CBSE Class X

Question 16:

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer:

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

Applying Pythagoras theorem in ΔABE, we obtain

AB² = AE² + BE²

Question 15:

In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9 AD² = 7 AB².

Answer:

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

And, AE = a√3/2

Given that, BD = 1/3 BC

∴ BD = a/3

DE = BE − BD = (a/2) – (a/3) = a/6

Applying Pythagoras theorem in ΔADE, we obtain

AD² = AE² + DE²

Question 14:

The perpendicular from A on side BC of a ;ABC intersect BC at D such that DB = 3 CD. Prove that 2 AB2 =2 AC² + BC²

Answer:

Applying Pythagoras theorem for ΔACD, we obtain

AC² = AD² + DC²

AD² = AC² – DC²     (1)

Applying Pythagoras theorem in ∆ABD, we obtain

AB² = AD² + DB²

AD² = AB² – DB²       (2)

From equation (1) and (2), we obtain

AC² – DC² = AB² – DB²    (3)

It is given that 3DC = DB

∴DC= BC/4 and DB = 3BC/4

Putting these values in equation (3), we obtain

Question 13:

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE²

Answer:

Applying Pythagoras theorem in ΔACE, we obtain

AC² + CE² = AE²       (1)

Applying Pythagoras theorem in ∆BCD, we obtain

BC² + CD² = BD²     (2)

Using equation (1) and equation (2), we obtain

AC² + CE² + BC² + CD² = AE² + BD²    (3)

Applying Pythagoras theorem in ∆CDE, we obtain

DE² = CD² + CE²

Applying Pythagoras theorem in ∆ABC, we obtain

AB² = AC² + CB²

Putting the values in equation (3), we obtain

DE² + AB² = AE² + BD²

Question 12:

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

Answer:

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

Therefore, CP = 11 − 6 = 5 m

From the figure, it can be observed that AP = 12m

Applying Pythagoras theorem for ΔAPC, we obtain

AP² + PC² = AC²

(12 m)² + (5 m)²= AC²

AC² = (144 + 25) m² = 169 m²

AC = 13 m

Therefore, the distance between their tops is 13 m.

Question 11:

An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the twoplanes after  hours?

Answer:

Distance travelled by the plane flying towards north in

Similarly, distance travelled by the plane flying towards west in

Let these distances be represented by OA and OB respectively.

Applying Pythagoras theorem,

Distance between these planes after

Question 10:

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be
driven so that the wire will be taut?

Answer:

Let OB be the pole and AB be the wire.

By Pythagoras theorem,

AB² = OB² + OA²

(24 m)² = (18 m)² + OA²

OA² = (576 – 324) m² = 252 m²

Therefore, the distance from the base is 6√7 m.

Question 9:

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladderfrom base of the wall.

Answer:

Let OA be the wall and AB be the ladder.

Therefore, by Pythagoras theorem,

AB² = OA² + BO²

(10 m)² = (8 m)² + OB²

100 m² = 64 m² + OB²

OB² = 36 m²

OB = 6 m

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

Question 8:

In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA² + OB² + OC² − OD² − OE² − OF² = AF² + BD² + CE²

(ii) AF² + BD² + CE² = AE² + CD² + BF²

Answer:

Join OA, OB, and OC.

(i) Applying Pythagoras theorem in ∆AOF, we obtain

OA² = OF² + AF²

Similarly, in ∆BOD,

OB² = OD² + BD²

Similarly, in ∆COE,

OC² = OE² + EC²

Adding these equations,

OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²

OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + EC²

(ii) From the above result,

AF² + BD² + EC² = (OA² – OE²) + (OC² –OD²⁾ + (OB² –OF²)

∴ AF² + BD² + EC² = AE² + CD² + BF²

Question 7:

Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Answer:

In ∆AOB, ∆BOC, ∆COD, ∆AOD,

Applying Pythagoras theorem, we obtain

Question 6:

ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer:

Let AD be the altitude in the given equilateral triangle, ∆ABC.

We know that altitude bisects the opposite side.

∴ BD = DC = a

In ∆ADB,

∠ADB = 90°

Applying  pythagoras theorem, we obtain

AD² + DB² = AB²

⇒ AD² + a² = (2a)²

⇒AD² + a² = 4a²

⇒ AD² = 3a²

⇒AD = a√3

In an equilateral triangle, all the altitudes are equal in length.

Therefore, the length of each altitude will be √3a.

Question 5:

ABC is an isosceles triangle with AC = BC. If AB² = 2 AC², prove that ABC is a right triangle.

Answer:

Given that,

AB² = 2AC²

⇒ AB² = AC² + AC²

⇒ AB² = AC² + BC² (As AC = BC)

The triangle is satisfying the pythagoras theorem.

Therefore, the given triangle is a right – angled traingle.

Question 4:

ABC is an isosceles triangle right angled at C. prove that AB² = 2 AC².

Answer:

Given that ∆ABC is an isosceles triangle

∴ AC = CB

Applying Pythagoras theorem in ∆ABC (i.e., right-angled at point C), we obtain

AC² + CB² = AB²

⇒ AC² + AC² = AB²    (AC = CB)

⇒ 2AC² = AB²

Question 3:

In the following figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that

(i) AB² = BC × BD
(ii) AC² = BC × DC
(iii) AD² = BD × CD

Answer:

(i) In ∆ADB and ∆CAB,

∠DCA = ∠ DAB (Each 90º)

∠CDA = ∠ ADB (Common angle)

∴∆DCA ~ ∆DAB

⇒ DC/DA = DA/DB

⇒ AD² = BD × CD

Question 2:

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM × MR.

Answer:

Question 1:

Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle,write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Answer:

(i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of these sides, we will obtain 49, 576, and 625.

49 + 576 = 625

or, 7² + 24² = 25²

The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.

We know that the longest side of a right triangle is the hypotenuse.

Therefore, the length of the hypotenuse of this triangle is 25 cm.

(ii) It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will obtain 9, 64, and 36.

However, 9 + 36 ≠ 64

Or, 3² + 6² ≠ 8²

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iii)Given that sides are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50² + 80² ≠ 100²

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iv)Given that sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will obtain 169, 144, and 25.

Clearly, 144 +25 = 169

Or, 12² + 5² = 13²

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

We know that the longest side of a right triangle is the hypotenuse.

Therefore, the length of the hypotenuse of this triangle is 13 cm.

Question 9:

Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio

(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81

Answer:

If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.

It is given that the sides are in the ratio 4:9.

Therefore, ratio between areas of these triangles = (4/9)² = 16/81

Hence, the correct answer is (D)

Question 8:

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is

(A) 2 : 1        (B) 1 : 2

(C) 4 : 1        (D) 1 : 4

Answer:

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

Let side of ΔABC = x

Therefore, side of ΔBDE = x/2

Hence, the correct answer is (C).

Question 7:

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Answer:

Let ABCD be a square of side a.

Therefore, its diagonal = √2a

Two desired equilateral triangles are formed as ∆ABE and ∆DBF.

Side of an equilateral triangle, ∆ABE, described on one of its sides = a

Side of an equilateral triangle, ∆DBF, described on one of its diagonals = √2a

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

Question 6:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer:

Let us assume two similar triangles as ∆ABC ∼ ∆PQR. Let AD and PS be the medians of these triangles.

∵∆ABC ~ ∆PQR