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# Maxima & Minima – A Semi-Circular Triangular Rectangular Affair

(8)
Tuition given in the topic of A-Maths Tuition Questions from the desk of at 5:54 pm (Singapore time)

Updated on

To aid those who have run out of questions to do (your kind actually exist???), and as a form of (ahem) public service for the cause against the LMBFH Syndrome, Miss Loi will post here from time to time questions that she found (either during her daily tuition or by her own creation) to be challenging, interesting, important, introspective, deep, meditative , sadistic (haha) etc. etc…

Enough of lame intros, let’s kick-off with the first question!

A semi-circle is shown with diameter AB. Rectangle DEFG is drawn within triangle ABC. Given that AC = 8 cm and BC = 6 cm, find

1. The perpendicular distance from C to AB.
2. Let DG be x. Show that the area of rectangle DEFG, A m2, is given by
A = 10x – (25/12)x2
3. What’s the value of x for A to have a stationary value? Is this value of A a maximum or a minimum?

### Revision Exercise

To show that you have understood what Miss Loi just taught you from the centre, you must:

1. john commented in tuition class

2007
Mar
23
Fri
11:07pm

1

1) h=4.25
the rest dunno so difficult maybe too tired ...

2. maple commented in tuition class

2007
Mar
26
Mon
10:34pm
3. bruce commented in tuition class

2007
Mar
29
Thu
6:31am

3

As a geometry teacher, I like this problem. It touches on properties of semicircles, area of triangles, coordinate geometry, and a little derivative at the end. Oh yes, and the height is NOT 4.25 (I don't want to spoil the answer, but it's a little taller).

4. Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class

2007
Mar
29
Thu
10:52am

4

Wah Level 10 guru here to comment! *Kneels down and bow*

John: someone's trying to pwn u!

I should put the answers up soon but some of my students still haven't try yet - slackers! (Just joking here la - actually they're also very busy with school at the moment :))

5. shihui commented in tuition class

2007
Mar
30
Fri
7:10pm

5

haha... i got h=4.8cm, the rest dunno how to do. =P

6. Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class

2007
Apr
1
Sun
11:14am

6

Part 1:

AB = √(82 + 62) = 10 cm
∴ Area of ABC = (1/2)(8)(6) = 24 cm2

h being height of triangle ABC, so solving this equation:
(1/2)(h)(AB) = 24 cm2

You should get h = 4.8 cm! 🙂 (shihui where u want Miss Loi to treat u???)

Part 2:

Using similar triangles,

(GF)/(AB) = (h-GD)/h
GF = AB(1 - GD/h)
= (10)(1 - x/h)
= 10 - 10x/h

Now using the value of h from part 1, 10/h = 10/4.8 = 25/12.

GF = 10 - (25/12)x
A = GD x GF
= x( 10 - (25/12)x )
= 10x - (25/12)x2 🙂

Part 3:

∴ A is max when x = 2.4 cm 🙂

Wah type until fingers also numb liao! Actually managed to get all these maths symbols out!

BTW thanks to all of you who tried and responded but why all only do part 1???? Next time don't 半途而废喔!

7. TYS commented in tuition class

2007
Sep
29
Sat
5:22pm

7

Anyway, I find that one of the steps in part 2 is wrong.

GF/ AB = GD / h

GF/ AB = (h - GD)/h

Other than that, I agree with you that this is a gd qn. In fact, there are a lot of concepts tested in this qn.

8. Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class

2007
Sep
29
Sat
6:03pm

8

Thanks so much TYS, for pointing out this typo all the way back from the pioneering days of this blog!

At the same time, Miss Loi has taken the liberty to 'enhance' the readability of the said working (just a little). She's feeling so noob now looking at her early posts 😛

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