Since MR = RO, Angle NOR = 21EC72 wrote:

Need help with this question. Thank you.

Therefore Angle ORQ = 42

Angle RMN + Angle RQO + Angle OPQ×2 = 180

Angle OPQ = (180-21-42)/2

Academic support for Primary 6 and PSLE

- Nebbermind
- KiasuGrandMaster
**Posts:**16632**Joined:**Tue Dec 22,**Total Likes:**168

Post
by **Nebbermind** » Wed Jul 01, 2015 11:55 am

Since MR = RO, Angle NOR = 21EC72 wrote:

Need help with this question. Thank you.

Therefore Angle ORQ = 42

Angle RMN + Angle RQO + Angle OPQ×2 = 180

Angle OPQ = (180-21-42)/2

Post
by **EC72** » Wed Jul 01, 2015 2:01 pm

Beautyful MindsBeautyful Minds wrote:

Thank you Beautyful Minds! Can you please explain to me how did you get the ratio in the distance travelled 2:3?

Speed is proportional to distance travelled

1.2 : 1.8 = 2 : 3

Beautyful Minds

Post
by **Ilye** » Wed Jul 01, 2015 3:04 pm

Hi, may I know how to see angle ASQ=20? Is there any quick way to see this? Thanks.Nebbermind wrote:Angle QCB = 70EC72 wrote:

Need help with this question. Thank you.

Angle ASQ = 20

Therefore Angle DSQ = 160

- Nebbermind
- KiasuGrandMaster
**Posts:**16632**Joined:**Tue Dec 22,**Total Likes:**168

Post
by **Nebbermind** » Wed Jul 01, 2015 3:14 pm

If you extend the line RC to tbe right, the angle below QCB is 20.

Post
by **EC72** » Wed Jul 01, 2015 5:22 pm

The distance between Town A and Town C is 825 km. Bob set off from Town A at 1.30 p.m., at an average speed of 60 km/h. At 2.45 p.m., Carl also set off from Town A to Town B at an average speed of 75 km/h. Carl arrived at Town B at 10.45 p.m.

(a) At what time did Carl overtake Bob?

(b) How long did Bob take to travel from Town B to Town C if he increased his speed by 50% starting from Town B?

(a) At what time did Carl overtake Bob?

(b) How long did Bob take to travel from Town B to Town C if he increased his speed by 50% starting from Town B?

Post
by **DearSonR** » Wed Jul 01, 2015 6:30 pm

a)EC72 wrote:The distance between Town A and Town C is 825 km. Bob set off from Town A at 1.30 p.m., at an average speed of 60 km/h. At 2.45 p.m., Carl also set off from Town A to Town B at an average speed of 75 km/h. Carl arrived at Town B at 10.45 p.m.

(a) At what time did Carl overtake Bob?

(b) How long did Bob take to travel from Town B to Town C if he increased his speed by 50% starting from Town B?

1.30p.m. --> 2.45p.m. 1h15min

1h15min*60km/h=75km

75km/h-60km/h=15km/h

75km/15km/h=5h

2.45p.m-->5h-->7.45p.m.

b)

2.45p.m. --> 10.45p.m. 8h

8h*75km/h=600km

825km-600km=225km

75km/h*150%=112.5kmh

225km/112.5kmh=2h

a)7.45p.m.

b)2h

Sorry if it is a little messy.

Post
by **DearSonR** » Wed Jul 01, 2015 6:44 pm

A few things you should note before my number statements:Ilye wrote:Hi, may I know how to see angle ASQ=20? Is there any quick way to see this? Thanks.Nebbermind wrote:Angle QCB = 70EC72 wrote:

Need help with this question. Thank you.

Angle ASQ = 20

Therefore Angle DSQ = 160

1. QCB+ABC=180

2. The angle at the extended line from QC in the parallelogram and QCB when added together is 180 too.

3. Angles on a straight line add up to 180.

4. Angles in a triangle add up to 180.

To get ASQ:

QCB-->180-110=70

Angle at the extended line from QC in the parallelogram-->180-70=110

Angle at the extended line from QC in the triangle (not the right angle)-->180-110=70

ASQ-->180-70-90=20

If you knew that angles which are opposite of one another in a parallelogram and rhombus are equal, you could save the first 2 steps and just go on to the 3rd.

Hope this helps you!!!

Post
by **Daddy** » Wed Jul 01, 2015 9:24 pm

Thank Nebbermind.

Why u square root 196,even if 196 is a triangle?And how is 14 the radius of the circle?Nebbermind wrote:Daddy wrote:Hi all, need help on these questions.

The shaded triangles form a square of side 14.

Radius of circle is 14

Perimeter = (3/4 * 2 * pi * 14) + 14 + 14

Post
by **DearSonR** » Wed Jul 01, 2015 9:38 pm

I believe that the radius of the circle should be square root of 98.Daddy wrote:Thank Nebbermind.Why u square root 196,even if 196 is a triangle?And how is 14 the radius of the circle?Nebbermind wrote:Daddy wrote:Hi all, need help on these questions.

The shaded triangles form a square of side 14.

Radius of circle is 14

Perimeter = (3/4 * 2 * pi * 14) + 14 + 14

Because the length of 1 shaded triangle is twice the height of 1 shaded triangle. So, if you let the height of 1 shaded triangle be x,

x*2x*0.5*2=196=2x^2

x^2=98

x=square root of 98

(square root of 98*2*3/4*22/7)+square root of 98+square root of 98=66.4680374315354673

Sorry if this did not solve your question... Did the question tell the student to round off to nearest whole number?

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