Pythagoras Theorem

ICSE Class 9 Mathematics · Selina Solutions Concise Mathematics Class 9 ICSE · 48 solved questions

Exercise

1(a)Exercise 12(A)

If the lengths of the sides of a triangle are in the ratio 5 : 12 : 13; then the triangle is :

(a)acute-angled triangle
(b)scalene triangle
(c)scalene right-angled triangle
(d)obtuse-angled triangle

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Given,

Lengths of the sides of a triangle are in the ratio 5 : 12 : 13.

Let length of the sides of triangle are 5x, 12x and 13x.

Squaring all the sides, we get :

⇒ (13x)² = 169x²

⇒ (12x)² = 144x²

⇒ (5x)² = 25x²

⇒ (12x)² + (5x)² = 144x² + 25x² = 169x².

Since,

⇒ (13x)² = (12x)² + (5x)².

∴ Triangle obeys pythagoras theorem.

∴ Triangle is a scalene right-angled triangle.

Hence, Option 3 is the correct option.

1(b)Exercise 12(A)

In a right-angled triangle, hypotenuse is 10 cm and the ratio of the other two sides is 3 : 4, the sides are :

(a)6 cm and 4 cm
(b)8 cm and 6 cm
(c)3 cm and 4 cm
(d)8 cm and 4 cm

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Given,

Hypotenuse is 10 cm and the ratio of the other two sides is 3 : 4.

Let other two sides be 3x and 4x.

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ 10² = (3x)² + (4x)²

⇒ 100 = 9x² + 16x²

⇒ 100 = 25x²

⇒ x² = $\frac{100}{25}$

⇒ x² = 4

⇒ x = $\sqrt{4} = \pm 2$.

Since, side cannot be negative.

∴ x = 2.

⇒ 3x = 3(2) = 6 cm and 4x = 4(2) = 8 cm.

Hence, Option 2 is the correct option.

1(c)Exercise 12(A)

ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. The area of the triangle is :

(a)$32\sqrt{2}$ cm²
(b)$16\sqrt{2}$ cm²
(c)$8\sqrt{2}$ cm²
(d)$12\sqrt{2}$ cm²

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Let AD be the altitude.

ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. The area of the triangle is : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

In an isosceles triangle, the altitude from the vertex bisects the base.

∴ BD = CD = $\frac{BC}{2} = \frac{8}{2}$ = 4 cm.

In right angle triangle ABD,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AD² + BD²

⇒ 12² = AD² + 4²

⇒ 144 = AD² + 16

⇒ AD² = 144 - 16

⇒ AD² = 128

⇒ AD = $\sqrt{128} = 8\sqrt{2}$ cm.

Area of right angle triangle = $\frac{1}{2}$ × base × height

From figure,

⇒ Area of △ ABC = Area of △ ABD + Area of △ ACD

⇒ Area of △ ABC = $\frac{1}{2} \times BD \times AD + \frac{1}{2} \times CD \times AD$

⇒ Area of △ ABC

$= \frac{1}{2} \times AD \times (BD + CD) = \frac{1}{2} \times 8\sqrt{2} \times (4 + 4) = 4\sqrt{2} \times 8 = 32\sqrt{2} \text{ cm}^2.$

Hence, Option 1 is the correct option.

1(d)Exercise 12(A)

In a rhombus, its diagonals are 30 cm and 40 cm, its perimeter is :

(a)20 cm
(b)10 cm
(c)60 cm
(d)100 cm

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Let ABCD be the rhombus, with diagonals AC and BD intersecting at O.

In a rhombus, its diagonals are 30 cm and 40 cm, its perimeter is : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

We know that,

Diagonals of rhombus intersect at right angles.

Let AC = 40 cm and BD = 30 cm.

∴ AO = $\frac{AC}{2} = \frac{40}{2} = 20 \text{ cm}, BO = \frac{BD}{2} = \frac{30}{2}$ = 15 cm.

In right angle triangle AOB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AO² + BO²

⇒ AB² = (20)² + (15)²

⇒ AB² = 400 + 225

⇒ AB² = 625

⇒ AB = $\sqrt{625}$ = 25 cm.

We know that,

All sides of rhombus are equal.

∴ Perimeter of rhombus = 4 × side = 4 × 25 = 100 cm.

Hence, Option 4 is the correct option.

1(e)Exercise 12(A)

In the given figure, AD = 13 cm, DC = 12 cm and BC = 3 cm, then AB is equal to :

(a)4 cm
(b)3 cm
(c)5 cm
(d)6 cm

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In right angled △ DCB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ BD² = DC² + BC²

⇒ BD² = (12)² + (3)²

⇒ BD² = 144 + 9

⇒ BD² = 153

⇒ BD = $\sqrt{153}$ cm.

In right angle △ ABD,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AD² = AB² + BD²

⇒ 13² = AB² + $(\sqrt{153})^2$

⇒ 169 = AB² + 153

⇒ AB² = 169 - 153

⇒ AB² = 16

⇒ AB = $\sqrt{16}$ = 4 cm.

Hence, Option 1 is the correct option.

2Exercise 12(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.

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Let AB be the ladder.

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. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

From figure,

In right angle △ ACB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AC² + BC²

⇒ 13² = AC² + 5²

⇒ 169 = AC² + 25

⇒ AC² = 169 - 25

⇒ AC² = 144

⇒ AC = $\sqrt{144}$ = 12 m.

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

3Exercise 12(A)

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

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Let A be the initial position of the man.

A man goes 40 m due north and then 50 m due west. Find his distance from the starting point. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

From figure,

ABC is a right-angled triangle.

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AC² = AB² + BC²

⇒ AC² = 40² + 50²

⇒ AC² = 1600 + 2500

⇒ AC² = 4100

⇒ AC = $\sqrt{4100}$

⇒ AC = 64.03 m.

Hence, distance from starting point is 64.03 meters.

4Exercise 12(A)

In the figure : ∠PSQ = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

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In right angled triangle PQS,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ PQ² = PS² + QS²

⇒ 10² = PS² + 6²

⇒ PS² = 10² - 6²

⇒ PS² = 100 - 36

⇒ PS² = 64

⇒ PS = $\sqrt{64}$ = 8 cm.

From figure,

RS = RQ + QS = 9 + 6 = 15 cm.

In right angled triangle PRS,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ PR² = PS² + RS²

⇒ PR² = 8² + 15²

⇒ PR² = 64 + 225

⇒ PR² = 289

⇒ PR = $\sqrt{289}$ = 17 cm.

Hence, PR = 17 cm.

5Exercise 12(A)

In a quadrilateral PQRS, ∠Q = ∠S = 90° then prove that 2PR² - QR² = PQ² + PS² + SR².

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In quadrilateral PQRS, since ∠Q = ∠S = 90°, triangles PQR and PSR are right-angled triangles.

In a quadrilateral PQRS, ∠Q = ∠S = 90° then prove that 2PR2 - QR2 = PQ2 + PS2 + SR2. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

According to Pythagoras theorem,

In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In triangle PQR,

⇒ PR² = PQ² + QR² ......................(1)

In triangle PSR,

⇒ PR² = PS² + SR² ..................(2)

Adding eq (1) and (2):

⇒ PR² + PR² = PQ² + QR² + PS² + SR²

⇒ 2PR² = PQ² + PS² + SR² + QR²

⇒ 2PR² - QR² = PQ² + PS² + SR²

Hence, proved that 2PR² - QR² = PQ² + PS² + SR².

6Exercise 12(A)

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.

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In △ ABD and △ ACD,

⇒ ∠ADB = ∠ADC (Both equal to 90°)

⇒ AD = AD (Common side)

⇒ AB = AC (Since, ABC is an equilateral triangle)

∴ △ ABD ≅ △ ACD (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BD = CD = $\frac{BC}{2} = \frac{10}{2}$ = 5 cm.

In right-angled triangle ABD,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AD² + BD²

⇒ 10² = AD² + 5²

⇒ AD² = 10² - 5²

⇒ AD² = 100 - 25

⇒ AD² = 75

⇒ AD = $\sqrt{75}$ = 8.7 cm.

Hence, AD = 8.7 cm.

7Exercise 12(A)

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

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Let length of OC be x cm.

In right angled triangle AOC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AC² = AO² + OC²

⇒ 3² = AO² + x²

⇒ AO² = 3² - x²

⇒ AO² = 9 - x²

⇒ AO = $\sqrt{9 - x^2}$ cm.

From figure,

BO = BC + CO = (6 + x) cm.

In right angled triangle AOB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AO² + BO²

⇒ 8² = $(\sqrt{9 - x^2})^2$ + (6 + x)²

⇒ 64 = 9 - x² + 36 + x² + 12x

⇒ 64 = 45 + 12x

⇒ 12x = 64 - 45

⇒ 12x = 19

⇒ x = $\frac{19}{12} = 1\frac{7}{12}$.

Hence, OC = $1\frac{7}{12}$ cm.

8Exercise 12(A)

In triangle ABC,

AB = AC = x; BC = 10 cm and the area of the triangle is 60 cm². Find x.

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In △ ABC,

Draw AD ⊥ BC

In triangle ABC, AB = AC = x; BC = 10 cm and the area of the triangle is 60 cm2. Find x. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

By formula,

Area of triangle = $\frac{1}{2} \times$ base × height

Given,

Area of triangle ABC = 60 cm²

$$\begin{aligned}\Rightarrow \frac{1}{2} \times BC \times AD = 60 \\[1em] \Rightarrow \frac{1}{2} \times 10 \times AD = 60 \\[1em] \Rightarrow 5 \times AD = 60 \\[1em] \Rightarrow AD = \frac{60}{5} = 12 \text{ cm}.\end{aligned}$$

We know that,

In an isosceles triangle, the altitude from the vertex bisects the base.

∴ BD = CD = $\frac{BC}{2} = \frac{10}{2}$ = 5 cm.

In right-angled triangle ADB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AD² + BD²

⇒ x² = 12² + 5²

⇒ x² = 144 + 25

⇒ x² = 169

⇒ x = $\sqrt{169}$ = 13 cm.

Hence, x = 13 cm.

9Exercise 12(A)

If the sides of a triangle are in the ratio 1 : $\sqrt{2}$ : 1, show that it is a right-angled triangle.

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Let ABC be the triangle.

If the sides of a triangle are in the ratio 1 : 2 : 1, show that it is a right-angled triangle. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

Given,

Sides of a triangle are in the ratio 1 : $\sqrt{2}$ : 1.

Let AB = x, BC = $\sqrt{2}x$ and AC = x.

Squaring both sides we get :

AB² = x², BC² = 2x² and AC² = x².

⇒ AB² + AC² = x² + x² = 2x² = BC².

Since,

⇒ BC² = AB² + AC².

∴ Triangle ABC satisfies pythagoras theorem.

Hence, proved that ABC is a right-angled triangle.

10Exercise 12(A)

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.

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Let AB and CD be two poles of height 6 m and 11 m respectively.

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. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

From figure,

ABDE is a rectangle.

∴ AE = BD = 12 m and ED = AB = 6 m.

From figure,

CE = CD - ED = 11 - 6 = 5 m.

In right-angled triangle,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AC² = CE² + AE²

⇒ AC² = 5² + 12²

⇒ AC² = 25 + 144

⇒ AC² = 169

⇒ AC = $\sqrt{169}$ = 13 m.

Hence, the distance between the tips of two poles is 13 m.

1(a)Exercise 12(B)

In △ ABC, ∠C = 90° and AC = BC, then AB² is equal to :

(a)AC²
(b)2AC²
(c)BC²
(d)2BC² - AC²

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ABC is a right-angled triangle at C.

In △ ABC, ∠C = 90° and AC = BC, then AB2 is equal to : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AC² + BC²

⇒ AB² = AC² + AC² (Since, AC = BC)

⇒ AB² = 2AC².

Hence, Option 2 is the correct option.

1(b)Exercise 12(B)

In the given diagram, AE² + BD² is equal to :

In the given diagram, AE2 + BD2 is equal to : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

(a)AB² - DE²
(b)DE² - AB²
(c)AB² + DE²
(d)DE × AB

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By Formula,

By pythagoras theorem,

(Hypotenuse)² = (Perpendicular)² + (Base)².

In right-angled triangle ABC,

⇒ AB² = AC² + BC² ........(1)

In right-angled triangle DEC,

⇒ DE² = CD² + EC² ........(2)

In right-angled triangle AEC,

⇒ AE² = AC² + EC² ........(3)

In right-angled triangle BDC,

⇒ BD² = CD² + BC² ........(4)

Adding equations (3) and (4), we get :

⇒ AE² + BD² = AC² + EC² + CD² + BC²

= (AC² + BC²) + (CD² + EC²)

= AB² + DE² [From equations (1) and (2)].

Hence, Option 3 is the correct option.

1(c)Exercise 12(B)

In the given figure, the value of AB × CD is :

(a)AC × BC
(b)AC × CD
(c)AC × AB
(d)AC² + BC²

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By formula,

⇒ Area of triangle = $\frac{1}{2}$ × base × height

From figure,

⇒ Area of △ ABC = $\frac{1}{2}$ × AB × CD .........(1)

⇒ Area of △ ABC = $\frac{1}{2}$ × BC × AC .........(2)

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

⇒ $\frac{1}{2}$ × AB × CD = $\frac{1}{2}$ × BC × AC

⇒ AB × CD = BC × AC.

Hence, Option 1 is the correct option.

1(d)Exercise 12(B)

ABC is an isosceles triangle right-angled at C. Then 2AC² is equal to :

(a)BC²
(b)AC²
(c)AC² - BC²
(d)AB²

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ABC is an isosceles triangle right-angled at C. Then 2AC2 is equal to : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

Since, ABC is an isosceles triangle right-angled at C.

∴ ∠A = ∠B

∴ BC = AC [Sides opposite to equal angles are equal] ........(1)

In right-angled △ ABC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AC² + BC²

⇒ AB² = AC² + AC² [From equation (1)]

⇒ AB² = 2AC².

Hence, Option 4 is the correct option.

2Exercise 12(B)

In the figure, given below, AD ⊥ BC. Prove that :

c² = a² + b² - 2ax.

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By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled △ ABD,

⇒ AB² = AD² + BD²

⇒ c² = h² + (a - x)² .............(1)

In right-angled △ ACD,

⇒ AC² = AD² + CD²

⇒ b² = h² + x²

⇒ h² = b² - x² ..........(2)

Substituting value of h² from equation (2) in (1), we get :

⇒ c² = b² - x² + (a - x)²

⇒ c² = b² - x² + a² + x² - 2ax

⇒ c² = a² + b² - 2ax.

Hence, proved that c² = a² + b² - 2ax.

3Exercise 12(B)

In equilateral △ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.

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In △ ABD and △ ACD,

⇒ ∠ADB = ∠ADC (Both equal to 90°)

⇒ AD = AD (Common side)

⇒ AB = AC (Since, ABC is an equilateral triangle)

∴ △ ABD ≅ △ ACD (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BD = CD = $\frac{BC}{2} = \frac{x}{2}$ cm.

In right-angled triangle ABD,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AD² + BD²

⇒ x² = AD² + $\Big(\frac{x}{2}\Big)^2$

⇒ AD² = x² - $\Big(\frac{x}{2}\Big)^2$

⇒ AD² = $x^2 - \frac{x^2}{4}$

⇒ AD² = $\frac{4x^2 - x^2}{4}$

⇒ AD² = $\frac{3x^2}{4}$

⇒ AD = $\sqrt{\frac{3x^2}{4}} = \frac{\sqrt{3}x}{2}$ cm.

Hence, AD = $\frac{\sqrt{3}x}{2}$ cm.

4Exercise 12(B)

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

AM² + BC² = AC² + BM².

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ABC is a triangle, right-angled at B. M is a point on BC. Prove that : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled △ ABM,

⇒ AM² = AB² + BM²

⇒ AB² = AM² - BM² ...........(1)

In right-angled △ ABC,

⇒ AC² = AB² + BC²

⇒ AB² = AC² - BC² ...........(2)

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

⇒ AM² - BM² = AC² - BC²

⇒ AM² + BC² = AC² + BM².

Hence, proved AM² + BC² = AC² + BM².

5Exercise 12(B)

M and N are the mid-points of the sides QR and PQ respectively of a △ PQR, right-angled at Q. Prove that :

(i) PM² + RN² = 5 MN²

(ii) 4 PM² = 4 PQ² + QR²

(iii) 4 RN² = PQ² + 4 QR²

(iv) 4 (PM² + RN²) = 5 PR²

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M and N are the mid-points of the sides QR and PQ respectively of a △ PQR, right-angled at Q. Prove that : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

Since, M and N are the mid-points of the sides QR and PQ respectively.

PN = NQ and QM = RM

(i) In △ MNQ,

By pythagoras theorem,

⇒ MN² = NQ² + QM² ..........(1)

In △ PQM,

By pythagoras theorem,

⇒ PM² = PQ² + QM²

⇒ PM² = (PN + NQ)² + QM²

⇒ PM² = PN² + NQ² + 2.PN.NQ + QM²

⇒ PM² = MN² + PN² + 2.PN.NQ [From equation (1)] ...........(2)

In △ RNQ,

By pythagoras theorem,

⇒ RN² = NQ² + RQ²

⇒ RN² = NQ² + (QM + RM)²

⇒ RN² = NQ² + QM² + RM² + 2.QM.RM

⇒ RN² = MN² + RM² + 2.QM.RM [From equation (1)] ...........(3)

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

⇒ PM² + RN² = MN² + PN² + 2.PN.NQ + MN² + RM² + 2.QM.RM

⇒ PM² + RN² = 2MN² + PN² + RM² + 2.PN.NQ + 2.QM.RM

Substituting PN = QN and RM = QM in above equation, we get :

⇒ PM² + RN² = 2MN² + QN² + QM² + 2.NQ.NQ + 2.QM.QM

⇒ PM² + RN² = 2MN² + QN² + QM² + 2 QN² + 2 QM²

⇒ PM² + RN² = 2MN² + (QN² + QM²) + 2 (QN² + QM²)

⇒ PM² + RN² = 2MN² + MN² + 2MN² [From equation (1)]

⇒ PM² + RN² = 5MN².

Hence, proved that PM² + RN² = 5MN².

(ii) In △ PQM,

By pythagoras theorem,

⇒ PM² = PQ² + QM²

Multiplying both sides of the above equation by 4, we get :

⇒ 4PM² = 4PQ² + 4QM²

⇒ 4PM² = 4PQ² + $4 \times \Big(\frac{1}{2}QR\Big)^2$

⇒ 4PM² = 4PQ² + $4 \times \frac{1}{4} \times QR^2$

⇒ 4PM² = 4PQ² + QR².

Hence, proved that 4PM² = 4PQ² + QR².

(iii) In △ RQN,

By pythagoras theorem,

⇒ RN² = NQ² + QR²

Multiplying both sides of the above equation by 4, we get :

⇒ 4RN² = 4NQ² + 4QR²

⇒ 4RN² = 4QR² + $4 \times \Big(\frac{1}{2}PQ\Big)^2$

⇒ 4RN² = 4QR² + $4 \times \frac{1}{4} \times PQ^2$

⇒ 4RN² = 4QR² + PQ².

Hence, proved that 4RN² = PQ² + 4QR².

(iv) Proved in part (i), we get :

⇒ PM² + RN² = 5MN²

Multiplying both side of the above equation by 4, we get :

⇒ 4(PM² + RN²) = 4 × 5 MN²

⇒ 4(PM² + RN²) = 4 × 5 (NQ² + MQ²)

⇒ 4(PM² + RN²) = 4 × 5 $\times \Big[\Big(\frac{1}{2}PQ\Big)^2 + \Big(\frac{1}{2}RQ\Big)^2\Big]$

⇒ 4(PM² + RN²) = 4 × 5 $\times \Big[\frac{1}{4}PQ^2 + \frac{1}{4}RQ^2\Big]$

⇒ 4(PM² + RN²) = 4 × 5 $\times \frac{1}{4}$ (PQ² + RQ²)

⇒ 4(PM² + RN²) = 5 (PQ² + RQ²) .....(4)

In right angled triangle PQR,

By pythagoras theorem,

⇒ PR² = PQ² + RQ²

Substituting above value of PR² in equation (4), we get :

⇒ 4(PM² + RN²) = 5 PR².

Hence, proved that 4(PM² + RN²) = 5 PR².

6Exercise 12(B)

In triangle ABC, ∠B = 90° and D is the mid-point of BC. Prove that : AC² = AD² + 3CD².

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Given,

D is the mid-point of BC.

∴ CD = BD ........(1)

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled △ ABC,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (BD + CD)²

⇒ AC² = AB² + (CD + CD)² ..........[From equation (1)]

⇒ AC² = AB² + (2CD)²

⇒ AC² = AB² + 4 CD² ...........(2)

In right-angled △ ABD,

⇒ AD² = AB² + BD²

⇒ AD² = AB² + CD² [From equation (1)] .......(3)

Subtracting equation (3) from (2), we get :

⇒ AC² - AD² = AB² + 4 CD² - (AB² + CD²)

⇒ AC² - AD² = AB² - AB² + 4 CD² - CD²

⇒ AC² - AD² = 3 CD²

⇒ AC² = AD² + 3 CD².

Hence, proved that AC² = AD² + 3 CD².

7Exercise 12(B)

In a rectangle ABCD, prove that :

AC² + BD² = AB² + BC² + CD² + DA².

Show SolutionHide Solution↓

In a rectangle ABCD, prove that : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

In rectangle the interior angles equal to 90° and opposite sides are equal.

∴ ∠A = ∠B = ∠C = ∠D = 90°, AB = DC and BC = AD.

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled △ ACD,

⇒ AC² = AD² + CD² ........(1)

In right-angled △ BCD,

⇒ BD² = BC² + CD²

⇒ BD² = BC² + AB² (As CD = AB) .........(2)

Adding equation (1) and (2), we get :

⇒ AC² + BD² = AD² + CD² + BC² + AB².

Hence, proved that AC² + BD² = AB² + BC² + CD² + DA².

8Exercise 12(B)

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°. Prove that :

2AC² - AB² = BC² + CD² + DA².

Show SolutionHide Solution↓

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°. Prove that : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

In right-angled triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AB² = AC² - BC² .........(1)

In right-angled triangle ADC,

By pythagoras theorem,

⇒ AC² = AD² + DC² ........(2)

To prove :

2AC² - AB² = BC² + CD² + DA².

Solving L.H.S. of the equation :

⇒ 2AC² - AB² = 2AC² - (AC² - BC²) [From equation (1)]

= 2AC² - AC² + BC²

= AC² + BC²

= AD² + DC² + BC² [From equation (2)]

= R.H.S.

Hence, proved that 2AC² - AB² = BC² + CD² + DA².

9Exercise 12(B)

O is any point inside a rectangle ABCD. Prove that :

OB² + OD² = OC² + OA².

Show SolutionHide Solution↓

O is any point inside a rectangle ABCD. Prove that : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

Through point O, draw PQ || BC so that P lies on AB and Q lies on DC.

Since, PQ || BC and AB ⊥ BC,

∴ AB ⊥ PQ and DC ⊥ PQ.

∴ BPQC and APQD are both rectangles.

In right-angled triangle OBP,

By pythagoras theorem,

⇒ OB² = OP² + BP² ..........(1)

In right-angled triangle OQD,

By pythagoras theorem,

⇒ OD² = OQ² + DQ² ..........(2)

In right-angled triangle OQC,

By pythagoras theorem,

⇒ OC² = OQ² + QC² ..........(3)

In right-angled triangle OAP,

By pythagoras theorem,

⇒ OA² = OP² + AP² ..........(4)

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

⇒ OB² + OD² = OP² + BP² + OQ² + DQ²

= OP² + CQ² + OQ² + AP² (As, BP = CQ and DQ = AP)

= CQ² + OQ² + OP² + AP²

= OC² + OA². [From equation (3) and (4)]

Hence, proved that OB² + OD² = OC² + OA².

10Exercise 12(B)

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

AR² + BP² + CQ² = AQ² + CP² + BR².

Show SolutionHide Solution↓

Join OA, OB and OC.

By formula,

By pythagoras theorem,

(Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled triangle AOR,

⇒ AO² = AR² + OR²

⇒ AR² = AO² - OR² ..........(1)

In right-angled triangle BOP,

⇒ BO² = BP² + OP²

⇒ BP² = BO² - OP² ..........(2)

In right-angled triangle COQ,

⇒ CO² = CQ² + OQ²

⇒ CQ² = CO² - OQ² ..........(3)

In right-angled triangle AOQ,

⇒ AO² = AQ² + OQ²

⇒ AQ² = AO² - OQ² ..........(4)

In right-angled triangle COP,

⇒ CO² = CP² + OP²

⇒ CP² = CO² - OP² ..........(5)

In right-angled triangle BOR,

⇒ BO² = BR² + OR²

⇒ BR² = BO² - OR² ..........(6)

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

⇒ AR² + BP² + CQ² = AO² - OR² + BO² - OP² + CO² - OQ² ..........(7)

Adding equations (4), (5) and (6), we get :

⇒ AQ² + CP² + BR² = AO² - OQ² + CO² - OP² + BO² - OR²

⇒ AQ² + CP² + BR² = AO² - OR² + BO² - OP² + CO² - OQ² ...........(8)

From equations (7) and (8), we get :

⇒ AR² + BP² + CQ² = AQ² + CP² + BR².

Hence, proved that AR² + BP² + CQ² = AQ² + CP² + BR².

1(a)Test Yourself

Angle AOB is:

(a)60°
(b)90°
(c)45°
(d)none of these

Show SolutionHide Solution↓

5, 12, and 13 are pythagorean triplets.

As,

13² = 12² + 5²

So, side BO = 13 units is the hypotenuse.

We know that,

Angle opposite to hypotenuse is a right-angle triangle.

So, ∠OAB = 90°

According to angle sum property,

⇒ ∠OAB + ∠ABO + ∠AOB = 180°

⇒ 90° + ∠ABO + ∠AOB = 180°

⇒ ∠ABO + ∠AOB = 180° - 90°

⇒ ∠ABO + ∠AOB = 90°

Thus, ∠AOB, ∠ABO cannot be defined individually.

Hence, option 4 is the correct option.

1(b)Test Yourself

Ranbeer runs 10 km due North and 24 km due West. The distance between his two position is:

(a)34 km
(b)17 km
(c)26 km
(d)none of these

Show SolutionHide Solution↓

Starting from point A, Ranbeer runs 10 km due north and reached at point B then 24 km due west, ending at point C.

Ranbeer runs 10 km due North and 24 km due West. The distance between his two position is 34 km 2. 17 km 3. 26 km 4. none of these. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

According to Pythagoras Theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 10² + 24²

⇒ AC² = 100 + 576

⇒ AC² = 676

⇒ AC = $\sqrt{676}$

⇒ AC = 26 km.

Hence, option 3 is the correct option.

1(c)Test Yourself

Angle AOB is:

(a)60°
(b)90°
(c)45°
(d)none of these

Show SolutionHide Solution↓

Since, AO = BO = x

Squaring all the sides we get :

AB² = $(x\sqrt{2})^2 = 2x^2$

OB = x²

OA = x²

In triangle AOB,

⇒ AO² + BO² = x² + x² = 2x² = AB².

Thus, AOB is a right angle triangle with AB as hypotenuse.

Since, AB is hypotenuse, then ∠AOB (angle opposite to hypotenuse) = 90°

Hence, option 2 is the correct option.

1(d)Test Yourself

The sides of a rectangle are 12 cm and 16 cm. The length of its diagonal is:

(a)28 cm
(b)4 cm
(c)$\sqrt{12^2 + 16^2}$ cm
(d)$\sqrt{16^2 - 12^2}$ cm

Show SolutionHide Solution↓

ABCD is a rectangle such that AB = 16 cm and BC = 12 cm.

The sides of a rectangle are 12 cm and 16 cm. The length of its diagonal is: Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

Join AC. AC is a diagonal of the rectangle ABCD.

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 16² + 12²

⇒ AC = $\sqrt{16^2 + 12^2}$ cm

Hence, option 3 is the correct option.

1(e)Test Yourself

Statement 1: ABCD is a rhombus, its diagonal AC = 16 cm and diagonal BD = 12 cm, perimeter of rhombus = 64 cm.

Statement 2: OA = 8 cm, OB = 6 cm. Then, AB = 10 cm

And, perimeter of rhombus = 40 cm

(a)Both the statements are true.
(b)Both the statements are false.
(c)Statement 1 is true, and statement 2 is false.
(d)Statement 1 is false, and statement 2 is true.

Show SolutionHide Solution↓

Given, ABCD is a rhombus, diagonal AC = 16 cm and diagonal BD = 12 cm.

The diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = 90°

Each diagonal is divided into two equal segments.

∴ AC = 16 cm ⇒ AO = OC = $\frac{16}{2}$ = 8 cm

∴ BD = 12 cm ⇒ BO = OD = $\frac{12}{2}$ = 6 cm

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In Δ AOB,

⇒ AB² = BO² + OA²

⇒ AB² = 6² + 8²

⇒ AB² = 36 + 64

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm

According to formula, perimeter of rhombus = 4 x length of side

= 4 x 10

= 40 cm.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

1(f)Test Yourself

Statement 1: Area of given triangle ABC = 6 x 5 cm².

Area of given triangle ABC = 6 x 5 cm2, Area of given triangle ABC = 1/2 x 6 x 4 cm2. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

Statement 2: Area of given triangle ABC = $\frac{1}{2}$ x 6 x 4 cm².

(a)Both the statements are true.
(b)Both the statements are false.
(c)Statement 1 is true, and statement 2 is false.
(d)Statement 1 is false, and statement 2 is true.

Show SolutionHide Solution↓

Draw a perpendicular bisector, AD.

The perpendicular bisector of the base also passes through the midpoint of the base, effectively dividing it into two equal segments.

BD = DC = $\frac{6}{2}$ = 3 cm

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In Δ ABD,

⇒ AB² = AD² + BD²

⇒ 5² = AD² + 3²

⇒ 25 = AD² + 9

⇒ AD² = 25 - 9

⇒ AD² = 16

⇒ AD = $\sqrt{16}$

⇒ AD = 4 cm.

Using formula, area of triangle = $\frac{1}{2}$ x base x height

Here, base, BC = 6 cm and height, AD = 4 cm

⇒ Area of triangle = $\frac{1}{2}$ x 6 x 4 cm²

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

1(g)Test Yourself

Assertion (A): Angle BOC = 90°.

Reason (R): OC² = 3² + 4² = 25

OB² = 6² + 8² = 100

OC² + OB² = 125 = BC²

Angle BOC = 90°. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

(a)A is true, but R is false.
(b)A is false, but R is true.
(c)Both A and R are true, and R is the correct reason for A.
(d)Both A and R are true, and R is the incorrect reason for A.

Show SolutionHide Solution↓

Given, OD = 3 cm and DC = 4 cm

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

In Δ ODC,

⇒ OC² = OD² + DC²

⇒ OC² = 3² + 4²

⇒ OC² = 9 + 16

⇒ OC² = 25

⇒ OC = $\sqrt{25}$

⇒ OC = 5 cm

Similarly, it it given that OA = 6 cm and AB = 8 cm

In Δ OAB,

⇒ OB² = OA² + AB²

⇒ OB² = 6² + 8²

⇒ OB² = 36 + 64

⇒ OB² = 100

⇒ OB = $\sqrt{100}$

⇒ OB = 10 cm

Squaring all sides of triangle BOC,

⇒ OB² = 10² = 100

⇒ OC² = 5² = 25

⇒ BC² = $(5\sqrt{5})^2$ = 125

Since,

⇒ BC² = OC² + OB²

Since, sides of triangle BOC, satisfy pythagoras theorem. So, BOC is a right angle triangle with BC as hypotenuse.

∴ ∠BOC = 90°.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

1(h)Test Yourself

Assertion (A): x = $5\sqrt{2}$

X = 5 root of 2, AC^2 = 8^2 + 6^2 = x + x^2: Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

Reason (R): AC² = 8² + 6² = x² + x²

(a)A is true, but R is false.
(b)A is false, but R is true.
(c)Both A and R are true, and R is the correct reason for A.
(d)Both A and R are true, and R is the incorrect reason for A.

Show SolutionHide Solution↓

In Δ ABC,

Since angle ABC = 90°.

According to Pythagoras theorem, in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

⇒ Hypotenuse² = Base² + Height²

⇒ AC² = AB² + BC²

⇒ AC² = 8² + 6² ........................(1)

In Δ ADC,

Since angle ADC = 90°.

Using pythagoras theorem,

⇒ AC² = AD² + DC²

⇒ AC² = x² + x² .....................(2)

From equation (1) and (2),

⇒ 8² + 6² = x² + x²

⇒ 64 + 36 = 2x²

⇒ 2x² = 100

⇒ x² = $\frac{100}{2}$

⇒ x² = 50

⇒ x = $\sqrt{50}$

⇒ x = 5 $\sqrt{2}$

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

2Test Yourself

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

Show SolutionHide Solution↓

In right-angled triangle ABD,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ BD² = AD² + AB²

⇒ 25² = AD² + 7²

⇒ 625 = AD² + 49

⇒ AD² = 625 - 49

⇒ AD² = 576

⇒ AD = $\sqrt{576}$ = 24 cm.

Draw BE perpendicular to CD.

From figure,

ABED is a rectangle. Since, opposite sides of rectangle are equal.

∴ ED = AB = 7 cm and BE = AD = 24 cm.

⇒ CE = CD - ED = 17 - 7 = 10 cm.

In right-angled triangle BEC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ BC² = BE² + CE²

⇒ BC² = 24² + 10²

⇒ BC² = 576 + 100

⇒ BC² = 676

⇒ BC = $\sqrt{676}$ = 26 cm.

Hence, BC = 26 cm.

3Test Yourself

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

Show SolutionHide Solution↓

Given,

AX : XB = 1 : 2

Let AX = x and XB = 2x.

From figure,

⇒ AX + XB = AB

⇒ x + 2x = 12

⇒ 3x = 12

⇒ x = $\frac{12}{3}$ = 4 cm.

⇒ AX = x = 4 cm and XB = 2x = 2(4) = 8 cm.

Given,

AY : YC = 1 : 2

$$\begin{aligned}\Rightarrow \frac{AY}{YC} = \frac{1}{2} \\[1em] \Rightarrow \frac{8}{YC} = \frac{1}{2} \\[1em] \Rightarrow YC = 8 \times 2 = 16.\end{aligned}$$

From figure,

⇒ AC = AY + YC = 8 + 16 = 24 cm.

In right-angled triangle ABC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ AC² = AB² + BC²

⇒ 24² = 12² + BC²

⇒ 576 = 144 + BC²

⇒ BC² = 576 - 144

⇒ BC² = 432

⇒ BC = $\sqrt{432} = 12\sqrt{3}$ = 20.78 cm.

Hence, AC = 24 cm and BC = 20.78 cm.

4Test Yourself

In △ ABC, ∠B = 90°. 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.

Show SolutionHide Solution↓

(i) In right-angled triangle ABC,

In △ ABC, ∠B = 90°. Find the sides of the triangle, if : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ AC² = AB² + BC²

⇒ (x + 6)² = (x - 3)² + (x + 4)²

⇒ x² + 6² + 12x = x² + 9 - 6x + x² + 16 + 8x

⇒ x² + 36 + 12x = x² + 9 - 6x + x² + 16 + 8x

⇒ x² + 36 + 12x = 2x² + 25 + 2x

⇒ 2x² - x² + 2x - 12x + 25 - 36 = 0

⇒ x² - 10x - 11 = 0

⇒ x² - 11x + x - 11 = 0

⇒ x(x - 11) + 1(x - 11) = 0

⇒ (x + 1)(x - 11) = 0

⇒ x + 1 = 0 or x - 11 = 0

⇒ x = -1 or x = 11.

Since, side cannot be negative.

∴ x = 11.

AB = x - 3 = 11 - 3 = 8 cm,

BC = x + 4 = 11 + 4 = 15 cm,

AC = x + 6 = 11 + 6 = 17 cm.

Hence, AB = 8 cm, BC = 15 cm and AC = 17 cm.

(ii) In right-angled triangle ABC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ AC² = AB² + BC²

⇒ (4x + 5)² = x² + (4x + 4)²

⇒ (4x)² + 5² + 2 × (4x) × 5 = x² + (4x)² + 4² + 2 × (4x) × 4

⇒ 16x² + 25 + 40x = x² + 16x² + 16 + 32x

⇒ 16x² + 25 + 40x = 17x² + 16 + 32x

⇒ 17x² - 16x² + 16 - 25 + 32x - 40x = 0

⇒ x² - 8x - 9 = 0

⇒ x² - 9x + x - 9 = 0

⇒ x(x - 9) + 1(x - 9) = 0

⇒ (x + 1)(x - 9) = 0

⇒ x + 1 = 0 or x - 9 = 0

⇒ x = -1 or x = 9.

Since, side cannot be negative.

∴ x = 9.

AB = x = 9 cm,

BC = 4x + 4 = 4(9) + 4 = 36 + 4 = 40 cm,

AC = 4x + 5 = 4(9) + 5 = 35 + 5 = 41 cm.

Hence, AB = 9 cm, BC = 40 cm and AC = 41 cm.

5Test Yourself

If a side of a rhombus is 10 cm and one of the diagonals is 16 cm, find the other diagonal.

Show SolutionHide Solution↓

Let ABCD be the rhombus and the diagonals intersect at point O.

Let diagonal AC = 16 cm.

If a side of a rhombus is 10 cm and one of the diagonals is 16 cm, find the other diagonal. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

We know that,

Diagonals of rhombus bisect each other at right angles.

∴ AO = OC = $\frac{AC}{2} = \frac{16}{2}$ = 8 cm and BO = OD = x cm (let).

In right angle triangle AOB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

⇒ AB² = AO² + OB²

⇒ 10² = 8² + x²

⇒ 100 = 64 + x²

⇒ x² = 100 - 64

⇒ x² = 36

⇒ x = $\sqrt{36}$ = 6 cm.

From figure,

⇒ BD = BO + OD = 6 + 6 = 12 cm.

Hence, length of other diagonal = 12 cm.

6Test Yourself

In the given figure, diagonals AC and BD intersect at right angle. Show that :

AB² + CD² = AD² + BC²

Show SolutionHide Solution↓

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle AOB,

⇒ AB² = OB² + OA² ..........(1)

In right angle triangle COD,

⇒ CD² = OD² + OC² ..........(2)

In right angle triangle AOD,

⇒ AD² = OD² + OA² ..........(3)

In right angle triangle BOC,

⇒ BC² = OB² + OC² ..........(4)

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

⇒ AB² + CD² = OB² + OA² + OD² + OC²

⇒ AB² + CD² = (OA² + OD²) + (OC² + OB²)

Substituting value from (3) and (4) in above equation, we get :

⇒ AB² + CD² = AD² + BC².

Hence, proved that AB² + CD² = AD² + BC².

7Test Yourself

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

OA² + OC² = 2AD² - $\frac{BD^2}{2}$

Show SolutionHide Solution↓

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle AOD,

Diagonals of rhombus ABCD intersect each other at point O. Prove that : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

By pythagoras theorem,

⇒ AD² = OA² + OD²

⇒ OA² = AD² - OD² ..........(1)

In right angle triangle COD,

By pythagoras theorem,

⇒ CD² = OC² + OD²

⇒ OC² = CD² - OD² ..........(2)

To prove :

OA² + OC² = 2AD² - $\frac{BD^2}{2}$

Solving L.H.S. of the above equation, we get :

⇒ OA² + OC² = AD² - OD² + CD² - OD²

⇒ OA² + OC² = AD² + CD² - 2OD² .........(3)

Since, all sides of a rhombus are equal and diagonals of rhombus bisect each other,

∴ CD = AD and OD = $\frac{BD}{2}$

Substituting value of CD and OD in equation (3), we get :

⇒ OA² + OC² = AD² + AD² - $2\Big(\frac{BD}{2}\Big)^2$

⇒ OA² + OC² = 2AD² - $2 \times \frac{BD^2}{4}$

⇒ OA² + OC² = 2AD² - $\frac{BD^2}{2}$ = R.H.S.

Hence, proved that OA² + OC² = 2AD² - $\frac{BD^2}{2}$.

8Test Yourself

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

AC² = 2.BC.DC.

Show SolutionHide Solution↓

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ AD² = AB² - BD² .........(1)

In right angle triangle ACD,

By pythagoras theorem,

⇒ AC² = AD² + DC²

= (AB² - BD²) + (DB + BC)² [From equation (1)]

= AB² - BD² + DB² + BC² + 2.DB.BC

= AB² + BC² + 2.DB.BC

= BC² + BC² + 2.DB.BC [As, AB = BC]

= 2BC² + 2.DB.BC

= 2BC(BC + DB)

= 2BC.DC

Hence, proved that AC² = 2BC.DC

9Test Yourself

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

AD² = AC² + BD.CD

Show SolutionHide Solution↓

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

Draw AE perpendicular to BC.

In an isosceles triangle ABC; AB = AC and D is a point on BC produced. Prove that : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

In right angle triangle AED,

By pythagoras theorem,

⇒ AD² = AE² + ED²

⇒ AD² = AE² + (EC + CD)² .........(1)

In right angle triangle AEC,

By pythagoras theorem,

⇒ AC² = AE² + EC²

⇒ AE² = AC² - EC² .........(2)

Substituting value of AE² from equation (2) in (1), we get :

⇒ AD² = AC² - EC² + (EC + CD)²

⇒ AD² = AC² - EC² + EC² + CD² + 2.EC.CD

⇒ AD² = AC² + CD(CD + 2EC) ..........(3)

Since, ABC is an isosceles triangle.

We know that,

In an isosceles triangle altitude from the vertex bisects the base.

∴ E is the mid-point of BC.

⇒ EC = $\frac{1}{2}$ BC

⇒ BC = 2EC.

From figure,

⇒ BD = BC + CD

⇒ BD = 2EC + CD

Substituting above value in equation (3), we get :

⇒ AD² = AC² + CD.BD

Hence, proved that AD² = AC² + BD.CD

10Test Yourself

In triangle ABC, angle A = 90°, CA = AB and D is a point on AB produced. Prove that :

DC² - BD² = 2AB.AD.

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By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle ACD,

By pythagoras theorem,

⇒ CD² = AC² + AD²

⇒ CD² = AC² + (AB + BD)²

⇒ CD² = AC² + AB² + BD² + 2.AB.BD .........(1)

In right angle triangle ABC,

By pythagoras theorem,

⇒ BC² = AC² + AB²

⇒ BC² = AB² + AB² (Since, CA = AB)

⇒ BC² = 2AB²

⇒ AB² = $\frac{1}{2}$ BC² .........(2)

Substituting value of AB² from equation (2) in (1), we get :

⇒ CD² = AC² + $\frac{1}{2}BC^2$ + BD² + 2.AB.BD

⇒ CD² - BD² = AB² + $\frac{1}{2}BC^2$ + 2.AB.BD [As, AC = AB]

⇒ CD² - BD² = AB² + AB² + 2.AB.(AD - AB) [From equation (2)]

⇒ CD² - BD² = 2AB² + 2.AB.AD - 2.AB²

⇒ CD² - BD² = 2.AB.AD

Hence, proved that CD² - BD² = 2.AB.AD

11Test Yourself

In triangle ABC, AB = AC and BD is perpendicular to AC. Prove that :

BD² - CD² = 2CD × AD

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By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In triangle ABC, AB = AC and BD is perpendicular to AC. Prove that : Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

In right-angled triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ AD² = AB² - BD² .......(1)

From figure,

⇒ AC = AD + DC

Squaring both sides, we get :

⇒ AC² = (AD + DC)²

⇒ AC² = AD² + DC² + 2AD.DC

⇒ AC² = AB² - BD² + DC² + 2AD.DC [From equation (1)]

Substituting AB = AC, in above equation :

⇒ AC² = AC² - BD² + DC² + 2AD.DC

⇒ AC² - AC² + BD² - DC² = 2AD.DC

⇒ BD² - DC² = 2CD × AD.

Hence, proved that BD² - DC² = 2CD × AD.

12Test Yourself

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

Prove that : 2AC² = 2AB² + BC².

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Given,

D divides BC in the ratio 1 : 3.

⇒ BD : DC = 1 : 3

Let BD = x and DC = 3x

From figure,

⇒ BC = BD + DC = x + 3x = 4x.

$$\begin{aligned}\Rightarrow \frac{BD}{BC} = \frac{x}{4x} \\[1em] \Rightarrow \frac{BD}{BC} = \frac{1}{4} \\[1em] \Rightarrow BD = \frac{1}{4}BC. \\[1em] \Rightarrow \frac{CD}{BC} = \frac{3x}{4x} \\[1em] \Rightarrow \frac{CD}{BC} = \frac{3}{4} \\[1em] \Rightarrow CD = \frac{3}{4}BC.\end{aligned}$$

In right angle triangle ADC,

By pythagoras theorem,

⇒ AC² = AD² + CD² ............(1)

In right angle triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD² ............(2)

Subtracting equation (2) from (1), we get :

⇒ AC² - AB² = AD² + CD² - (AD² + BD²)

⇒ AC² - AB² = CD² - BD²

Substituting value of BD and CD in above equation, we get :

$$\begin{aligned}\Rightarrow AC^2 - AB^2 = \Big(\frac{3}{4}BC\Big)^2 - \Big(\frac{1}{4}BC\Big)^2 \\[1em] \Rightarrow AC^2 - AB^2 = \frac{9}{16}BC^2 - \frac{1}{16}BC^2 \\[1em] \Rightarrow AC^2 - AB^2 = \frac{8}{16}BC^2 \\[1em] \Rightarrow AC^2 - AB^2 = \frac{1}{2}BC^2 \\[1em] \Rightarrow 2(AC^2 - AB^2) = BC^2 \\[1em] \Rightarrow 2AC^2 - 2AB^2 = BC^2 \\[1em] \Rightarrow 2AC^2 = 2AB^2 + BC^2.\end{aligned}$$

Hence, proved that 2AC² = 2AB² + BC².

13Test Yourself

In the given figure, AB = 16 cm, BC = 12 cm and CA = 6 cm; find the length of CD.

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By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

Let CD be x cm.

In right angle triangle ACD,

By pythagoras theorem,

⇒ AC² = AD² + CD²

⇒ 6² = AD² + x²

⇒ 36 = AD² + x²

⇒ AD² = 36 - x² ........(1)

In right angle triangle ABD,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ 16² = AD² + (BC + CD)²

⇒ 256 = 36 - x² + (12 + x)² [From equation (1)]

⇒ 256 = 36 - x² + 12² + x² + 2(12)x

⇒ 256 = 36 + 144 + 24x

⇒ 256 = 180 + 24x

⇒ 24x = 256 - 180

⇒ 24x = 76

⇒ x = $\frac{76}{24} = \frac{19}{6} = 3\frac{1}{6}$ cm.

Hence, CD = $3\frac{1}{6}$ cm.

14Test Yourself

In a quadrilateral ABCD, ∠A + ∠D = 90°, prove that : AC² + BD² = AD² + BC².

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Produce AB and DC such that they meet at point E.

In a quadrilateral ABCD, ∠A + ∠D = 90°, prove that : AC2 + BD2 = AD2 + BC2. Pythagoras Theorem, Concise Mathematics Solutions ICSE Class 9.

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In △ AED,

⇒ ∠A + ∠D + ∠E = 180°

⇒ 90° + ∠E = 180°

⇒ ∠E = 180° - 90° = 90°.

By pythagoras theorem,

⇒ AD² = AE² + DE² .........(1)

In △ BEC,

By pythagoras theorem,

⇒ BC² = BE² + CE² .........(2)

In △ AEC,

By pythagoras theorem,

⇒ AC² = AE² + CE² .........(3)

In △ BED,

By pythagoras theorem,

⇒ BD² = BE² + DE² .........(4)

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

⇒ AD² + BC² = AE² + DE² + BE² + CE²

⇒ AD² + BC² = (AE² + CE²) + (BE² + DE²)

⇒ AD² + BC² = AC² + BD².

Hence, proved that AC² + BD² = AD² + BC².

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