Math Is Fun Forum

Monox D. I-Fly

Member

From: Indonesia

Registered: 2015-12-02

Posts: 2,000

[ASK] About Triangle

The triangle ABC is a right triangle with A as the right angle and BD is the bisector of angle B. If AB = 12 cm and BC = 15 cm, the length of AD is ....
A. 3 cm
B. 4 cm
C. 5 cm
D. 6 cm
It was a question for a 9th grader and the book hasn't covered trigonometry by name yet (As in, they don't know about the term sine, cosine, and tangent, but the books do explain the length ratio of triangle which has 45°-90°-45° angle or 30°-60°-90° angle. How to do it and explain it to them?

---

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: [ASK] About Triangle

hi Monox D. I-Fly

This is a tricky one. The method I have found only works because of the specific measurements in the problem.

Start by marking point E, on BA, 9cm from B.  Note that EA = 3cm.

Complete the rectangle EFCA which is 9 by 3.

Mark point G on EF so that EG = 3cm.  Note If H is on AC so that AH = 3cm then a circle centred on G with radius 3 will go through both E and H because BA and AC are tangents to this circle.

So the circle is the inscribed circle for triangle BAC and G will lie on the angle bisectors of angle B (and angle A).  So BG extended will meet AC at D.

Triangles BEG and BAD  are similar; BE:EG = 9:3 so BA:AD = 12:4  Therefore AD = 4cm.

Bob

ps. Just looked at my post and I see the diagram is a bit large.  Sorry.

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Monox D. I-Fly

Member

From: Indonesia

Registered: 2015-12-02

Posts: 2,000

Re: [ASK] About Triangle

ps. Just looked at my post and I see the diagram is a bit large.  Sorry.

It's okay, bob. Thanks for your help.

---

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: [ASK] About Triangle

hi

After I've posted I like to think over what I've done to check it is correct.  I have realised I've made an incorrect assumption with the above.  But it is easily corrected so I'll do that now:

I have said that the circle is the inscribed one for ABC.  But I should prove that BC is also a tangent before we can be sure this is true.

Construct a perpendicular  from G to the line BC meeting that line at point H.

Equation of BC is 12x + 9y = 108   (1)

GH will have equation y = 3x/4 + k and goes through (3,3) so k = 3/4

Rearrange this to 16y = 12x + 12    (2)

Eliminating 12x from these gives y = 24/5 and hence x = 27/5. So H is (27/5, 24/5)

Using Pythagoras GH = Sqroot[( 3 - 27/5)^2 + (3-24/5)^2] = Sqroot[  (1/5)^2 . 3^2 . (4^2 + 3^2)  ]

Therefore GH = 3.

So H is on the circle and GH is perpendicular to BC.

So the circle has all three sides as tangents so it is the inscribed circle.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Monox D. I-Fly

Member

From: Indonesia

Registered: 2015-12-02

Posts: 2,000

Re: [ASK] About Triangle

Okay, thanks for your correction bob!

---

Actually I never watch Star Wars and not interested in it anyway, but I choose a Yoda card as my avatar in honor of our great friend bobbym who has passed away.
May his adventurous soul rest in peace at heaven.