Question

Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that:
(i) angle SAQ = angle ABC
(ii) SQ=AC​

Answer 1

Answer:

∠SAQ  = ∠ABC

SQ = AC

Step-by-step explanation:

Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that:

(i) angle SAQ = angle ABC

(ii) SQ=AC​

At Point A

∠BAQ  + ∠BAD + ∠DAS + ∠SAQ = 360°

∠BAQ = 90° ( angle of square)

∠DAS = 90° (Angle of Square)

=> 90 + ∠BAD  + 90 + ∠SAQ = 360°

=> ∠SAQ  + ∠BAD = 180°

in Parallelogram

∠BAD + ∠ABC = 180°  

Equating both

∠SAQ  + ∠BAD = ∠BAD + ∠ABC

=> ∠SAQ  = ∠ABC

Now SQ² = AQ² + AS² - 2AQ.ASCos∠SAQ

AQ = AB  (sides of square)  AS = AD (sides of square)

AD = BC (opposite side of parallelogram) => AS = BC

∠SAQ  = ∠ABC  , Using all these

=> SQ² = AB² + BC² - 2AB* BC Cos ∠ABC

=> SQ² = AC²

=> SQ = AC

Answer 2

Step-by-step explanation:

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