Question

ABC is a triangle where, angle C = 90°. Let BC = a, CA= b, AB = c and let p be the length of perpendicular from C on AB. prove that 1/p²=1/a²+1/b²​

Answer 1

Question :-

ABC is a triangle where, angle C = 90°. Let BC = a, CA= b, AB = c and let p be the length of perpendicular from C on AB. prove that 1/p²=1/a²+1/b²

$\large\underline{\sf{Solution-}}$

Given that,

• ABC is a right angle triangle right-angled at C.

• BC = a, CA = b, AB = c and p be the length of perpendicular from C on AB.

Let assume that

• Perpendiculars drawn from C intersect AB at D

Now, we know

$\rm \: Area_{(\triangle\:ABC)} = \dfrac{1}{2} \times AC \times BC = \dfrac{1}{2} \times CD \times AB$

$\rm \: \dfrac{1}{2} \times AC \times BC = \dfrac{1}{2} \times CD \times AB$

$\rm \: AC \times BC = CD \times AB$

$\rm \: b \times a = p \times c$

$\rm\implies \:c = \dfrac{ab}{p} \cdots \cdots \: (1)$

Now, In right triangle ABC,

Using Pythagoras Theorem, we have

$\rm \: {AB}^{2} = {BC}^{2} + {AC}^{2}$

$\rm \: {c}^{2} = {a}^{2} + {b}^{2}$

$\rm \: {\bigg(\dfrac{ab}{p} \bigg) }^{2} = {a}^{2} + {b}^{2}$

$\rm \: \dfrac{ {a}^{2} {b}^{2} }{ {p}^{2} } = {a}^{2} + {b}^{2}$

$\rm \: \dfrac{1}{ {p}^{2} } = \dfrac{{a}^{2}}{{a}^{2}{b}^{2}} + \dfrac{{b}^{2}}{{a}^{2}{b}^{2}}$

$\rm \: \dfrac{1}{ {p}^{2} } = \dfrac{1}{{b}^{2}} + \dfrac{1}{{a}^{2}}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{1}{ {p}^{2} } = \dfrac{1}{{a}^{2}} + \dfrac{1}{{b}^{2}} \: \: }}$

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Additional Information :-

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem,

This theorem states that : If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

Answer 2

Question :-

ABC is a triangle where, angle C = 90°. Let BC = a, CA= b, AB = c and let p be the length of perpendicular from C on AB. prove that $\displaystyle\rm\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$

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Since $\rm \triangle A B C$ is a right-angled triangle with $\rm \angle C=90^{\circ}$

$\[ \begin{array}{l} \displaystyle\rm \therefore A B^{2}=B C^{2}+A C^{2} \\\\ \displaystyle\rm \Rightarrow c^{2}=a^{2}+b^{2} \\\\ \displaystyle\rm \Rightarrow\left(\frac{a b}{p}\right)^{2}=a^{2}+b^{2} \\\\ \displaystyle\rm \therefore c p=a b \\\\ \displaystyle\rm \Rightarrow c=\frac{a b}{p} \\ \\ \displaystyle\rm\Rightarrow \frac{1}{p^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}} \\ \\ \displaystyle\rm\Rightarrow \frac{1}{p^{2}}=\frac{1}{b^{2}}+\frac{1}{a^{2}} \\\\ \displaystyle\rm \text { Thus } \frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} \end{array} \]$