
ABC is a right-angled triangle, right-angled at A and AD is the altitude on BC. If $AB:AC=3:4$, then what is the ratio of $BD:DC$?
A. $3:4$
B. $9:16$
C. $2:3$
D. $1:2$
Answer
509.4k+ views
Hint: For solving this question you should know about the general properties of triangle and right triangle also. Here, we will find the side of the triangle and the common angle from that and then we will compare those angles. We will then compare again with the sides for calculating the new ratio of $BD:DC$.
Complete step-by-step solution:
According to our question we have ABC, a right-angled triangle, right angled at A and AD is the altitude on BC. If $AB:AC=3:4$, then we have to find the ratio of $BD:DC$. So, if we see our triangles $\Delta ABC$ and $\Delta DBA$, we can see that angle B is common in both the triangles. So, we can write it here as:
In $\Delta ABC$ and $\Delta DBA$, $\angle B$ is common. Then,
$\begin{align}
& \angle CAB=\angle BDA={{90}^{\circ }} \\
& \Rightarrow \Delta ABC\sim \Delta DBA \\
\end{align}$
And we know that if two right-angled triangles are similar to each other then the ratio of their sides will be $\dfrac{3}{4}$. So, we can write it as,
$\begin{align}
& \Rightarrow \dfrac{AB}{AC}=\dfrac{DB}{DA} \\
& \because \dfrac{DB}{DA}=\dfrac{3}{4} \\
& \Rightarrow \dfrac{AB}{AC}=\dfrac{DB}{DA}=\dfrac{3}{4} \\
& \Rightarrow AD=\dfrac{4}{3}DB\ldots \ldots \ldots \left( i \right) \\
\end{align}$
If we see in $\Delta ABD$ and $\Delta ADC$, then,
$\angle DAB=\angle ACD$
Since the third angles of similar triangles $ABC$ and $DBA$ is,
$\begin{align}
& \angle ADB=\angle ADC={{90}^{\circ }} \\
& \therefore \dfrac{AB}{AC}=\dfrac{AD}{CD}=\dfrac{3}{4} \\
& \Rightarrow AD=\dfrac{3}{4}CD\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
Now, from equations (i) and (ii), we get,
$\begin{align}
& \dfrac{4}{3}DB=\dfrac{3}{4}CD \\
& \Rightarrow \dfrac{BD}{CD}=\dfrac{3}{4}\times \dfrac{3}{4}=\dfrac{9}{16} \\
\end{align}$
So, the correct answer is option B, $9:16$.
Note: While solving this type of questions you should compare both the triangles to get the values of ratios of the sides of the triangles with each other. So, if we have been given any one of the angles or sides of the triangle, then we can calculate the other if they are similar to each other.
Complete step-by-step solution:
According to our question we have ABC, a right-angled triangle, right angled at A and AD is the altitude on BC. If $AB:AC=3:4$, then we have to find the ratio of $BD:DC$. So, if we see our triangles $\Delta ABC$ and $\Delta DBA$, we can see that angle B is common in both the triangles. So, we can write it here as:
In $\Delta ABC$ and $\Delta DBA$, $\angle B$ is common. Then,
$\begin{align}
& \angle CAB=\angle BDA={{90}^{\circ }} \\
& \Rightarrow \Delta ABC\sim \Delta DBA \\
\end{align}$
And we know that if two right-angled triangles are similar to each other then the ratio of their sides will be $\dfrac{3}{4}$. So, we can write it as,
$\begin{align}
& \Rightarrow \dfrac{AB}{AC}=\dfrac{DB}{DA} \\
& \because \dfrac{DB}{DA}=\dfrac{3}{4} \\
& \Rightarrow \dfrac{AB}{AC}=\dfrac{DB}{DA}=\dfrac{3}{4} \\
& \Rightarrow AD=\dfrac{4}{3}DB\ldots \ldots \ldots \left( i \right) \\
\end{align}$
If we see in $\Delta ABD$ and $\Delta ADC$, then,
$\angle DAB=\angle ACD$
Since the third angles of similar triangles $ABC$ and $DBA$ is,
$\begin{align}
& \angle ADB=\angle ADC={{90}^{\circ }} \\
& \therefore \dfrac{AB}{AC}=\dfrac{AD}{CD}=\dfrac{3}{4} \\
& \Rightarrow AD=\dfrac{3}{4}CD\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
Now, from equations (i) and (ii), we get,
$\begin{align}
& \dfrac{4}{3}DB=\dfrac{3}{4}CD \\
& \Rightarrow \dfrac{BD}{CD}=\dfrac{3}{4}\times \dfrac{3}{4}=\dfrac{9}{16} \\
\end{align}$
So, the correct answer is option B, $9:16$.
Note: While solving this type of questions you should compare both the triangles to get the values of ratios of the sides of the triangles with each other. So, if we have been given any one of the angles or sides of the triangle, then we can calculate the other if they are similar to each other.
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