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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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On a triangle ABC, a circle with diameter BC is drawn, intersecting A...
Let us draw the diagram according to the available information.
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We can see that triangle BPC and BQC are inscribed inside a semicircle. Hence, we can say that
Therefore, we can say that BQ ⊥ AC and CP ⊥ AB.
In triangle ABC,
Area of triangle = (1/2)*Base*Height = (1/2)*AB*CP = (1/2)*AC*BQ
=> BQ = AB *CP/AC = 30 * 20/25 = 24 cm.
Most Upvoted Answer
On a triangle ABC, a circle with diameter BC is drawn, intersecting A...
To find the length of BQ, we can use the property of circles that states that angles formed by intersecting lines and a circle are equal. Let's break down the solution into steps:
1. Draw the triangle ABC with AB = 30 cm, AC = 25 cm, and CP = 20 cm.
2. Draw the circle with diameter BC. Let's label the center of the circle as O.
3. Let's label the points of intersection of the circle with AB and AC as P and Q, respectively.
4. Since the circle has diameter BC, angle BOC is a right angle.
5. Let's denote the length of BQ as x.
6. Using the property mentioned earlier, we can say that angle BPC is equal to angle BQC.
7. Since angle BOC is a right angle, angles BPC and BQC are both 90 degrees.
8. This means that triangle BPQ is a right-angled triangle with BP as the hypotenuse.
9. Using the Pythagorean theorem, we can find the length of BP:
- BP^2 = BQ^2 + PQ^2
- BP^2 = x^2 + (30 - x)^2 [Since BP = AB - AP = 30 - x]
- BP^2 = x^2 + 900 - 60x + x^2
- BP^2 = 2x^2 - 60x + 900
10. We also know that triangle CPQ is a right-angled triangle with CP as the hypotenuse.
11. Using the Pythagorean theorem, we can find the length of CP:
- CP^2 = CQ^2 + PQ^2
- CP^2 = (25 - x)^2 + x^2 [Since CP = AC - AQ = 25 - x]
- CP^2 = 625 - 50x + x^2 + x^2
- CP^2 = 2x^2 - 50x + 625
12. Since CP = 20 cm, we can equate the expressions for CP^2 from steps 11 and 12:
- 2x^2 - 50x + 625 = 400
- 2x^2 - 50x + 225 = 0
13. Solving this quadratic equation, we find that x = 24 or x = 4.5.
- We discard x = 4.5 since it is not a feasible length for BQ.
14. Therefore, the length of BQ is 24 cm, as given in the correct answer.
1. Draw the triangle ABC with AB = 30 cm, AC = 25 cm, and CP = 20 cm.
2. Draw the circle with diameter BC. Let's label the center of the circle as O.
3. Let's label the points of intersection of the circle with AB and AC as P and Q, respectively.
4. Since the circle has diameter BC, angle BOC is a right angle.
5. Let's denote the length of BQ as x.
6. Using the property mentioned earlier, we can say that angle BPC is equal to angle BQC.
7. Since angle BOC is a right angle, angles BPC and BQC are both 90 degrees.
8. This means that triangle BPQ is a right-angled triangle with BP as the hypotenuse.
9. Using the Pythagorean theorem, we can find the length of BP:
- BP^2 = BQ^2 + PQ^2
- BP^2 = x^2 + (30 - x)^2 [Since BP = AB - AP = 30 - x]
- BP^2 = x^2 + 900 - 60x + x^2
- BP^2 = 2x^2 - 60x + 900
10. We also know that triangle CPQ is a right-angled triangle with CP as the hypotenuse.
11. Using the Pythagorean theorem, we can find the length of CP:
- CP^2 = CQ^2 + PQ^2
- CP^2 = (25 - x)^2 + x^2 [Since CP = AC - AQ = 25 - x]
- CP^2 = 625 - 50x + x^2 + x^2
- CP^2 = 2x^2 - 50x + 625
12. Since CP = 20 cm, we can equate the expressions for CP^2 from steps 11 and 12:
- 2x^2 - 50x + 625 = 400
- 2x^2 - 50x + 225 = 0
13. Solving this quadratic equation, we find that x = 24 or x = 4.5.
- We discard x = 4.5 since it is not a feasible length for BQ.
14. Therefore, the length of BQ is 24 cm, as given in the correct answer.
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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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