Question

10) Triangle ABC is right angled at A. AD is perpendicular to
BC. If AB = 8 cm, BC = 10 cm and AC = 6 cm.
Find the area of Triangle ABC. Also, find the length of AD.​

Answer 1

Answer:

perimeter of the triangle is (8+10+6)cm=24cm

half of the perimeter is 24/2cm=12cm

area of the triangle ABC is

√{12(12-10)(12-8)(12-6)}sqcm

=√(12.2.4.6)sqcm

=√(12.12.2.2)sqcm

=24sqcm

let AD =x cm

so,

1/2.BC.AD=24

1/2.10.x=24

x=24/5=4.8

so length of AD is 4.8cm

Answer 2

Answer:

Answer:

We can see this is an Right Angle Triangle.

⠀

Here ∠ A = 90°

▪ Longest Side [ BC = 10 cm ] is Hypotenuse.

▪ AC = 6 cm [ Base ]

▪ AB = 8 cm [ Height ]

⠀

From another view, where ∠ D = 90°

▪ BC = 10 cm [ Base ]

▪ AD = ? [ Height ]

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf Area\:of\:\triangle ABC_{\angle A \:is\:90} = Area\:of\:\triangle ABC_{\angle D \:is \:90}\\\\\\:\implies\sf \dfrac{1}{2} \times Height \times Base=\dfrac{1}{2} \times Height \times Base\\\\\\:\implies\sf \dfrac{1}{2} \times AB \times AC = \dfrac{1}{2} \times AD \times BC\\\\\\:\implies\sf AB \times AC = AD \times BC\\\\\\:\implies\sf 8 \:cm \times 6 \:cm = AD \times 10 \:cm\\\\\\:\implies\sf \dfrac{8 \:cm \times 6\:cm}{10\:cm} = AD\\\\\\:\implies\sf \dfrac{48 \:cm}{10} = AD\\\\\\:\implies\underline{\boxed{\sf AD = 4.8 \:cm}}$

⠀

$\therefore\:\underline{\textsf{Hence, length of AD is \textbf{4.8 cm}}}.$