Question

One of the angles of a parallelogram is ${150}^{\circ }$. Altitudes are drawn from the vertex of this angle. If these altitudes measure $6cm$ and $8cm$, then find the perimeter of the parallelogram

• A. $50$
• B. $56$
• C. $60$
• D. $66$

Answer 1

Answer: (B)

$56$

Here, $ABCD$ is a parallelogram, where $DE\perp AB$ and $DF\perp BC$.

$\angle ADC={150}^o$

$\Rightarrow$ $\angle A+\angle D={180}^o$ [ Sum o adjacent angles of parallelogram is ${180}^o$. ]

$\Rightarrow$ $\angle A+{150}^o={180}^o$

$\Rightarrow$ $\angle A={30}^o$

$\Rightarrow$ $\angle A=\angle C={30}^o$ [ Opposite angles are equal in parallelogram ]

In $△AED,$

$\Rightarrow$ $\angle AED+\angle EDA+\angle DAE={180}^o$

$\Rightarrow$ ${90}^o+\angle EDA+{30}^o={180}^o$

$\therefore$ $\angle EDA={60}^o.$

$\therefore$ $△AED$ is ${30}^o-{60}^o-{90}^o$ triangle.

$\Rightarrow$ $DE=AD\sin {30}^o$

$\Rightarrow$ $6=AD\times \frac{1}{2}$

$\therefore$ $AD=12$

$\Rightarrow$ $AD=BC=12$ ---- ( 1 ) [ Opposite sides of parallelogram are equal ]

In $△CFD$,

$\Rightarrow$ $\angle CFD+\angle FDC+\angle DCF={180}^o$

$\Rightarrow$ ${90}^o+\angle FDC+{30}^o={180}^o$

$\Rightarrow$ $\angle FDC={60}^o$

$\therefore$ $△CFD$ is ${30}^o-{60}^o-{90}^o$ triangle.

$\Rightarrow$ $DF=CD\sin {30}^o$

$\Rightarrow$ $8=CD\times \frac{1}{2}$

$\therefore$ $CD=16$

$\Rightarrow$ $CD=AB=16$ ---- ( 2 ) [ Opposite sides of parallelogram are equal ]

$\Rightarrow$ Now, perimeter of parallelogram $ABCD=AB+BC+CD+DA$

$=16+12+16+12$ [ From ( 1 ) and ( 2 ) ]

$=56$