
If the height of a triangle is decreased by \[40\% \] and its base is increased by \[40\% \] what will be the effect on its area?
A) No change
B) \[8\% \] decrease
C) \[16\% \] decrease
D) \[16\% \] increase
E) None of these
Answer
592.2k+ views
Hint: The area of the triangle is given by \[A = \dfrac{1}{2} \times b \times h\] where b is the base and h is the height of the triangle get the area of the triangle by reducing the height by \[40\% \] and increasing the base by \[40\% \] then compare with the known formula to get the final results.
Complete Step by Step Solution:
We know that the area of a triangle is given by \[A = \dfrac{1}{2} \times b \times h\] where b is the base and h is the height of the triangle.
Let us assume that the new triangle formed by reducing the height and increasing the base is given by \[A' = \dfrac{1}{2} \times b' \times h'\] Where \[b'\] is the new base, \[h'\] is the new height and \[A'\] is the new area.
Now using this let us try to get the height when it will be reduced by \[40\% \]
\[\begin{array}{l}
\therefore b' = b + 40\% \times b\\
= b + \dfrac{{40}}{{100}}b\\
= 1.4b
\end{array}\]
Similarly for height we will have
\[\begin{array}{l}
\therefore h' = h - 40\% \times h\\
= h - \dfrac{{40}}{{100}}h\\
= 0.6h
\end{array}\]
Now putting all of this in the area formula i.e.,
\[\begin{array}{l}
A' = \dfrac{1}{2} \times b' \times h'\\
= \dfrac{1}{2} \times \left( {1.4b} \right) \times \left( {0.6h} \right)\\
= 0.84 \times \dfrac{1}{2} \times b \times h
\end{array}\]
Now we know that \[A = \dfrac{1}{2} \times b \times h\]
Putting this we get it as
\[A' = 0.84A\]
Clearly \[A > A'\]
Therefore if we subtract A’ from A we will get that how much less it is
\[\begin{array}{l}
\therefore A - A' = A - 0.84A\\
= 0.16A
\end{array}\]
So if we convert it into percent we will get that \[16\%\] of A
Therefore A’ is reduced by 16% of A
Which means option C is correct. It is a 16% decrease.
Note: Do note that getting a relation between the old area and the new area is a key step. The ultimate goal in these types of questions must be to establish a relation between the new and the old value.
Complete Step by Step Solution:
We know that the area of a triangle is given by \[A = \dfrac{1}{2} \times b \times h\] where b is the base and h is the height of the triangle.
Let us assume that the new triangle formed by reducing the height and increasing the base is given by \[A' = \dfrac{1}{2} \times b' \times h'\] Where \[b'\] is the new base, \[h'\] is the new height and \[A'\] is the new area.
Now using this let us try to get the height when it will be reduced by \[40\% \]
\[\begin{array}{l}
\therefore b' = b + 40\% \times b\\
= b + \dfrac{{40}}{{100}}b\\
= 1.4b
\end{array}\]
Similarly for height we will have
\[\begin{array}{l}
\therefore h' = h - 40\% \times h\\
= h - \dfrac{{40}}{{100}}h\\
= 0.6h
\end{array}\]
Now putting all of this in the area formula i.e.,
\[\begin{array}{l}
A' = \dfrac{1}{2} \times b' \times h'\\
= \dfrac{1}{2} \times \left( {1.4b} \right) \times \left( {0.6h} \right)\\
= 0.84 \times \dfrac{1}{2} \times b \times h
\end{array}\]
Now we know that \[A = \dfrac{1}{2} \times b \times h\]
Putting this we get it as
\[A' = 0.84A\]
Clearly \[A > A'\]
Therefore if we subtract A’ from A we will get that how much less it is
\[\begin{array}{l}
\therefore A - A' = A - 0.84A\\
= 0.16A
\end{array}\]
So if we convert it into percent we will get that \[16\%\] of A
Therefore A’ is reduced by 16% of A
Which means option C is correct. It is a 16% decrease.
Note: Do note that getting a relation between the old area and the new area is a key step. The ultimate goal in these types of questions must be to establish a relation between the new and the old value.
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