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Parallelogram Perimeter and Area

XAT 2010 · Quantitative Ability & Data Interpretation · Medium · Quantitative Ability

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ABCD is a parallelogram with ∠ABC=60∘\angle ABC = 60^\circ. If the longer diagonal is of length 7 cm and the area of the parallelogram ABCD is 153215\frac{\sqrt{3}}{2} sq.cm, then the perimeter of the parallelogram (in cm) is

  1. A.

    15

  2. B.

    15315\sqrt{3}

  3. C.

    16

  4. D.

    16316\sqrt{3}

  5. E.

    None of the above

Answer

C

Explanation

Area = absin⁡60∘=ab32=1532  ⟹  ab=15a b \sin 60^\circ = a b \frac{\sqrt{3}}{2} = 15 \frac{\sqrt{3}}{2} \implies a b = 15. Adjacent angle ∠DAB=180∘−60∘=120∘\angle DAB = 180^\circ - 60^\circ = 120^\circ. Longer diagonal BD=7 cmBD = 7\text{ cm}. By law of cosines in △ABD\triangle ABD: BD2=a2+b2−2abcos⁡120∘  ⟹  49=a2+b2−2(15)(−1/2)=a2+b2+15BD^2 = a^2 + b^2 - 2ab \cos 120^\circ \implies 49 = a^2 + b^2 - 2(15)(-1/2) = a^2 + b^2 + 15 a2+b2=34a^2 + b^2 = 34 (a+b)2=a2+b2+2ab=34+30=64  ⟹  a+b=8(a + b)^2 = a^2 + b^2 + 2ab = 34 + 30 = 64 \implies a + b = 8 Perimeter =2(a+b)=2×8=16 cm= 2(a + b) = 2 \times 8 = 16\text{ cm}.

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