Tutor MathsGuru: Ask me for your burning Maths questions!

[Daddy]

Hi starlight1968sg,


"Area of unshaded triangle = 1/2 * 4 * 12 = 24cm2".
Why the area’s formula can applied on the triangle which is not a right angle?

24*3 ??? not understand.

thanks alot.

[ruyu]

Daddy:

Dear Maths guru,


Please help. Thanks.

1. Triangle ABC is a right-angle triangle. The area of the unshaded portion is 1/3 the area of the shaded portion. Find the area of triangle ABC.

http://www.postimage.org/image.php?v=aV9Uz90

Thanks

1/2 x 4 x 12 = 24

24 x 3 = 72

[ruyu]

Daddy:

Hi starlight1968sg,


\"Area of unshaded triangle = 1/2 * 4 * 12 = 24cm2\".
Why the area's formula can applied on the triangle which is not a right angle?

24*3 ??? not understand.

thanks alot.

it can be applied even if it is not an right angled triangle.

[starlight1968sg]

Daddy:

Hi starlight1968sg,


\"Area of unshaded triangle = 1/2 * 4 * 12 = 24cm2\".
Why the area's formula can applied on the triangle which is not a right angle?

24*3 ??? not understand.

thanks alot.

In any triangle, the formula half base * height holds. Note the height must be the vertical height ie it is perpendicular to the base. So in this case for the unshaded triangle, the base is 4 and the vertical height is 12. Thus the area is half of 48 = 24cm2.

Since area of unshaded is 1/3 of the area of shaded triangle ie 1 portion of unshaded and 3 portions of shaded.

Thus shaded triangle is 24 * 3 = 72cm2.

I hope my ans is correct!

[Daddy]

Thanks starlight1968sg for good explaination…

Thanks ruyu

[Dharma]

firebird:

Dear Dharma


Good morning.

Thank you very much for you help. I clearly understand now.

With best regards
Firebird

You're welcome, firebird.

[firebird]

Dear maths guru


Please help me on the following maths which were taken from Paper 2 CHIJ St Nicholas Girl’s School Preliminary Exam 2009:

Question 9,
I apologise for not inserting the diagram, which I am unable to.
The figure below is not drawn to scale. PQRS is a square and CBRD is a rectangle. BRE is a straight line. Given that angle DRF = 136 degree and angle SHD = 129 degree. Find the ratio of angle HRD to angle ERF.

Question 13,
Korey’s allowance to Skyler’s allowance was in the ratio 10:11. Both of them spent an equal amount. Then Korey’s remaining allowance was 7/9 of Skyler’s remaining allowance. What percentage of Skyler’s allowance did she spend? Leave your answer correct to 2 decimal places.

Question 17
At carpark P, the number of lorries to that of vans was in the ratio 3:7.
At carpark Q, the number of lorries to that of vans was in the ratio 8:9.
When 40% more lorries from an industrial estate entered carpark P and 20% of the vans at Q moved to carpark P, there were 76 fewer lorries at carpark P than Carpark Q. How many vehicles were there althogether at the two carparks finally?

Thank you
firebird

[CoffeeCat]

firebird:

Dear maths guru


Please help me on the following maths which were taken from Paper 2 CHIJ St Nicholas Girl's School Preliminary Exam 2009:

Question 13,
Korey's allowance to Skyler's allowance was in the ratio 10:11. Both of them spent an equal amount. Then Korey's remaining allowance was 7/9 of Skyler's remaining allowance. What percentage of Skyler's allowance did she spend? Leave your answer correct to 2 decimal places.

Question 17
At carpark P, the number of lorries to that of vans was in the ratio 3:7.
At carpark Q, the number of lorries to that of vans was in the ratio 8:9.
When 40% more lorries from an industrial estate entered carpark P and 20% of the vans at Q moved to carpark P, there were 76 fewer lorries at carpark P than Carpark Q. How many vehicles were there althogether at the two carparks finally?

Thank you
firebird

Q13, ratio question, ratio method...
Before Korey : Skyler = 10:11 = 20 : 22 (difference now 2)
After Korey : Skyler = 7 : 9 (difference 2 units)
The key here is equal amount spent, so the difference is constant throughout
Percentage of skyler's allowance spent = (22-9) / 22 = 59.09%

[Vanilla Cake]

firebird:

from Paper 2 CHIJ St Nicholas Girl's School Preliminary Exam 2009:

Question 9,
I apologise for not inserting the diagram, which I am unable to.
The figure below is not drawn to scale. PQRS is a square and CBRD is a rectangle. BRE is a straight line. Given that angle DRF = 136 degree and angle SHD = 129 degree. Find the ratio of angle HRD to angle ERF.

Hi firebird,
There is a typo error in your question (see highlighted in red colour). The original question asked for the ratio of angle HRB to angle ERF and not angle HRD to angle ERF.

Angle DHR=180⁰-129⁰ =51⁰
Angle HRB=Angle DHR=51⁰ (alternate angles)

Angle DRB=Angle DRE=90⁰ (BRE is a straight line)
Angle ERF=136⁰-90⁰ = 46⁰

Ratio of angle HRB to angle ERF = 51:46

firebird:

Question 17
At carpark P, the number of lorries to that of vans was in the ratio 3:7.
At carpark Q, the number of lorries to that of vans was in the ratio 8:9.
When 40% more lorries from an industrial estate entered carpark P and 20% of the vans at Q moved to carpark P, there were 76 fewer lorries at carpark P than Carpark Q. How many vehicles were there althogether at the two carparks finally?

Before
Carpark P
L : V
3 : 7

Carpark Q
L : V
8 : 9

1.4x3u=4.2u (40% more lorries from an industrial estate entered carpark P)
0.2x9u=1.8u (20% of the vans at Q moved to carpark P)

After
Carpark P
L : V
4.2 : 8.8

Carpark Q
L : V
8 : 7.2

(8-4.2)u = 3.8u
3.8u = 76
u = 20
(4.2+8.8+8+7.2)u = 28.2u
28.2u = 28.2x20=564

Finally, there were 564 vehicles at the two carparks.

[Maths Monster]

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