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Question
primary 5 | Maths
| Area & Volume
One Answer Below
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Pls help
Date Posted: 4 years ago
Views: 444
The answer to this question can be solved using Heron's formula which applies to finding areas of triangles where only the length of sides are known. The problem is that the formula's proof and why it works is not intuitive at the primary level.
The formula is as such: Area of triangle=
√(Sx(S-A)x(S-B)x(S-C)) where A, B and C are all the respective lengths of triangle's sides and S is the perimeter divided by 2. √ is the symbol for square root
Hence the answer is:
√(21x(21-13)x(21-14)x(21-15))
=√(21x8x7x6)
=84 cm squared
The formula is as such: Area of triangle=
√(Sx(S-A)x(S-B)x(S-C)) where A, B and C are all the respective lengths of triangle's sides and S is the perimeter divided by 2. √ is the symbol for square root
Hence the answer is:
√(21x(21-13)x(21-14)x(21-15))
=√(21x8x7x6)
=84 cm squared
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