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 cm2 – 25 cm2
⇒ (perpendicular)2 = 144 cm2
⇒ (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 cm2
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 cm2 + 49 cm2
⇒ (hypotenuse)2 = 625 cm2
⇒ (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 cm2 – 256 cm2
On further calculation, we get,
(x cm)2 = 3969 cm2
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 cm2
4. A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.
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,
OB2 = OA2 + AB2
⇒ OB2 = (10m)2 + (24 m)2
⇒ OB2 = 100 m2 + 576 m2
⇒ OB2 = 676 m2
⇒ OB2 = (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.
Let AC be the ladder and A be the position of the window
Then,
AC = 25 m and AB = 20 m
Using Pythagoras theorem,
AC2 = AB2 + BC2
By substituting AC = 25 and AB = 20
We get,
(25 m)2 = (20 m)2 + BC2
⇒ BC2 = (25 m)2 – (20 m)2
⇒ BC2 = 625 m2 – 400 m2
⇒ BC2 = 225 m2
⇒ BC2 = (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,
x2 = p2 – q2
⇒ x2 = (p + q) (p – q)
Substituting (p – q) = 1
We get,
x2 = (p + q) (1)
⇒ x2 = (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.
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,
BO2 = AO2 – AB2
⇒ BO2 = (15 m)2 – (9 m)2
⇒ BO2 = 225 m2 – 81 m2
On simplification, we get,
BO2 = 144 m2
⇒ BO2 = (12 m)2
We get,
BO = 12 m
Using Pythagoras theorem,
In △COD,
DO2 = CO2 – CD2
⇒ DO2 = (15 m)2 – (12 m)2
⇒ DO2 = 225 m2 – 144 m2
We get,
DO2 = 81 m2
⇒ 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?
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,
AC2 = AB2 + BC2
⇒ AC2 = (8 m)2 + (6 m)2
⇒ AC2 = 64 m2 + 36 m2
⇒ AC2 = 100 m2
We get,
AC = 10 m
We know that, AC = DE = 10 m
Using Pythagoras theorem,
In △DBE,
BE2 = DE2 – BD2
⇒ BE2 = (10 m)2 – (8 m)2
⇒ BE2 = 100 m2 – 64 m2
⇒ BE2 = 36 m2
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.
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,
AC2 = AE2 + CE2
⇒ AC2 = (5m)2 + (12 m)2
⇒ AC2 = 25 m2 + 144 m2
⇒ AC2 = 169 m2
⇒ AC2 = 13 m2
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,
d1 = 24 cm and d2 = 10 cm
The diagonals of a rhombus bisect each other
Hence,
(d1/2)2 + (d2/2)2 = side2
⇒ side2 = 122 + 52
⇒ side2 = 144 + 25
⇒ side2 = 169
⇒ side2 = 132
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, d1 = 16 cm
Let d2 be the other diagonal of the rhombus
The diagonals of a rhombus bisect each other
Therefore,
(d1/2)2 + (d2/2)2 = side2
On further calculation, we get,
82 + (d2/2)2 = 102
⇒ (d2/2)2 = 102 – 82
⇒ (d2/2)2 = 100 – 64
We get,
(d2/2)2 = 62
⇒ (d2/2) = 6
⇒ d2 = 6×2
⇒ d2 = 12
Therefore, the length of the other diagonal of rhombus is 12 cm
12. In △ABC, AD is perpendicular to BC. Prove that
AB2 + CD2 = AC2 + BD2
Answer
AB2 = AD2 + BD2 …(i)
AC2 = AD2 + CD2 …(ii)
Subtracting (ii) from (i), we get,
AB2 – AC2 = BD2 – CD2
We get,
AB2 + CD2 = AC2 + BD2
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) AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
(b) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Answer
We have,
OA2 = AF2 + OF2
OB2 = BD2 + OD2
OC2 = CE2 + OE2
By adding all these results, we get,
OA2 + OB2 + OC2 = AF2 + BD2 + CE2 + OF2 + OD2 + OE2
⇒ AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
Hence, proved
(b) In right triangles ODB and ODC,
We have,
OB2 = OD2 + BD2
OC2 = OD2 + CD2
On subtracting, we get,
OB2 – OC2 = (OD2 + BD2) – (OD2 + CD2)
We get,
OB2 – OC2 = BD2 – CD2 …(1)
Similarly, we have,
OC2 – OA2 = CE2 – AE2 …(2)
OA2 – OB2 = AF2 – BF2 …(3)
Adding equations (1), (2) and (3) we get,
(OB2 – OC2) + (OC2 – OA2) + (OA2 – OB2) = (BD2 – CD2) + (CE2 – AE2) + (AF2 – BF2)
On further calculation, we get,
(BD2 + CE2 + AF2) – (AE2 + CD2 + BF2) = 0
⇒ AF2 + BD2 + CE2 = AE2 + CD2 + BF2
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 OB2+ OD2= OC2 + OA2
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,
OA2 = OE2 + AE2 and
OC2 = OF2 + CF2
On adding, we get,
OA2 + OC2 = (OE2 + AE2) + (OF2 + CF2)
⇒ OA2 + OC2 = OE2 + OF2 + AE2 + CF2 …(1)
Now,
In right triangles △OFB and △ODE,
We have,
OB2 = OF2 + FB2 and
OD2 = OE2 + DE2
On adding, we get,
OB2 + OD2 = (OF2 + FB2) + (OE2 + DE2)
⇒ OB2 + OD2 = OE2 + OF2 + DE2 + BF2
⇒ OB2 + OD2 = OE2 + OF2 + CF2 + AE2 …(2)
∵ DE = CF and AE = BF
From equations (1) and (2) we get,
OA2 + OC2 = OB2 + OD2
Hence, proved
15. ABCD is a rhombus. Prove that AB2+ BC2+ CD2 + DA2 = AC2 + BD2
Answer
Applying Pythagoras theorem
AB2 = AO2 + OB2
BC2 = BO2 + OC2
CD2 = CO2 + OD2
AD2 = AO2 + OD2
On adding all these equations, we get,
AB2 + BC2 + CD2 + AD2 = 2 (AO2 + OB2 + OC2 + OD2)
= 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 AD2 = 7 AB2.
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) AF2 + BD2 + CE2 = OA2 + OB2 + OC2 – OD2 – OE2 – OF2
(b) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Answer
18. In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that :
(a) AC2 = AD2 + BC×DE + 1/4BC2
(b) AB2 = AD2 – BC×DE + 1/4.BC2
(c) AB2 + AC2 = 2AD2 + 1/2.BC2
(d) AC2 – AB2 = 2BC×ED
(e) AB2 + AC2 = 2(AD2 + CD2)
Answer
21. The perpendicular AD on the base BC of a △ABC intersects BC at D such that DB = 3 CD. Prove that 2AB2 = 2AC2 + BC2
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) 9AQ2 = 9AC2 + 4BC2
(ii) 9BP2 = 9BC2 + 4AC2
(iii) 9(AQ2 + BP2) = 13 AB2
Answer
23. In the given figure, PQ = RS/3 = 8 cm, 3ST = 4QT = 48 cm.
Show that ∠RTP = 90°.
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
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
