Quadrilaterals Test - 28

From the above figure, its is clear that APCQ is a parallelogram.

Given, $$\angle AEB  = 70^{\circ}$$
$$\angle AEB + \angle AEC = 180$$
$$70 + \angle AEC = 180$$
$$\angle AEC = 110^{\circ}$$

Now, in quadrilateral AECD,
$$\angle AEC+ \angle ADC + \angle DCE + \angle EAD = 360$$
$$110^\circ + \angle ADC + 90^\circ + 90^\circ = 360^\circ$$
$$\angle ADC = 70^{\circ}$$

From question, 
In $$\Delta$$ ABC, D is the mid-point of AB and DE is drawn parallel to BC and it meets at E on AC. Since, DE $$\parallel$$ BC (By B.P. Theorem)
$$\displaystyle \dfrac{AD}{DB} = \dfrac{AE}{EC}$$ ......(i)
also, AD = DB  ($$\because D\ is\ mid - point$$)
$$\therefore \dfrac{AD}{DB} = 1$$
Now $$\displaystyle \frac{AE}{EC} = 1$$  ....(From  ...(i))
$$\Rightarrow$$  AE = EC
Hence, the line biscets the third line

A rectangle is made up of $$4$$ sides. 

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

The given angles are, $$70^o, 60^o, 90^o$$.

Let the fourth angle be $$x^o$$.

Then, $$70^o+60^o+90^o+x^o=360^o$$.

$$\Rightarrow$$ $$360^o - (70^o + 60^o + 90^o ) =x^{\circ}$$

$$\Rightarrow$$ $$x^o=360^o - 220^o  =140^{\circ}$$.

Therefore, the fourth angle is $$x^o=140^o$$.

Hence, option $$D$$ is correct.

The sum of all interior angles of a quadrilateral is $$360^{\circ}$$.

$$\textbf{Step-1: Comparing with trapezium}$$ 

                $$\text{The definition of a square does not comply with the}$$

                $$\text{definition of a trapezoid. The definition of a trapezoid}$$

                $$\text{is a quadrilateral (a closed plane figure with 4 sides) with exactly one pair of parallel sides.}$$

$$\textbf{Step-2: Comparing with other option}$$ 

                $$\text{On the other hand, a square is a very special kind of}$$

                $$\text{quadrilateral; A square is also a parallelogram because}$$

                $$\text{it has two sets of parallel sides and four right angles. A}$$

                $$\text{square is also a parallelogram because opposite sides}$$

                $$\text{are parallel. A square is always a rhombus. A rhombus}$$

                $$\text{is a quadrilateral with four congruent sides. If the rhombus has 4}$$

                $$\text{right angles it may also be called a square. So every square is a rhombus.}$$ 

$$\textbf{Hence , Square is not a (C) trapezium}$$

So, we can say that the four sides of a square are equal.

                $$\text{ Now in}$$  $$\Delta ABC, $$

                $$\begin{align}
  & \angle CAB+\angle ABC+\angle BCA={{180}^{\circ }}........(1) \\
 
 & \text{In}\Delta ACD, \\
 
 & \angle CAD+\angle ADC+\angle DCA={{180}^{\circ }}.......(2) \\
\end{align}$$

                $$\text{Adding  (1) and (2), we get,}$$