ACS Primary 2009 PSLE Math Prelim Paper 2 Q10

Given that the length of DB is 14 cm and the shaded region is 115 square cm, find the lenght of AC, correct to 2 decimal places.

Solution

Area of Triangle = Half base x height.
Area of Triangle ABC --> (1/2)(AC)(DX + 14)
Area of Triangle ACD --> (1/2)(AC)(DX)

Area of shaded area
--> Area of Triangle ABC - Area of Triangle ACD = 115

(1/2)(AC)(DX + 14) - (1/2)(AC)(DX) = 115

(1/2)(AC)(DX) + (1/2)(AC)(14) - (1/2)(AC)(DX) = 115

(1/2)(AC)(14) = 115
7(AC) =115
AC = 115 divided by 7
= 16.428
~ 16.43 (nearest to 2 decimal places)

Answer: 16.43 cm

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