In parallelogram ABCD, point S is the intersection point of its diagonals. The perimeter of ∆CDS is 5.6 cm smaller than the perimeter of ∆BCS. The angle bisector of angle BAD intersects segment BC in point M. Find the lengths of sides of
the parallelogram if
BM: MC = 7:4.
Answers
Answer:
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Step-by-step explanation:
From the diagram, OD−→−+DP−→−=OP−→−
OA−→−+AP−→−=OP−→−
OB−→−+BP−→−=OP−→−
OC−→−+CP−→−=OP−→−
∴OD−→−+DP−→−+OA−→−+AP−→−+OB−→−+BP−→−+OC−→−+CP−→−=4OP−→−
⇒OA−→−+OB−→−+OC−→−+OD−→−+AP−→−+CP−→−+BP−→−+DP−→−=4OP−→−→(1)
Also, from the diagram, we can see that,
PA−→−+PC−→−=0
PD−→−+PB−→−=0
So, equation (1) becomes,
⇒OA−→−+OB−→−+OC−→−+OD−→−+0+0=4OP−→−
⇒OA−→−+OB−→−+OC−→−+OD−→−=4OP−→−
Answer:
Step-by-step explanation:
Draw MN parallel to AB and CD (as shown)
=> AB = NM = CD ………… (1)
Consider Parallelogram ABMN. AM is a diagonal and also an angular bi-sector of angle A (given) hence ABMN is a rhombus
Hence AB = BM
But from the given condition, BM = (7/4) BC.
=> AB = (7/4) BC ………………………. (2)
Again from the question we know that —
According to the properties of a Parallelogram, Diagonals bisect each other,
=> DS = SB = x (suppose) and AS = SC = y (suppose)
The perimeter of ∆CDS is 5.6 cm smaller than the perimeter of ∆BCS.
x + y + CD + 5.6 = x + y + BC
=> CD = BC - 5.6
=> AB - BC = 5.6 ………….. (3)
From (2) and (3)
=. 5.6 + BC = (7/4) BC
=> ( 7/4 - 1 ) BC = 5.6
=> BC = 5.6 * (4/3) = 22.4/3 = 7.466 cm
And AB = (7/4) BC = (7/4) = 13.07 cm
Hence Length of sides are 7.466 and 13.07 cm ………. Answer
