Tutor MathsGuru: Ask me for your burning Maths questions!

[elkniwt]

firebird:

ABCD is a parallelogram. Line AB is on top and line DC is bellow. AB is parallel to DC and AD is parallel to BC. There is a point E inside the parallelogram. This point E is in between and nearer to D and C. AE is equal to DE. Therefore triangle AED is an isosceles triangle. Given Angle ADE is 48 degrees and angle AEB is 62 degrees. The question is find angle EBC.


Please help. I uncertain after about 2 hours of trying.

Thank you
Firebird

Hi Firebird,

Try to draw a line parallel to AD (and BC) through the point E. Extend this line all the way such that it cuts line AB at F.

Angle FEA = 48 deg (alternate angles)
Angle FEB = 62 - 48 = 14 deg
Angle EBC = Angle FEB = 14 deg (alternate angles)

Regards.

[firebird]

Dear elkniwt


Good morning.

Many many many and a lot of thanks.

With best regards
Firebird

[clblinym]

Dear Dharma,


Many many thanks.

With best regards
Amy

[clblinym]

Dear parents,


Please help with the following question.


http://www.postimage.org/image.php?v=PqMocRJ


Many thanks.
Amy

[Dharma]

clblinym:

Dear parents,


Please help with the following question.


http://www.postimage.org/image.php?v=PqMocRJ


Many thanks.
Amy

Area of triangle BEF = 8cm2

Since BE : AE = 3 : 6 = 1 : 2
Area of triangle AEF = 2 x Area of BEF = 2 x 8cm2 = 16cm2

Area of triangle ABF = 8cm2 + 16cm2 = 24cm2

Since BF : FC = 4 : 5
Area of triangle AFC = 24cm2 x (5/4) = 30cm2

Area of quadrilateral AEFC
= Area of triangle AEF + Area of triangle AFC
= 16cm2 + 30cm2
= 46cm2

[clblinym]

Dharma:

clblinym:

Dear parents,


Please help with the following question.


http://www.postimage.org/image.php?v=PqMocRJ


Many thanks.
Amy

Area of triangle BEF = 8cm2

Since BE : AE = 3 : 6 = 1 : 2
Area of triangle AEF = 2 x Area of BEF = 2 x 8cm2 = 16cm2

Area of triangle ABF = 8cm2 + 16cm2 = 24cm2

Since BF : FC = 4 : 5
Area of triangle AFC = 24cm2 x (5/4) = 30cm2

Area of quadrilateral AEFC
= Area of triangle AEF + Area of triangle AFC
= 16cm2 + 30cm2
= 46cm2

~~~~~~~~

Thanks Dharma!!!
I really appreciate your great help.

[Almighty]

Dharma:

tiger262:

Can anyone post a solution for this speed question?


Kumar and Clara started jogging along a circular track.
Kumar started at Point M while Clara started at Point N where the line MN formed the diameter of the circle.
Kumar and Clara jogged toward each other along the circular track from their respective starting point and first met at Point P which was 80 m from Point M.
After they met for the first time, they continued jogging along the track and finally met again for the second time at Point Q which was 60 m from Point N. Find the distance of the circular track.

Thank you so much.

When Kumar met Clara for the 1st time at P ; he had jogged for 80m from M.
When Kumar met Clara for the 2nd time at Q; he would have jogged another 160m (2 x 80m) from P.

So, distance from P ->N -> Q = 160m
Distance PN = 160m – 60m = 100m
Circumference of the jogging track = 2 x (80m + 100m) = 360m

Hi Dharma,
Can you please explain how did u do the thing marked in red?? :stupid:

[Dharma]

Almighty:

Dharma:

[quote=\"tiger262\"]Can anyone post a solution for this speed question?


Kumar and Clara started jogging along a circular track.
Kumar started at Point M while Clara started at Point N where the line MN formed the diameter of the circle.
Kumar and Clara jogged toward each other along the circular track from their respective starting point and first met at Point P which was 80 m from Point M.
After they met for the first time, they continued jogging along the track and finally met again for the second time at Point Q which was 60 m from Point N. Find the distance of the circular track.

Thank you so much.

When Kumar met Clara for the 1st time at P ; he had jogged for 80m from M.
When Kumar met Clara for the 2nd time at Q; he would have jogged another 160m (2 x 80m) from P.

So, distance from P ->N -> Q = 160m
Distance PN = 160m – 60m = 100m
Circumference of the jogging track = 2 x (80m + 100m) = 360m

Hi Dharma,
Can you please explain how did u do the thing marked in red?? :stupid:[/quote]Hi Almighty,

When Kumar met Clara for the 1st time at P, he had jogged 80m from M. When they met at P, in total they both would covered a total distance of half the circumference of the track (M to N).

Now when Kumar and Clara met for the 2nd time at Q, in total they would have covered the circumference of the track ( P to Q).

M to P => 80m (Kumar) when Kumar & Clara covered ½ the circumference
P to Q => 80m x 2 = 160m (Kumar) when Kumar & Clara covered the whole circumference.

[clblinym]

Dear parents,


My son has two questions to ask. We tried to solve them, but failed.
Hope parents here can help us get solutions. Thanks in advance.

http://www.postimage.org/image.php?v=gxBAuPA

http://www.postimage.org/image.php?v=PqN6UjA

Amy

[Dharma]

clblinym:

Dear parents,


My son has two questions to ask. We tried to solve them, but failed.
Hope parents here can help us get solutions. Thanks in advance.

http://www.postimage.org/image.php?v=gxBAuPA

http://www.postimage.org/image.php?v=PqN6UjA

Amy

Hi Amy,

Q1

Area of quadrlateral ABCD
= Area of triangle ABC + Area of triangle ADC
= (1/2 x 3cm x 4cm) + (1/2 x 2cm x 2cm)
= 6cm2 + 2cm2
= 8cm2


Q2

Shaded area
= (Area of triangle GBE + Area of triangle HBD) - Area of quad. ABCD
= (1/2 x 10cm x 10cm) - 7cm2
= 50cm2 - 7cm2
= 43cm2