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primary 5 | Maths
| Area & Volume
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Please help. Thanks
Date Posted: 5 years ago
Views: 336
area ABC = 0.5 x 20 x 12 = 120 cm²
area BDC = 0.5 x 5 x 12 = 30 cm²
area ADC = 120 - 30 = 90 cm²
area BDC = 0.5 x 5 x 12 = 30 cm²
area ADC = 120 - 30 = 90 cm²
As assumption has to be made here, that the unshaded triangle is a right angled one.
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Actually triangle with dimensions 5cm, 12 cm and 13 cm is right angle triangle. .. but primary level haven’t teach. Hope this helps
Date Posted:
5 years ago
This diagram is mathematically incorrect as the unshaded triangle is supposed be right angled as well, but its dimensions don't satisfy the Pythagoras' theorem since 15² + 12² = 369 ≠ 20²
still make use of info that angle ADC is right angle
To be fair to a P5 student, the value of angle BDC and ADC aren't even given. Amd neither are stated as right angles.
Hence my earlier comment that it has to be assumed that BDC is right angled (which then would imply that ADC is right angled since AB is taken to be a straight line), since they're not expected to infer that the shaded triangle satisfies Pythagoras' Theorem and thus conclude BDC is 90°.
Or, assume that ADC is right angled.
Hence my earlier comment that it has to be assumed that BDC is right angled (which then would imply that ADC is right angled since AB is taken to be a straight line), since they're not expected to infer that the shaded triangle satisfies Pythagoras' Theorem and thus conclude BDC is 90°.
Or, assume that ADC is right angled.
How about use AD : BD = 3:1... are we able to find area for ABC... ?
Will still get 120 cm² for ABC
Meaning we still need to assume ADC is a right angle.Really mathematically error question;(
Yup. The teacher/set probably expects assumption to be done. Otherwise it could be that they forgot to mark either as a right angle.
If we let AC = AB be x instead,
Then x² = 12² + (x - 5)²
x² = 144 + x² - 10x + 25
10x = 169
x = 16.9
So what the teacher/setter should have done is let AB = AC = 16.9 cm
Then x² = 12² + (x - 5)²
x² = 144 + x² - 10x + 25
10x = 169
x = 16.9
So what the teacher/setter should have done is let AB = AC = 16.9 cm
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