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### Triangles

`☼` Pythagoras Theorem .

● You are already familiar with the Pythagoras Theorem from your earlier classes. You had verified this theorem through some activities and made use of it in solving certain problems.

● Now, we shall prove this theorem using the concept of similarity of triangles. In proving this, we shall make use of a result related to similarity of two triangles formed by the perpendicular to the hypotenuse from the opposite vertex of the right triangle.

● Now, let us take a right triangle ABC, right angled at B. Let BD be the perpendicular to the hypotenuse AC (see Fig. 6.45).

You may note that in `Δ ADB` and `Δ ABC`

`∠ A = ∠ A` and `∠ ADB = ∠ ABC`

So,` Δ ADB ~ Δ ABC`....................... (1)

Similarly, `Δ BDC ~ Δ ABC`...................... (2)

● So, from (1) and (2), triangles on both sides of the perpendicular BD are similar to the whole triangle ABC.

Also, since `Δ ADB ~ Δ ABC` and `Δ BDC ~ Δ ABC`

So, `Δ ADB ~ Δ BDC` (From Remark in Section 6.2)

The above discussion leads to the following theorem :

`color{red}{"Theorem 6.7 :"}` If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Let us now apply this theorem in proving the Pythagoras Theorem:

`color{red}{"Theorem 6.8 :"}` In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Proof : We are given a right triangle ABC right angled at B.

We need to prove that `AC^2 = AB^2 + BC^2`

● Let us draw `BD ⊥ AC` (see Fig.)

●Now, `Δ ADB ~ Δ ABC` (Theorem 6.7)

so `(AD)/(AB) = (AB)/(AC)` (Sides are proportional)

or, `AD . AC = AB^2` ........ (1)

Also, `Δ BDC ~ Δ ABC` (Theorem 6.7)

so `(CD)/(BC) = (BC)/(AC)`

or, `CD . AC = BC^2` .................(2)

Adding (1) and (2),

`AD . AC + CD . AC = AB^2 + BC^2`

or, `AC (AD + CD) = AB^2 + BC^2`

or, `AC . AC = AB^2 + BC^2`

or, `color{orange}{AC^2 = AB^2 + BC^2}`

●The above theorem was earlier given by an ancient Indian mathematician Baudhayan (about 800 B.C.E.) in the following form :

● The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e., length and breadth).

For this reason, this theorem is sometimes also referred to as the Baudhayan Theorem.

● `color{red}{"Theorem 6.9 :"}` In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Proof : Here, we are given a triangle `ABC` in which `AC^2 = AB^2 + BC^2.` We need to prove that `∠ B = 90°.`

To start with, we construct a `Δ PQR` right angled at `Q` such that `PQ = AB` and

`QR = BC` (see Fig. 6.47).

Now, from `Δ PQR`, we have :

`PR^2 = PQ^2 + QR^2` (Pythagoras Theorem, as `∠ Q = 90°`)

or, `PR^2 = AB^2 + BC^2` (By construction) ......................................... (1)

But `AC^2 = AB^2 + BC^2` (Given) .............................. (2)

So, `AC = PR` [From (1) and (2)] ...................................... (3)

Now, in `Δ ABC` and `Δ PQR`,

`AB = PQ` (By construction)

`BC = QR` (By construction)

`AC = PR` [Proved in (3) above]

So, `Δ ABC ≅ Δ PQR` (SSS congruence)

Therefore, `∠ B = ∠ Q `(CPCT)

But `∠ Q = 90°` (By construction)

So,` ∠ B = 90°`

Note : Also see Appendix 1 for another proof of this theorem.

● Now, we shall prove this theorem using the concept of similarity of triangles. In proving this, we shall make use of a result related to similarity of two triangles formed by the perpendicular to the hypotenuse from the opposite vertex of the right triangle.

● Now, let us take a right triangle ABC, right angled at B. Let BD be the perpendicular to the hypotenuse AC (see Fig. 6.45).

You may note that in `Δ ADB` and `Δ ABC`

`∠ A = ∠ A` and `∠ ADB = ∠ ABC`

So,` Δ ADB ~ Δ ABC`....................... (1)

Similarly, `Δ BDC ~ Δ ABC`...................... (2)

● So, from (1) and (2), triangles on both sides of the perpendicular BD are similar to the whole triangle ABC.

Also, since `Δ ADB ~ Δ ABC` and `Δ BDC ~ Δ ABC`

So, `Δ ADB ~ Δ BDC` (From Remark in Section 6.2)

The above discussion leads to the following theorem :

`color{red}{"Theorem 6.7 :"}` If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Let us now apply this theorem in proving the Pythagoras Theorem:

`color{red}{"Theorem 6.8 :"}` In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Proof : We are given a right triangle ABC right angled at B.

We need to prove that `AC^2 = AB^2 + BC^2`

● Let us draw `BD ⊥ AC` (see Fig.)

●Now, `Δ ADB ~ Δ ABC` (Theorem 6.7)

so `(AD)/(AB) = (AB)/(AC)` (Sides are proportional)

or, `AD . AC = AB^2` ........ (1)

Also, `Δ BDC ~ Δ ABC` (Theorem 6.7)

so `(CD)/(BC) = (BC)/(AC)`

or, `CD . AC = BC^2` .................(2)

Adding (1) and (2),

`AD . AC + CD . AC = AB^2 + BC^2`

or, `AC (AD + CD) = AB^2 + BC^2`

or, `AC . AC = AB^2 + BC^2`

or, `color{orange}{AC^2 = AB^2 + BC^2}`

●The above theorem was earlier given by an ancient Indian mathematician Baudhayan (about 800 B.C.E.) in the following form :

● The diagonal of a rectangle produces by itself the same area as produced by its both sides (i.e., length and breadth).

For this reason, this theorem is sometimes also referred to as the Baudhayan Theorem.

● `color{red}{"Theorem 6.9 :"}` In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Proof : Here, we are given a triangle `ABC` in which `AC^2 = AB^2 + BC^2.` We need to prove that `∠ B = 90°.`

To start with, we construct a `Δ PQR` right angled at `Q` such that `PQ = AB` and

`QR = BC` (see Fig. 6.47).

Now, from `Δ PQR`, we have :

`PR^2 = PQ^2 + QR^2` (Pythagoras Theorem, as `∠ Q = 90°`)

or, `PR^2 = AB^2 + BC^2` (By construction) ......................................... (1)

But `AC^2 = AB^2 + BC^2` (Given) .............................. (2)

So, `AC = PR` [From (1) and (2)] ...................................... (3)

Now, in `Δ ABC` and `Δ PQR`,

`AB = PQ` (By construction)

`BC = QR` (By construction)

`AC = PR` [Proved in (3) above]

So, `Δ ABC ≅ Δ PQR` (SSS congruence)

Therefore, `∠ B = ∠ Q `(CPCT)

But `∠ Q = 90°` (By construction)

So,` ∠ B = 90°`

Note : Also see Appendix 1 for another proof of this theorem.

Q 3119480319

In Fig. 6.48, `∠ ACB = 90°` and `CD ⊥ AB`. Prove that `(BC^2)/(AC^2)

Class 10 Chapter 6 Example 10

Class 10 Chapter 6 Example 10

`Δ ACD ~ Δ ABC` (Theorem 6.7)

so `(AC)/(AB) = (AD)/(AC)`

or `AC^2 = AB . AD` .............(1)

Similarly, `Δ BCD ~ Δ BAC` (Theorem 6.7)

so `(BC)/(BA) = (BD)/(BC)`

or, `BC^2 = BA . BD` .................................... (2)

Therefore, from (1) and (2),

`(BC^2)/(AC^2) = (BA . BD)/(AB . AD) = (BD)/(AD)`

Q 3119580410

A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.

Class 10 Chapter 6 Example 11

Class 10 Chapter 6 Example 11

Let AB be the ladder and CA be the wall with the window at A (see Fig. 6.49).

Also, `BC = 2.5 m` and `CA = 6 m`

From Pythagoras Theorem, we have:

`AB^2 = BC^2 + CA^2`

`= (2.5)^2 + (6)^2`

`= 42.25`

So, `AB = 6.5`

Thus, length of the ladder is 6.5 m.

Q 3139580412

In Fig. 6.50, if `AD ⊥ BC`, prove that `AB^2 + CD^2 = BD^2 + AC^2.`

Class 10 Chapter 6 Example 12

Class 10 Chapter 6 Example 12

From `Δ ADC`, we have

`AC^2 = AD^2 + CD^2`

(Pythagoras Theorem) (1) From `Δ ADB`, we have

`AB^2 = AD^2 + BD^2`

(Pythagoras Theorem) (2)

Subtracting (1) from (2), we have

`AB^2 – AC^2 = BD^2 – CD^2`

or, `AB^2 + CD^2 = BD^2 + AC^2`

Q 3119680510

`BL` and `CM` are medians of a triangle `ABC` right angled at A. Prove that `4 (BL^2 + CM^2) = 5 BC^2.`

Class 10 Chapter 6 Example 13

Class 10 Chapter 6 Example 13

`BL` and `CM` are medians of the `Δ ABC` in which `∠ A = 90°` (see Fig. 6.51).

From `Δ ABC,`

`BC^2 = AB^2 + AC^2` (Pythagoras Theorem) ..................... (1)

From `Δ ABL,`

`BL^2 = AL^2 + AB^2`

or `BL^2 = ((AC)/2)^2 + AB^2 ` (L is the mid-point of AC)

or `BL^2 = (AC^2)/4 + AB^2`

or, `4 BL^2 = AC^2 + 4 AB^2` ....................(2)

From `Δ CMA,`

`CM^2 = AC^2 + AM^2`

`CM^2 = AC^2 + ((AB)/2)^2` (M is the mid-point of AB)

`CM^2 = AC^2 +(AB^2)/4`

or `4 CM^2 = 4 AC^2 + AB^2` ..............(3)

Adding (2) and (3), we have

`4 (BL^2 + CM^2) = 5 (AC^2 + AB^2)`

i.e., `4 (BL^2 + CM^2) = 5 BC^2`

Q 3139680512

`O` is any point inside a rectangle `ABCD` (see Fig. 6.52). Prove that

`OB^2 + OD^2 = OA^2 + OC^2`.

Class 10 Chapter 6 Example 14

`OB^2 + OD^2 = OA^2 + OC^2`.

Class 10 Chapter 6 Example 14

Through `O,` draw `PQ ∥ BC` so that `P` lies on `AB` and `Q` lies on `DC.`

Now `PQ ∥ BC`

Therefore, `PQ ⊥ AB` and `PQ ⊥ DC` (`∠ B = 90°` and `∠ C = 90°`)

So, `∠ BPQ = 90°` and `∠ CQP = 90°`

Therefore, `BPQC` and `APQD` are both rectangles.

Now, from Δ OPB,

`OB^2 = BP^2 + OP^2` .......................... (1)

Similarly, from `Δ OQD,`

`OD^2 = OQ^2 + DQ^2` .............................. (2)

From `Δ OQC`, we have

`OC^2 = OQ^2 + CQ^2` ...................................... (3)

and from `Δ OAP`, we have

`OA^2 = AP^2 + OP^2` ................................................ (4)

Adding (1) and (2),

`OB^2 + OD^2 = BP^2 + OP^2 + OQ^2 + DQ^2`

`= CQ^2 + OP^2 + OQ^2 + AP^2`

(As BP = CQ and DQ = AP)

`= CQ^2 + OQ^2 + OP^2 + AP^2`

`= OC^2 + OA^2` [From (3) and (4)]