Question

ABCD is a parallelogram bc = 10cms and angle a = 30° . find :dc​

Answer 1

Question :

ABCD is a parallelogram , BC = 10cm and angle A= 30° . Find DE

Answer:

The value of DE = 5cm

and AE =5√3cm

Step by step explanation

Given ABCD is a parallelogram

Therefore ,

AB = CD and

AD = BC = 10cm

Now from triangle

$\triangle AED$

By Pythagorean triplet we get

${AD}^{2} = {AE}^{2} + {DE}^{2} \\ \implies ({10cm)}^{2} = {AE}^{2} + {DE}^{2} \\ \implies {AE}^{2} + {DE}^{2} = 100 {cm}^{2}$

Again given

$\angle A = 30 \degree$

So By trigonometric functions we get

$\sin A = \frac{DE}{AD} \\ \implies \sin30 \degree = \frac{DE}{10cm} \\ \implies \frac{1}{2} = \frac{DE}{10cm} \\ \implies DE = 5cm$

Now using the value of DE in previous equation we have

$AE {}^{2} + {DE}^{2} = 100 {cm}^{2} \\ \implies {AE}^{2} + ({5cm)}^{2} = 100cm {}^{2} \\ \implies {AE}^{2} =( 100 - 25) {cm}^{2} \\ \implies {AE}^{2} = 75 {cm}^{2} \\ \implies {AE} = 5 \sqrt{3} cm$