Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Fig $1$ and Fig $2$ show an equilateral triangle divided into thirds and quarters respectively. They are combined in Fig $3$ . Calculate the fraction of Fig $3$ that is shaded.\begin{align} & \left( A \right)\dfrac{7}{12} \\ & \left( B \right)\dfrac{5}{12} \\ & \left( C \right)\dfrac{5}{24} \\ & \left( D \right)\dfrac{11}{24} \\ \end{align}

Last updated date: 11th Aug 2024
Total views: 390.9k
Views today: 5.90k
Verified
390.9k+ views
Hint: At first, we need to break down the given shaded area into simpler areas which are easier to find out from $Fig1$ and $Fig2$ . We then add and subtract these fundamental areas from one another to find out the shaded area of $Fig3$ . As we can see, these fundamental areas are one out of the three areas in $Fig1$ and its half, and one out of the four areas in $Fig2$ and its half.

We are given the question that in $Fig1$ , the triangle is divided into three equal parts. The shaded area in $Fig4$ is nothing but one of the individual three areas in $Fig1$ . Thus, if we assume the total area of the triangle to be $A$ , then this shaded area will be one third of the total area, that is $\dfrac{A}{3}$ .
In the shaded part in $Fig3$ , we can clearly see that the shaded area in $Fig4$ has not been entirely taken. Therefore, we need to take away one half of the shaded area in $Fig4$ . This area which needs to be taken away is half that of the shaded area in $Fig4$ which is $\dfrac{1}{2}\times \dfrac{A}{3}=\dfrac{A}{6}$ .
After the above two operations, we see that the shaded area in $Fig3$ has still not been reached. So, we now add the shaded area of $Fig6$ . In $Fig2$ , we have been told that the triangle has been divided equally into four equal areas. Again, if we observe closely, we can see that the shaded area in $Fig6$ is one third of the individual triangle of $Fig2$ . So, we can say that the shaded area of $Fig6$ is $\dfrac{1}{3}\times \dfrac{A}{4}=\dfrac{A}{12}$ .
Again, we need to subtract half the area added previously as more than required had been taken. Half of the previous area means $\dfrac{1}{2}\times \dfrac{A}{12}=\dfrac{A}{24}$ .
After all these operations, we can see that the final area is equal to the area which was required to be found out. This area is equal to $\dfrac{A}{3}-\dfrac{A}{6}+\dfrac{A}{12}-\dfrac{A}{24}=\dfrac{A}{6}+\dfrac{A}{24}=\dfrac{5A}{24}$ . If we want to find out what fraction of area is the shaded one out of the whole triangle, then we need to divide them, which is $\dfrac{\dfrac{5A}{24}}{A}=\dfrac{5}{24}$ .
Therefore, we can conclude that the shaded area in $Fig3$ is $\dfrac{5}{24}$ of the triangle.