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RHS Criteria for Congruency

ABC is a righ...

Question

$△ A B C$ is a right angled in which $L C = 90^{o}$ and $C D ⊥ A B$ . If $B C = a, C A = b, A B = c$ and $C D = p$ then prove that
$c p = a b$

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Solution

Area of triangle is $b a s e \times h e i g h t$
When $c$ is taken as base Then, $p$ becomes height so the area is $\frac{1}{2} c p$
When $b$ is taken as base Then, $a$ becomes height so the area is $\frac{1}{2} a b$
Since area is same
$⟹ a b = c p$

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Similar questions

△ A B C

is a right angled in which

L C = 90^{o}

C D ⊥ A B

B C = a, C A = b, A B = c

C D = p

then prove that

\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

A B C

is a right angled triangle with

∠ C = 90^{o}, B C = a, A C = b, C D ⊥ A B

C D = P

\frac{1}{P^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

Q. In a right angled

△ A B C

, right angled at

A

A D ⊥ B C

A D = p

B C = a, C A = b

A B = c

A B C

is a right triangle, right-angled at

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B C = a, C A = b, A B = c

p

be the length of perepndicular from

C

A B

, prove that
i)

c p = a b

\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

Q. ABC is a right triangle, right-angled at C. If BC = a,CA = b, AB = c and p be the length of perpendicular from C on AB, then
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c p = a b

\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

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