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11 Aug 2025, 09:26
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A rhombus has a perimeter of 40 cm. The line
joining the midpoints of two adjacent sides is 6 cm
long. Find the area of the rhombus.
(A) 48 sq.cm. (B) 84 sq.cm.
(C) 96 sq.cm. (D) 372 sq.cm.
joining the midpoints of two adjacent sides is 6 cm
long. Find the area of the rhombus.
(A) 48 sq.cm. (B) 84 sq.cm.
(C) 96 sq.cm. (D) 372 sq.cm.
23 Sep 2025, 03:08
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Let each side of the Rhombus be "a"
4a=40, a=10
Hence each side of the rhombus is equal to 10
Now look at the diagram
E & F are mid-points, therefore AE=ED=5 & AF=FD=5
also we have EF=6
Using similarity,
Tr(AEF)~Tr(ADB)
AE/ED=EF/BD
5/10=6/BD
BD=12
Now apply Heron's formula on Tr(ABD)
s=(10+10+12)/2=16
a=10, b=10, c=12
Ar(ABD)= Root[s(s-a)(s-b)(s-c)]
Substitute values and you'll get Ar(ABD)=48
Ar(ABCD)=2*Ar(ABD)
Ar(ABCD)=96
GMAT-Club-Forum-3k79w7b2.png [ 3.89 KiB | Viewed 155 times ]
GMAT-Club-Forum-rzrabni5.png [ 2.44 KiB | Viewed 154 times ]
4a=40, a=10
Hence each side of the rhombus is equal to 10
Now look at the diagram
E & F are mid-points, therefore AE=ED=5 & AF=FD=5
also we have EF=6
Using similarity,
Tr(AEF)~Tr(ADB)
AE/ED=EF/BD
5/10=6/BD
BD=12
Now apply Heron's formula on Tr(ABD)
s=(10+10+12)/2=16
a=10, b=10, c=12
Ar(ABD)= Root[s(s-a)(s-b)(s-c)]
Substitute values and you'll get Ar(ABD)=48
Ar(ABCD)=2*Ar(ABD)
Ar(ABCD)=96
Attachment:
GMAT-Club-Forum-3k79w7b2.png [ 3.89 KiB | Viewed 155 times ]
Attachment:
GMAT-Club-Forum-rzrabni5.png [ 2.44 KiB | Viewed 154 times ]
