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In triangle $A B C$, point $D$ divides the side $A B$ such that $\frac
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06 Aug 2025, 07:09
06 Aug 2025, 07:09
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In triangle $\(A B C\)$, point $D$ divides the side $\(A B\)$ such that $\(\frac{A D}{D B}=\frac{1}{3}\)$. If the area of the triangle $\(A B C\)$ is ' $\(r\)$ ', then what is the area of the triangle $\(A D C\)$ in terms of ' $\(r\)$ '?
(A) $\(\frac{r}{2}\)$
(B) $\(\frac{r}{3}\)$
(C) $\(\frac{2 r}{3}\)$
(D) $\(\frac{r}{4}\)$
(E) $\(\frac{3 r}{4}\)$
(A) $\(\frac{r}{2}\)$
(B) $\(\frac{r}{3}\)$
(C) $\(\frac{2 r}{3}\)$
(D) $\(\frac{r}{4}\)$
(E) $\(\frac{3 r}{4}\)$
Re: In triangle $A B C$, point $D$ divides the side $A B$ such that $\frac
[#permalink]
15 Aug 2025, 10:43
15 Aug 2025, 10:43
Expert Reply
Given:
- Triangle $A B C$ with area $r$
- Point $D$ divides side $A B$ such that $\(\frac{A D}{D B}=\frac{1}{3}\)$
Step 1: Express lengths
Since $\(\frac{A D}{D B}=\frac{1}{3}\)$, the length $A B$ can be split as:
$$
\(A B=A D+D B\)
$$
If $\(A D=x\)$, then $\(D B=3 x\)$, so:
$$
\(A B=x+3 x=4 x\)
$$
Thus,
$$
\(A D=\frac{1}{4} A B, \quad D B=\frac{3}{4} A B\)
$$
Step 2: Use area ratios based on base segments
The triangles $A D C$ and $D B C$ share the same height from vertex $C$ onto $A B$.
The area of a triangle is proportional to its base length when the height is the same.
- Area of $\(\triangle A D C\)$ corresponds to base $\(A D=\frac{1}{4} A B\)$
- Area of $\(\triangle D B C\)$ corresponds to base $\(D B=\frac{3}{4} A B\)$
Step 3: Calculate area of $\(\triangle A D C\)$
Area of $\(\triangle A D C\)$ is:
$$
\(\operatorname{Area}(\triangle A D C)=\frac{A D}{A B} \times \operatorname{Area}(\triangle A B C)=\frac{1}{4} \times r=\frac{r}{4}\)
$$
Answer:
The area of $\(\triangle A D C\)$ is $\(\frac{r}{4}\)$, which corresponds to option (D).
- Triangle $A B C$ with area $r$
- Point $D$ divides side $A B$ such that $\(\frac{A D}{D B}=\frac{1}{3}\)$
Step 1: Express lengths
Since $\(\frac{A D}{D B}=\frac{1}{3}\)$, the length $A B$ can be split as:
$$
\(A B=A D+D B\)
$$
If $\(A D=x\)$, then $\(D B=3 x\)$, so:
$$
\(A B=x+3 x=4 x\)
$$
Thus,
$$
\(A D=\frac{1}{4} A B, \quad D B=\frac{3}{4} A B\)
$$
Step 2: Use area ratios based on base segments
The triangles $A D C$ and $D B C$ share the same height from vertex $C$ onto $A B$.
The area of a triangle is proportional to its base length when the height is the same.
- Area of $\(\triangle A D C\)$ corresponds to base $\(A D=\frac{1}{4} A B\)$
- Area of $\(\triangle D B C\)$ corresponds to base $\(D B=\frac{3}{4} A B\)$
Step 3: Calculate area of $\(\triangle A D C\)$
Area of $\(\triangle A D C\)$ is:
$$
\(\operatorname{Area}(\triangle A D C)=\frac{A D}{A B} \times \operatorname{Area}(\triangle A B C)=\frac{1}{4} \times r=\frac{r}{4}\)
$$
Answer:
The area of $\(\triangle A D C\)$ is $\(\frac{r}{4}\)$, which corresponds to option (D).
gmatclubot
Re: In triangle $A B C$, point $D$ divides the side $A B$ such that $\frac [#permalink]
15 Aug 2025, 10:43
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