J's answer to Eve lim's Primary 6 Maths Algebra Singapore question.
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Perimeter of square = 4 sides = 12 cm
1 side = 12 cm ÷ 4 = 3cm
Perimeter of ABCDE = 3 sides of the square + 2 longer sides of the triangle = (10p + 11) cm
Perimeter of BCD = 1 side of the square + 2 longer sides of the triangle
So to get this value, you can subtract two sides of the square from ABCDE's perimeter
10p + 11 - 3 - 3 = 10p + 5
Perimeter of ABCD = (10p + 5) cm
b) Perimeter of ABCDE = (10p + 11) cm
If the perimeter is equal to 31 cm,
Then 10p + 11 = 31
10p = 31 - 11
10p = 20
p = 20 ÷ 10 = 2
1 side = 12 cm ÷ 4 = 3cm
Perimeter of ABCDE = 3 sides of the square + 2 longer sides of the triangle = (10p + 11) cm
Perimeter of BCD = 1 side of the square + 2 longer sides of the triangle
So to get this value, you can subtract two sides of the square from ABCDE's perimeter
10p + 11 - 3 - 3 = 10p + 5
Perimeter of ABCD = (10p + 5) cm
b) Perimeter of ABCDE = (10p + 11) cm
If the perimeter is equal to 31 cm,
Then 10p + 11 = 31
10p = 31 - 11
10p = 20
p = 20 ÷ 10 = 2
Date Posted:
3 years ago
Correction :
It should be perimeter of BCD = (10p + 5)cm and not perimeter of ABCD
It should be perimeter of BCD = (10p + 5)cm and not perimeter of ABCD
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