If each quadrilateral below is a rectangle, find the missing measure.

please help :(

m<JMK = 54

m<JKH = 63

m<HLK = 90

m<HJL = 27

m<LHK = 63

m<JLK = 27

Step-by-step explanation:

1. Solve for m<JMK

It is given that (<HMJ) and (<JMK)  form line (HK), meaning that they are supplementary angles. Thus, their degree measures add up to (180), using this one can form the following equation,

(<HMJ) + (<JMK) = 180

126 + (<JMK) = 180

(<JMK) = 54

2. Solve for m<JKH

It is given that the figure is a rectangle, one of the properties of a rectangle is that diagonals are congruent, and the diagonals bisect each other. This means that segments (JM) and (MK) are congruent. Hence, the triangle (JMK) is an isoceles triangle. Therefore, one can use the base-angles theorem, which states that (<MJK) is congruent to (<JKH). Since the sum of angle measures in any triangle is (180), one can form the following equation,

(<JMK) + (<MJK) + (<JKH) = 180

54 + 2(<JKH) = 180

<JKH = 63

3. Solve for m<HLK

As per its definition, the corner angles in a rectangle are always (90) degrees, this property is applicable for the given rectangle. Thus, (<HLK) = 90 degrees.

4. Solve for m<HJL

To solve for this angle, one can use similar logic as they did to solve for (<JKH).  Form the following equation and solve,

(<HMJ) + (<HJL) + (<JHK) = 180

126 + 2(<HJL) = 180

(<HJL) = 27

5. Solve for m<LHK

The vertical angles theorem states that when two lines intersect, the opposite angles are congruent. One can apply this theorem here and say (<JMK) = (<HML). Then, to solve for this angle, one can use the same logic that they used to solve for (<JKH) and (<HJL). Form the equation and solve,

(<HML) +(<LHK) + (<HLJ) = 180

54 + 2(<LHK) = 180

<LHK = 63

6.Solve for m<JLK

To solve for this angle, one can use the same logic that they used to solve for (<LHK). Form the equation and solve,

(<LMK) + (<JLK) + (HKL) = 180

126 + 2(<JLK) = 180

<JLK = 27

step-by-step explanation:

86

step-by-step explanation:

find the area of the triangle then multiply by 2 then add the area of the parollelegram multiplied by 2