Math, asked by nandlalchaudhary4990, 4 days ago

area of a parallelogram ABCD is
 {120cm}^{2}
. If AB = 15cm , then find the distance between AB and DC. IF angle CDA = 150°, find the perimeter of the parallelogram.

Answers

Answered by mathdude500
5

Answer:

\:\boxed{\begin{aligned}& \qquad \:\sf \:Distance \: between \: AB \: and \: DC=8 \: cm \qquad \: \\ \\& \qquad \:\sf \: Perimeter\:of\:parallelogram=62 \: cm\end{aligned}} \qquad  \\  \\

Step-by-step explanation:

Given that, ABCD is a parallelogram such that AB = 15 cm and  \angleCDA = 150°.

We know, in a parallelogram, sum of adjacent angles is 180°.

So,

\sf \: \angle CDA + \angle DAB = 180 \degree \:  \\  \\

\sf \: 150\degree + \angle DAB = 180 \degree \:  \\  \\

\sf \: \angle DAB = 180 \degree - 150\degree \:  \\  \\

\sf\implies \sf \: \angle DAB = 30\degree \:  \\  \\

Let from D, draw DE perpendicular to AB intersecting AB at E.

So, Distance between AB and CD is DE.

Given that,

\sf \: Area\:of\:parallelogram = 120 \:  {cm}^{2}  \\  \\

\sf \: AB \times DE = 120 \:   \\  \\

\sf \: 15 \times DE = 120 \:   \\  \\

\sf \:  DE = \dfrac{120}{15}  \:   \\  \\

\sf\implies \sf \:  \boxed{ \sf{ \:DE = 8 \: cm \: }}  \:   \\  \\

So, Distance between AB and CD is 8 cm.

Now, In right-angle triangle ADE

\sf \: sin30\degree = \dfrac{DE}{DA}  \\  \\

On substituting the value of sin30° and DE, we get

\sf \: \dfrac{1}{2}  = \dfrac{8}{DA}  \\  \\

\sf\implies \sf \: DA \:  =  \: 16 \: cm \\  \\

Now,

\sf \: Perimeter\:of\:parallelogram \\  \\

\sf \:  =  \: 2(AB + DA) \\  \\

\sf \:  =  \: 2(15 + 16) \\  \\

\sf \:  =  \: 2 \times 31 \\  \\

\sf \:  =  \: 62 \: cm \\  \\

Hence,

\bf\implies \: Perimeter\:of\:parallelogram \:  =  \: 62 \: cm \\  \\

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