Pythagoras Theorem — Multiple Choice Questions

The value of x is:

1. 10 m

2. 9 m

3. 16 m

4. none of these

Answer

In the given figure,

Base = x, Perpendicular = 12 m, Hypotenuse = 15 m

According to Pythagoras theorem :

⇒ Hypotenuse² = Base² + Perpendicular²

⇒ Base² = Hypotenuse² - Perpendicular²

⇒ x² = 15² - 12²

⇒ x² = 225 - 144

⇒ x² = 81

⇒ x = $\sqrt{81}$ = 9 m

Hence, option 2 is the correct option.

A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. The original height of the tree is:

1. 13 m

2. 27 m

3. 18 m

4. 20 m

Answer

The standing part of the tree is 5 m and the broken part (from the break to the top) touches the ground 12 m from the base, forming the hypotenuse of a right-angled triangle.

Let the broken part = h.

According to Pythagoras theorem :

⇒ h² = 5² + 12²

⇒ h² = 25 + 144

⇒ h² = 169

⇒ h = $\sqrt{169}$ = 13 m

Original height of the tree = standing part + broken part = 5 m + 13 m = 18 m

Hence, option 3 is the correct option.

From a point P, a boy travels 12 km due east and then travels 9 km due north to point Q. The shortest distance between points P and Q is:

1. 15 km

2. 3 km

3. 21 km

4. 20 km

Answer

Since east and north are perpendicular directions, the shortest distance PQ is the hypotenuse of a right-angled triangle with legs 12 km and 9 km.

According to Pythagoras theorem :

⇒ PQ² = 12² + 9²

⇒ PQ² = 144 + 81

⇒ PQ² = 225

⇒ PQ = $\sqrt{225}$ = 15 km

Hence, option 1 is the correct option.

The value of x is:

1. $5\sqrt{2}$

2. 5

3. 10

4. 6

Answer

The given triangle is an isosceles right-angled triangle in which the two equal sides (legs) are each x and the hypotenuse is $10\sqrt{2}$.

According to Pythagoras theorem :

⇒ Hypotenuse² = x² + x²

⇒ $\left(10\sqrt{2}\right)$² = 2x²

⇒ 200 = 2x²

⇒ x² = 100

⇒ x = $\sqrt{100}$ = 10

Hence, option 3 is the correct option.

The value of x is:

1. 15

2. 21

3. 23

4. 17

Answer

The altitude from the top vertex divides the base into two parts, 6 and 15, and is common to both right-angled triangles.

First, find the altitude h using the left right-angled triangle :

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ 10² = 6² + h²

⇒ 100 = 36 + h²

⇒ h² = 100 - 36

⇒ h² = 64

⇒ h = $\sqrt{64}$ = 8

Now, in the right-angled triangle on the right, x is the hypotenuse with base 15 and height (h = 8) :

⇒ x² = 15² + 8²

⇒ x² = 225 + 64

⇒ x² = 289

⇒ x = $\sqrt{289}$ = 17

Hence, option 4 is the correct option.

The value of x is:

1. 21

2. 15

3. 17

4. 27

Answer

The altitude from the top vertex is 8 and is common to both right-angled triangles. AB = 10 and AD = 17. Base BD = x

Let BC be a units and CD be b units.

In right-angled triangle ACB,

⇒ AB² = AC² + BC²

⇒ 10² = 8² + a²

⇒ 100 = 64 + a²

⇒ a² = 36

⇒ a = 6

In right-angled triangle ACD,

⇒ AD² = AC² + CD²

⇒ 17² = 8² + b²

⇒ 289 = 64 + b²

⇒ b² = 225

⇒ b = $\sqrt{225}$

⇒ b = 15

x = a + b = 6 + 15 = 21

Hence, option 1 is the correct option.

The perimeter of quadrilateral (rhombus) ABCD is:

1. 56 cm

2. 40 cm

3. 28 cm

4. none of these

Answer

In a rhombus, the diagonals bisect each other at right angles at O.

AO = 8 cm and OB = 6 cm

In right-angled triangle AOB, AB is the hypotenuse :

⇒ AB² = AO² + OB²

⇒ AB² = 8² + 6²

⇒ AB² = 64 + 36

⇒ AB² = 100

⇒ AB = $\sqrt{100}$ = 10 cm

Since all sides of a rhombus are equal,

Perimeter = 4 × 10 = 40 cm

Hence, option 2 is the correct option.

The perimeter of given rectangle ABCD is:

1. 32 cm

2. 36 cm

3. 28 cm

4. 38 cm

Answer

In rectangle ABCD, the diagonal AC = 10 cm and the side AB = 8 cm.

In right-angled triangle ABC, AC is the hypotenuse :

⇒ AC² = AB² + BC²

⇒ 10² = 8² + BC²

⇒ 100 = 64 + BC²

⇒ BC² = 100 - 64

⇒ BC² = 36

⇒ BC = $\sqrt{36}$ = 6 cm

Perimeter = 2 × (AB + BC) = 2 × (8 + 6) = 2 × 14 = 28 cm

Hence, option 3 is the correct option.

Length of BC is:

1. 16 cm

2. 28 cm

3. 10 cm

4. 20 cm

Answer

In triangle ABC, angle B = 40° and angle C = 50°.

Angle A = 180° - (40° + 50°) = 90°

So the triangle is right-angled at A and BC is the hypotenuse.

According to Pythagoras theorem :

⇒ BC² = AB² + AC²

⇒ BC² = 16² + 12²

⇒ BC² = 256 + 144

⇒ BC² = 400

⇒ BC = $\sqrt{400}$ = 20 cm

Hence, option 4 is the correct option.