In this figure ABCD is a quadrilateral whose vertices are on a
semicircle such that AB=BC=CD=10cm and AD is a diameter of circle having centre at
O. Find the perimeter of the quadrilateral ABCD.
A. 30
B. 100
C. 50
c. 70
Answers
Answer:
your answer is 50. for sloution please see the attached images..
Step-by-step explanation:
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The perimeter of the quadrilateral ABCD is 70.
the answer is (C) 70.
First, observe that triangle ABD is a right-angled triangle, with hypotenuse AD being the diameter of the circle.
Thus, by Pythagoras' Theorem, we have:
AB² + BD² = AD²
Substituting AB = BC = CD = 10, we get:
10² + BD² = (2r)², where r is the radius of the circle
Simplifying, we get:
BD² = 4r² - 100
Next, observe that quadrilateral ABCD can be divided into two congruent right-angled triangles ABD and BCD, with a rectangle ABCD in between them.
Thus, the perimeter of ABCD is:
AB + BD + BC + CD + AD
= 10 + √(4r² - 100) + 10 + 10 + 2r
= √(4r² - 100) + 2r + 30
To find r, we use the fact that the perpendicular bisectors of AB and CD intersect at the center of the circle O.
Thus, O is the midpoint of both AB and CD. Since AB = 10, we have OA = OB = 5.
Similarly, CD = 10, so we have OC = OD = 5.
Thus, triangle OAD is an isosceles right-angled triangle, with hypotenuse AD being 2r.
Thus, by Pythagoras' Theorem, we have:
OA² + AD² = OD²
Substituting OA = OD = 5 and AD = 2r, we get:
5² + (2r)² = 5²
Simplifying, we get:
4r² = 75
Thus, r = √(75/4) = (5/2)√3
Substituting this value of r into the expression we derived for the perimeter of ABCD, we get:
Perimeter of ABCD = √(4r² - 100) + 2r + 30
= √(4(5/2)²(3) - 100) + 5√3 + 30
= √(75) + 5√3 + 30
= 5√3 + 45
Thus, the answer is (C) 70.
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