Question

${\underline {\frak{\pink{~Question :-~}}}}$

$\triangle$ABC is right angled at A. AD is perpendicular to BC. If AB = 5cm, BC = 13cm and AC = 12cm. Find the area of $\triangle$ABC. Also find the Length of AD.


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Answer 1

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Answer 2

Given -

• ABC is a triangle.

• AB = 5 cm

• BC = 13 cm

• AC = 12 cm

To find -

• Area of the triangle ABC and length of AD

Formula used -

• Area of triangle

Solution -

In the question, we are provided with the length of AB, BC & AC, and we need to find the area if the triangle + length of AD, for that first by applying the formula of area of triangle, we will find the area of ∆ABC, After that we will find the length of AD, by again applying the same formula as given above.

According to question -

• Length of AB (h) = 5 cm

• Length of BC = 13 cm

• Length of AC (b) = 12 cm

Area of Triangle -

• $\bf\dfrac{1}{2} \: \times b \: \times h$

On substituting the values -

$\bf \longrightarrow \: A = \: \dfrac{1}{ \cancel2^{(1)} } \: \times \cancel{12}^{(6)} \: cm \: \times 5 \: cm \\ \\ \bf \longrightarrow \: A \: = 6 \: cm \: \times 5 \: cm \\ \\ \bf \longrightarrow \: A \: = 30 { \: cm}^{2}$

Now -

We will find the length of AD, by again applying the formula of area of Triangle. In this we will take AD as base and AC as height.

$\bf \longrightarrow \: A \: = \dfrac{1}{2} \: \times \: b \: \times \: h \\ \\ \bf \longrightarrow \: 30 { \: cm}^{2} \: = \dfrac{1}{2} \: \times \: AD \: \times 13 \: cm \\ \\ \bf \longrightarrow \: AD \: = \frac{30 \: cm \: \times 2}{13 \: cm} \\ \\ \bf \longrightarrow \: AD \: = \dfrac{60}{13}$

$\therefore$ The area of ∆ABC is 30cm² and length of AD is 60/13 cm

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