Question

PQRS is a quadirilateral . show that PQ+QR+RS + Sp​

Answer 1

We know that in any triangle the sum of any two sides is greater than the third side.

In △PQO,

PQ+QO>PQ→(1)

In △SOP,

SO+PO>PS→(2)

In △SOR,

SO+OR>RS→(3)

In △QOR,

QO+OR>QR→(4)

Adding equations (1) , (2) , (3) & (4) , we get :

=> 2(PO+OQ+SO+OR)>PQ+QR+RS+SP

= > PQ+QR+RS+SP<2(PQ+OR+SO+OQ)

=> PQ+QR+RS+SP<2(PR+SQ)

TERMS USED:

WHAT IS A QUARDILATERAL?

A quadrilateral is a polygon in Euclidean plane geometry with four edges and four vertices. Other names for quadrilateral include quadrangle and tetragon. A quadrilateral with vertices A, B, C and D is sometimes denoted as

${\displaystyle \square ABCD}.$

Answer 2

Answer:

The converse of this theorem is also true; that is, if two lines kk and ll are cut by a transversal so that the alternate interior angles are congruent, then k∥lk∥l .Proof.

Since k∥lk∥l , by the Corresponding Angles Postulate ,

∠1≅∠5∠1≅∠5 .

Therefore, by the definition of congruent angles ,

m∠1=m∠5m∠1=m∠5 .

Since ∠1∠1 and ∠2∠2 form a linear pair , they are supplementary , so

m∠1+m∠2=180°m∠1+m∠2=180° .

Also, ∠5∠5 and ∠8∠8 are supplementary, so

m∠5+m∠8=180°m∠5+m∠8=180° .

Substituting m∠1m∠1 for m∠5m∠5 , we get

m∠1+m∠8=180°m∠1+m∠8=180° .

Subtracting m∠1m∠1 from both sides, we have

m∠8=180°−m∠1           =m∠2m∠8=180°−m∠1           =m∠2 .

Therefore, ∠2≅∠8∠2≅∠8 .

You can prove that ∠3≅∠5∠3≅∠5 using the same method.So, in the figure below, if k∥lk∥l , then ∠2≅∠8∠2≅∠8 and ∠3≅∠5∠3≅∠5 .