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Date Posted: 5 years ago
Views: 308
DAE = BCE = 90° (angles in a semicircle)
DAC = CBD
(angles in the same segment DC)
By AA similarity, △DAE and△CBE are similar.
AE/DE = BE/EC
36/DE = BE/28
DE = 36 x 28 ÷ BE
DE = 1008/BE ①
DB bisects ABC.
So BC/CE = AB/AE (angle bisector theorem)
BC/28 = AB/36
9/7 BC = AB
81/49 BC² = AB² ②
AC² + BC² = AB² (Pythagoras' theorem)
(36 + 28)² + BC² = AB²
Using ②,
64² + BC² = 81/49 BC²
64² = 32/49 BC²
BC² = 64² x 49/32 = 6272
Now, BCE is also right angled.
So BC² + CE² = BE²
6272 + 28² = BE²
BE² = 7056
BE = √7056 = 84
Sub this into ①,
DE = 1008/84 = 12
DAC = CBD
(angles in the same segment DC)
By AA similarity, △DAE and△CBE are similar.
AE/DE = BE/EC
36/DE = BE/28
DE = 36 x 28 ÷ BE
DE = 1008/BE ①
DB bisects ABC.
So BC/CE = AB/AE (angle bisector theorem)
BC/28 = AB/36
9/7 BC = AB
81/49 BC² = AB² ②
AC² + BC² = AB² (Pythagoras' theorem)
(36 + 28)² + BC² = AB²
Using ②,
64² + BC² = 81/49 BC²
64² = 32/49 BC²
BC² = 64² x 49/32 = 6272
Now, BCE is also right angled.
So BC² + CE² = BE²
6272 + 28² = BE²
BE² = 7056
BE = √7056 = 84
Sub this into ①,
DE = 1008/84 = 12
Thank you very much J
Welcome. Do clarify if there's any doubts about the working
I can fully understand your working. It is very detailed and clear. Tqvm!
Most welcome. Glad it was clear and understandable
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