Understanding Quadrilaterals Test - 18

$$AD\parallel  BC$$, $$DB$$ is the transversal, therefore,
$$\angle CBD=\angle ADB={55}^{o}$$(alt$$\angle s$$)
In $$\triangle BOC$$,
$$\angle BOC={180}^{o}-(\angle OCB+\angle OBC)$$
$$={180}^{o}-({35}^{o}+{55}^{o})={!80}^{o}-{90}^{o}={90}^{o}$$
$$\Rightarrow$$ $$BD\bot AC$$
Also, in $$\triangle ABC$$, $$\angle BAC={180}^{o}-({110}^{o}-{35}^{o})$$
$$={180}^{o}-{145}^{o}={35}^{o}$$
Hence, $$\angle ACB=\angle BAC\Rightarrow AB=BC$$
$$\therefore$$ $$ABCD$$ is a parallelogram, whose adjacent sides are equal and diagonals are perpendicular to each other
$$\Rightarrow$$ $$ABCD$$ is a rhombus.

Perimeter of $$\parallel gm $$ $$=40cm$$

We have $$\angle SPQ=60^{\circ}$$
$$\angle P+\angle Q=180^{\circ}$$ (adj. $$\angle s$$ of a || gm are supp)
$$\angle Q=180^{\circ}-60^{\circ}=120^{\circ}$$
Now $$PQ||SR$$ and $$AP$$ is the transversal.
$$\therefore SAP=\angle APQ=30^{\circ}$$  ($$\because AP$$ bisects $$\angle SPQ$$)
$$\angle SPA=\angle SAP\Rightarrow SA=SP$$ (isos. $$\Delta $$ property)  ...(i)
Also $$\angle RAQ=\angle AQP=60^{\circ}$$
($$PQ||SR,\:AQ$$ is transversal, alt $$\angle s$$)
and $$\angle AQP=\dfrac{1}{2}\angle PQR=60^{\circ}$$
$$\Rightarrow \angle RQA=\angle RAQ$$  (AQ bisect $$\angle PQR$$)
$$\Rightarrow RA=RQ$$  (isos. $$\Delta $$ property)  ....(ii)
$$\therefore $$ from eqn. (i) and (ii)
$$AS=AR$$   ($$\because SP=RQ$$, opp. side of a ||gm)
Also, in $$\Delta ARQ,\:\angle ARQ=60^{\circ}$$
$$\Rightarrow \Delta ARQ$$ is equilateral
$$\Rightarrow AR=RQ\Rightarrow AR=SP$$
Therefore, the incorrect statement is $$AQ=PQ$$.

$$Let\quad ABCD\quad a\quad parallelogram\quad with\quad four\quad angles\quad as\quad \angle A,\angle B,\angle C\quad and\quad \angle D\\ \angle A+\angle C\quad =\quad 130(Given)\\ Also\quad \angle A\quad =\angle C\quad (Opposite\quad angles\quad of\quad \parallel gram\quad are\quad equal).......(1)\\ \Longrightarrow \angle A\quad +\quad \angle A\quad =130\\ \Longrightarrow 2\angle A\quad =\quad 130\\ \Longrightarrow \angle A\quad =\quad \frac { 130 }{ 2 } \quad =\quad 65\\ \Longrightarrow \angle A\quad =\quad \angle C\quad =\quad 65....(2)\\ Also\quad \angle A\quad +\quad \angle B\quad =180(Sum\quad of\quad cointerior\quad angles\quad is\quad 180)\\ \Longrightarrow 65\quad +\quad \angle B\quad =\quad 180\\ \Longrightarrow \angle B\quad =\quad 180\quad -\quad 65\quad =\quad 115\\ Also\quad \angle B\quad =\quad \angle D\quad =115(Opposite\quad angles\quad of\quad \parallel gram\quad are\quad equal)\\ Hence\quad all\quad the\quad angles\quad are\quad 65, 115, 65\quad and\quad 115$$

$$\because$$ In a parallelogram, opposite angles are equal. 

Given: $$ABCD$$ is a parallelogram

We know that the diagonals of a parallelogram bisect each other.

$$\Rightarrow $$ The diagonals $$BD$$ and $$AC$$ bisect each other.

The sum of all the four angles of a parallelogram is $$ 360^{\circ}$$ , Hence option (D) is false.

In $$\parallel gm$$ ABCD, AC and BD are diagonals