9. The figure shows a rhombus ABCD where
the diagonals AC and BD intersect at E. Find the
value ofr.
(3x + 7°
D
C С
E
A
(2x + 53)
B
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In rhombus ABCD, the diagonals AC and BD intersect at E. If AE=x, BE=(x+7), and AB=(x+8), find the lengths of the diagonals AC and BD.
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ANSWER
It is given that AE=x,BE=(x+7) and AB=(x+8).
In △AEB, ∠E=90
0
Using pythagoras theorem, we have
AB
2
=AE
2
+BE
2
⇒(x+8)
2
=x
2
+(x+7)
2
⇒x
2
+16x+64=x
2
+x
2
+14x+49(∵(a+b)
2
=a
2
+b
2
+2ab)
⇒x
2
−2x
2
+16x−14x+64−49=0
⇒−x
2
+2x+15=0
⇒x
2
−2x−15=0
⇒x
2
−5x+3x−15=0
⇒x(x−5)+3(x−5)=0
⇒(x+3)=0,(x−5)=0
⇒x=−3,x=5
Since the length of the rhombus cannot be negative thus, x=5.
Therefore, AE=5 cm, BE=5+7=12 cm,
AC=2×AE=2×5=10 cm and
BD=2×BE=2×12=24 cm
Hence, the length of the diagonals are AC=10 cm and BD=24 cm.
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