Question

ΔABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ΔABC. Also find the length of AD.

Answer 1

I hope you got your solution

Answer 2

Given:

• ΔABC is a right angled triangle at A.
• AD is perpendicular to BC means AD is Height and BC is base.
• AB = 5cm, BC = 13cm, AC = 12cm

To Attain:

• Area of Triangle
• Length of AD

☀️ Explanation:

In this Question, We are provided with that ΔABC is a right angled Triangle. In this right angled Triangle, AD is Height and BC is base. The value of AB is 5cm means it's the base and value of BC is 13cm means it's the height of Triangle. then we are asked to find the Area of Triangle. by applying the formula of Triangle, we will know the area of Triangle. after it that is also asked to find the Length of AD. So, to find the length of AD we will divide the area of Triangle with the base if Triangle!

Solution:

60/13cm

Step by Step Explanation:

$: \implies \sf \: Area \: of \: Δ = \dfrac{1}{2} \times B \times H \\ \\ \\ : \implies \sf \: Area \: of \: Δ = \dfrac{1}{ \cancel{2}} \times 5 \times \cancel{12} \\ \\ \\ : \implies \sf {30cm}^{2}$

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$\therefore \underline {\sf{Hence, \: area \: of \: triangle \: is \: \pmb {30cm}^{2} }}$

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Now, We have to find the length of AD!

⠀⠀

$: \implies \sf \: Area \: of \: Δ = \sf\dfrac{1}{2} \times B \times H \\ \\ \\ : \implies \sf \: \dfrac{1}{2} \times bc \times ad = {30cm}^{2} \\ \\ \\ : \implies \sf \: \dfrac{1}{2} \times 13 \times ad = {30cm}^{2} \\ \\ \\ : \implies \sf \: ad \: = 30 \times \dfrac{2}{13} \\ \\ \\ : \implies \sf \dfrac{60}{13} cm \: or \: 4.61cm$

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$\therefore \underline{\sf{Length \: of \: AD \: is \: {\pmb \: \dfrac{60}{13} cm}}}$