# Frank Solutions for Chapter 17 Pythagoras Theorem Class 9 Mathematics ICSE

**Exercise 17.1**

**1. Find the length of the perpendicular of a triangle whose base is 5 cm and the hypotenuse is 13 cm. Also, find its area.**

**Answer**

Given

Base = 5 cm

Hypotenuse = 13 cm

By Pythagoras theorem,

(perpendicular)^{2} = (13 cm)^{2} – (5 cm)^{2}

⇒ (perpendicular)^{2} = 169 cm^{2} – 25 cm^{2}

⇒ (perpendicular)^{2} = 144 cm^{2}

⇒ (perpendicular)^{2} = (12 cm)^{2}

Hence,

perpendicular = 12 cm

Now,

Area of triangle = (1/2)× (Base × Perpendicular)

= (1/2)×5×12

= 5×6

We get,

= 30 cm^{2}

**2. ****Find the length of the hypotenuse of the triangle whose other two sides are 24 cm and 7 cm.**

**Answer**

Given

The two sides (excluding hypotenuse) of a right angled triangle are 24 cm and 7 cm

(hypotenuse)^{2} = (24 cm)^{2} + (7 cm)^{2}

⇒ (hypotenuse)^{2} = 576 cm^{2} + 49 cm^{2}

⇒ (hypotenuse)^{2} = 625 cm^{2}

⇒ (hypotenuse)^{2} = (25 cm)^{2}

⇒ (hypotenuse) = 25 cm

Therefore, the length of the hypotenuse of the triangle is 25 cm

**3. ****Calculate the area of a right – angled triangle whose hypotenuse is 65 cm and one side is 16 cm**

**Answer**

Given

Hypotenuse = 65 cm

One side = 16 cm

Let the length of the other side = x cm

By Pythagoras theorem,

(65 cm)^{2} = (16 cm)^{2} + (x cm)^{2}

⇒ (x cm)^{2} = (65 cm)^{2} – (16 cm)^{2}

⇒ (x cm)^{2} = 4225 cm^{2} – 256 cm^{2}

On further calculation, we get,

(x cm)^{2} = 3969 cm^{2}

We get,

(x cm)^{2} = (63 cm)^{2}

Hence,

x = 63 cm

Area of the triangle = (1/2)×(Base×Height)

= (1/2) × 16 cm × 63 cm

= 8 cm × 63 cm

We get,

= 504 cm^{2}

**4. ****A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.**

**Answer**

Let us take ‘O’ as the original position of the man

From the figure, we come to know that B is the final position of the man

Here,

△AOB is right angled at A

By Pythagoras theorem,

OB^{2} = OA^{2} + AB^{2}

⇒ OB^{2} = (10m)^{2} + (24 m)^{2}

⇒ OB^{2} = 100 m^{2} + 576 m^{2}

⇒ OB^{2} = 676 m^{2}

⇒ OB^{2} = (26 m)^{2}

We get,

OB = 26 m

Therefore, the man is at a distance of 26 m from the starting point

**5.** **A ladder 25 m long reaches a window of a building 20 m above the ground. Determine the distance of the foot of the ladder from the building.**

**Answer**

Let AC be the ladder and A be the position of the window

Then,

AC = 25 m and AB = 20 m

Using Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

By substituting AC = 25 and AB = 20

We get,

(25 m)^{2} = (20 m)^{2} + BC^{2}

⇒ BC^{2} = (25 m)^{2} – (20 m)^{2}

⇒ BC^{2} = 625 m^{2} – 400 m^{2}

⇒ BC^{2} = 225 m^{2}

⇒ BC^{2} = (15 m)^{2}

We get,

BC = 15 m

Hence,

The distance of the foot of the ladder from the building is 15 m

**6. ****A right triangle has hypotenuse p cm and one side q cm. If p – q = 1, find the length of the third side of the triangle.**

**Answer**

Given

Hypotenuse = p cm

One side = q cm

Let the length of the third side = x cm

Using Pythagoras theorem,

x^{2} = p^{2} – q^{2}

⇒ x^{2} = (p + q) (p – q)

Substituting (p – q) = 1

We get,

x^{2} = (p + q) (1)

⇒ x^{2} = (p + q)

⇒ x = √p + q

Therefore, the length of the third side of the triangle is √p + q cm

**7. ****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 other side of the street to reach a window 12 m high. Find the width of the street.**

**Answer**

Let ‘O’ be the foot of the ladder

Let AO be the position of the ladder when it touches the window at A which is 9 m high and CO be the position of the ladder when it touches the window at C which is 12 m high.

Using Pythagoras theorem,

In △AOB,

BO^{2} = AO^{2} – AB^{2}

⇒ BO^{2} = (15 m)^{2} – (9 m)^{2}

⇒ BO^{2} = 225 m^{2} – 81 m^{2}

On simplification, we get,

BO^{2} = 144 m^{2}

⇒ BO^{2} = (12 m)^{2}

We get,

BO = 12 m

Using Pythagoras theorem,

In △COD,

DO^{2} = CO^{2} – CD^{2}

⇒ DO^{2} = (15 m)^{2} – (12 m)^{2}

⇒ DO^{2} = 225 m^{2} – 144 m^{2}

We get,

DO^{2} = 81 m^{2}

⇒ DO = 9 m

Width of the street = DO + BO

= 9 m + 12 m

= 21 m

Therefore, the width of the street is 21 m

**8. ****The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall to what height does its top reach?**

**Answer**

Let AC be the ladder and A be the position of the window which is 8 m above the ground

Now,

The ladder is shifted such that its foot is at point D which is 8 m away from the wall

Hence,

BD = 8 m

At this instance, the position of the ladder is DE

Hence,

AC = DE

Using Pythagoras theorem,

In △ABC,

AC^{2} = AB^{2} + BC^{2}

⇒ AC^{2} = (8 m)^{2} + (6 m)^{2}

⇒ AC^{2} = 64 m^{2} + 36 m^{2}

⇒ AC^{2} = 100 m^{2}

We get,

AC = 10 m

We know that, AC = DE = 10 m

Using Pythagoras theorem,

In △DBE,

BE^{2} = DE^{2} – BD^{2}

⇒ BE^{2} = (10 m)^{2} – (8 m)^{2}

⇒ BE^{2} = 100 m^{2} – 64 m^{2}

⇒ BE^{2} = 36 m^{2}

We get,

BE = 6 m

Hence,

The required height up to which the ladder reaches is 6 m above the ground

**9. ****Two poles of height 9m and 14 m stand on a plane ground. If the distance between their feet is 12m, find the distance between their tops.**

**Answer**

Let AB and CD be the two poles of 9 m and 14 m respectively

Given that,

BD = 12 m

Thus,

CE = 12 m

Now,

AE = AB – EB

⇒ AE = 14 m – 9 m

We get,

AE = 5 m

Using Pythagoras theorem in △ACE,

AC^{2} = AE^{2} + CE^{2}

⇒ AC^{2} = (5m)^{2} + (12 m)^{2}

⇒ AC^{2} = 25 m^{2} + 144 m^{2}

⇒ AC^{2} = 169 m^{2}

⇒ AC^{2} = 13 m^{2}

We get,

AC = 13 m

Therefore, the distance between the tops of their poles is 13 m

**10. ****The length of the diagonals of rhombus are 24 cm and 10 cm. Find each side of the rhombus.**

**Answer**

Given

The length of the diagonals of rhombus are 24 cm and 10 cm respectively

Therefore,

d_{1} = 24 cm and d_{2} = 10 cm

The diagonals of a rhombus bisect each other

Hence,

(d_{1}/2)^{2} + (d_{2}/2)^{2} = side^{2}

⇒ side^{2} = 12^{2} + 5^{2}

⇒ side^{2} = 144 + 25

⇒ side^{2} = 169

⇒ side^{2} = 13^{2}

We get,

side = 13

Therefore, each side of the rhombus is of length 13 cm

**11.** **Each side of rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal.**

**Answer**

Given

Side of the rhombus = 10 cm

Length of one diagonal, d_{1} = 16 cm

Let d_{2} be the other diagonal of the rhombus

The diagonals of a rhombus bisect each other

Therefore,

(d_{1}/2)^{2} + (d_{2}/2)^{2} = side^{2}

On further calculation, we get,

8^{2} + (d_{2}/2)^{2} = 10^{2}

⇒ (d_{2}/2)^{2} = 10^{2} – 8^{2}

⇒ (d_{2}/2)^{2} = 100 – 64

We get,

(d_{2}/2)^{2} = 6^{2}

⇒ (d_{2}/2) = 6

⇒ d_{2} = 6×2

⇒ d_{2} = 12

Therefore, the length of the other diagonal of rhombus is 12 cm

**12**. **In ****△****ABC, AD is perpendicular to BC. Prove that**

**AB ^{2} + CD^{2} = AC^{2} + BD^{2}**

**Answer**

AB^{2} = AD^{2} + BD^{2} **…(i)**

AC^{2} = AD^{2} + CD^{2} **…(ii)**

Subtracting (ii) from (i), we get,

AB^{2} – AC^{2} = BD^{2} – CD^{2}

We get,

AB^{2} + CD^{2} = AC^{2} + BD^{2}

Hence, proved

**13. ****From a point O in the interior of a ****△****ABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that:**

**(a) AF ^{2} + BD^{2} + CE^{2} = OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}**

**(b) AF ^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}**

**Answer**

**(a)**In right triangles OFA, ODB and OEC,

We have,

OA^{2} = AF^{2} + OF^{2}

OB^{2} = BD^{2} + OD^{2}

OC^{2} = CE^{2} + OE^{2}

By adding all these results, we get,

OA^{2} + OB^{2} + OC^{2} = AF^{2} + BD^{2} + CE^{2} + OF^{2} + OD^{2} + OE^{2}

⇒ AF^{2} + BD^{2} + CE^{2} = OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

Hence, proved

**(b)** In right triangles ODB and ODC,

We have,

OB^{2} = OD^{2} + BD^{2}

OC^{2} = OD^{2} + CD^{2}

On subtracting, we get,

OB^{2} – OC^{2} = (OD^{2} + BD^{2}) – (OD^{2} + CD^{2})

We get,

OB^{2} – OC^{2} = BD^{2} – CD^{2} **…(1)**

Similarly, we have,

OC^{2} – OA^{2} = CE^{2} – AE^{2} **…(2)**

OA^{2} – OB^{2} = AF^{2} – BF^{2} **…(3)**

Adding equations (1), (2) and (3) we get,

(OB^{2} – OC^{2}) + (OC^{2} – OA^{2}) + (OA^{2} – OB^{2}) = (BD^{2} – CD^{2}) + (CE^{2} – AE^{2}) + (AF^{2} – BF^{2})

On further calculation, we get,

(BD^{2} + CE^{2} + AF^{2}) – (AE^{2} + CD^{2} + BF^{2}) = 0

⇒ AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

Hence, proved

**14. ****A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that OB ^{2}+ OD^{2}= OC^{2} + OA^{2}**

**Answer**

Let ABCD is the given rectangle and ‘O’ be a point within it

Join OA, OB, OC and OD

Then, ABFE is a rectangle

In right triangles △OEA and △OFC,

We have,

OA^{2} = OE^{2} + AE^{2} and

OC^{2} = OF^{2} + CF^{2}

On adding, we get,

OA^{2} + OC^{2} = (OE^{2} + AE^{2}) + (OF^{2} + CF^{2})

⇒ OA^{2} + OC^{2} = OE^{2} + OF^{2} + AE^{2} + CF^{2} **…(1)**

Now,

In right triangles △OFB and △ODE,

We have,

OB^{2} = OF^{2} + FB^{2} and

OD^{2} = OE^{2} + DE^{2}

On adding, we get,

OB^{2} + OD^{2} = (OF^{2} + FB^{2}) + (OE^{2} + DE^{2})

⇒ OB^{2} + OD^{2} = OE^{2} + OF^{2} + DE^{2} + BF^{2}

⇒ OB^{2} + OD^{2} = OE^{2} + OF^{2} + CF^{2} + AE^{2} **…(2)**

∵ DE = CF and AE = BF

From equations (1) and (2) we get,

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

Hence, proved

**15. ****ABCD is a rhombus. Prove that AB ^{2}+ BC^{2}+ CD^{2} + DA^{2} = AC^{2} + BD^{2}**

**Answer**

Applying Pythagoras theorem

AB^{2} = AO^{2} + OB^{2}

BC^{2} = BO^{2} + OC^{2}

CD^{2} = CO^{2} + OD^{2}

AD^{2} = AO^{2} + OD^{2}

On adding all these equations, we get,

AB^{2} + BC^{2} + CD^{2} + AD^{2} = 2 (AO^{2} + OB^{2} + OC^{2} + OD^{2})

= 2 [(AC/2)^{2} + (BD/2)^{2} + (AC/2)^{2} + (BD/2)^{2}]

Since diagonals bisect each other

= 2 [(AC)^{2}/2 + (BD)^{2}/2]

= (AC)^{2} + (BD)^{2}

Hence, proved

**16. In an equilateral triangle ABC, the sides BC is trisected at D. Prove that 9 AD ^{2} = 7 AB^{2}. **

**Answer**

**17. From a point O in the interior of a ****△****ABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: **

**(a) AF ^{2} + BD^{2} + CE^{2} = OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} **

**(b) AF ^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2} **

**Answer**

**(a)**

**(b)**

**18. In a triangle ABC, AC > AB, D is the midpoint BC, and AE ****⊥**** BC. Prove that :**

**(a) AC ^{2 }= AD^{2} + BC×DE + 1/4BC^{2} **

**(b) AB ^{2} = AD^{2} – BC**

**×DE + 1/4.BC**

^{2}**(c) AB ^{2 }+ AC^{2} = 2AD^{2} + 1/2.BC^{2} **

**(d) AC ^{2} – AB^{2} = 2BC×ED **

**(e) AB ^{2} + AC^{2} = 2(AD^{2} + CD^{2}) **

**Answer**

**21. The perpendicular AD on the base BC of a ****△****ABC intersects BC at D such that DB = 3 CD. Prove ****that 2AB ^{2} = 2AC^{2} + BC^{2} **

**Answer**

**22. In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides are ratio 2 : 1. **

**Prove that: **

**(i) 9AQ ^{2} = 9AC^{2 }+ 4BC^{2} **

**(ii) 9BP ^{2 }= 9BC^{2} + 4AC^{2} **

**(iii) 9(AQ ^{2} + BP^{2}) = 13 AB^{2 }**

**Answer**

**23. In the given figure, PQ = RS/3 = 8 cm, 3ST = 4QT = 48 cm. **

**Show that ****∠****RTP = 90****°. **

**Answer**

**24. In a right-angled triangle ABC, ****∠****ABC = 90****°****, AC = 10 cm, BC = 6 cm and BC produced to D such CD = 9 cm. Find the length of AD. **

**Answer**

**25. In the given figure, PQ = PS, ****∠****P = ****∠****R = 90****°.**** RS = 20 cm** **and QR = 21 cm. Find the length of PQ correct to two decimal places. **

**Answer**

**26. In a right angled triangle PQR, right-angled at Q, S and T are points on PQ and QR respectively such as PT = SR = 13 cm, QT = 5 cm and PS = TR. Find the length of PQ and PS.**

**Answer**

**27. PQR is an isosceles triangle with PQ = PR = 10 cm and QR = 12 cm. Find the length of the perpendicular from P to QR. **

**Answer**

**28. In a square PQRS of side 5 cm, A, B, C and D are points on sides PQ, QR, RS and SP respectively such as PA = PD = RB = RC = 2 cm. Prove that ABCD is a rectangle. Also, find the area and perimeter of the rectangle. **

**Answer**