Pythagoras Theorem — Exercise 12(A)

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

1. acute-angled triangle

2. scalene triangle

3. scalene right-angled triangle

4. obtuse-angled triangle

Answer

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.

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

1. 6 cm and 4 cm

2. 8 cm and 6 cm

3. 3 cm and 4 cm

4. 8 cm and 4 cm

Answer

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 = $\dfrac{100}{25}$

⇒ 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.

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

1. $32\sqrt{2}$ cm²

2. $16\sqrt{2}$ cm²

3. $8\sqrt{2}$ cm²

4. $12\sqrt{2}$ cm²

Answer

Let AD be the altitude.

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

∴ BD = CD = $\dfrac{BC}{2} = \dfrac{8}{2}$ = 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 = $\dfrac{1}{2}$ × base × height

From figure,

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

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

⇒ Area of △ ABC

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

Hence, Option 1 is the correct option.

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

1. 20 cm

2. 10 cm

3. 60 cm

4. 100 cm

Answer

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

We know that,

Diagonals of rhombus intersect at right angles.

Let AC = 40 cm and BD = 30 cm.

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

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

1. 4 cm

2. 3 cm

3. 5 cm

4. 6 cm

Answer

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.

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.

Answer

Let AB be the ladder.

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.

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

Answer

Let A be the initial position of the man.

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.

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

Answer

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.

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

Answer

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

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.

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.

Answer

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 = $\dfrac{BC}{2} = \dfrac{10}{2}$ = 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.

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

Answer

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 = $\dfrac{19}{12} = 1\dfrac{7}{12}$.

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

In triangle ABC,

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

Answer

In △ ABC,

Draw AD ⊥ BC

By formula,

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

Given,

Area of triangle ABC = 60 cm²

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

We know that,

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

∴ BD = CD = $\dfrac{BC}{2} = \dfrac{10}{2}$ = 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.

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

Answer

Let ABC be the triangle.

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.

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.

Answer

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

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.