In the given trapezium PQRS , A and B are the mid - points of the diagonals QS and PR respectively. Prove that : 1. AB SR

2. AB = ( SR - PQ )

[ Hint : Join Q and B and produce QB to meet SR at C ].

~Thanks in advance !

Step-by-step explanation:jj

Given: In trapezium PQRS , A and B are midpoints of diagonals PR and QS.

To Show: AB=1/2(PQ-SR).

Construction:draw midpoints C and D of SP and RQ respectively and join CA and BD

Proof : In triangle SPQ ,C and B are midpoints of SP and SQ

Therefore,by midpoint theorem

CB || PQ and 2CB= PQ 1

In triangle QSR B and D are midpoints of QS and RQ

Therefore, by midpoint theorem

BD || SR and 2BD =SR2

In triangle SPR C and A are midpoints of SP and RP

Therefore by midpoint theorem

CA = 1/2 SR 3

From 2

2 BD = SR

BD =1/2 SR 4

From 3 and 4

CA = BD 5

From 1 and 2

PQ - SR = 2 CB - 2 BD

PQ - SR = 2(CB - BD)

1/2(PQ - SR) =CB - BD

1/2(PQ - SR)= CA+AB - BD

From 5

CA = BD , therefore they cancel each other

Therefore ,

1/2(PQ-SR) =AB

AB= 1/2(PQ-SR)

third option: 6x+8

step-by-step explanation:

we can see that the given question is a sum of series and the values of n vary from 0 to 3

the first term will be:

when n=0, [(0)*x+2] => (0+2) => 2

when n=1, [(1)*x+2] => (x+2) => x+2

when n=2, [(2)*x+2] => 2x+2

when n=4, [(3)*x+2] => 3x+2

so adding up all the terms will give us:

=2+x+2+2x+2+3x+2

=x+2x+3x+2+2+2+2

=6x+8

so,

third option is the correct answer ..

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)