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Ask QuestionPosted by Shivang Sharma 2 years, 1 month ago
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Posted by Charu Singh 2 years ago
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Preeti Dabral 2 years ago
Given: Let ABCD is a quadrilateral.
Let its diagonal AC and BD bisect each other at right angle at point O.
∴OA = OC, OB = OD
And ∠AOB = ∠BOC = ∠COD = ∠AOD = 90∘
To prove: ABCD is a rhombus.
Proof: In △AOD and △ BOC,
OA = OC [Given]
∠AOD = ∠BOC [Given]
OB = OD [Given]
∴ △AOD ≅ △COB [By SAS congruency]
⇒ AD = CB [By C.P.C.T.]……….(i)
Again, In △AOB and △COD,
OA = OC [Given]
∠ AOB = ∠COD [Given]
OB = OD [Given]
∴ △AOB ≅ △COD [By SAS congruency]
⇒ AB = CD [By C.P.C.T.]……….(ii)
Now In △AOB and △BOC,
OA = OC [Given]
∠ AOB = ∠BOC [Given]
OB = OB [Common]
∴ △AOB ≅ △COB [By SAS congruency]
⇒ AB = BC [By C.P.C.T.]……….(iii)
From eq. (i), (ii) and (iii),
AD = BC = CD = AB
And the diagonals of quadrilateral ABCD bisect each other at right angle.
Therefore, ABCD is a rhombus.
Posted by Bhavani S. K 2 years, 1 month ago
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Sehajpreet Kaur 2 years, 1 month ago
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Akshadeep Meshram 2 years, 1 month ago
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Posted by Lavesh Sahu 2 years, 1 month ago
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Shivang Sharma 2 years, 1 month ago
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Preeti Dabral 2 years, 1 month ago
Angle HYD = 120°
Angle BTG = 60°
Angle YXC = 70°
Angle SXF = 70°
Angle ASE = 70°
Posted by Garvit Chawla 2 years, 1 month ago
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Shivang Sharma 2 years, 1 month ago
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Posted by Anushka Awasthi 2 years, 1 month ago
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Preeti Dabral 2 years, 1 month ago
Let the radius of the sphere be r cm.
Surface area = 154 cm2
⇒ 4πr2 = 154
⇒ 4×227×r2=154
⇒ r2=154×74×22=494
⇒ r = √494=72
⇒ r = 3.5 cm
∴ the radius of the sphere is 3.5 cm.
Siddhi Gupta 2 years, 1 month ago
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Posted by Nithin S 2 years, 1 month ago
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Poojith K Gowda 2 years, 1 month ago
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Dhairya Sharma 2 years, 1 month ago
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Pooja Arya 2 years, 1 month ago
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