# SELINA Solutions for Class 9 Maths Chapter 13 - Pythagoras Theorem [Proof and Simple Applications with Converse]

Explore Selina Solutions for ICSE Class 9 Mathematics Chapter 13 Pythagoras theorem [Proof and Simple Applications with Converse] online on TopperLearning. Revise the textbook solutions on Pythagoras theorem to understand applications of the theorem in solving real-life problems. For instance, learn how to calculate the distance from point A to point B based on given data using Pythagoras theorem.

The Selina solutions Maths textbook questions for this chapter also helps you with calculating the length of sides of right-angled triangles in quadrilaterals. Other ICSE Class 9 Maths resources that you can browse on TopperLearning are video lessons, most important questions, sample question papers and more.

## Chapter 13 - Pythagoras Theorem [Proof and Simple Applications with Converse] Exercise Ex. 13(A)

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.

The pictorial representation of the given problem is given below,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

(i)Here, AB is the hypotenuse. Therefore applying the Pythagoras theorem we get,

_{}

Therefore, the distance of the other end of the ladder from the ground is 12m

A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.

Here , we need to measure the distance AB as shown in the figure below,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Therefore , in this case

_{}

Therefore the required distance is 64.03 m.

In the figure: _{}PSQ = 90^{o}, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras theorem we get,

_{}

The length of PR_{}

The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and _{}ABD = _{}BCD = 90^{o}. Calculate the length of AB.

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras theorem we get,

_{}

The length of AB is 4 cm.

AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.

Since ABC is an equilateral triangle therefore, all the sides of the triangle are of same measure and the perpendicular AD will divide BC in two equal parts.

Here, we consider the _{}and applying Pythagoras theorem we get,

_{}Therefore, the length of AD is 8.7 cm

In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC= 3 cm. Calculate the length of OC.

We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, from (i) and(ii),

_{}

_{}

In triangle ABC,

AB = AC = x, BC = 10 cm and the area of the triangle is 60 cm^{2}. Find x.

Here, the diagram will be,

We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since, ABC is an isosceles triangle, therefore perpendicular from vertex will cut the base in two equal segments.

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

_{}

Therefore, x is 13cm

If the sides of triangle are in the ratio 1 : _{}: 1, show that is a right-angled triangle.

Let, the sides of the triangle be, _{}

Now, _{}

Here, in (i) it is shown that, square of one side of the given triangle is equal to the addition of square of other two sides. This is nothing but Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Therefore, the given triangle is a right angled triangle.

Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m; find the distance between their tips.

The diagram of the given problem is given below,

We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

_{}

In the given figure, AB//CD, AB = 7 cm, BD = 25 cm and CD = 17 cm; find the length of side BC.

Take M be the point on CD such that AB = DM.

So DM = 7cm and MC = 10 cm

Join points B and M to form the line segment BM.

So BM || AD also BM = AD.

_{}

In the given figure, ∠B = 90^{°}, XY || BC, AB = 12cm, AY = 8cm and AX: XB = 1: 2 = AY: YC. Find the lengths of AC and BC.

Given that AX: XB = 1: 2 = AY: YC.

Let x be the common multiple for which this proportion gets satisfied.

So, AX = 1x and XB = 2x

AX + XB = 1x + 2x = 3x

⇒ AB = 3x .….(A - X - B)

⇒ 12 = 3x

⇒ x = 4

AX = 1x = 4 and XB = 2x = 2 × 4 = 8

Similarly,

AY = 1y and YC = 2y

AY = 8…(given)

⇒ 8 = y

∴ YC = 2y = 2 × 8 = 16

∴ AC = AY + YC = 8 + 16 = 24 cm

∆ABC is a right angled triangle. …. Given

∴ By Pythagoras Theorem, we get

⇒ AB^{2} + BC^{2} = AC^{2}

⇒ BC^{2 }= AC^{2} - AB^{2}

⇒ BC^{2 }= (24)^{2} - (12)^{2}

⇒ BC^{2 }= 576 - 144

⇒ BC^{2 }= 432

⇒ BC= cm

∴ AC = 24 cm and BC =cm

In ΔABC, _{} Find the sides of the triangle, if:

(i) AB = (x - 3) cm, BC = (x + 4) cm and AC = (x + 6) cm

(ii) AB = x cm, BC = (4x + 4) cm and AC = (4x + 5) cm

_{}

_{}

## Chapter 13 - Pythagoras Theorem [Proof and Simple Applications with Converse] Exercise Ex. 13(B)

In the figure, given
below, AD _{} BC. Prove
that: c^{2} = a^{2} + b^{2} - 2ax.

First, we consider the _{} and applying
Pythagoras theorem we get,

_{}

First,
we consider the _{} and applying
Pythagoras theorem we get,

_{}

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

_{}

Hence Proved.

In
equilateral Δ ABC, AD _{} BC and BC = x
cm. Find, in terms of x, the length of AD.

In
equilateral Δ ABC, AD _{} BC.

Therefore, BD = DC = x/2 cm.

_{}

ABC is a triangle, right-angled at B. M is a point on BC. Prove that:

AM^{2} + BC^{2} = AC^{2} + BM^{2}.

The pictorial form of the given problem is as follows,

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras theorem we get,

_{}

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

_{}

M andN are the mid-points of the sides QR and PQ respectively of a _{}PQR, right-angled at Q. Prove that:

(i) PM^{2} + RN^{2} = 5 MN^{2}

(ii) 4 PM^{2} = 4 PQ^{2} + QR^{2}

(iii) 4 RN^{2} = PQ^{2} + 4 QR^{2}

(iv) 4 (PM^{2} + RN^{2}) = 5 PR^{2}

We draw , PM,MN,NR

Since, M andN are the mid-points of the sides QR and PQ respectively, therefore, PN=NQ,QM=RM

(i)

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras

theorem we get,

_{}

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

_{}

(ii)

We consider the _{}and applying Pythagoras theorem we get,

_{}

_{}^{.}

(iii)

We consider the _{}and applying Pythagoras theorem we get,

_{}

(iv)

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Now, we consider the _{}and applying Pythagoras theorem we get,

_{}

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

_{}

In triangle ABC, _{}B = 90^{o} and D is the mid-point of BC. Prove that: AC^{2} = AD^{2} + 3CD^{2}.

In triangle ABC,_{ }B = 90^{o} and D is the mid-point of BC. Join AD. Therefore, BD=DC

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Similarly, we get from rt. angle triangles ABC we get,

_{}

From (i) and (ii) ,

_{}

In a rectangle ABCD, prove that:

AC^{2} + BD^{2} = AB^{2} + BC^{2} + CD^{2} + DA^{2}.

Since, ABCD is a rectangle angles A,B,C and D are rt. angles.

First, we consider the _{}and applying Pythagoras theorem we get,

_{}

Similarly, we get from rt. angle triangle BDC we get,

_{}

Adding (i) and (ii) ,

_{}

In a
quadrilateral ABCD, _{}B = 90^{0} and _{}D = 90^{0}. Prove that: 2AC^{2} - AB^{2}
= BC^{2} + CD^{2} + DA^{2}

_{}

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

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH.

Using Pythagorean theorem we have from the above diagram:

OA^{2 }= AH^{2 }+ OH^{2 }= AH^{2 }+ AE^{2 }

OC^{2 }= CG^{2 }+ OG^{2 }= EB^{2 }+ HD^{2 }

OB^{2 }= EO^{2 }+ BE^{2 }= AH^{2 }+ BE^{2 }

OD^{2 }= HD^{2 }+ OH^{2 }= HD^{2 }+ AE^{2 }

Adding these equalities we get:

OA^{2 }+ OC^{2 }= AH^{2 }+ HD^{2 }+ AE^{2 }+ EB^{2 }

OB^{2 }+ OD^{2 }= AH^{2 }+ HD^{2 }+ AE^{2 }+ EB^{2 }

From which we prove that for any point within the rectangle there is the relation

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

Hence Proved.

In the following figure, OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC. Prove that:

AR^{2} + BP^{2} + CQ^{2} = AQ^{2} + CP^{2} + BR^{2}

Here, we first need to join OA, OB, and OC after which the figure becomes as follows,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the _{}and applying Pythagoras theorem we get,

_{}

_{Similarly, from triangles, BPO,COQ,AOQ,CPO and BRO we get the following results,}

_{}

Adding (*i*), (*ii*) and (*iii*),we get

_{}

Adding (*iv*), (*v*) and (*vi*),we get ,

_{}

From (*vii*) and (*viii*), we get,

AR^{2} + BP^{2} + CQ^{2} = AQ^{2} + CP^{2} + BR^{2}

Hence proved.

Diagonals of rhombus ABCD intersect each other at point O. Prove that:

OA^{2}
+ OC^{2} = 2AD^{2} - _{}

Diagonals of the rhombus are perpendicular to each other.

_{}

In the figure AB = BC and AD is perpendicular to CD. Prove that:

AC^{2} = 2BC. DC.

We consider the _{}and applying Pythagoras theorem we get,

_{}

In an isosceles triangle ABC; AB = AC and D is point on BC produced. Prove that:

AD^{2} = AC^{2} + BD.CD.

In an isosceles triangle ABC; AB = AC and D is point on BC produced. Construct AE perpendicular BC.

We consider the rt. angled _{}and applying Pythagoras theorem we get,

_{}

_{}

In triangle ABC, angle A = 90^{o}, CA = AB and D is point on AB produced. Prove that DC^{2} -BD^{2} = 2AB.AD.

We consider the rt. angled _{}and applying Pythagoras theorem we get,

_{}

Similarly, in _{,}

_{}

Putting, _{}from (ii) in (i) we get,

_{}

Hence Proved.

In
triangle ABC, AB = AC and BD is perpendicular to AC. Prove that: BD^{2}
- CD^{2} = 2CD × AD.

_{}

In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1: 3.

Prove that : 2AC^{2} = 2AB^{2} + BC^{2}

Here,

_{}

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