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[iFruit]

YLH88:

acehkr3009:

Hi,


Here is a area of trangle question in MCQ...

Triangle ABC is an isosceles triangle with sides AB = 12cm AC = BC = 10cm.
Find the area of the triangle ABC?

Pls help...thks

Attached is a picture of the triangle.

http://postimage.org/image/tr6pyybo/

If you join 2 of the same isosceles triangles together, as shown in the diagram below, it will become a square.

Not necessarily. It will become a rhombus but not necessarily a square.

The perpendicular line shown in the picture bisects AB. So in the right angle triangle, base = 6, hypotenuse = 10. So height = 8 (sqrt of 100-36)

So area of the triangle is 1/2 x 12 x 8 = 48 cm²

[Dharma]

YLH88:

Dharma:

[quote=\"Brenda10\"]
Hi,

According to dd, the height is 5cm.

Area of 1/2 of the triangle : 1/2 x 5 x 10 = 25CM²
Total : 25CM² x 2 = 50CM²


Hope the answer is correct.

Perpendicular height is 8cm

Hi Dharma,

How do you get the height = 8 cm ?[/quote]Pythagoras theorem but it is not primary maths stuff

[YLH88]

iFruit:

YLH88:

[quote=\"acehkr3009\"]Hi,


Here is a area of trangle question in MCQ...

Triangle ABC is an isosceles triangle with sides AB = 12cm AC = BC = 10cm.
Find the area of the triangle ABC?

Pls help...thks

Attached is a picture of the triangle.

http://postimage.org/image/tr6pyybo/

If you join 2 of the same isosceles triangles together, as shown in the diagram below, it will become a square.

Not necessarily. It will become a rhombus but not necessarily a square.

The perpendicular line shown in the picture bisects AB. So in the right angle triangle, base = 6, hypotenuse = 10. So height = 8 (sqrt of 100-36)

So area of the triangle is 1/2 x 12 x 8 = 48 cm²[/quote]Hi Dharma, iFruit,

Thanks for pointing out my mistake and I have already deleted my solution posted earlier. But the kids don't learn Pythagoras theorem in primary school ....

acehkr3009,
May I know the source of this question ? is it for primary level ?

[iFruit]

amberzbon:

Please help to solve these problems. thanks in advance

\t
Q1) Ali, Ben and Col each had a certain number of cards. Ali gave to Ben and Col some cards which is equal to the amount Ben and Col each originally had. Next, Ben gave to Ali an Col some cards which is equal to the amount Ali and Col each already had and finally Col gave to Ali and Ben in the similar manner. At the end, each had 32 cards each. Find the number of cards each had originally.

This is a backwards problem..

At the end,

A=B=C=32

1 step back,

A = 32/2 = 16
B = 32/2 = 16
C = 32 + 16 +16 = 64

2 steps back

A = 16/2 = 8
C = 64/2 = 32
B = 16 + 8 + 32 = 56

3 steps back (at the beginning)

C = 32/2 = 16
B = 56/2 = 28
A = 8+28+16 = 52

Draw this in a table..

[iFruit]

amberzbon:

Please help to solve these problems. thanks in advance



Q5) A box contains both pink and blue pens. 1/2 of the pink pens and 1/3 of the blue pens make up a total of 13 pens. 1/3 of the pink pens and 1/2 of the blue pens make up a total of 12 pens. How many pens are there of each colour?

Blue pens = B
Pink Pens = P

P/2 + B/3 = 13----> 3P +2B = 78 -----> 6P+4B = 156-------(1)
P/3 + B/2 = 12----> 2P + 3B = 72-----> 6P+9B = 216-------(2)

From (1) and (2)
5B = 60---> B=12

P = 18

HTH

[amberzbon]

iFruit:

amberzbon:

Please help to solve these problems. thanks in advance



Q5) A box contains both pink and blue pens. 1/2 of the pink pens and 1/3 of the blue pens make up a total of 13 pens. 1/3 of the pink pens and 1/2 of the blue pens make up a total of 12 pens. How many pens are there of each colour?

Blue pens = B
Pink Pens = P

P/2 + B/3 = 13----> 3P +2B = 78 -----> 6P+4B = 156-------(1)
P/3 + B/2 = 12----> 2P + 3B = 72-----> 6P+9B = 216-------(2)

From (1) and (2)
5B = 60---> B=12

P = 18

HTH

Thanks alot

[peapot]

Hi everyone!


I need to post here a P1 question which was from my DD recent test paper.

Please help to solve.

Jason uses some sticks to form some squares as shown below. How many such squares can he make if he has 24 sticks?

[/img][/list]

[peapot]

sorry, I miss out the picture as I can’t seems to upload it. Exasperating!!!

[acehkr3009]

YLH88:

iFruit:

[quote=\"YLH88\"]
If you join 2 of the same isosceles triangles together, as shown in the diagram below, it will become a square.

Not necessarily. It will become a rhombus but not necessarily a square.

The perpendicular line shown in the picture bisects AB. So in the right angle triangle, base = 6, hypotenuse = 10. So height = 8 (sqrt of 100-36)

So area of the triangle is 1/2 x 12 x 8 = 48 cm²

Hi Dharma, iFruit,

Thanks for pointing out my mistake and I have already deleted my solution posted earlier. But the kids don't learn Pythagoras theorem in primary school ....

acehkr3009,
May I know the source of this question ? is it for primary level ?[/quote]Hi everyone,

I saw this question in P5 test paper 2 Section A Question 5 MCQ, from Advanced Learners Final Exam paper (grey colour loose plastic pack in Popular). 4 options here-(1) 40 cm sq (2) 48 cm sq (3) 50 cm sq (4) 60 cm sq & Ans is 48 cm sq. But dont know how to explain to my son if using Pythagoras theorem.

My solution to my son....I told him that the height can never be longer than slanted side which is 10cm. With this, use guess & check to do 6 (half of 12cm) x 9 or 8 or 7... until he gets one of the answer in the MCQ, which is 48 cm sq. I think that will be the simplest option in the context of Primary level.

Thanks all for trying..

[atutor2001]

acehkr3009:

..


My solution to my son....I told him that the height can never be longer than slanted side which is 10cm. With this, use guess & check to do 6 (half of 12cm) x 9 or 8 or 7... until he gets one of the answer in the MCQ, which is 48 cm sq. I think that will be the simplest option in the context of Primary level.

Thanks all for trying..

I believe your approach is the same as what setter of the question has in mind. To find the area without knowledge of Pythagoras theorem is equivalent to the discovery of Pythagoras theorem. Was googling and found this site very interesting on how Pythagoras theorem can be proven.

http://www.1728.com/pytproof.htm