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primary 5 | Maths
| Angles & Geometry
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This is only what I know ADE is 2 cm2 while ABE is 6cm2 ..............How do I solve this problem ?? Please help me!!thanks
Date Posted: 3 years ago
Views: 433
Triangles ABC and BDC have the same height, and same base BC.
So their areas are equal.
Since triangle BEC is common to both of them and contained inside them,
this tells you that triangle DCE's area is equal to triangle ABE's area (6cm)
(Area of ABC = Area of BEC + ABE
(Area of BDC = Area of BEC + DCE)
So their areas are equal.
Since triangle BEC is common to both of them and contained inside them,
this tells you that triangle DCE's area is equal to triangle ABE's area (6cm)
(Area of ABC = Area of BEC + ABE
(Area of BDC = Area of BEC + DCE)
Now ADE and ABE also have a same height (if you take DE and BE to be their respective bases)
Since the area of ADE is 3 times of ABE, this means the length of base BE is 3 times the length of DE.
Since the area of ADE is 3 times of ABE, this means the length of base BE is 3 times the length of DE.
Now DEC and BEC also have the same height (if you take their bases to be DE and BE respectively)
Since BE is 3 times the length of DE, then the area of BEC must be 3 times of DEC
3 × 6cm² = 18cm²
Total area of trapezium
= 18cm² + 6cm² + 6cm² + 2cm²
= 32cm²
Since BE is 3 times the length of DE, then the area of BEC must be 3 times of DEC
3 × 6cm² = 18cm²
Total area of trapezium
= 18cm² + 6cm² + 6cm² + 2cm²
= 32cm²
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Thank you so much
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