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25 Nov 2019, 06:34
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Difficulty:
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38% (02:29) correct
63%
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wrong
based on 48
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The sum of the lengths of the diagonals of a rhombus of side 5 cm is 12 cm. What is its area?
a) \(9 cm^2\)
b) \(11 cm^2\)
c) \(20 cm^2\)
d) \(24 cm^2\)
e) \(32 cm^2\)
a) \(9 cm^2\)
b) \(11 cm^2\)
c) \(20 cm^2\)
d) \(24 cm^2\)
e) \(32 cm^2\)
26 Nov 2019, 18:26
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CaptainLevi
The sum of the lengths of the diagonals of a rhombus of side 5 cm is 12 cm. What is its area?
a) \(9 cm^2\)
b) \(11 cm^2\)
c) \(20 cm^2\)
d) \(24 cm^2\)
e) \(32 cm^2\)
a) \(9 cm^2\)
b) \(11 cm^2\)
c) \(20 cm^2\)
d) \(24 cm^2\)
e) \(32 cm^2\)
Now we can draw a rhombus as shown..
Sum of diagonals = 2a+2b=12...a+b=6
\(a^2+b^2=5^2......(a+b)^2-2ab=25........6^2-25=2ab=36-25=9\)
Now each of the FOUR triangles formed have an area of \(\frac{1}{2}*a*b\), so all four combined will have an area of \(4*\frac{1}{2}*a*b=2ab=9\)
A
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Untitled9.png [ 7.75 KiB | Viewed 6091 times ]
globaldesi
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16 Dec 2019, 07:54
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chetan2u
CaptainLevi
The sum of the lengths of the diagonals of a rhombus of side 5 cm is 12 cm. What is its area?
a) \(9 cm^2\)
b) \(11 cm^2\)
c) \(20 cm^2\)
d) \(24 cm^2\)
e) \(32 cm^2\)
a) \(9 cm^2\)
b) \(11 cm^2\)
c) \(20 cm^2\)
d) \(24 cm^2\)
e) \(32 cm^2\)
Now we can draw a rhombus as shown..
Sum of diagonals = 2a+2b=12...a+b=6
\(a^2+b^2=5^2......(a+b)^2-2ab=25........6^2-25=2ab=36-25=9\)
Now each of the FOUR triangles formed have an area of \(\frac{1}{2}*a*b\), so all four combined will have an area of \(4*\frac{1}{2}*a*b=2ab=9\)
A
chetan2u
In you same image i considered diagonals as \(d_1, d_2\)
so i did \(d_1+d_2 = 12\)
and \((d_1 +d_2)^2 = 144\)
given
\((\frac{d_1}{2})^2 +(\frac{d_2}{2})^2 = 25\)
or \((d_1)^2 +(d_2)^2 = 100\)
sb 2 equations
gives \(2 *d_1 *d_2\) = 44
or Area of rhombus =\( \frac{d_1 *d_2}{2}\) = 11
where did I go wrong?
