In a rectangle KLMN, the diagonals KM and LN intersect at O. (i) If KO = 4y + 6 and ON = 3y + 11, find the lengths of its diagonals. (ii) If ∠NKM = 32°, find ∠KON and ∠KMN. GUYS, IT'S VERY URGENT. I HAVE TO SUBMIT TOMORROW. PLS ANS. PLS DON'T GIVE WRONG ANSWERS

1). Length of diagonals = 52

2). m∠KON = 103°, m∠KMN = 45°

Step-by-step explanation:

KLMN is a rectangle.

Diagonals KM and LN intersect at point O.

1). Since, diagonals of the rectangle equally bisect each other.

   KM = 2(KO)

   KM = 2(4y + 6)

   And LN = 2(ON)

   LN = 2(3y + 11)

   Since, length of diagonals of a rectangle are equal in measure.

   2(4y + 6) = 2(3y + 11)

   4y + 6 = 3y + 11

   4y - 3y = 11 - 6

   y = 5

   Therefore, length of diagonals = 2(3y + 11)

                                                       = 2(15 + 11)

                                                       = 52 units

2). m∠NKM = 32°

    Since, m∠KNM = 90° [internal angle of a rectangle]

    And diagonal LN bisects this angle,

    m∠KNO = 45°

    In ΔKNO,

    m(∠NKM) + m(∠KNO) + m(KON) = 180°

    32° + 45° + m∠KON = 180°

    m∠KON = 180° - 77°

                   = 103°

    Since, m∠KMN = (m∠LMN)

    m∠KMN = = 45°

$450 i think. hope this ☺

hi,

the area of abcd is 245.

3*24=230

4*94=1039

2*12=24

add all of them up, you get 2450