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secondary 3 | E Maths
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last qn’s answer is 540. pls help :)
Date Posted: 3 years ago
Views: 183
Looks like 810 to me and not 540...
Because the two triangles of interest, ARE and BRC, are similar triangles with length ratio 1 : 3 and an area ratio 1 : 9.
Because the two triangles of interest, ARE and BRC, are similar triangles with length ratio 1 : 3 and an area ratio 1 : 9.
No. AR and RC are not the corresponding sides.
Oh ya, good point. I linked the triangles wrongly in my visualisation.
Will update if I can find a way
Found an idea. Updating soon.
△ARE and △BRC are similar.
(You can prove using AA similarity.
Notice the vertically opposite angles AREA and BRC.
Either pair of the other two pairs of angles can be shown to be equal by the property 'angles in the same segment are equal in measure' )
AR = ¼ AC
Looking at △ARE, by sine rule,
AR / sin∠REA = RE/sin∠EAR
¼AC / sin45° = RE / sin60°
RE = ¼AC x sin60°/sin45°
= ¼AC x (√3/2)/(1/√2)
= ¼AC x √2√3/2
= √6 / 8 AC
Since AR = ¼ AC, and AC = AR + RC,
then RC = AC - AR
= AC - ¼AC
= ¾ AC
Now RE and RC are corresponding sides of the two similar triangles.
RE/RC = (√6 /8)AC / ¾AC
= √6 / 8 x 4/3
= √6/6
= 1/√6
so RE : RC = 1 : √6
Since the ratio of their sides is 1 : √6, the ratio of the area of △ARE to △BRC
= 1² : (√6)²
= 1 : 6
Therefore △BRC's area
= 6 x △ARE'S area
= 6 x 90cm²
= 540cm²
(Shown)
(You can prove using AA similarity.
Notice the vertically opposite angles AREA and BRC.
Either pair of the other two pairs of angles can be shown to be equal by the property 'angles in the same segment are equal in measure' )
AR = ¼ AC
Looking at △ARE, by sine rule,
AR / sin∠REA = RE/sin∠EAR
¼AC / sin45° = RE / sin60°
RE = ¼AC x sin60°/sin45°
= ¼AC x (√3/2)/(1/√2)
= ¼AC x √2√3/2
= √6 / 8 AC
Since AR = ¼ AC, and AC = AR + RC,
then RC = AC - AR
= AC - ¼AC
= ¾ AC
Now RE and RC are corresponding sides of the two similar triangles.
RE/RC = (√6 /8)AC / ¾AC
= √6 / 8 x 4/3
= √6/6
= 1/√6
so RE : RC = 1 : √6
Since the ratio of their sides is 1 : √6, the ratio of the area of △ARE to △BRC
= 1² : (√6)²
= 1 : 6
Therefore △BRC's area
= 6 x △ARE'S area
= 6 x 90cm²
= 540cm²
(Shown)
Weird. I got an entirely different answer and could not find a flaw in my workings thus far. I obtained 90 x sqrt 12.
I have attached my workings in the posting section, but I avoided similar triangles. I find it shocking that I did not get 540, but I could not find a mistake after reading again and again.
When I wake up, I go check again. I’m pretty sure I went wrong somewhere. Either I am wrong somewhere and I am probably too tired to spot it now, or the stipulation AR : AC = 1 : 4 is not even valid in the first place (a data “overload”). If I am not too lazy and can remember or spare the time, I will do a scale diagram for this to see whether AR : AC is valid.
When I wake up, I go check again. I’m pretty sure I went wrong somewhere. Either I am wrong somewhere and I am probably too tired to spot it now, or the stipulation AR : AC = 1 : 4 is not even valid in the first place (a data “overload”). If I am not too lazy and can remember or spare the time, I will do a scale diagram for this to see whether AR : AC is valid.
J, I just realised something.
AB / sin 45 = AE / sin 30
AB
= AE sin 45 / sin 30
= AE x sqrt 2 / 2 divided by 1/2
= AE x sqrt 2
Since AB = BC,
BC = AE x sqrt 2
Since triangles ARE and BRC are similar, AE and BC are corresponding sides.
This means that the ratio BC : AE must be sqrt 2 : 1, so the ratio of the areas of both triangles must be the square of this, or 2 : 1.
Therefore, area of ARE is twice that of BRC.
Now I am almost 100% certain of the overload. The data AR = 0.25 AC is definitely wrong.
540 cm2 is definitely not the answer.
180 cm2 is definitely the correct answer. Moreover, the triangle BRC does not even appear six times as big as triangle ARE.
AB / sin 45 = AE / sin 30
AB
= AE sin 45 / sin 30
= AE x sqrt 2 / 2 divided by 1/2
= AE x sqrt 2
Since AB = BC,
BC = AE x sqrt 2
Since triangles ARE and BRC are similar, AE and BC are corresponding sides.
This means that the ratio BC : AE must be sqrt 2 : 1, so the ratio of the areas of both triangles must be the square of this, or 2 : 1.
Therefore, area of ARE is twice that of BRC.
Now I am almost 100% certain of the overload. The data AR = 0.25 AC is definitely wrong.
540 cm2 is definitely not the answer.
180 cm2 is definitely the correct answer. Moreover, the triangle BRC does not even appear six times as big as triangle ARE.
thanks all !!!
The rest of the info fits well, i.e all the angles are correct. The answer given by then is based on AR = ¼AC.
Clare, where did you get this question from?
Clare, where did you get this question from?
See 3 Answers
done
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clear
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Last question use sin rule
Date Posted:
3 years ago
The working can be improved:
① 1.22 is premature approximation. Can be left as exact form.
② it should be AR instead of 1. ER ≠ 1.22 as that is mathematically incorrect.
Writing like that will result in loss of marks.
By sine rule,
sin 45° / AR = sin 60° / RE
Cross multiply,
RE sin 45° = AR sin 60°
RE = AR sin 60° / sin 45°
RE = AR (√3/2) / (1/√2)
RE = √3√2 / 2 AR = √6 / 2 AR
AR = ¼AC → 4AR = AC
RC = AC - AR
= 4AR - AR
= 3AR
RC/RE = (3AR)/(√6 / 2 AR)
= 6/√6
= √6
Since they are similar, and RC and RE are their corresponding sides,
area△BRC / area △ARE
= RC ² / RE ²
= (√6)²
= 6
area of △ BRC
= 6 x area of △ARE
= 6 x 90cm²
= 540 cm²
① 1.22 is premature approximation. Can be left as exact form.
② it should be AR instead of 1. ER ≠ 1.22 as that is mathematically incorrect.
Writing like that will result in loss of marks.
By sine rule,
sin 45° / AR = sin 60° / RE
Cross multiply,
RE sin 45° = AR sin 60°
RE = AR sin 60° / sin 45°
RE = AR (√3/2) / (1/√2)
RE = √3√2 / 2 AR = √6 / 2 AR
AR = ¼AC → 4AR = AC
RC = AC - AR
= 4AR - AR
= 3AR
RC/RE = (3AR)/(√6 / 2 AR)
= 6/√6
= √6
Since they are similar, and RC and RE are their corresponding sides,
area△BRC / area △ARE
= RC ² / RE ²
= (√6)²
= 6
area of △ BRC
= 6 x area of △ARE
= 6 x 90cm²
= 540 cm²
done
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clear
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I have a different answer altogether, but could not find anything wrong with my working. I did not use similar triangles however.
Date Posted:
3 years ago
done
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Clare, I have confirmed that there is an overload of data which is in conflict with each other. J's and A Math God's solutions, as well as my previous solutions, are therefore incorrect. Or, all are correct, but because of the conflict of data, two different, but logically sound, answers are possible.
Here are my workings for this question. There can only be one correct answer for triangle BRC, and that is 180 cm2.
Here are my workings for this question. There can only be one correct answer for triangle BRC, and that is 180 cm2.
Date Posted:
3 years ago
thank you for the abundant analysis ! appreciate it !
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