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Ask QuestionPosted by Shivang Sharma 1 year, 9 months ago
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Posted by Charu Singh 1 year, 8 months ago
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Preeti Dabral 1 year, 8 months ago
Given: Let ABCD is a quadrilateral.
Let its diagonal AC and BD bisect each other at right angle at point O.
{tex}\therefore {/tex}OA = OC, OB = OD
And {tex}\angle {/tex}AOB = {tex}\angle {/tex}BOC = {tex}\angle {/tex}COD = {tex}\angle {/tex}AOD = {tex}{90^ \circ }{/tex}
To prove: ABCD is a rhombus.
Proof: In {tex}\triangle{/tex}AOD and {tex}\triangle{/tex} BOC,
OA = OC [Given]
{tex}\angle {/tex}AOD = {tex}\angle {/tex}BOC [Given]
OB = OD [Given]
{tex}\therefore {/tex} {tex}\triangle{/tex}AOD {tex} \cong{/tex} {tex}\triangle{/tex}COB [By SAS congruency]
{tex} \Rightarrow {/tex} AD = CB [By C.P.C.T.]……….(i)
Again, In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}COD,
OA = OC [Given]
{tex}\angle {/tex} AOB = {tex}\angle {/tex}COD [Given]
OB = OD [Given]
{tex}\therefore {/tex} {tex}\triangle{/tex}AOB {tex} \cong {/tex} {tex}\triangle{/tex}COD [By SAS congruency]
{tex} \Rightarrow {/tex} AB = CD [By C.P.C.T.]……….(ii)
Now In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}BOC,
OA = OC [Given]
{tex}\angle {/tex} AOB = {tex}\angle {/tex}BOC [Given]
OB = OB [Common]
{tex}\therefore {/tex} {tex}\triangle{/tex}AOB {tex} \cong{/tex} {tex}\triangle{/tex}COB [By SAS congruency]
{tex} \Rightarrow {/tex} 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 1 year, 9 months ago
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Sehajpreet Kaur 1 year, 9 months ago
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Akshadeep Meshram 1 year, 8 months ago
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Posted by Lavesh Sahu 1 year, 9 months ago
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Shivang Sharma 1 year, 8 months ago
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Posted by Anjali Mattu 1 year, 9 months ago
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Preeti Dabral 1 year, 8 months ago
Angle HYD = 120°
Angle BTG = 60°
Angle YXC = 70°
Angle SXF = 70°
Angle ASE = 70°
Posted by Garvit Chawla 1 year, 9 months ago
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Shivang Sharma 1 year, 8 months ago
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Posted by Anushka Awasthi 1 year, 9 months ago
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Preeti Dabral 1 year, 9 months ago
Let the radius of the sphere be r cm.
Surface area = 154 cm2
⇒ {tex}4\pi r^2{/tex} = 154
⇒ {tex}4\times{22\over7}\times r^2=154{/tex}
⇒ {tex}r^2={{154\times7}\over4\times22}={49\over4}{/tex}
⇒ r = {tex}{\sqrt{49\over4}}={7\over2}{/tex}
⇒ r = 3.5 cm
∴ the radius of the sphere is 3.5 cm.
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Dhairya Sharma 1 year, 9 months ago
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Pooja Arya 1 year, 8 months ago
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