## Triangles | Pythagoras Theorem

**Theorem:
**

**In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
**

**Given:** A right triangle *ABC*, right angled at *B
*

**To prove:**

**Construction: **Draw *BD *^ *AC*

**Proof: **

*In D*

*ADB*and D

*ABC*, we have

Ð*ADB *= Ð*ABC* [Each equal to 90°]

Ð*A* = Ð*A *[Common]

\

DADB ~ DABC [By AA similarity]

Þ [Corresponding sides of similar triangles are proportional]

Þ …(i)

In DBCD and DACB, we have

ÐCDB = ÐCBA [Each equal to 90°]

ÐC = ÐC [Common]

By AA similarity criterion

D*BCD* ~ D*ACB*

\

Þ …(ii)

Adding equations (i) and (ii), we get

Hence,

**Theorem (Converse of Pythagoras theorem):
**

** In a triangle if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.
**

**Given:** A triangle *ABC* such that

**To prove:**

**Construction:** Draw a D*PQR* right angled at *Q* such that *PQ *= *AB *and *QR *= *BC*.

**Proof:** In right triangle *PQR*, we have

[By Pythagoras Theorem]

Þ …(i)

[

*PQ* = *AB* and *QR* = *BC* (by construction)]

But [Given] …(ii)

From equation (i) and equation (ii), we get

Þ …(iii)

Now, in D*ABC *and D*PQR*, we have

*AB *= *PQ *and *BC *= *QR* [By construction]

and [From (iii)]

\ D*ABC *@

D*PQR* [By *SSS *congruency]

\ Ð*B* = Ð*Q* = 90° [c.p.c.t*.*]

Hence, Ð*B* = 90°.

## Solved Questions based on Pythagoras Theorem

### Question:

#### The sides of some triangles are given below. Determine which of them are right triangles.

#### (i) 7 cm, 24 cm, 25 cm

#### (ii) 7 cm, 8 cm, 6 cm

### Solution:

If the sum of the squares of the smaller sides of a triangle is equal to the square of the larger side, then the triangle is right angled.

**(i)
**

(7 cm)^{2} + (24 cm)^{2} = (49 + 576) cm^{2} = 625 cm^{2} and (25 cm)^{2} = 625 cm^{2}

Hence, the given triangle is a right triangle.

**(ii)
**

(7 cm)^{2} + (6 cm)^{2} = (49 + 36) cm^{2} = 85 cm^{2} and (8 cm)^{2} = 64 cm^{2}

Hence, the given triangle is not a right triangle.

### Question:

#### A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its top reaches a window 12 m above the ground. Determine the length of the ladder.

### Solution:

Let *AB *be the ladder, *B* be the window and *BC *be the wall

Then *BC* = 12 m, *AC* = 5m and Ð*ACB* = 90°

In right triangle *ACB*, we have

[By Pythagoras Theorem]

Þ

Þ

Hence, the length of the ladder is 13 m.

### Question:

#### A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.

### Solution:

Let *AB *be the street and let *C* and *D* be the windows at heights of 9 m and 12 m respectively from the ground.

Let *E* be the foot of the ladder. Then *EC* and *ED* are the two positions of the ladder.

Clearly *AC* = 9 m, *BD* = 12 m, *EC* = *ED* = 15 m and Ð*CAE* = Ð*DBE* = 90°

In right triangle *CAE*, we have

[By Pythagoras Theorem]

Þ (15)^{2} = (9)^{2} + *AE*^{2}

Þ *AE*^{2} = (15)^{2} – (9)^{2} = (225 – 81) = 144

Þ *AE* = 12 m …(i)

In right triangle *DBE*, we have

[By Pythagoras Theorem]

Þ (15)^{2} = (12)^{2} + *EB*^{2}

Þ *EB*^{2} = (15)^{2} – (12)^{2} = (225 – 144) = 81

Þ *EB* = 9 m …(ii)

Adding equations (i) and (ii), we get

*AE* + *EB* = (12 + 9) m Þ *AB* = 21 m

Hence, the width of the street is 21 m.

### Question:

#### In figure, Ð*ACB* = 90° and *CD*

^

*AB*. Prove that

### Solution:

**Given:**

Ð*ACB* = 90° and *CD *^ *AB
*

**To prove: **

**Proof:*** *In D*ACD *and D*ABC
*

*
*Ð

*A*= Ð

*A*[Common]

Ð*ADC *= Ð*ACB* [Both 90°]

*\ *D*ACD *~ D*ABC* [By *AA *similarity]

So,

or, …(i)

Similarly, D*BCD *~ D*BAC*

So,

or, …(ii)

Therefore, from (i) and (ii),

### Question:

*P* and *Q* are the midpoints of the sides *CA* and *CB* respectively of D*ABC*, right angled at *C*. Prove that

#### (i)

#### (ii)

#### (iii)

### Solution:

**Given:*** P* and *Q* are the midpoints of the sides *CA* and *CB* respectively of D*ABC*, right angled at *C*

**To Prove:
**

(i)* *

(ii)

(iii)

**Proof:
**

**(i)
**

From right D*ACQ*, we have

* *[By Pythagoras Theorem]

Þ [** ***Q* is the midpoint of *CB*]

Þ

Þ …(i)

**(ii)
**

From right D*BCP*

[By Pythagoras Theorem]

Þ [** ***P* is the mid-point of side *CA*]

Þ

Þ … (ii)

**(iii)
**

Adding equation (i) and equation (ii), we get

Þ

Þ [ In D*ABC, *]

Hence,

### Question:

*O* is any point inside a rectangle *ABCD* in figure. Prove that *OB*^{2} + *OD*^{2} = *OA*^{2} + *OC*^{2}.

### Solution:

**Given:*** O* is any point inside a rectangle *ABCD
*

**To prove:**

*OB*^{2} + *OD*^{2} = *OA*^{2} + *OC*^{2}

**Construction:***
*Through

*O*, draw

*PQ*||

*BC*so that

*P*lies on

*AB*and

*Q*lies on

*DC*.

**Proof:*** PQ* || *BC *[By construction]

Therefore, *PQ *^ *AB* and *PQ *^ *DC* (Ð*B* = 90° and Ð*C* = 90°)

So, Ð*BPQ* = 90° and Ð*CQP* = 90°

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

Now, from D*OPB*,

…(i)

Similarly, from D*OQD*,

…(ii)

From D*OQC*, we have

…(iii)

and from D*OAP*, we have

…(iv)

Adding (i) and (ii),

(As *BP* = *CQ* and *DQ* = *AP*)

[From (iii) and (iv)]

### Question:

*ABC *is a triangle right-angled at *C* and *p* is the length of the perpendicular from *C* to *AB*. Show that

#### (a) *pc *=* ab*

#### (b) where *a* = *BC*, *b* = *AC* and *c* = *AB*.

### Solution:

**(a)
**

Taking *c* as the base and *p* as the altitude, we have

area of D*ABC*

…(i)

Taking *b* as the base and a as the altitude, we have

area D*ABC* = …(ii)

\ [From (i) and (ii)]

Þ *pc = ab* Hence Proved.

**(b)
**

*ABC* is a right triangle-angled at *C*.

\ …(iii)

[By Pythagoras Theorem]

*pc* = *ab * [proved above]

\

or

Þ [From equation (iii)]

Þ

### Question:

#### In D*ABC*, Ð*C* > 90° and side *AC* is produced to *D* such that segment *BD* is perpendicular to segment *AD*. Prove that

### Solution:

**Given:**

D*ABC* in which Ð*ACB* > 90°.

**To prove: **

**Construction:** Draw *BD *^ *AC* (produced).

**Proof:*** *In right-triangle D*BDA*, we get

…(i) [By Pythagoras theorem]

[*AD* = *AC* + *CD*]

[ In right-angled D*BDC*, *BD*^{2} + *CD*^{2} = *BC*^{2}] [By Pythagoras theorem]

Hence,

### Question:

#### In D*ABC*, Ð*B* < 90° and *AD* is drawn perpendicular to *BC*. Prove that

### Solution:

**Given:**

D*ABC* in which Ð*B* < 90° and AD is perpendicular to *BC*

**To prove: **

**Proof:*** *In right-angled D*ABD*, we have

[By Pythagoras theorem]…(i)

In right-angled D*ADC*, we have

[*DC *= *BC *– *BD*]

[In D

*ABD*, *AB ^{2 }*=

*AD*+

^{2}*BD*]

^{2}

Þ

### Question:

#### The perpendicular from *A* on the side *BC* of a D*ABC* intersects *BC* at *D* such that *DB* = 3 *CD*. Prove that

### Solution:

**Given:** In D*ABC, **AD *^ *BC* and *BD* = 3 *CD*.

**To prove: **

**Proof: **

*(given)*

Adding *CD* on both sides

*BC* = 4*CD*

…(i)

In D*ABD*, we have,

[Pythagoras Theorem]

Þ

[ In D*ADC*, ]

[From (i)]

Þ

or

### Question:

#### In an equilateral triangle *ABC*, *D* is a point on *BC* such that Prove that

### Solution:

**Given:** In D*ABC *is an equilateral triangle and *D* is a point on *BC* such that

**To prove:**

**Construction:*** *Draw and join *A* to *D*.

**Proof:** In D*AEB* and D*AEC*, we have

[Common]

Ð*AEB* = Ð*AEC* [Each = 90°]

[*ABC* is an equilateral triangle]

\ D*AEB *@ D*AEC* [R.H.S. congruency]

\ *BE = EC* [c.p.c.t.]

Now, and

In D*ABE
*

[By Pythagoras theorem]

=

= [In D*ADE*

]

[]

[ and ]

### Question:

#### Prove that three times the square of any side of an equilateral-triangle is equal to four times the square of the altitude.

### Solution:

**Given:**

*ABC* be an equilateral triangle and let *AD *^ *BC*.

**To prove:** 3*AB ^{2}* = 4

*AD*

^{2}

**Proof:***
*In D

*ADB*and D

*ADC*, we have

*AB* = *AC* [Given]

Ð*B* = Ð*C* [Each equal to 60°]

and, Ð*ADB* = Ð*ADC *[Each equal to 90°]

\ D*ADB *@ D*ADC *[By AAS congruency]

Þ *BD* = *DC* [c.p.c.t.]

Þ

Since D*ADB* is a right triangle right-angled at *D*.

\

Þ

Þ [

*BC* = *AB*]

Þ [

*BC* = *AB*]

Þ

Þ

### Question:

*ABC *is a right triangle right-angled at *C* and Prove that Ð*ABC *= 60°.

### Solution:

**Given:**

*ABC *is a right triangle right-angled at *C* and

**To prove:**

Ð*ABC *= 60°

**Construction:** Let *D *be the mid-point of *AB*. Join *CD*.

**Proof:**

*ABC* is a right triangle right-angled at *C*.

\ [By Pythagoras theorem]

Þ [

(Given)]

Þ

But, i.e., *AB* = 2*BD
*

\ *BD* = *BC
*

But mid-point of the hypotenuse of a right triangle is equidistant from the vertices.

\ *CD = AD = BD*

Þ *CD = BC* [** ***BD* = *BC*]

Thus, in D*BCD*, we have

*BD = CD = BC
*

Þ D*BCD* is equilateral Þ Ð*ABC* = 60°

### Question:

#### The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

### Solution:

**Given:** A D*ABC* in which *AD* is the internal bisector of Ð*A* and meets *BC* in *D*.

**To Prove:*** *

**Construction:** Draw *CE* || *DA* to meet *BA* produced at *E*.

**Proof:** Ð2 = Ð3 …(i)

[Alternate angles]

and, Ð1 = Ð4 …(ii)

[Corresponding angles]

But, Ð1 = Ð2 [

*AD* is the bisector of Ð*A*]

From (i) and (ii), we get

Ð3 = Ð4

Þ *AE* = *AC* …(iii)

[Sides opposite to equal angles are equal]

Now, in D*BCE*, we have

[By construction]

Þ [Using Basic Proportionality Theorem]

Þ [(From (iii)]

Hence,

### Question:

*ABCD *is a quadrilateral in which *AB *= *AD*. The bisector of Ð*BAC *and Ð*CAD *intersect the sides *BC *and *CD *at the points *E *and *F *respectively. Prove that *EF *|| *BD*.

### Solution:

**Given:** A quadrilateral *ABCD *in which *AB *= *AD *and the bisectors of Ð*BAC *and Ð*CAD *meet the sides *BC *and *CD *at *E *and *F *respectively.

**To Prove:**

*EF || BD*

**Construction: **Join *AC*, *BD* and *EF*.

**Proof: ** In D*CAB*, *AE* is the bisector of Ð*BAC*.

\ …(i)

In D*ACD*, *AF* is the bisector of Ð*CAD*.

\

Þ [

*AD* = *AB*] …(ii)

From (i) and (ii), we get

Þ

Therefore, by the converse of BPT Theorem, we have

*EF* || *BD*.

### Question:

#### Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out in figure? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

### Solution:

**(i)
**

In right angled DABC, AB = 1.8 cm, BC = 2.4 cm

\

Þ

Hence the original length of the string *AC* is 3 m.

**(ii)
**

When Nazima pulls in the string at the rate of 5 cm/sec, then the length of the string decreases = 5 ´ 12 cm = 60 cm = 0.60 m in 12 seconds.

\ Remaining length of the string (AD) after 12 seconds = (3 – 0.60) = 2.40 m

Now in right angled D*ABD*,

Þ

m Þ

= 1.587 m

Þ Horizontal distance (*DE*) of the fly from Nazima

= (1.587 + 1.2) m = 2.787 m = 2.79 m.