Question

solve the attachment!!!✌✌​

Answer 1

Let ABC be an Isosceles triangle having perimeter 44 cm.

The ratio of the equal side to its base is 4:3.

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ Let sides of triangle be in form of 'x'

⠀

Therefore,

⠀

Equal sides, AB = AC = 4x

Base of triangle, BC = 3x

⠀

We know that,

⠀

$\star\;{\boxed{\sf{\purple{Perimeter_{\triangle} = Sum\;of\;sides}}}}$

$:\implies\sf 4x + 4x + 3x = 44$

$:\implies\sf 11x = 44$

$:\implies\sf x = \cancel{ \dfrac{44}{11}}$

$:\implies{\boxed{\frak{\pink{x = 4}}}}\;\bigstar$

Therefore, measure of sides are,

⠀

AB = a = 4 × 4 = 16 cm

AC = b = 4 × 4 = 16 cm

BC = c = 3 × 4 = 12 cm

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now, for finding the height of a triangle we will use pythagoras theorem in 1 half of the given isosceles triangle as for pythagoras theorem we need right angled triangle, Then the base of the triangle will be $\sf {\cancel\frac{12}{2}} = 6 cm.$

$:\implies\sf (hypotenuse)^2 = base^2 + height^2$

⠀

$:\implies\sf 16 \times 16 = 6 \times 6 + {(height)}^{2}$

$:\implies\sf 256 = 36 + {(height)}^{2}$

$:\implies\sf 256 - 36 = {(height)}^{2}$

$:\implies\sf 220 = {(height)}^{2}$

$:\implies\sf \sqrt{220} = height$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ Also, we can find Area of given Isosceles triangle:

$\star\;{\boxed{\sf{\purple{Perimeter_{\triangle} = \frac{1}{2} \times base \times \: height}}}}$

$:\implies\sf \frac{1}{2} \times 12 \times \sqrt{220}$

$:\implies\sf 6 \times 2 \sqrt{55}$

$:\implies{\boxed{\frak{\pink{12 \sqrt{55}\:sq.\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;area\;of\; Isosceles\;triangle\;is\; \bf{12\sqrt{55}\;sq.\;cm}.}}}$

Answer 2

Answer:

Let ABC be an Isosceles triangle having perimeter 44 cm.

The ratio of the equal side to its base is 4:3.

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ Let sides of triangle be in form of 'x'

⠀

Therefore,

⠀

Equal sides, AB = AC = 4x

Base of triangle, BC = 3x

⠀

We know that,

⠀

\begin{gathered}\star\;{\boxed{\sf{\purple{Perimeter_{\triangle} = Sum\;of\;sides}}}}\\ \\\end{gathered}

⋆

Perimeter

△

=Sumofsides

\begin{gathered}:\implies\sf 4x + 4x + 3x = 44\\ \\\end{gathered}

:⟹4x+4x+3x=44

\begin{gathered}:\implies\sf 11x = 44\\ \\\end{gathered}

:⟹11x=44

\begin{gathered}:\implies\sf x = \cancel{ \dfrac{44}{11}}\\ \\\end{gathered}

:⟹x=

11

44

\begin{gathered}:\implies{\boxed{\frak{\pink{x = 4}}}}\;\bigstar\\ \\\end{gathered}

:⟹

x=4

★

Therefore, measure of sides are,

⠀

AB = a = 4 × 4 = 16 cm

AC = b = 4 × 4 = 16 cm

BC = c = 3 × 4 = 12 cm

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Now, for finding the height of a triangle we will use pythagoras theorem in 1 half of the given isosceles triangle as for pythagoras theorem we need right angled triangle, Then the base of the triangle will be \sf {\cancel\frac{12}{2}} = 6 cm.

2

12

=6cm.

\begin{gathered}:\implies\sf (hypotenuse)^2 = base^2 + height^2 \\ \\\end{gathered}

:⟹(hypotenuse)

2

=base

2

+height

2

⠀

\begin{gathered}:\implies\sf 16 \times 16 = 6 \times 6 + {(height)}^{2} \\ \\\end{gathered}

:⟹16×16=6×6+(height)

2

\begin{gathered}:\implies\sf 256 = 36 + {(height)}^{2} \\ \\\end{gathered}

:⟹256=36+(height)

2

\begin{gathered}:\implies\sf 256 - 36 = {(height)}^{2} \\ \\\end{gathered}

:⟹256−36=(height)

2

\begin{gathered}:\implies\sf 220 = {(height)}^{2} \\ \\\end{gathered}

:⟹220=(height)

2

\begin{gathered}:\implies\sf \sqrt{220} = height \\ \\\end{gathered}

:⟹

220

=height

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

☯ Also, we can find Area of given Isosceles triangle:

\begin{gathered}\star\;{\boxed{\sf{\purple{Perimeter_{\triangle} = \frac{1}{2} \times base \times \: height}}}}\\ \\\end{gathered}

⋆

Perimeter

△

=

2

1

×base×height

\begin{gathered}:\implies\sf \frac{1}{2} \times 12 \times \sqrt{220}\\ \\\end{gathered}

:⟹

2

1

×12×

220

\begin{gathered}:\implies\sf 6 \times 2 \sqrt{55} \\ \\\end{gathered}

:⟹6×2

55

\begin{gathered}:\implies{\boxed{\frak{\pink{12 \sqrt{55}\:sq.\;cm}}}}\;\bigstar\\ \\\end{gathered}

:⟹

12

55

sq.cm

★

\therefore\;{\underline{\sf{Hence,\;the\;area\;of\; Isosceles\;triangle\;is\; \bf{12\sqrt{55}\;sq.\;cm}.}}}∴

Hence,theareaofIsoscelestriangleis12

55

sq.cm.