Question

ABCD is a parallelogram with sides AB = $12$ cm BC = $10$ cm and diagonal AC = $16$ cm The area of the parallelogram and the distance between the shorter sides are respectively

• A. $120.5$ $c{m}^2$ ; $12.05$ cm
• B. $119.8$ $c{m}^2$ ; $11.98$ cm
• C. $118.71$ $c{m}^2$ ; $11.87$ cm
• D. $117.9$ $c{m}^2$ ; $11.79$ cm

Answer 1

The correct option is B $119.8$ $c{m}^2$ ; $11.98$ cm

Area of Parallelogram ABCD$=2\times ar△ABC$

$ar△ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$s=\frac{a+b+c}{2}=\frac{16+10+12}{2}=19cm$

$ar△ABC=\sqrt{19(19-16)\times (19-10)(19-12)}$

$=\sqrt{169\times 3\times 9\times 7}$

$=59.9c{m}^2$

$\therefore$ area of parallelogram ABCD$=2\times 59.9=119.8c{m}^2$

Now, area of parallelogram ABCD=base X height

$119.8=10\times h\Rightarrow h=11.98cm$

$\therefore$ Distance between the shortest sides is $11.98cm$