Question

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Answer 1

Proof:

See fig. (1).

Suppose the triangle is not a right triangle. Take it as ΔABC.

There are two possibilities for ΔABC: one is of an acute triangle (left) and the other is of an obtuse triangle (right).

Each sides are labelled as a, b, c. So let,

a² + b² = c²

Draw CD of length b perpendicular to BC.

Form ΔBCD which is right triangle.

In ΔBCD, by Pythagoras' theorem,

BD² = a² + b², which is equal to c².

∴ BD² = c²

∴ BD = c

Okay, also we get two isosceles triangles ACD and ABD.

In ΔACD,

AC = CD

∠ADC = ∠CAD

In ΔABD,

AB = BD

∴ ∠ADB = ∠BAD

From the acute triangle at left, it seems that ∠ADB < ∠ADC.

As ∠ADB = ∠BAD and ∠ADC = ∠CAD, it should also seem that ∠BAD < ∠CAD.

But it seems that ∠BAD > ∠CAD.

So here it is a contradiction.

Also, from the obtuse triangle at right, it seems that ∠ADB > ∠ADC.

So it should seem that ∠BAD > ∠CAD, as ∠ADB = ∠BAD and ∠ADC = ∠CAD.

But it seems that ∠BAD < ∠CAD.

So here it is also a contradiction.

So these mean that ΔABC can't be either acute or obtuse; only right.

Hence it's proved!!!

Question:

Determine the length of an altitude of an equilateral triangle of side 2 cm.

Answer:

√3 cm

Step-by-step explanation:

See fig. (2).

We have to prove it by the converse of the above explained proof.

I.e., Finding the answer by the concept that, "If one angle of a triangle is right angle, then the square of the side opposite to this angle is equal to the sum of the squares of the other two sides."

Okay, let's find.

Here, AB = BC = AC = 2 cm

ΔABC is an equilateral triangle. So altitude from any point of this triangle is the perpendicular bisector of the opposite side.

∴ The altitude AD is the perpendicular bisector of BC.

BD = CD = 2 / 2 = 1 cm

Consider ΔACD.

∠ADC = 90°

∴ AC² = AD² + CD²

⇒ 2² = AD² + 1²

⇒ 4 = AD² + 1

⇒ AD² = 4 - 1

⇒ AD² = 3

⇒ AD = √3

(As AD² = 3, AD = - √3 too. But as the altitude is asked, AD can't be - √3.)

∴ √3 is the answer.

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Hope this may be helpful.

Thank you. Have a nice day. :-)

#adithyasajeevan