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An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

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Length of the equal sides = 12cm Perimeter of the triangle = 30cm Length of the third side = 30 - (12+12) cm = 6cm Semi perimeter of the triangle(s) = 30/2 cm = 15cm Using heron's formula, Area of the triangle = √s (s-a) (s-b) (s-c) = √15(15...
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Length of the equal sides = 12cm
Perimeter of the triangle = 30cm
Length of the third side = 30 - (12+12) cm = 6cm
Semi perimeter of the triangle(s) = 30/2 cm = 15cm
Using heron's formula,
Area of the triangle = √s (s-a) (s-b) (s-c)                    
                             = √15(15 - 12) (15 - 12) (15 - 6)    
                             = √15 × 3 × 3 × 9                     
                             = 9√15 
 
 
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Let a be the third side. ∴ a + 12cm + 12cm = 30cm ⇒ a + 24cm = 30cm ∴ a = 30cm - 24cm = 6cm. ∴ the area of the triangle = √ 15 ( 15 - 12 ) ( 15 - 12 ) ( 15 - 6 ) = √ 15 × 3 × 3 × 9 = √ 15 × 3 × 3 × 3 × 3 = 9 &rad...
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Let a be the third side.

∴ a + 12cm + 12cm = 30cm

 ⇒ a + 24cm = 30cm 

∴ a = 30cm - 24cm = 6cm.

∴ the area of the triangle

= √ 15 ( 15 - 12 ) ( 15 - 12 ) ( 15 - 6 )

= √ 15 × 3 × 3 × 9 = √ 15 × 3 × 3 × 3 × 3 

= 9 √15cm².

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8 years of experience as home tutor

Triangle ABC is an isosceles triangle with two equal sides AB= 12 cm, AC= 12 cm, BC = x cm perimeter of an triangle= sum of three sides since it is an isosceles triangle, so in this case 30= 12+12+x ⇒x= 6cm so the length of the side BC is 6 cm. so in this way we find out the the length of all sides...
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Triangle ABC is an isosceles triangle with two equal sides

AB= 12 cm, AC= 12 cm, BC = x cm

perimeter of an triangle= sum of three sides

since it is an isosceles triangle, so in this case

30= 12+12+x

⇒x= 6cm

so the length of the side BC is 6 cm.

so in this way we find out the the length of all sides of triangle ABC, where AB= 12 cm, AC= 12 cm, and BC= 6 cm. Now draw a perpinduclar from point A in the line BC, let the middle point is P so in this way the length of BP and BC= 3 cm.

Now according to the pythagorus theorm we will find the length of AP

AB²= AP²+BP²

⇒144= AP²+9

⇒ AP= √135

so now area of triangle ABP= 1/2∗base*height

=1/2*3* √135

=3√135/2

so area of triangle ABC= 2* area of triangle ABP

= 2* 3√135/2

= 3√135

= 3*11.61

= 34.83 cm²

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Professional teacher with 5 years of experiance.

The third side of the triangle will be 30-(12+12)cm = 30-24 = 6cm. Therefore the area of the triangle = √S(s-a)(s-b)(s-c) √15(15-12)(15-12)(15-6) √15×3×3×9 9√15cm² Answer : Area of the triangle is 9√15cm²
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The third side of the triangle will be 30-(12+12)cm

= 30-24 = 6cm. 

Therefore the area of the triangle 

 

= √S(s-a)(s-b)(s-c)

√15(15-12)(15-12)(15-6)

√15×3×3×9

9√15cm²

Answer : Area of the triangle is 9√15cm²

 

 

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Maths Teacher for classes 6th to 10th with 6 years experience and having my own YouTube Channel for clearing doubts

The perimeter of the isosceles triangle is the sum of all its three sides. Each of the equal sides is 12cm. So, the third side of that isosceles triangle is = 30cm - (12cm + 12cm) = 30cm - 24cm = 6cm. The area of that isosceles triangle, using Heron's Formula is, √15 (15-12) (15-12) (15-6) =...
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The perimeter of the isosceles triangle is the sum of all its three sides. Each of the equal sides is 12cm.

So, the third side of that isosceles triangle is = 30cm - (12cm + 12cm) = 30cm - 24cm = 6cm.

The area of that isosceles triangle, using Heron's Formula is, √15 (15-12) (15-12) (15-6) = √15x3x3x9 = 3x3x√15 = 9√15cm²  

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Science Teacher with an Experience of 2 years in North India schools

34.85
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34.86
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8 years of experience as home tutor

34.83 cm²
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