Mathematics
Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that:
(i) ∠SAQ = ∠ABC
(ii) SQ = AC.
Triangles
3 Likes
Answer
(i) Given,
ABPQ and ADRS are square.
Each angle of a square = 90°
From figure,
⇒ ∠SAQ + ∠SAD + ∠BAD + ∠BAQ = 360°
⇒ ∠SAQ + 90° + ∠BAD + 90° = 360°
⇒ ∠SAQ + ∠BAD + 180° = 360°
⇒ ∠SAQ + ∠BAD = 360° - 180°
⇒ ∠SAQ = 180° - ∠BAD …(1)
In parallelogram ABCD,
⇒ ∠ABC + ∠BAD = 180° (Sum of adjacent angles of a // gm = 180°)
⇒ ∠ABC = 180° - ∠BAD ….(2)
From eq.(1) and (2), we have:
⇒ ∠SAQ = ∠ABC
Hence, proved that ∠SAQ = ∠ABC.
(ii) In square ADRS,
AS = SR = RD = AD
⇒ AD = BC (Opposite sides of a parallelogram ABCD are equal)
∴ AS = BC
In △SAQ and △CBA,
⇒ ∠SAQ = ∠ABC (Proved above)
⇒ AS = BC (Proved above)
⇒ AQ = AB (Sides of a square ABPQ)
∴ △SAQ ≅ △CBA (By S.A.S axiom)
⇒ SQ = AC (Corresponding parts of congruent triangles are equal)
Hence, proved that SQ = AC.
Answered By
3 Likes
Related Questions
In the given figure, ABCD is a square and △PAB is an equilateral triangle.
(i) Prove that △APD ≅ △BPC.
(ii) Show that ∠DPC = 15°.
In the given figure, in △ABC, ∠B = 90°. If ABPQ and ACRS are squares, prove that:
(i) △ACQ ≅ △ABS
(ii) CQ = BS.
In the given figure, ABCD is a parallelogram, E is the mid-point of BC. DE produced meets AB produced at L. Prove that:
(i) AB = BL
(ii) AL = 2DC
Equilateral triangle ABD and ACE are drawn on the sides AB and AC of △ABC as shown in the figure. Prove that :
(i) ∠DAC = ∠EAB
(ii) DC = BE
