Question

ABCD is a parallelogram with AB=8cm, AD=4cm, $$\mathrm{\angle }B=120{}^{\circ }$$
(a) What is $\mathrm{\angle }$A?
(b) What is the perpendicular distance from D to AB?
(c) What is the area of ABCD?

Answer 1

Hint: Use the property of parallelogram that opposite angles of a parallelogram are equal. Then use the property that the sum of all angles of a quadrilateral is $$360{}^{\circ }$$ . For the second part, use cosine of angle A found from the first part to find perpendicular distance. For the third part, use the formula for Area of parallelogram = Length of base $$\times$$ Height .

Complete step-by-step answer:

We are given that ABCD is a parallelogram with AB=8cm, AD=4cm, $$\mathrm{\angle }B=120{}^{\circ }$$
We need to find the following
(a) $\mathrm{\angle }$A
(b) the perpendicular distance from D to AB
(c) the area of ABCD
Let us first solve (a).
We know that the opposite angles of a parallelogram are equal.
Using this, $$\mathrm{\angle }B=\mathrm{\angle }D=120{}^{\circ }$$
And, $$\mathrm{\angle }A=\mathrm{\angle }C$$
Also, we know that the sum of all angles of a quadrilateral is $$360{}^{\circ }$$ .
$$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D=360{}^{\circ }$$
Substituting the relations between the angles in this equation, we will get the following:
$$2\mathrm{\angle }A+2\mathrm{\angle }B=360{}^{\circ }$$
$$2\mathrm{\angle }A+2\times 120{}^{\circ }=360{}^{\circ }$$
$$2\mathrm{\angle }A+240{}^{\circ }=360{}^{\circ }$$
$$2\mathrm{\angle }A=120{}^{\circ }$$
So, $$\mathrm{\angle }A=60{}^{\circ }$$
Now, let us solve (b)
For this, draw DE as perpendicular from D to AB as shown in the figure.
We need to find the length of DE.

 Now, we found in the previous part that $$\mathrm{\angle }A=60{}^{\circ }$$
We also know that $$\sin 60{}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$$
So, $$\sin \mathrm{\angle }A={\displaystyle \frac{DE}{AD}}={\displaystyle \frac{\sqrt{3}}{2}}$$
$$\sin \mathrm{\angle }A={\displaystyle \frac{DE}{4}}={\displaystyle \frac{\sqrt{3}}{2}}$$
This gives us: $$DE=2\sqrt{3}$$ cm
So, the perpendicular distance from D to AB is $$2\sqrt{3}$$ cm.
Now, we will solve (c).
We need to find the area of ABCD.
We know that:
Area of parallelogram = Length of base $$\times$$ Height
Here, we have length of base, AB = 8 cm
Height, DE = $$2\sqrt{3}$$ cm
Putting these values in the formula, we will get the following:
Area of ABCD = Length of AB $$\times$$ Length of DE
Area of ABCD = $$8\times 2\sqrt{3}$$ $$=16\sqrt{3}c{m}^2$$
So, Area of ABCD is $$16\sqrt{3}c{m}^2$$

Note: You have to remember the various geometrical properties to solve this kind of question. For example, in this question too, you need to use the properties like opposite angles of a parallelogram are equal and that the sum of all angles of a quadrilateral is $$360{}^{\circ }$$