Question

The angle bisector bisects the opposite side into the two lengths, 2 and 4 units long.
The length of the height on that side is 15 units. Determine the lengths of the other
two sides of the triangle.​

Answer 1

Solution :-

given that,

• AD is angle bisector of ∠A .
• BD = 2 units .
• DC = 4 units .
• AE ⟂ BC and √15 units .

So,

→ AB / AC = BD / DC { By angle bisector theorem }

→ AB / AC = 2/4

→ AB / AC = 1/2 ----------- Eqn.(1)

also,

→ BC = BD + DC = 2 + 4 = 6 units .

now let BE = x units .

So, in right angled ∆AEB ,

→ AB = √(15 + x²) { By pythagoras theorem }

and, in right angled ∆AEC ,

→ AC = √{15 + (x - 6)²} { By pythagoras theorem }

Putting both values in Eqn.(1),

→ AB/AC = 1/2

→ √(15 + x²)/√(15 + x² + 36 - 12x) = 1/2

squaring both sides,

→ (15 + x²) / (x² - 12x + 51) = 1/4

→ 4(15 + x²) = x² - 12x + 51

→ 60 + 4x² = x² - 12x + 51

→ 4x² - x² + 12x + 60 - 51 = 0

→ 3x² + 12x + 9 = 0

→ 3x² + 3x + 9x + 9 = 0

→ 3x(x + 1) + 9(x + 1) = 0

→ (3x + 9)(x + 1) = 0

→ x = (-3) and (-1) .

taking x = (-1),

→ AB = √(15 + x²) = √(15 + 1) = 4 units .

So,

→ AC = 4 * 2 = 8 units .

taking x = (-3),

→ AB = √(15 + 9) = √24 = 4√6 units.

So,

→ AC = 2 * 4√6 = 8√6 units .

Note :- since x is negative, D lies on BC when produced .

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Answer 2

The lengths of the other two sides of the triangle is  8√6 units .

Does an angle bisector bisect the opposite side into the two lengths?

• The angle bisector theorem is commonly used when the angle bisectors and side lengths are known.
• It can be used in a calculation or in a proof.
• An immediate consequence of the theorem is that the angle bisector of the vertex angle of an isosceles triangle will also bisect the opposite side.

How do you write an angle bisector?

• An angle bisector is a line or ray that divides an angle into two congruent angles .
• In the figure, the ray →KM bisects the angle ∠JKL .
• The angles ∠JKM and ∠LKM are congruent.
• So, m∠JKM = m∠LKM .

According to the question:

Given that,

AD is angle bisector of ∠A .

BD = 2 units .

DC = 4 units .

AE ⟂ BC and √15 units .

So, AB / AC = BD / DC { By angle bisector theorem }

AB / AC = 2/4

AB / AC = 1/2........................... Eqn.(1)

Also, BC = BD + DC = 2 + 4 = 6 units .

now let BE = x units .

So, in right angled ∆AEB ,

AB = √(15 + x²) { By pythagoras theorem }

and, in right angled ∆AEC ,

AC = √{15 + (x - 6)²} { By pythagoras theorem }

Putting both values in Eqn.(1),

AB/AC = 1/2

√(15 + x²)/√(15 + x² + 36 - 12x) = 1/2

Squaring both sides,

(15 + x²) / (x² - 12x + 51) = 1/4.

4(15 + x²) = x² - 12x + 51.

60 + 4x² = x² - 12x + 51.

4x² - x² + 12x + 60 - 51 = 0.

3x² + 12x + 9 = 0.

3x² + 3x + 9x + 9 = 0.

3x(x + 1) + 9(x + 1) = 0.

(3x + 9)(x + 1) = 0.

x = (-3) and (-1) .

Taking x = (-1),

AB = √(15 + x²) = √(15 + 1) = 4 units .

So, AC = 4 * 2 = 8 units .

taking x = (-3),

AB = √(15 + 9) = √24 = 4√6 units.

So, AC = 2 * 4√6

= 8√6 units .

since, x is negative, D lies on BC when produced .

Hence, The lengths of the other two sides of the triangle is  8√6 units .

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