Question

pls solve it very easy

Answer 1

$\huge{\bold{\underline{Answer\: \colon}}}$

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$\bold{\underline{Given\: \colon}}$

The given triangle ABC is a isosceles having two equal side AB = AC and base BC = 24 cm and equal sides are 4 cm greater than height.

$\bold{\underline{To\: Find\: \colon}}$

Perimeter and area of triangle ABC

$\bold{\underline{Solution\: \colon}}$

In attachment we proved that perpendicular of isosceles triangle in which AB = AC in perpendicular trim the vertex A to BC bisect the base BC into two equal parts which means the BC bisect by AO in two equal parts.

So,

$\bold{\boxed{BO\: =\: OC}}$

$\therefore BO = \frac{BC}{2} \\\\ = \:\frac{24}{2} \\\\ =\: 12\: cm$

We have a statement from the question that the two equal sides are 4 cm greater than the height of triangle

Let AB = AC = a {two equal sides}

Let AO = h { height of triangle }

By statement

a = ( h + 4 ) cm

h = ( a - 4 ) cm

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AB = (h + 4) cm

AO = h cm

BO = 12 cm

Now in triangle ABO

By Pythagoras :

$AB^{2} = AO^{2} + BO^{2}\\\\ a^{2} = h^{2} + 12^{2} \\\\ (h+4)^{2} - h^{2} = 12^{2} \\\\ h^{2} + 16 +8h -h^{2} = 12^{2} \\\\ 8 h = 12\times 12 - 16 \\\\ 8 h = 144-16 \\\\ 8 h = 128 \\\\ h = \frac{128}{8} \\\\ h = 16 \: cm$

So,

h = 16 cm

Now

AB = BC = a = ( h + 4) = ( 16 + 4 ) cm

AB = BC = a = 20 cm

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$\bold{\underline{Perimeter \: of \: triangle \: \colon}}$

Sum of all three side

$P_{\triangle{ABC}} = AB + BC + CA \\\\ = 20 + 24 + 20 \\\\ = 64 \: \: cm$

$\bold{\underline{Area\: of \: \triangle{ABC} \colon}}$

$A_{\triangle{ABC}} = \frac{1}{2} \times{Base} \times{Height} \\\\\implies \frac{1}{2}\times 24 \times 16 \\\\\implies 12 \times 16 \\\\ \implies 192 cm^{2}$

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Perimeter of triangle ABC = 64 cm

Area of triangle ABC = 192 cm²

Thanks