Kite E F G H is inscribed in a rectangle. Points F and H are midpoints of sides of the rectangle, and creates a side length of x. Lines are drawn from point F to point H and from point E to point G and intersect at a point. Line E G is parallel to a rectangle side. The distance from point F to the intersection is 2, and the distance from H to the intersection is 5. Kite EFGH is inscribed in a rectangle where F and H are midpoints of parallel sides. The area of EFGH is 35 square units. What is the value of x? 4 units 5 units 6 units 7 units

5 units

Step-by-step explanation:

Let point O be the point of intersection of the kite diagonals.

|OF| = 2, |OH| = 5

|FH| = |OF| + |OH| = 2 + 5 = 7

FH and EG are the diagonals of the kite. Hence the area of thee kite is:

Area of kite EFGH = (FH * EG) / 2

Substituting:

35 = (7 * |EG|) / 2

|EG| * 7 = 70

|EG| = 10 units

The longer diagonal of a kite bisects the shorter one, therefore |GO| = |EO| = 10 / 2 = 5 units

x = |GO| = |EO| = 5 units

343t464564435rri76i6u

step-by-step explanation:

because i said it is

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐