Question

7. 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

Answer:

$\huge\tt{\underline{Question}}$

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.

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Given that:-

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

To find:-

• The area of triangle ABC
• Length of AD

Answer:-

From the question it is given that,

AB = 5 cm, BC = 13 cm, AC = 12 cm

We know that,

Area of the ΔABC = ½ × base × height

= ½ × AB × AC

= ½ × 5 × 12

= 1 × 5 × 6

= 30 cm2

Now,

Area of ΔABC = ½ × base × height

30 = ½ × AD × BC

30 = ½ × AD × 13

(30 × 2)/13 = AD

AD = 60/13

AD = 4.6 cm

• Therefore area of triangle ABC = 30cm² and length of AD = 4.6cm

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Answer 2

$\huge{\underline{\mathtt{\red{QU}\pink{E}\green{S}\blue{T}\purple{I}\orange{ON}}}}$

✿ΔABC is right-angled at A. AD is perpendicular to BC, AB=5cm, BC=13 cm and AC=12 cm. Find the area of ΔABC. Also, find the length of AD.

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$\huge{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

$\boxed{\rm{\orange{Given \longrightarrow }}}$

ΔABC is right-angled at "A".

AD is perpendicular to BC.

BC=13 cm

AB=5cm

AC=12 cm

$\boxed{\rm{\red{To\:Find\longrightarrow }}}$

Area of ΔABC

Length of AD

$\boxed{\rm{\pink{Explanation\longrightarrow }}}$

In right angles triangle BAC, AB=5cm and AC=12cm

Area of triangle,

$\sf\purple{↬} \dfrac{1}{2}×base×height$

$\sf\green{↬} \dfrac{1}{2}×AB×AC$

$\sf\orange{↬}\dfrac{1}{2}×5×12$ =30cm²

Now, in ΔABC,

Area of triangle ABC,

$\sf\pink{↬} \dfrac{1}{2}×BC×AD$

$\sf\blue{↬} \dfrac{1}{2}×30×AD$

$\sf\red{↬} AD=\dfrac{13×2}{13}=1360\:cm$

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