Question

the perimeter of rhombus is 40 cm . if 1 diagnil is 16 cm . find : (i) its other side (ii) area

Answer 1

☆ԹՌՏաᴇՐ:☆

ɪᴛ's ᴏɴᴇ ᴅɪᴀɢᴏɴᴀʟ ɪs 16 ᴄᴍ. sɪɴᴄᴇ ,ᴅɪᴀɢᴏɴᴀʟ ᴏғ ʀʜᴏᴍʙᴜs ɪɴᴛᴇʀsᴇᴄᴛ ᴇᴀᴄʜ ᴏᴛʜᴇʀ ᴀᴛ ʀɪɢʜᴛ ᴀɴɢʟᴇs. ʟᴇɴɢᴛʜ ᴏғ ᴏᴛʜᴇʀ ᴅɪᴀɢᴏɴᴀʟ ᴡɪʟʟ ʙᴇ: ʟᴇɴɢᴛʜ ᴏғ ᴏᴛʜᴇʀ ᴅɪᴀɢᴏɴᴀʟ= 2(10²-8²)=2×6=12 ᴄᴍ.

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Answer 2

Answer:

each side is 40/4= 10 cm

ALSO CONSIDER DIAGONAL AC= 16CM

we know that diagonals of a rhombus intersect at 90 degrees.

when we consider triangle ABE-    angle AEB =90 degrees

IN TRIANGLE ABE -hyp^2= base^2 + height^2

                               -AB^2= [AC/2]^2 + EB^2                        

                               -10^2=[16/2]^2 + EB^2

                               -100=64 + EB^2

                               -EB^2=100-64

                               -EB^2=36

                               -EB^2=6cm

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[i] THE OTHER DIAGONAL=EB+EB=12CM

[ii] area of rhombus = 1/2*product of diagonals

                                 = 1/2*16*12

                                 = 96cm^2