Question

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In a ∆ABC, ∠A = 90°, AB = 5 cm and AC = 12 cm. If AD⊥BC, then AD=
(a) 13/2 cm
(b) 60/13 cm
(c) 13/60 cm
(d) 2√15/13 cm

$\sf \underline {Explain \: Your \: Answer \: Properly}$
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Answer 1

Answer:-

(b) $\frac {60}{13}$

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Given:-

In ΔABC,

✯ ∠A = 90º

✯ AD is perpendicular to BC.

✯ AC = 12 cm

✯ AB = 5 cm.

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To Find:-

✯ length of AD

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Solution:-

In ΔACB and ΔADC

∠C = ∠C (common angle)

∠A = ∠ADC (90° because AD is perpendicular to BC)

Therefore, ΔACB ~ ΔADC (by AA similarity criterion)

Therefore, by the property of similar triangles, we get:-

$\frac{AD}{AB} = \frac{AC}{BC}$

$\frac{AD}{5} = \frac{12}{13}$

$AD = \frac{12 \times 5}{13}$

$AD = \frac{60}{13}$

Therefore, the correct answer is $\frac {60}{13}$.

Answer 2

Explanation:

The given side of triangle are 5, 12 and 13.

The semi-perimeter s of given triangle with sides

a = 5, b = 12 and c = 13 are given as

s = a+b+c/2

= 5+12+13/2

= 30/2

= 15.

Now, using Heron's formula, the area Δ of given triangle is given as,

Δ = √s(s-a)(s-b)(s-c)

= √15(15-5)(15-12)(15-13)

= 30.

If h is the length of perpendicular to the side c=13

drawn from opposite vertex then the area of given triangle.

Δ = 1/2×(13)×(h)

30 = 13/2 × h

h = 60/13.

Hope it may helps you...