Question

Somebody help me pleaseee?

Answer 1

`\sf\\\textsf{Here, since IJKL is a rhombus, all of its sides are equal. That is, IJ=JK = KL=IL.}\\\textsf{This means }\triangle IJL\textsf{ is an isosceles triangle because two of it's sides IJ and IL are}\\\textsf{equal.}`

`\textsf{Now we know that diagonals of a parallelogram bisect each other. Rhombus is a }\\\textsf{type of parallelogram whose all sides are equal. So in rhombus IJKL, JM = ML}\\\textsf{and KM = MI}`

`\textsf{This means JM = MI. So MI is the line in triangle IJL which joins the midpoint of}\\\textsf{it's base and the opposite vertex. The line that joins the midpoint of base and the}\\\textsf{opposite vertex in an isosceles triangle is perpendicular to the base and bisects}\\\textsf{the vertex angle.}`

`\sf\\\textsf{Hence, }IM\perp JL,\textsf{ which means }\angle IMJ=90^o. \textsf{ Additionally, }\angle MIJ = \angle MIL.`

Hope this helps! Let me know if you have any questions.

Answer 2

• 90°

Given :

• A Rhombus IJKL

To find :

• m∠IMJ

Solution :

we know that,

• A rhombus's diagonals bisect each other at 90°.

Thus,

• m∠IMJ = 90°

`\:`