Questions & Answers

Question

Answers

A. \[{75^0}\]

B. \[{90^0}\]

C. \[{120^0}\]

D. \[{135^0}\]

Answer

Verified

128.4k+ views

Let us consider a \[\Delta ABC\] with side lengths \[a,b,c\] units.

Given that the altitudes of the \[\Delta ABC\] are 12,15 and 20 units as shown in the below figure:

We know that if \[h\] is the height of the triangle and \[x\] is the length of base of the triangle then the area of that triangle is given by \[\dfrac{1}{2} \times x \times h\].

So, area of \[\Delta ABC\] \[ = \dfrac{1}{2} \times a \times 12 = \dfrac{1}{2} \times b \times 15 = \dfrac{1}{2} \times c \times 20\]

Let \[\dfrac{1}{2} \times a \times 12 = \dfrac{1}{2} \times b \times 15 = \dfrac{1}{2} \times c \times 20 = k\]

Taking first and last term we have

\[ \Rightarrow \dfrac{1}{2} \times a \times 12 = k \\

\therefore a = \dfrac{{2k}}{{12}} = \dfrac{{2k}}{{12}} \times \dfrac{5}{5} = \dfrac{{10k}}{{60}} \]

Taking the second and last term we have

\[ \Rightarrow \dfrac{1}{2} \times b \times 15 = k \\

\therefore b = \dfrac{{2k}}{{15}} = \dfrac{{2k}}{{15}} \times \dfrac{4}{4} = \dfrac{{8k}}{{60}} \]

Taking the third and the last term we have

\[ \Rightarrow \dfrac{1}{2} \times c \times 20 = k \\

\therefore c = \dfrac{{2k}}{{20}} = \dfrac{{2k}}{{20}} \times \dfrac{3}{3} = \dfrac{{6k}}{{60}} \]

Let us consider \[{b^2} + {c^2}\]

\[ \Rightarrow {b^2} + {c^2} = {\left( {\dfrac{{8k}}{{60}}} \right)^2} + {\left( {\dfrac{{6k}}{{60}}} \right)^2} \\

\Rightarrow {b^2} + {c^2} = \dfrac{{64{k^2}}}{{3600}} + \dfrac{{36{k^2}}}{{3600}} \\

\therefore {b^2} + {c^2} = \dfrac{{100{k^2}}}{{3600}}...............................\left( 1 \right) \]

And consider the value of \[{a^2}\]

\[ \Rightarrow {a^2} = {\left( {\dfrac{{10k}}{{60}}} \right)^2} \\

\therefore {a^2} = \dfrac{{100{k^2}}}{{3600}}...................................\left( 2 \right) \]

From equation (1) and (2), we have

\[{b^2} + {c^2} = {a^2}\]

We know that Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides” i.e., \[{\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Adjacent side}}} \right)^2} + {\left( {{\text{Opposite side}}} \right)^2}\].

So, clearly the \[\Delta ABC\] with sides \[a,b,c\] forms a right-angle triangle and it can be redrawn as:

We know that in a right-angle triangle the greatest angle is \[{90^0}\].

Thus, the greatest angle in \[\Delta ABC\] is \[{90^0}\].