Question

In a quadrilateral PQRS; ANGLE P : ANGLE Q : ANGLE R : ANGLE S=3:4:6:7.
Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other.
Also find if PS is also parallel to QR.

Answer 1

Step-by-step explanation:

let the angle be x

3x+4x+6x+7x=360 by the quad theorem

20x=360

x=360/20

x=18

so 3*x=3*18=54

4*x=4*18=72

6*x=6*18=108

7*x=7*18=126

the answer is

p=54

q=72

r=108

s=126

Answer 2

Answer:

The angles of the quadrilateral are $54^{o},~ 72^{o}, ~108^{o},~ 126^{o}$ and $PQ \parallel SR~~ \&~~PS \not\parallel QR$

Step-by-step explanation:

Given the ratio of angles in a quadrilateral PQRS as

$\angle P : \angle Q : \angle R :\angle S=3:4:6:7$

Let the common factor of all the angles be $x$.

The sum of angles in a quadrilateral $=360^{o}$

Therefore, $\angle P+ \angle Q +\angle R +\angle S=3x+4x+6x+7x=360^{o}$

$\implies 20x=360$

$\implies x=\frac{360}{20} =18$

Therefore, $\angle P=3x=3\times 18=54^{o}$

$\angle Q=4x=4\times 18=72^{o}$

$\angle R=6x=6\times 18=108^{o}$

$\angle S=7x=7\times 18=126^{o}$

The angles of the quadrilateral are $54^{o},~ 72^{o}, ~108^{o},~ 126^{o}$.

Two sides in a quadrilateral are parallel, if the sum of interior angles on the same side of transversal is $180^{o}$

For PQ and SR, if we take transversal along QR,

then $\angle Q+\angle R=72+108=180^{o}$

Therefore, PQ and SR are parallel to each other.

Similarly for PS and QR, if we take transversal along PQ,

then $\angle P+\angle Q=54+72=126^{o} \neq180^{o}$

Therefore, PS and QR are not parallel to each other.