NCERT Solutions Class 10 Maths Chapter 6 Triangles

Question

NCERT Solutions for Class 10 Maths Chapter 6 Triangles are provided here which is one of the most important study materials for the students preparing for the CBSE board examination. Our NCERT Solutions are based on the latest syllabus of the CBSE. Prepared and designed by subject matter experts these solutions are on to the point and explained in a point-wise method to make it easier for the students to learn, study, and last minute revision. We have also covered all the shortcuts, equations, rules, theorems, and axioms.

In the NCERT Solution Class 10 provided by us, we have explained all the topics in detail which will surely help students clear all the required concepts. Important topics discussed here are

• Pythagoras theorem – Pythagoras theorem states that “In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of squares of the other two sides.
• Similar figures – Two triangles are said to be similar when they have the same shape, irrespective of their size.
• Congruent figures – two triangles are said to be congruent if and only if they are an exact replica of each other, same size and same shape.
• Congruency and Similarity of triangles and their differences.
• Basic Proportionality Theorem (Thales Theorem) – It provides the relationship between the sides of two equiangular triangles.
• Angle-Angle-Angle (AAA) similarity – Two triangles that have all the angles equal to each other have AAA similarity.
• Comparison between AA and AAA, are they the same?
• Side-Side-Side (SSS) similarity – Two triangles that have all the sides equal to each other have SSS similarity.
• Side-Angle-Side (SAS) similarity – Two triangles that have two sides and the angle between them equal to each other are said to have SAS similarity.
• Ratios of areas of two similar triangles – In two similar triangles ratio of the area of two triangles is proportional to the square of the ratio of their corresponding sides.
• Perpendicular bisector of the right angle in a right angle triangle created who similar triangle in which smaller triangles are also similar triangles the larger triangle.

Our NCERT Solutions Class 10 Maths is the best resource to practice and learn about all the important concepts. Regular practice of our solutions will help students develop problem-solving skills related to the subject matter. We have solutions for all kinds of queries including exercise and in-text questions.

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Answer 1

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

Answer:

Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.

By applying Pythagoras Theorem in ∆DEA, we get,

DE² + EA² = DA² …………… (i)

By applying Pythagoras Theorem in ∆DEB, we get,

DE² + EB² = DB²

DE² + (EA + AB) ² = DB²

(DE² + EA²) + AB² + 2EA × AB = DB²

DA² + AB² + 2EA × AB = DB² ……………. (ii)

By applying Pythagoras Theorem in ∆ADF, we get,

AD² = AF² + FD²

Again, applying Pythagoras theorem in ∆AFC, we get,

AC² = AF² + FC² = AF² + (DC − FD) ²

= AF² + DC² + FD² − 2DC × FD

= (AF² + FD²) + DC² − 2DC × FD AC²

AC²= AD² + DC² − 2DC × FD …………… (iii)

Since ABCD is a parallelogram,

AB = CD ………….…(iv)

And BC = AD …………. (v)

In ∆DEA and ∆ADF,

∠DEA = ∠AFD (Each 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common Angles)

∴ ∆EAD ≅ ∆FDA (AAS congruence criterion)

⇒ EA = DF …………… (vi)

Adding equations (i) and (iii), we get,

DA² + AB² + 2EA × AB + AD² + DC² − 2DC × FD = DB² + AC²

DA² + AB² + AD² + DC² + 2EA × AB − 2DC × FD = DB² + AC²

From equation (iv) and (vi),

BC² + AB² + AD² + DC² + 2EA × AB − 2AB × EA = DB² + AC²

AB² + BC² + CD² + DA² = AC² + BD²

62. In Figure, two chords AB and CD intersect each other at the point P. Prove that :

(i) ∆APC ~ ∆ DPB

(ii) AP . PB = CP . DP

Answer:

Firstly, let us join CB, in the given figure.

(i) In ∆APC and ∆DPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

Therefore,

∆APC ∼ ∆DPB (AA similarity criterion)

(ii) In the above, we have proved that ∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.

∴ AP/DP = PC/PB = CA/BD

⇒AP/DP = PC/PB

∴AP. PB = PC. DP

Hence, proved.

63. In Figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(i) ∆ PAC ~ ∆ PDB

(ii) PA . PB = PC . PD.

Answer:

(i) In ∆PAC and ∆PDB,

∠P = ∠P (Common Angles)

As we know, exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is opposite interior angle, which are both equal.

∠PAC = ∠PDB

Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)

(ii) We have already proved above,

∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.

Therefore,

AP/DP = PC/PB = CA/BD

AP/DP = PC/PB

∴ AP. PB = PC. DP

64. In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC.

Answer:

In the given figure, let us extend BA to P such that;

AP = AC.

Now join PC.

Given, BD/CD = AB/AC

⇒ BD/CD = AP/AC

By using the converse of basic proportionality theorem, we get,

AD || PC

∠BAD = ∠APC (Corresponding angles) …………….. (i)

And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)

By the new figure, we have;

AP = AC

⇒ ∠APC = ∠ACP ………………. (iii)

On comparing equations (i), (ii), and (iii), we get,

∠BAD = ∠APC

Therefore, AD is the bisector of the angle BAC.

Hence, proved.

65. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Answer:

Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the horizontal distance of the fly from the tip of the fishing rod. 

Therefore, AC is now the length of the string.

To find AC, we have to use Pythagoras theorem in ∆ABC, is such way;

AC² = AB²+ BC²

AB² = (1.8 m) ² + (2.4 m) ²

AB² = (3.24 + 5.76) m²

AB² = 9.00 m²

⟹ AB = √9 m = 3m

Thus, the length of the string out is 3 m.

As its given, she pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m

Let us say now, the fly is at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds

= (3.00 − 0.6) m

= 2.4 m

In ∆ADB, by Pythagoras Theorem,

AB² + BD² = AD²

(1.8 m) ² + BD² = (2.4 m) ²

BD² = (5.76 − 3.24) m² = 2.52 m²

BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m = 2.787 m

= 2.79 m