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secondary 3 | E Maths
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if i physically use satay sticks construct the triangle , i think it won’t be 90 degrees
Date Posted: 4 years ago
Views: 693
The length XR is in the same rectangular “plane” as the rectangular side WVRS.
The length QR is in the same rectangular “plane” as the rectangular side UVRQ.
Both rectangular planes in the cuboid are perpendicular to each other. Hence, any line drawn along the first plane must be perpendicular to the any line drawn along the second plane.
So, that triangle is a right angled triangle, but then the right angled triangle seems elevated.
The length QR is in the same rectangular “plane” as the rectangular side UVRQ.
Both rectangular planes in the cuboid are perpendicular to each other. Hence, any line drawn along the first plane must be perpendicular to the any line drawn along the second plane.
So, that triangle is a right angled triangle, but then the right angled triangle seems elevated.
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i find it hard to visualise even after construction . Am i right to say that this triangle involves 2 planes ?
Date Posted:
4 years ago
Yes, imagining it would be difficult. At least this 3D illustration makes things simpler, but in theory the triangle is right angled.
Because the triangle still involves two perpendicular planes of the cuboid, the angle there is still 90 degrees.
The point directly below X (let’s call it Y) will make triangle QRY right angled. But lines XR and YR are on the same plane, so if triangle QRY is right angled, then so does triangle QRX.
The point directly below X (let’s call it Y) will make triangle QRY right angled. But lines XR and YR are on the same plane, so if triangle QRY is right angled, then so does triangle QRX.
Another way to see it :
Extend RV and name the end of the extended line as Z. Let RZ = √45 cm
(√(6² + 3²) = √45)
Draw triangle QRZ, which is right angled.
Now, treating QR as a hinge, rotate the whole QRZ anticlockwise until Z coincides with X.
(Notice the circular rotation, where RZ and RX are actually radii of this circle)
Now you have triangle QRX. But notice the right angle QRZ (Which is now QRX) didn't change.
Extend RV and name the end of the extended line as Z. Let RZ = √45 cm
(√(6² + 3²) = √45)
Draw triangle QRZ, which is right angled.
Now, treating QR as a hinge, rotate the whole QRZ anticlockwise until Z coincides with X.
(Notice the circular rotation, where RZ and RX are actually radii of this circle)
Now you have triangle QRX. But notice the right angle QRZ (Which is now QRX) didn't change.
Technically if someone views the picture from a very large height (seeing triangle QRX as the cross sectional surface, something like an aerial view of an NDP parade), the person would observe a right angled triangle.
Or you draw the rectangle QRXY, where Y is a point 3cm from U and along TU.
Then XQR is just this rectangle's diagonal.
It follows that triangle XQR is right angled since QR is perpendicular to RX, meaning XRQ is a right angle (90°)
Then XQR is just this rectangle's diagonal.
It follows that triangle XQR is right angled since QR is perpendicular to RX, meaning XRQ is a right angle (90°)
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thanks for all the explanation . Would this be the physical cut out ?
Date Posted:
4 years ago
Yup, kind of
i understood ur explanation earlier . It’s just that i feel safer to test out physically , in case my visualisation is wrong
First or second explanation? Second one I guess?
yup 2nd one about the rect . the 1st one u mentioned extending 1 point , i was wondering do u mean extend higher up , then hinge , rotate ... due to naturally poor visualisation , i m unable to imagine lol
I'll try to illustrate, later will post
actually it’s ok !! hinge is a bit physics related ... too cheem for me .
can i conclude that , as long i am able to PHYSICALLY fit a right angled triangle into the situation , means it is indeed right angled ?
In a sense yes.
btw i’m a novice tutor , not a student. Going forward , it is okay if u don’t feel like replying me . cos u don’t benefit from answering my qn (i’m not a prospective student )
Doesn't matter if you're a novice tutor or student. We're all here to learn :) anyway I've uploaded the 'sketch', feel free to clarify doubts should you have any
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Something like this.
Date Posted:
4 years ago
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