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Q.

In a right triangle ABC, right-angled at C, a point 'D' is taken on AB such that CD is perpendicular to AB. Is the condition $\frac{1}{A C^{2}} + \frac{1}{BC} = \frac{1}{C D^{2}}$ true or false?

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a

True

b

False

answer is A.

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Detailed Solution

It is given that in a right triangle ABC, right-angled at C, a point 'D' is taken on AB such that CD is perpendicular to AB.
Let us draw the figure,
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(i) cp =

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(i) cp =

We know that the formula for area of a triangle is given by ar.=

\frac{1}{2} \times base \times height

.
So, for ΔABC,

a r Δ A B C = 12 × B C × A C a r Δ A B C = 12 a b

For ΔABC,

a r Δ A B C = 12 × A B × C D a r Δ A B C = 12 c p

We know that the areas must be same for the same triangle.
So,

12 a b = 12 c p a b = c p ...... 1

In the right-angled triangle ABC,

A B 2 = B C 2 + A C 2 a b p 2 = a 2 + b 2

Δ A B C

Hence, the given statement is true i.e,

\frac{1}{A C^{2}} + \frac{1}{BC} = \frac{1}{C D^{2}}

.
Hence, the option 1 is correct.

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