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05 Jul 2021, 08:45
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C
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Difficulty:
5%
(low)
Question Stats:
84% (01:42) correct
16%
(01:51)
wrong
based on 68
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Triangle ABC has area 36. If side AC is twice as long as side CB, what is the length of side AB?
(A) 6
(B) 12
(C) \(6\sqrt{5}\)
(D) 18
(E) \(12\sqrt{5}\)
Attachment:
2021-06-18_00-59-49.png [ 17.91 KiB | Viewed 2235 times ]
05 Jul 2021, 10:12
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Solution:
We know area od triangle \(= \frac{1}{2} \times base \times height\)
Looking at the triangle we can say: area \(= \frac{1}{2} ]times AC \times BC\)
\(⇒ \frac{1}{2} \times AC \times BC= 36 \)
\(⇒ AC \times BC = 36\times 2\)
\(⇒ AC \times BC = 72\)
We are given that \(AC = 2BC\)
\(⇒ 2BC \times BC = 72\)
\(⇒ BC^2 = 36\)
\(⇒ BC = 6.\) (cannot be -6 because its a side)
Now we know \(BC = 6\).
\(AC = 2BC = 2\times 6 = 12. \)
Using Pythagoras Theorem,\( AB^2 = BC^2 + CA^2\)
\(⇒ AB^2 = 36 + 144\)
\(⇒ AB^2 = 180\)
\(⇒ AB = \sqrt{180}\)
\(⇒ AB = 6\sqrt{5}\)
Hence the right answer is Option C.
We know area od triangle \(= \frac{1}{2} \times base \times height\)
Looking at the triangle we can say: area \(= \frac{1}{2} ]times AC \times BC\)
\(⇒ \frac{1}{2} \times AC \times BC= 36 \)
\(⇒ AC \times BC = 36\times 2\)
\(⇒ AC \times BC = 72\)
We are given that \(AC = 2BC\)
\(⇒ 2BC \times BC = 72\)
\(⇒ BC^2 = 36\)
\(⇒ BC = 6.\) (cannot be -6 because its a side)
Now we know \(BC = 6\).
\(AC = 2BC = 2\times 6 = 12. \)
Using Pythagoras Theorem,\( AB^2 = BC^2 + CA^2\)
\(⇒ AB^2 = 36 + 144\)
\(⇒ AB^2 = 180\)
\(⇒ AB = \sqrt{180}\)
\(⇒ AB = 6\sqrt{5}\)
Hence the right answer is Option C.
DHRJ0032
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05 Jul 2021, 10:45
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let side BC be x
then side AC will be 2x.
now, for right angles triangle,
1/2(x)(2x)=36
x comes out to be 6.
So,BC=6,
AC=12
applying pythagoras theorum,
AB=root180
i.e 6root5
C it is.
then side AC will be 2x.
now, for right angles triangle,
1/2(x)(2x)=36
x comes out to be 6.
So,BC=6,
AC=12
applying pythagoras theorum,
AB=root180
i.e 6root5
C it is.
