Question

The perimeter of an isosceles triangle is 64cm. The ratio of the equal side to its base is 3:2. Find the area of the triangle.

Answer 1

Answer:

Let the sides of the triangle be

3x,3x and 2x

perimeter=3x+3x,+2x=8x=64

x=64/8=8

So sides are 24,24 and 16

Area=√s(s-a)(s-b)(s-c)

s=8x/2=4x

Area=√(4x)(4x-3x)(4x-3x)((4x-2x)

=x^2√(4×2)=8×8×2√2=128√2

Answer 2

Given

➙ The perimeter of an isosceles triangle is 64 cm .The ratios of the equal sides to its base is 3:2 .

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To Find :

➙ Find the area of Triangle .

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Solution :

⚘ Concept :

As we have been already given in the question the perimeter and the ratio of sides using the formula perimeter of triangle we can derive the value of x in the ratios .And after finding the sides we can easily calculate the area of triangle by applying Heron's Formula . So,Let's solve :

$\qquad{━━━━━━━━━━━━}$

⚘ Let the Ratios :

We know that in a isosceles triangle two sides are equal therefore the ratio of equal sides is 3x and ratio of base is 2x .Hence,

➳ 1st side = 3x

➳ 2nd side = 3x

➳ 3rd side = 2x

⚘ Calculating the value of x :

Formula Used :

$\large{\gray{\bigstar}} \: \: \: {\underline{\boxed{\red{\sf{ Perimeter{\small_{(Triangle)}} = a + b + c}}}}}$

Calculation starts :

$\qquad{\longmapsto{\sf{ a + b + c = 64 \: cm}}}$

$\qquad{\longmapsto{\sf{ 3x + 3x + 2x = 64 \: cm}}}$

$\qquad{\longmapsto{\sf{ 8x = 64 \: cm}}}$

$\qquad{\longmapsto{\sf{ x = \cancel\dfrac{64}{8} }}}$

${\sf{ Value \: of \: x \: = {\green{\sf{ 8}}}}}$

⚘ Sides of Triangle :

➳ 1st side = 3x = 3 × 8 = 24 cm

➳ 2nd side = 3x = 3 × 8 = 24 cm

➳ 3rd side = 2x = 2 × 8 = 16 cm

⚘ Calculating the Area of Triangle :

Formula Used :

$\large{\gray{\bigstar}} \: \: \: {\underline{\boxed{\red{\sf{ Area{\small_{(Triangle)}} = \sqrt{s(s - a)(s - b)(s - c)}}}}}}$

Calculation starts :

Semi - Perimeter :

$\qquad{\longmapsto{\sf{S = \dfrac{Perimeter}{2}}}}$

$\qquad{\longmapsto{\sf{S = \cancel\dfrac{64}{2}}}}$

${\sf{ Semi \: Perimeter \: = {\pink{\sf{ 32 \: cm}}}}}$

Area :

$\qquad{\longmapsto{\sf{ Area =\sqrt{s(s - a)(s - b)(s - c)} }}}$

$\qquad{\longmapsto{\sf{ Area =\sqrt{32(32 - 24)(32 - 24)(32 - 16)} }}}$

$\qquad{\longmapsto{\sf{ Area =\sqrt{32 \times 8 \times 8 \times 16} }}}$

$\qquad{\longmapsto{\sf{ Area =\sqrt{32 \times 8 \times 8 \times 16} }}}$

${\sf{ Area \: of \: Triangle \:= {\orange{\sf{ 128 \sqrt{2} \: cm²}}}}}$

$\qquad{━━━━━━━━━━━━}$

Therefore :

$❝ {\sf{Area \: of \:the \:given \:triangle \: is \:{\blue{\underline{\red{\sf{ 128 \sqrt{2} \: cm². }}}}}}}❞$

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