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Post a LessonAnswered on 01 Jul Learn CBSE/Class 9/Mathematics/Geometry/Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 03 Jul Learn CBSE/Class 9/Mathematics/Geometry/Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Given: In square ABCD, AK = BL = CM = DN.
To prove: KLMN is a square.
In square ABCD,
AB = BC = CD = DA
And, AK = BL = CM = DN
(All sides of a square are equal.) (Given)
So, AB - AK = BC - BL = CD - CM = DA - DN
⇒ KB = CL = DM = AN.......... (1)
In △NAKand△KBL
∠NAK=∠KBL=900 (Each angle of a square is a right angle.)
AK=BL (Given)
AN=KB [From (1)]
So, by SAS congruence criteria,
\(\triangle NAK ≅ \triangle KBL \)
⇒NK=KL (Cpctc ...(2)
Similarly,
\(\triangle MDN≅ \triangle NAK \)
\(\triangle DNM≅ \triangle CML \)
\(\triangle MCL≅ \triangle LBK \)
\(\rightarrow MN = NK and \angle DNM=\angle KNA \) (Cpctc )… 3)
MN = JM and ∠DNM=∠CML (Cpctc )… 4)
ML = LK and ∠CML=∠BLK (Cpctc )… (5)
From (2), (3), (4) and (5), we get
NK = KL = MN = ML…….....(6)
And, ∠DNM=∠AKN=∠KLB=∠LMC
Now,
In △NAK
∠NAK=900
Let ∠AKN=x0
So, ∠DNK=900+x0 (Exterior angles equals sum of interior opposite angles.)
⇒∠DNM+∠MNK=900+x0
⇒x0+∠MNK=900+x0
⇒ ∠MNK=900
Similarly,
∠NKL=∠KLM=∠LMN=900 ...(7)
Using (6) and (7), we get
All sides of quadrilateral KLMN are equal and all angles are \ ( 90^0 \)
So, KLMN is a square.
Answered on 16 Jul Learn CBSE/Class 9/Mathematics/Geometry/Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 16 Jul Learn CBSE/Class 9/Mathematics/Geometry/Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Answered on 16 Jul Learn CBSE/Class 9/Mathematics/Geometry/Quadrilaterals
Deepika Agrawal
"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."
Let ABCD be a parallelogram.
∴∠A=∠C and ∠B=∠D (Opposite angles)
Let ∠A=x0 and ∠B=4x50
Now, ∠A+∠B=1800
(Adjacent angles are supplementary)
⇒x+4x50=1800
⇒9x5=1800
⇒x=20×5
⇒x=1000
Now, ∠A=1000 and ∠B=45×1000=800
Hence, ∠A=∠C=1000
∠B=∠D=800
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