Question

Find the area of isosceles triangle whose equal side is 6 cm. 6 cm and 8 cm
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Answer 1

Answer:

Let the lengths of the sides be a,b,c

Given:

a = 6cm

b = 6cm

c = 8cm

Applying Heron's formula:

Area = √{s(s - a)(s - b)(s - c)}

where, s = perimeter/2 = (6+6+8)/2 = 10cm

Therefore,

Area = √{10(10 - 6)(10 - 6)(10 - 8)}

=> Area = √{10*4*4*2} = √320 = 8√5 cm^2

= 17.8885 cm^2

Answer 2

Required Answer:-

$\underline {\underline{\bigstar \: \rm Given:-}}$

An isosceles triangle with

• 1st side = 6 cm
• 2nd side = 6 cm
• 3rd side = 8 cm

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$\underline {\underline{\bigstar \: \rm What \: To \: Find:-}}$

We have to find the area of the isosceles triangle.

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$\underline {\underline{\bigstar \: \rm Formula \: For \: Finding:-}}$

$\sf Area \: of \: triangle = \sqrt{s(s - a) \times (s -b) \times (s - c)}$

Where

• a = 1st side
• b = 2nd side
• c = 3rd side
• s = semi-perimeter

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$\underline {\underline{\bigstar \: \rm How \: To \: Find:-}}$

To find the area,

• First, find the perimeter of the triangle and divide it by 2 (semi means half and half means into 2 parts of a whole) to find the semi-perimeter.
• Then, use the formula for finding the area of the isosceles triangle.

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$\underline {\underline{\bigstar \: \rm Solution:-}}$

$\bf \bullet \: Finding \: the \: semi \: perimeter.$

First, find the perimeter by adding the sides,

$\sf \Rightarrow a + b + c = Perimeter$

Substitute the values,

$\sf \Rightarrow 6 \: cm + 6 \: cm + 8 \: cm = Perimeter$

Add the values,

$\sf \Rightarrow 20 \: cm = Perimeter$

Now, find the semi perimeter,

$\sf \Rightarrow \dfrac{20 \: cm}{2} = Semi \: Perimeter$

Divide the perimeter by 2,

$\sf \Rightarrow 10 \: cm = Semi \: Perimeter \: (s)$

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$\bf \bullet \: Finding \: the \: area.$

Using the formula,

$\sf \Rightarrow Area \: of \: triangle = \sqrt{s(s - a) \times (s -b) \times (s - c)}$

Substitute the values,

$\sf \Rightarrow Area \: of \: triangle = \sqrt{10(10 - 6) \times (10 - 6) \times (10 - 8)}$

Solve the brackets,

$\sf \Rightarrow Area \: of \: triangle = \sqrt{10 \times 4 \times 4 \times 2}$

Multiply 10, 4, 4, and 2,

$\sf \Rightarrow Area \: of \: triangle = \sqrt{320}$

Find the square root of 320,

$\sf \Rightarrow Area \: of \: triangle = 17.88 \: cm^2 approx.$

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$\underline {\underline{\bigstar \: \rm Verification:-}}$

Using the formula,

$\sf \Rightarrow Area \: of \: triangle = \sqrt{s(s - a) \times (s -b) \times (s - c)}$

Substitute the values,

$\sf \Rightarrow 17.88 \: cm^2 \: approx. = \sqrt{10(10 - 6) \times (10 - 6) \times (10 - 8)}$

Solve the brackets in RHS,

$\sf \Rightarrow 17.88 \: cm^2 \: approx. = \sqrt{10 \times 4 \times 4 \times 2}$

Multiply 10, 4, 4, and 2 in RHS,

$\sf \Rightarrow 17.88 \: cm^2 \: approx. = \sqrt{320}$

Find the square root of 320,

$\sf \Rightarrow 17.88 \: cm^2 approx. = 17.88 \: cm^2 approx.$

$\underbrace { \rm \because RHS = LHS , Hence \: verified.}$

$\underbrace {\boxed{ \rm \therefore Thus, \: the \: area \: of \: triangle \: 17.88 \: cm^2 approx.}}$

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