BP = CP - CB = 20 - sqrt(1625)/2
BQ = 20 - sqrt(1625)/2 + 5(sqrt(3) - 1)
BQ = 5sqrt(3) - sqrt(1625)/2 + 15
BQ = 24 (approx.)
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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, isCorrect answer is '24'. Can you explain this answer?
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