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In triangle $A B C$ shown in the above figure, $B C=3$ and $A C=4$. Wh
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07 Jan 2025, 02:21
07 Jan 2025, 02:21
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In triangle $A B C$ shown in the above figure, $B C=3$ and $A C=4$. What is the length of the line segment $\(C D\)$ ?
(A) 3
(B) $\(\frac{15}{4}\)$
(C) 5
(D) $\(\frac{16}{3}\)$
(E) $\(\frac{20}{3}\)$
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Re: In triangle $A B C$ shown in the above figure, $B C=3$ and $A C=4$. Wh
[#permalink]
18 Feb 2025, 01:53
18 Feb 2025, 01:53
Expert Reply
OFFICIAL EXPLANATION
In
right angle
$\(\triangle \mathrm{ABC}\)$,
$$
\(\mathrm{AB}=\sqrt{3^2+4^2}=\sqrt{25}=5\)
$$
(Using Pythagoras theorem - Hypotenuse $\(^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ )
Let the length of $\(C D=x\)$, so in right triangle $\(A C D\)$, we get $\(A D=\sqrt{4^2+x^2}=\sqrt{16+x^2}\)$
Now, applying Pythagoras theorem in triangle $\(A B D\)$, we get $\(A B^2+A D^2=B D^2\)$ i.e. $\(5^2+\left(\sqrt{16+x^2}\right)^2=(3+x)^2\)$ which when simplified gives $\(41=9+6 x \Rightarrow x=\frac{32}{6}=\frac{16}{3}\)$
Hence the answer is (D).
In
right angle
$\(\triangle \mathrm{ABC}\)$,
$$
\(\mathrm{AB}=\sqrt{3^2+4^2}=\sqrt{25}=5\)
$$
(Using Pythagoras theorem - Hypotenuse $\(^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ )
Let the length of $\(C D=x\)$, so in right triangle $\(A C D\)$, we get $\(A D=\sqrt{4^2+x^2}=\sqrt{16+x^2}\)$
Now, applying Pythagoras theorem in triangle $\(A B D\)$, we get $\(A B^2+A D^2=B D^2\)$ i.e. $\(5^2+\left(\sqrt{16+x^2}\right)^2=(3+x)^2\)$ which when simplified gives $\(41=9+6 x \Rightarrow x=\frac{32}{6}=\frac{16}{3}\)$
Hence the answer is (D).
gmatclubot
Re: In triangle $A B C$ shown in the above figure, $B C=3$ and $A C=4$. Wh [#permalink]
18 Feb 2025, 01:53
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