Question

triangle ABC is right angled at A. AD is perpendicular to BC. if ab is 5 cm, BC is 13 cm , ac is 12 cm , find area of ABC, also find length of ad​

Answer 1

Answer is 30cm^2 and 4.61 cm.

Step by step explain :)

First we find area of ∆ABC from Heron's formula.

Cemi perimeter of ∆ABC :)

$\: s= \frac{a + b + c}{2} \\ = \frac{5 + 13 + 12}{2} = 15cm$

area of ∆ABC :)

$= \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{15(15 - 5)(15 - 13)(15 - 12)} \\ = \sqrt{15 \times 10 \times 2 \times 3} \\ = \sqrt{3 \times 5 \times 5 \times 2 \times 2 \times 3} \\ = 3 \times 5 \times 2 \\ = 30 {cm}^{2}$

Second we find length of AD :)

We know that area of triangle is

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

Here we have area of ∆ABC =

$= 30 {cm}^{2}$

Then AD :)

$30 {cm}^{2} = \frac{1}{2} \times 13 \times height \\ = \frac{30 \times 2}{13} \\ = \frac{60 }{13} \\ = 4.61cm$

There fore AD =[【 4.61 cm 】]

OR

Take BD = x than DC = 13 - x

In ∆ABD,

AD^2 = AB^2 - BD^2 =

$= 25 - {x}^{2}$

In ∆ADC,

AD^2 =.AC^2 - DC^2

$= 144 - {(13 - x)}^{2} \\ = 144 - (169 - 26x + {x}^{2} ) \\ = 144 - 169 + 26x - {x}^{2}$

AD = AD

$25 - {x}^{2} = 144 - 169 + 26x - {x}^{2} \\ 25 = - 25 + 26x \\ 26x = 50 \\ x = \frac{50}{26} = 1.92$

AD^2 = 25 - x^2

put values of x

$= \sqrt{25 - {(1.92)}^{2} } \\ = \sqrt{25 - 3.6864} \\ = \sqrt{21.3134 } \\ = 4.61cm$

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Answer 2

Answer:

Step-by-step explanation:

Answer is 30cm^2 and 4.61 cm.

Step by step explain :)

First we find area of ∆ABC from Heron's formula.

Cemi perimeter of ∆ABC :)

area of ∆ABC :)

Second we find length of AD :)

We know that area of triangle is

Here we have area of ∆ABC =

Then AD :)

There fore AD =[【 4.61 cm 】]

OR

Take BD = x than DC = 13 - x

In ∆ABD,

AD^2 = AB^2 - BD^2 =

In ∆ADC,

AD^2 =.AC^2 - DC^2

AD = AD

AD^2 = 25 - x^2

put values of x

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