Question

Two sides of a triangle are equal in length. The base of the triangle (the side that isn’t equal in length to the other two) measures \[7.6\,cm\]. The two equal sides both measure \[5\,cm\].What is the area of the triangle?

Answer 1

Hint: In order to find the area, we first need to find the height of the triangle. For that, we will draw a perpendicular from a vertex to the base. After drawing the altitude, we will get two right triangles. We will apply Pythagoras Theorem in one of these right triangles and find the altitude of the triangle. After finding the altitude, we will find the area of the triangle using the formula \[\dfrac{1}{2} \times base \times height\].

Complete step by step answer:
Let us first draw the figure of the given question.

We have,
\[ \Rightarrow Base = BC = 7.6cm\]
\[ \Rightarrow AB = AC = 5cm\]
\[ \Rightarrow Height = AD\]
Now, since two sides of the triangle are equal, the triangle is an isosceles triangle. In an isosceles triangle, the altitude (Height) bisects the base of the triangle.Hence, we get
\[ \Rightarrow BD + DC = BC = 7.6cm - - - - - - (1)\]

Now, since \[AD\] bisects the base \[BC\], we have \[BD = DC\]
Using \[BD = DC\] in (1), we get
\[ \Rightarrow BD + BD = 7.6cm\]
\[ \Rightarrow 2BD = 7.6cm\]
Dividing both the sides by \[2\], we get
\[ \Rightarrow BD = 3.8cm\]
Hence, we get \[BD = DC = 3.8cm\]

Now, considering the diagram,

Now, we have two right triangles, namely, \[\vartriangle ABD\] and \[\vartriangle ADC\].
In \[\vartriangle ADC\], we have
\[ \Rightarrow Base = DC = 3.8cm\]
\[ \Rightarrow Hypotenuse = AC = 5cm\]
\[ \Rightarrow Perpendicular = AD\]
According to Pythagoras Theorem, \[{\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Perpendicular} \right)^2}\]
Now, applying Pythagoras Theorem in \[\vartriangle ADC\], we get
\[ \Rightarrow {\left( {AC} \right)^2} = {\left( {DC} \right)^2} + {\left( {AD} \right)^2}\]
Now, substituting the values, we get
\[ \Rightarrow {\left( 5 \right)^2} = {\left( {3.8} \right)^2} + {\left( {AD} \right)^2}\]

Now, finding the values of squared terms, we get
\[ \Rightarrow 25 = 14.44 + {\left( {AD} \right)^2}\]
Reshuffling the terms, we get
\[ \Rightarrow 25 - 14.44 = {\left( {AD} \right)^2}\]
Now, solving the Left Hand Side, we get
\[ \Rightarrow 10.56 = {\left( {AD} \right)^2}\]
Now, taking square root both the sides, we get
\[ \Rightarrow \pm 3.25 = AD\]
Since the side of a triangle cannot be negative, we get
\[ \Rightarrow AD = 3.25cm\]
Hence, in \[\vartriangle ABC\], we get
\[ \Rightarrow Base = BC = 7.6cm\]
\[ \Rightarrow Height = AD = 3.25cm\]

Now, using the formula \[\dfrac{1}{2} \times base \times height\] for the area of the triangle, we get
\[ \Rightarrow \]Area of \[\vartriangle ABC\]\[ = \dfrac{1}{2} \times BC \times AD\]
Now, substituting the values, we get
\[ \Rightarrow \]Area of \[\vartriangle ABC\]\[ = \dfrac{1}{2} \times \left( {7.6cm} \right) \times \left( {3.25cm} \right)\]
Now, on solving, we get
\[ \Rightarrow \]Area of \[\vartriangle ABC\]\[ = \left( {3.8cm} \right) \times \left( {3.25cm} \right)\]
Now, multiplying the numeric values, we get
\[ \therefore \]Area of \[\vartriangle ABC\]\[ = 12.35c{m^2}\]

Hence, we get, area of \[\vartriangle ABC\]\[ = 12.35c{m^2}\].

Note: We can also solve this question using Heron’s Formula. According to Heron’s Formula,
Area of the Triangle \[ = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \], where \[a,b,c\] are the sides of the triangle and \[s = \dfrac{{a + b + c}}{2}\]. We call \[s = \dfrac{{a + b + c}}{2}\] to be the semi-perimeter of the triangle. Also, we can use the direct formula for the area of the isosceles triangle. The formula is given as:
Area of an isosceles triangle \[ = \left( {\dfrac{1}{2} \times b \times \sqrt {{a^2} - \dfrac{{{b^2}}}{4}} } \right)uni{t^2}\], where \[a\] is the length of the equal sides and \[b\] is the base of the isosceles triangle.