ICSE Class 9 Maths Chapter 13 Pythagoras Theorem

Pythagoras Theorem

Points To Remember

1. Pythagoras Theorem: In a right-angled triangle the square on the hypotenuse is equal to the sum of squares on the other two sides.

In \(\triangle ABC\), \(\angle B = 90°\)

\(\therefore AC^2 = AB^2 + BC^2\)

2. Converse of Pythagoras Theorem: In a triangle, if the square of one side is equal to the sum of the square of other two sides, then the triangle is a right-angled.

Teacher's Note

The Pythagoras Theorem is fundamental to construction and engineering. Carpenters use the 3-4-5 triangle rule to ensure corners are perfectly square when building furniture or houses.

Exercise 13

Q. 1. In \(\triangle ABC\), \(\angle C = 90°\)

If \(BC = a\), \(AC = b\) and \(AB = c\), find:

(i) \(c\) when \(a = 8\) cm and \(b = 6\) cm.

(ii) \(a\) when \(c = 25\) cm and \(b = 7\) cm.

(iii) \(b\) when \(c = 13\) cm and \(a = 5\) cm.

Sol. In \(\triangle ABC\), \(\angle C = 90°\)

\(\therefore c^2 = a^2 + b^2\) (Pythagoras Theorem)

(i) If \(a = 8\) cm, \(b = 6\) cm, then

\(c^2 = a^2 + b^2 = (8)^2 + (6)^2 = 64 + 36 = 100\)

\(\therefore c = \sqrt{100} = 10\) cm.

(ii) \(c = 25\) cm, \(b = 7\) cm

But \(c^2 = a^2 + b^2 \Rightarrow (25)^2 = a^2 + (7)^2\)

\(\Rightarrow 625 = a^2 + 49 \Rightarrow a^2 = 625 - 49\)

\(\Rightarrow a^2 = 576 \Rightarrow c = \sqrt{576} = 24\) cm.

(iii) \(c = 13\) cm, \(a = 5\) cm

But \(c^2 = a^2 + b^2 \Rightarrow (13)^2 = (5)^2 + b^2\)

\(\Rightarrow 169 = 25 + b^2\)

\(\Rightarrow b^2 = 169 - 25 = 144\)

\(\therefore b = \sqrt{144} = 12\) cm.

Q. 2. A rectangular field is 40 m long and 30 m broad. Find the length of its diagonal.

Sol. In rectangular field ABCD, \(AB = 40\) m and \(BC = 30\) m

AC is its diagonal.

\(\therefore\) In \(\triangle ABC\), \(\angle B = 90°\)

\(AC^2 = AB^2 + BC^2\) (Pythagoras Theorem)

\(= (40)^2 + (30)^2 = 1600 + 900 = 2500\)

\(\therefore AC = \sqrt{2500} = 50\) m

Hence diagonal \(AC = 50\) m Ans.

Teacher's Note

Finding diagonals is essential in real estate and construction. Property surveyors use this principle to calculate diagonal measurements of rectangular plots without physically measuring across.

Q. 3. A man goes 15 m due west and then 8 m due north. How far is he from the starting point?

Sol. Let A is starting point.

\(\therefore\) In right \(\triangle ABC\), \(\angle B = 90°\)

\(AB = 15\) m and \(BC = 8\) m

\(\therefore AC^2 = AB^2 + BC^2\) (Pythagoras Theorem)

\(\Rightarrow AC^2 = (15)^2 + (8)^2 = 225 + 64 = 289\)

\(\therefore AC = \sqrt{289} = 17\) m

Hence he is 17 m far from the starting point.

Q. 4. A ladder 17 m long reaches the window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

Sol. Let AB the ladder, CB be the building and B is window, then

\(AB = 17\) m, \(BC = 15\) m

Now in right \(\triangle ACB\), \(\angle C = 90°\)

\(AB^2 = AC^2 + BC^2\) (Pythagoras Theorem)

\(\Rightarrow (17)^2 = AC^2 + (15)^2\)

\(\Rightarrow 289 = AC^2 - 225\)

\(\Rightarrow AC^2 = 289 - 225 = 64 = (8)^2\)

\(\therefore AC = 8\)

Hence the distance of the foot of the ladder from the building = 8 m Ans.

Q. 5. A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

Sol. Let AC be the ladder, BC be the building and AB be the distance from the foot of the ladder from the building.

\(\therefore AB = 5\) m, \(AC = 13\) m.

Now in right \(\triangle ABC\), \(\angle B = 90°\)

\(AC^2 = AB^2 + BC^2\) (Pythagoras Theorem)

\(\Rightarrow (13)^2 = (5)^2 + BC^2\)

\(\Rightarrow 169 = 25 + BC^2\)

\(\Rightarrow BC^2 = 169 - 25 = 144\)

\(\therefore BC = \sqrt{144} = 12\) m

Hence the distance of the other end of the ladder from the ground = 12 m Ans.

Teacher's Note

Ladder safety is critical in construction work. Workers use this calculation to determine safe angles and distances when placing ladders against walls to prevent slipping and accidents.

Q. 6. A ladder 15 m long reaches a window which is 9 m above the ground on one side of the 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.

Sol. Let CD and CE be two positions of the ladder, D is the window on one side and E is the window on the other side and AB be the width of the street.

\(\therefore CD = CE = 15\) m, \(AD = 9\) m, and \(BE = 12\) m

Now in right \(\triangle CAD\), by Pythagoras Theorem,

\(CD^2 = AC^2 + AD^2\)

\(\Rightarrow (15)^2 = AC^2 + (9)^2\)

\(\Rightarrow 225 = AC^2 + 81\)

\(\Rightarrow AC^2 = 225 - 81 = 144 = (12)^2\)

\(\therefore AC = 12\)

Similarly, in right \(\triangle ABD\),

\(AD^2 = BD^2 + AB^2\)

\(\Rightarrow (13)^2 = 153 + AB^2\)

\(\Rightarrow 169 = 153 + AB^2\)

\(\Rightarrow AB^2 = 169 - 153 = 16 = (4)^2\)

\(\therefore AB = 4\) cm Ans.

Q. 7. In the given figure, ABCD is a quadrilateral in which \(BC = 3\) cm, \(AD = 13\) cm, \(DC = 12\) cm and \(\angle ABD = \angle BCD = 90°\). Calculate the length of AB.

Sol. In quadrilateral ABCD,

\(BC = 3\) cm, \(AD = 13\) cm, \(DC = 12\) cm

\(\angle ABD = \angle BCD = 90°\)

In right \(\triangle BCD\),

\(BD^2 = CB + CD^2\) (Pythagoras Theorem)

\(= (12)^2 + (3)^2 = 144 + 9 = 153\)

Similarly, in right \(\triangle ABD\),

\(AD^2 = BD^2 + AB^2\)

\(\Rightarrow (13)^2 = 153 + AB^2\)

\(\Rightarrow 169 = 153 + AB^2\)

\(\Rightarrow AB^2 = 169 - 153 = 16 = (4)^2\)

\(\therefore AB = 4\) cm Ans.

Q. 8. In the given figure, \(\angle PSR = 90°\), PQ = 10 cm, QS = 6 cm and RQ = 9 cm, calculate the length of PR.

Sol. In the figure,

\(QS = 6\) cm, \(RQ = 9\) cm, \(PQ = 10\) cm.

Now in right \(\triangle PQS\), \(\angle S = 90°\)

\(PQ^2 = QS^2 + PS^2\) (Pythagoras Theorem)

\(\Rightarrow (10)^2 = (6)^2 + PS^2 \Rightarrow 100 = 36 + PS^2\)

\(\Rightarrow PS^2 = 100 - 36 = 64 = (8)^2\)

\(\therefore PS = 8\) cm

Similarly, in right \(\triangle PRS\),

\(PR^2 = RS^2 + PS^2 = (RO + OS)^2 + PS^2 = (9 + 6)^2 + (8)^2 = (15)^2 + (8)^2 = 225 + 64 = 289\)

\(\therefore PR = \sqrt{289} = 17\)

Hence PR = 17 cm Ans.

Teacher's Note

Navigation and surveying use the Pythagoras theorem extensively to calculate distances between points that cannot be directly measured, such as islands or mountain peaks.

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