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PQRST is a pentagon in which all the interior angles are unequal. A circle of radius ‘r’ is inscribed in each of the vertices. Find the area of portion of circles falling inside the pentagon.
- a)πr2
- b)1.5πr2
- c)2πr2
- d)1.25πr2
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
PQRST is a pentagon in which all the interior angles are unequal. A ci...
Since neither angles nor sides are given in the question, immediately the sum of angles of pentagon should come in mind. To use it,
We know the area of the sectors of a circle is given as,
We know the area of the sectors of a circle is given as,
Note => The above concept is applicable for a polygon of n sides.
Choice (B) is therefore, the correct answer.
Correct Answer: 1.5πr2
Choice (B) is therefore, the correct answer.
Correct Answer: 1.5πr2
Most Upvoted Answer
PQRST is a pentagon in which all the interior angles are unequal. A ci...
R is inscribed in pentagon PQRST. The circle is tangent to sides PQ, QR, RS, ST, and TP.
Let O be the center of the inscribed circle. Since the circle is tangent to sides PQ, QR, RS, ST, and TP, the line segments joining O to the midpoints of sides PQ, QR, RS, ST, and TP are radii of the circle. Let A, B, C, D, and E be the midpoints of sides PQ, QR, RS, ST, and TP, respectively.
Since PQRST is a pentagon in which all the interior angles are unequal, the sides of the pentagon are also unequal. Therefore, the line segments joining O to the midpoints of the sides are also unequal in length. Let r1, r2, r3, r4, and r5 be the lengths of the radii of the circle corresponding to sides PQ, QR, RS, ST, and TP, respectively.
Since the line segments joining O to the midpoints of the sides are radii of the circle, we have OA = OB = OC = OD = OE = r.
Consider triangle OAB. Since OA = OB = r, triangle OAB is isosceles. Therefore, angle OAB = angle OBA. Similarly, in triangles OBC, OCD, ODE, and OEA, we have angle OBC = angle OCB, angle OCD = angle ODC, angle ODE = angle OED, and angle OEA = angle OAE, respectively.
Since the interior angles of pentagon PQRST are unequal, the exterior angles at vertices P, Q, R, S, and T are also unequal. Let these exterior angles be a, b, c, d, and e, respectively.
Since the sum of the exterior angles of any polygon is 360 degrees, we have a + b + c + d + e = 360 degrees.
Since angle OAB = angle OBA, angle OBC = angle OCB, angle OCD = angle ODC, angle ODE = angle OED, and angle OEA = angle OAE, we have a + b = angle OAB, b + c = angle OBC, c + d = angle OCD, d + e = angle ODE, and e + a = angle OEA.
Since the angles at vertices P, Q, R, S, and T are exterior angles, we have angle OAB = angle P, angle OBC = angle Q, angle OCD = angle R, angle ODE = angle S, and angle OEA = angle T.
Therefore, a + b = angle P, b + c = angle Q, c + d = angle R, d + e = angle S, and e + a = angle T.
Substituting these equations into the equation a + b + c + d + e = 360 degrees, we have (angle P) + (angle Q) + (angle R) + (angle S) + (angle T) = 360 degrees.
Therefore, the sum of the exterior angles of pentagon PQRST is 360 degrees.
Let O be the center of the inscribed circle. Since the circle is tangent to sides PQ, QR, RS, ST, and TP, the line segments joining O to the midpoints of sides PQ, QR, RS, ST, and TP are radii of the circle. Let A, B, C, D, and E be the midpoints of sides PQ, QR, RS, ST, and TP, respectively.
Since PQRST is a pentagon in which all the interior angles are unequal, the sides of the pentagon are also unequal. Therefore, the line segments joining O to the midpoints of the sides are also unequal in length. Let r1, r2, r3, r4, and r5 be the lengths of the radii of the circle corresponding to sides PQ, QR, RS, ST, and TP, respectively.
Since the line segments joining O to the midpoints of the sides are radii of the circle, we have OA = OB = OC = OD = OE = r.
Consider triangle OAB. Since OA = OB = r, triangle OAB is isosceles. Therefore, angle OAB = angle OBA. Similarly, in triangles OBC, OCD, ODE, and OEA, we have angle OBC = angle OCB, angle OCD = angle ODC, angle ODE = angle OED, and angle OEA = angle OAE, respectively.
Since the interior angles of pentagon PQRST are unequal, the exterior angles at vertices P, Q, R, S, and T are also unequal. Let these exterior angles be a, b, c, d, and e, respectively.
Since the sum of the exterior angles of any polygon is 360 degrees, we have a + b + c + d + e = 360 degrees.
Since angle OAB = angle OBA, angle OBC = angle OCB, angle OCD = angle ODC, angle ODE = angle OED, and angle OEA = angle OAE, we have a + b = angle OAB, b + c = angle OBC, c + d = angle OCD, d + e = angle ODE, and e + a = angle OEA.
Since the angles at vertices P, Q, R, S, and T are exterior angles, we have angle OAB = angle P, angle OBC = angle Q, angle OCD = angle R, angle ODE = angle S, and angle OEA = angle T.
Therefore, a + b = angle P, b + c = angle Q, c + d = angle R, d + e = angle S, and e + a = angle T.
Substituting these equations into the equation a + b + c + d + e = 360 degrees, we have (angle P) + (angle Q) + (angle R) + (angle S) + (angle T) = 360 degrees.
Therefore, the sum of the exterior angles of pentagon PQRST is 360 degrees.
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Community Answer
PQRST is a pentagon in which all the interior angles are unequal. A ci...
Since neither angles nor sides are given in the question, immediately the sum of angles of pentagon should come in mind. To use it,
We know the area of the sectors of a circle is given as:
We know the area of the sectors of a circle is given as:
Note: The above concept is applicable for a polygon of n sides.
Choice (B) is therefore, the correct answer.
Correct Answer: 1.5πr2
Choice (B) is therefore, the correct answer.
Correct Answer: 1.5πr2
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