Question

ABC is a right triangle with AB = AC . if bisector of angle A meets BC at D and AD = 2 root2 cm , then the perimeter of triangle ABC is

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Answer 1

$\underline{\underline{\sf{\pink{Given :} }}}$

∆ABC is a right Angled triangle

AB = AC

AD is bisector of ∠A

$\sf AD = 2\sqrt{2}$

⠀ ⠀⠀ ⠀ ⠀⠀ ⠀

$\underline{\underline{\sf{\pink{To\:Find :} }}}$

Perimeter of ∆ABC

⠀ ⠀⠀ ⠀ ⠀⠀ ⠀

$\underline{\underline{\sf{\pink{Solution :} }}}$

In ∆ABC,

AB = AC

$\underline{\tt{As\:we\:know\:that,}}$

$\star\:{\boxed{\sf{\pink{Angle \: opposite \: to \: equal \: sides \: are \: equal }}}}$

$: \: \implies \sf{\angle C = \angle B - (i)}$

$\underline{\tt{Now, \:by \: angle\:sum \: property,}}$

∠A + ∠B + ∠C = 180°

From (i)

∠A + ∠B + ∠B = 180°

$: \: \implies \sf{90^{\circ} + 2 \angle B = 180^{\circ}}$

$: \: \implies \sf{2 \angle B = 180^{\circ} - 90^{\circ}}$

$: \: \implies \sf{2 \angle B = 90^{\circ}}$

$: \: \implies \sf{ \angle B = \dfrac{90^{\circ}}{2}}$

$: \: \implies \sf{\angle B = 45^{\circ}}$

$\underline{\tt{Now, \: \angle B = \angle C}}$

$: \: \implies \sf{\angle C = 45^{\circ}}$

_______________________________

In ∆ABD

As, AD bisects BC,

Therefore, ∠ADB = 90°

$\underline{\tt{Now, \: by \: angle\:sum \: property,}}$

∠B + ∠DAB + ∠ADB = 180°

45° + ∠DAB + 90° = 180°

45° + ∠DAB = 180° - 90°

∠DAB = 90° - 45°

∠DAB = 45°

As, ∠B = ∠DAB (both 45°) and we know that sides opposite to equal angles are equal

Therefore, BD = AD - (ii)

_______________________________

In ∆ACD

As, AD bisects BC,

Therefore, ∠ADC = 90°

$\underline{\tt{Now, \: by \: angle\:sum \: property,}}$

∠C + ∠DAC + ∠ADC = 180°

45° + ∠DAC + 90° = 180°

45° + ∠DAC = 180° - 90°

∠DAC = 90° - 45°

∠DAC = 45°

As, ∠C = ∠DAC (both 45°) and we know that sides opposite to equal angles are equal

Therefore, CD = AD -(iii)

_______________________________

From (ii) and (iii),

BD + CD = AD + AD

Now, we have BD + CD = BC

$: \: \implies \sf{BC = 2AD}$

Now, we are given AD = $\sf 2\sqrt{2}cm$

$: \: \implies \sf{BC = 2 \times 2\sqrt{2}}$

$: \: \implies \sf{BC = 4\sqrt{2}cm }$

_______________________________

In ∆ABC

$\underline{\tt{By \: Pythagoras' \: theorem,}}$

$\star\:{\boxed{\sf{\pink{H^{2} = B^{2} + P^{2}}}}}$

$: \: \implies \sf{BC^{2} = AB^{2} + AC^{2}}$

$: \: \implies \sf{{(4\sqrt{2})}^{2} = AB^{2} + AC^{2}}$

$\underline{\tt{Now, \: we \: are \: given \: AB = AC}}$

$: \: \implies \sf{{(4\sqrt{2})}^{2} = AB^{2} + AB^{2}}$

$: \: \implies \sf{16 \times 2 = 2AB^{2}}$

$: \: \implies \sf{32 = 2AB^{2}}$

$: \: \implies \sf{\dfrac{32}{2} = AB^{2}}$

$: \: \implies \sf{16 = AB^{2}}$

$: \: \implies \sf{\sqrt{16} = AB}$

$: \: \implies \sf{AB = 4 cm}$

As, AB = AC (given)

$: \: \implies \sf{AC = 4 cm}$

Now, We have

AB = 4 cm

AC = 4 cm

$\sf{BC = 4\sqrt{2}cm }$

_______________________________

$\underline{\tt{Now, \:we \: have \: to \: find \: perimeter,}}$

$\star\:{\boxed{\sf{\pink{Perimeter \: of \: triangle = a + b + c}}}}$

In ∆ABC,

a = AB = 4 cm

b = AC = 4 cm

c = $\sf{BC = 4\sqrt{2}cm }$

$: \implies \sf{Perimeter = 4 cm + 4 cm + 4\sqrt{2}cm }$

$: \implies \sf{Perimeter = 8 + 4\sqrt{2}cm}$

$\pink{\sf \therefore \: Perimeter \: of \: \triangle ABC = 8 + 4\sqrt{2}cm}$

Answer 2

Explanation:

Step-by-step explanation:

Given Equation:-

⠀⠀⠀⠀ $\sf{\bigg[ \dfrac{5 {x}^{2} - 10}{12 } \bigg] }$

To find:-

value of x

Solution:-

use factor theorem

take the value of equation =0

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow \left [\dfrac {5x^2-10}{12}\right]=0$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow \dfrac {5x^2-10}{12}=0$

using cross multiplication

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow 5x^2-10=0$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow 5x^2=10$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x^2=\dfrac {10}{5}$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x^2=2$

$\\\qquad\quad\displaystyle\sf{:}\longrightarrow x=\sqrt {2}$

$\\\\\therefore\sf x=\sqrt {2}.$