Pythagoras Theorem — Competency Focused Questions

The hypotenuse of a right triangle is 25 cm. If out of the two legs, one is longer than the other by 5 cm, then the sum of the lengths of the legs is:

1. 30 cm

2. 35 cm

3. 40 cm

4. 45 cm

Answer

Let one leg = x cm and the other = (x + 5) cm

From figure,

AB = (x + 5) cm, BC = x cm and AC = 25 cm, ∠B = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AC² = AB² + BC²

⇒ 25² = (x + 5)² + x²

⇒ 625 = x² + 25 + 10x + x²

⇒ 2x² + 10x + 25 - 625 = 0

⇒ 2x² + 10x - 600 = 0

⇒ 2(x² + 5x - 300) = 0

⇒ x² + 5x - 300 = 0

⇒ x² - 15x + 20x - 300 = 0

⇒ x(x - 15) + 20(x - 15) = 0

⇒ (x + 20)(x - 15) = 0

⇒ x = -20 or x = 15

Since, length cannot be negative.

x = 15 cm.

Sum of the lengths = x + (x + 5) = 15 + (15 + 5) = 15 + 20 = 35 cm.

Hence, Option 2 is the correct option.

An aeroplane leaves an airport and flies due North at 300 km/h. At the same time, another plane leaves the same airport and flies due West at 400 km/h. After 90 minutes, the distance between the two planes would be:

1. 1000 km

2. 900 km

3. 800 km

4. 750 km

Answer

Let aeroplane flying in north direction at 300 km/h be at point P after 1.5 hours and aeroplane flying in west direction at 400 km/h be at point Q after 1.5 hours.

Given, 90 minutes = 1.5 hours

Speed of aeroplane P = 300 km/h

As we know,

Distance traveled = Speed × Time taken

AP = 300 × 1.5 = 450 km.

Speed of aeroplane Q = 400 km/h

As we know,

Distance traveled = Speed × Time taken

AQ = 400 × 1.5 = 600 km.

From figure,

Let ∠A = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle APQ,

⇒ PQ² = AP² + AQ²

⇒ PQ² = 450² + 600²

⇒ PQ² = 202500 + 360000

⇒ PQ² = 562500

⇒ PQ = $\sqrt{562500}$

⇒ PQ = 750 km.

Hence, Option 4 is the correct option.

For going to city B from city A, there is a route via city C such that AC ⟂ CB, AC = 2x km and CB = 2(x + 7) km.

It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.

Answer

From figure,

Distance between two cities A and B = AB = 26 km

Since, AC ⟂ CB

∠A = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AB² = AC² + CB²

⇒ 26² = (2x)² + [2(x + 7)]²

⇒ 676 = 4x² + 4(x + 7)²

⇒ 676 = 4x² + 4(x² + 49 + 14x)

⇒ 676 = 4x² + 4x² + 196 + 56x

⇒ 8x² + 196 + 56x - 676 = 0

⇒ 8x² + 56x - 480 = 0

⇒ 8(x² + 7x - 60) = 0

⇒ x² + 7x - 60 = 0

⇒ x² + 12x - 5x - 60 = 0

⇒ x(x + 12) - 5(x + 12) = 0

⇒ (x - 5)(x + 12) = 0

⇒ x = 5 or x = -12.

Since, length cannot be negative.

x = 5

Distance traveled via city C = 2x + 2(x + 7) = 2x + 2x + 14 = 4x + 14 = 4 × 5 + 14 = 20 + 14 = 34 km.

Difference in distance traveled via city C and after construction of highway = 34 - 26 = 8 km.

Hence, 8 km distance will be saved in reaching city B from city A after the construction of the highway.

A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

Answer

First ladder reaches at A, thus AB = 4 m and AC = 5 m.

Given,

Foot of the ladder is moved 1.6 m towards the wall

The distance by which the top of the ladder would slide upwards on the wall = AE

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AC² = AB² + BC²

⇒ 5² = 4² + BC²

⇒ 25 = 16 + BC²

⇒ BC² = 25 - 16

⇒ BC² = 9

⇒ BC = $\sqrt{9}$

⇒ BC = 3 m

BD = BC - CD = 3 - 1.6 = 1.4 m

In triangle EBD,

Ladder ED = 5 m

⇒ ED² = EB² + BD²

⇒ 5² = EB² + (1.4)²

⇒ 25 = EB² + 1.96

⇒ EB² = 25 - 1.96

⇒ EB² = 23.04

⇒ EB = $\sqrt{23.04}$

⇒ EB = 4.8 m

From figure,

AE = EB - AB = 4.8 - 4 = 0.8 m

Hence, the distance by which the top of the ladder would slide upwards on the wall is 0.8 m.

The diagram shows a nest of 4 squares set one within another. The side of the outer most square is 20 cm. The midpoints of the sides are joined to give a second square, and the process is repeated to give the third and fourth squares. Find the length of a side of the smallest square.

Answer

Given, squares are formed joining mid-points of square ABCD.

AE = AH = $\dfrac{1}{2}$ × AB = $\dfrac{1}{2}$ × 20 = 10 cm.

In square ABCD,

∠A = 90°

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle AEH,

⇒ EH² = AE² + AH²

⇒ EH² = 10² + 10²

⇒ EH² = 100 + 100

⇒ EH² = 200

⇒ EH = $\sqrt{200} = \sqrt{2 \times 100}$

⇒ EH = $10\sqrt{2}$ cm.

In square EFGH,

∠E = 90°

EM = EN = $\dfrac{1}{2}$ × EH = $\dfrac{1}{2} \times 10 \sqrt{2} = 5\sqrt{2} \text{ cm.}$

By Pythagoras theorem,

In triangle EMN,

⇒ MN² = EN² + EM²

⇒ MN² = $(5\sqrt{2})^2 + (5\sqrt{2})^2$

⇒ MN² = 50 + 50

⇒ MN² = 100

⇒ MN = $\sqrt{100}$

⇒ MN = 10 cm.

In square MNOP,

∠M = 90°

IM = ML = $\dfrac{1}{2} × MN = \dfrac{1}{2} \times 10 = 5 \text{ cm.}$

By Pythagoras theorem,

In triangle IML,

⇒ IL² = IM² + ML²

⇒ IL² = 5² + 5²

⇒ IL² = 25 + 25

⇒ IL² = 50

⇒ IL = $\sqrt{25 \times 2}$

⇒ IL = $5\sqrt{2}$ cm.

Hence, the length of a side of the smallest square is $5\sqrt{2}$ cm.