RS Aggarwal Solutions for Class 7 Maths Chapter 15 – Properties of Triangles are available here. Maths is a subject which requires understanding and reasoning skills accompanied by logic. Along with this, it also requires students to practice Maths regularly. Students of Class 7 are suggested to solveÂ RS Aggarwal Class 7 Solutions to strengthen their fundamentals and be able to solve questions that are usually asked in the examination.

The primary aim is to help students understand and crack these problems. We at BYJU’S have prepared the RS Aggarwal Solutions for Class 7 Maths Chapter 15 wherein problems are solved step by step with in-depth explanations. Download pdf of Class 7 Chapter 15 in their respective links.

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### Also, access RS Aggarwal Solutions for Class 7 Chapter 15 Exercises

### Exercise 15A Page: 183

**1. In a Î”ABC, if âˆ A = 72 ^{o} and âˆ B = 63^{o}, find âˆ C.**

**Solution:-**

We know that the sum of the angles of a triangle is 180^{o}.

âˆ´ âˆ A + âˆ B + âˆ C = 180^{o}

= 72^{o}+ 63^{o}+ âˆ C = 180^{o}

= 135^{o} + âˆ C = 180^{o}

= âˆ C = 180^{o}– 135^{0}

= âˆ C = 45^{o}

Hence, the measures of âˆ C is 45^{o}.

**2. In a Î”DEF, if âˆ E = 105 ^{o} and âˆ F = 40^{o}, find âˆ D.**

**Solution:-**

We know that the sum of the angles of a triangle is 180^{o}.

âˆ´ âˆ D + âˆ E + âˆ F = 180^{o}

= âˆ D + 105^{o}+ 40^{o} = 180^{o}

= 145^{o} + âˆ D = 180^{o}

= âˆ D = 180^{o}– 145^{0}

= âˆ D = 35^{o}

Hence, the measures of âˆ D is 35^{o}.

**3. In a Î”XYZ, if âˆ X = 90 ^{o} and âˆ Z = 48^{o}, find âˆ Y.**

**Solution:-**

We know that the sum of the angles of a triangle is 180^{o}.

âˆ´ âˆ X + âˆ Y + âˆ Z = 180^{o}

= 90^{o} + âˆ Y + 48^{o} = 180^{o}

= 138^{o} + âˆ Y = 180^{o}

= âˆ Y = 180^{o}– 138^{0}

= âˆ Y = 42^{o}

Hence, the measures of âˆ Y is 42^{o}.

**4. Find the angles of a triangle which are in the ratio 4: 3: 2.**

**Solution:-**

Let the measures of the given angles of the triangles be (4x)^{o}, (3x)^{o }and (2x)^{o} respectively.

Then,

= 4x + 3x + 2x = 180^{o} â€¦ [âˆµsum of the angles of a triangle is 180^{o}]

= 9x = 180^{o}

= x = 180/9

= x = 20

So, the angle measures (4 Ã— 20)^{o}, (3 Ã— 20)^{o}, (2 Ã— 20)^{o},

i.e., 80^{o}, 60^{o}, 40^{o}.

Hence, the angles of the triangles are 80^{o}, 60^{o}, 40^{o}.

**5. One of the acute angles of right triangle is 36 ^{o}, find the other.**

**Solution:-**

Let the three angles be âˆ A, âˆ B, âˆ C

And,

âˆ A = 36^{o}

âˆ B = 90^{o}

^{o}]

âˆ C = x

Now,

= âˆ A + âˆ B + âˆ C = 180^{o} â€¦ [âˆµsum of the angles of a triangle is 180^{o}]

= 36^{o} + 90^{o} + âˆ C = 180^{o}

= 126^{o} + âˆ C = 180^{0}

= âˆ C = 180^{o} â€“ 126^{o}

= âˆ C = 54^{o}

Hence, the other angle is 54^{o}.

**6. The acute angle of a right triangle are in the ratio 2: 1. Find each of these angles.**

**Solution:-**

Let the three angles be âˆ A, âˆ B, âˆ C

And,

âˆ A = 2x^{o}

âˆ B = 1x^{o}

âˆ C = 90^{o}

^{o}]

Now,

= âˆ A + âˆ B + âˆ C = 180^{o} â€¦ [âˆµsum of the angles of a triangle is 180^{o}]

= 2x^{o} + x^{o} + 90^{o} = 180^{o}

= 3x^{o} = 180^{0}– 90^{o}

= 3x^{o} = 90^{o}

= x^{o} = 90^{o}/3

= x^{o} = 30

Now, substitute the value of x in the given angles.

âˆ A = 2x^{o} = 2 Ã— 30 = 60^{o}

âˆ B = 1x^{o}= 1 Ã— 60 = 30^{o}

Hence, the other angles are 60^{o} and 30^{o}.

**7. One of the angles of a triangle is 100 ^{o} and the other two angles are equal. Find each of the equal angles.**

**Solution:-**

Let the other two equal angles be x.

Then,

= x + x + 100^{o} = 180^{o} â€¦ [âˆµsum of the angles of a triangle is 180^{o}]

= 2x = 180^{o} â€“ 100^{o}

= 2x = 80^{o}

= x = 80^{o}/2

= x = 40^{o}

Hence, the other two equal angles are 40^{o} and 40^{o}.

Exercise 15B Page: 186

**1. In the figure given alongside, find the measure of âˆ ACD.**

**Solution:-**

In the given figure, side BC of Î”ABC is produced to D.

Consider the Î”ABC,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

âˆ´ âˆ ACD = âˆ ABC + âˆ BAC

âˆ ACD = 45^{o} + 75^{o}

âˆ ACD = 120^{o}

Hence, the measures of âˆ ACD is 120^{o}.

**2. In the figure given alongside, find the values of x and y.**

**Solution:-**

In the given figure, side BC of Î”ABC is produced to D.

Consider the Î”ABC,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

âˆ´ âˆ ABC + âˆ BAC = âˆ ACD

= 68^{o} + x = 130^{o}

= x = 130^{o} â€“ 68^{o}

= x = 62^{o}

Also, we know that the sum of all the angles of a triangles is 180^{o}.

âˆ´ x + y + 68^{o} = 180^{o}

= 62^{o} + y + 68^{o} = 180^{o}

= y + 130^{o} = 180^{o}

= y = 180^{o} â€“ 130^{o}

= y = 50^{o}

Hence, the value of x is 62^{o} and value of y is 50^{o}.

**3. In the figure given alongside, find the values of x and y.**

**Solution:-**

In the given figure, side BC of Î”ABC is produced to D.

Consider the Î”ABC,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

âˆ´ âˆ ABC + âˆ BAC = âˆ ACD

= x^{o} + 32^{o} = 65^{o}

= x = 65^{o} â€“ 32^{o}

= x = 33^{o}

Also, we know that the sum of all the angles of a triangles is 180^{o}.

âˆ´ A + B + C = 180^{o}

= 32^{o} + x^{o} + y^{o} = 180^{o}

= 32^{o} + 33^{o} + y^{o} = 180^{o}

= y + 65^{o} = 180^{o}

= y = 180^{o} â€“ 65^{o}

= y = 115^{o}

Hence, the value of x is 33^{o} and value of y is 115^{o}.

**4. An exterior angle of a triangle measures 110 ^{o} and its interior opposite angles are in the ratio 2: 3. Find the angles of the triangle.**

**Solution:-**

Let the given interior opposite angles be (2x)^{o} and (3x)^{o}.

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

âˆ´ 2x + 3x = 110^{o}

= 5x = 110

= x =110/5

= x = 22

âˆ´ âˆ A = 2x = 2 Ã— 22 = 44^{o}

âˆ B = 3x = 3 Ã— 22 = 66^{o}

But, âˆ A + âˆ B + âˆ C = 180^{o}

âˆ´ 44^{o} + 66^{o}+ âˆ C = 180^{o}

= 110 + âˆ C = 180^{o}

= âˆ C = 180 â€“ 110

= âˆ C = 70^{o}

âˆ´âˆ A = 44^{o}, âˆ B = 66^{o}, âˆ C = 70^{o}

Exercise 15C Page: 188

**1. Is it possible to draw a triangle, the lengths of whose sides are given below?**

**(i). 1 cm, 1cm, 1cm **

**Solution:-**

Consider the number1, 1, 1

It is clear that the sum of any two of these numbers is greater than the third.

Hence, it is possible to draw a triangle whose sides are 1cm, 1cm and 1cm.

**(ii). 2 cm, 3 cm, 4 cm **

**Solution:-**

Clearly, we have:

(2 + 3) > 4

(3 + 4) > 2

(2 + 4) > 3

Thus, the sum of any two of these numbers is greater than the third.

Hence, it is possible to draw a triangle whose sides are 2 cm, 3 cm and 4 cm.

**(iii). 7 cm, 8 cm, 15 cm **

**Solution:-**

Clearly, we have:

(7 + 8) = 15

Thus, the sum of any two of these numbers is not greater than the third.

Hence, it is not possible to draw a triangle whose sides are 7 cm, 8 cm and 15 cm.

**(iv). 3.4 cm, 2.1 cm, 5.3 cm **

**Solution:-**

Clearly, we have:

(3.4 + 2.1) > 5.3

(2.1 + 5.3) > 3.4

(3.4 + 5.3) > 2.1

Thus, the sum of any two of these numbers is greater than the third.

Hence, it is possible to draw a triangle whose sides are 3.4 cm, 2.1 cm and 5.3 cm.

**(iv). 6 cm, 7 cm, 14 cm **

**Solution:-**

Clearly, we have:

(6 + 7) < 14

Thus, the sum of any two of these numbers is less than the third.

Hence, it is not possible to draw a triangle whose sides are 6 cm, 7 cm and 14 cm.

**2. Two sides of a triangle are 5 cm and 9 cm long. What can be the length of its third side?**

**Solution:-**

Let us Assume the length of the third side be x.

Then,

(5 + 9) > x

âˆ´the length of its third side is less than 14 cm

**3. If P is appoint in the interior of Î”ABC then fill in the blanks with > or < or =.**

**(i). PA + PBâ€¦â€¦. AB**

**Solution:-**

PA + PB > AB

Because, the sum of any two side of triangle is greater than the third side.

**(ii). PB + PCâ€¦â€¦ BC**

**Solution:-**

PB + PC > BC

Because, the sum of any two side of triangle is greater than the third side.

**(iii). AC â€¦â€¦ PA + PC**

**Solution:-**

AC < PA + PC

Because, the sum of any two side of triangle is greater than the third side.

Exercise 15D Page: 193

**1. Find the length of the hypotenuse of a right triangle, the other two sides of which measures 9 cm and 12 cm.**

**Solution:-**

Let Î”ABC be right angled at B.

Let AB = 12 cm and BC = 9cm,

Hypotenuse (AC) =?

Then, by Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

AC^{2 }= 12^{2} + 9^{2}

AC^{2 }= 144 + 81

AC^{2} = 225

Sending power2 from LHS to RHS it becomes square root

AC = âˆš(225)

AC = 15 cm

âˆ´The length of the hypotenuse of a triangle is 15 cm.

**2. The hypotenuse of a right triangle is 26 cm long. If one of the remaining two sides is 10 cm long, find the length of the other side.**

**Solution:-**

Let Î”ABC be right angled at B.

Let AB = 10 cm

Hypotenuse (AC) = 26 cm

BC =?

Then, by Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

BC^{2 }= (AC^{2} – AB^{2})

BC^{2 }= (26^{2} â€“ 10^{2})

BC^{2} = (676 â€“ 100)

BC^{2} = 576

Sending power2 from LHS to RHS it becomes square root

BC = âˆš(576)

BC = 24 cm

âˆ´The length of other the side of a triangle is 24 cm.

**3. The length of one side of a right triangle is 4.5cm and the length of its hypotenuse is 7.5 cm. Find the length of its third side. **

**Solution:-**

From the question,

Let Î”ABC be right angled at B.

Let AB = 4.5 cm

Hypotenuse (AC) = 7.5 cm

BC = x

Then, by Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

BC^{2 }= (AC^{2} – AB^{2})

x^{2 }= (7.5^{2} â€“ 4.5^{2})

x^{2} = (56.25 â€“ 20.25)

x^{2} = 36

Sending power2 from LHS to RHS it becomes square root

x = âˆš(36)

x = 6 cm

âˆ´The length of other the side of a triangle is 6 cm.

**4. The two legs of a right triangle are equal and the square of its hypotenuse is 50. Find the length of its third side.**

**Solution:-**

Let the two legs of a right triangle be x.

Then,

= x^{2} + x^{2 }= 50

= 2x^{2} = 50

= x^{2 }= (50/2)

= x^{2} = 25

Sending power2 from LHS to RHS it becomes square root

= x = âˆš(25)

= x = 5

âˆ´The length of two legs of a right triangle is 5 cm.

**5. The sides of a triangle is measures 15 cm. 36 cm and 39 cm. Show that it is a right-angled triangle.**

**Solution:-**

Let us assume the largest value is the hypotenuse side i.e. 39 cm.

Then, by Pythagoras theorem,

= 39^{2} = 36^{2} + 15^{2}

= 1521 = 1296 + 225

= 1521 = 1521

The sum of square of two side of triangle is equal to the square of third side,

âˆ´The given triangle is right â€“angled triangle.

**6. In right Î”ABC, the lengths of its legs are given as a = 6 cm and b = 4.5 cm. Find the length of its hypotenuse.**

**Solution:-**

Let Î”ABC be right angled at B.

Let AB = 10 cm

Hypotenuse (AC) = 26 cm

BC =?

Then, by Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2}

c^{2 }= (a^{2} + b^{2})

c^{2 }= (6^{2} + 4.5^{2})

c^{2} = (36 + 20.25)

c^{2} = 56.25

Sending power2 from LHS to RHS it becomes square root

c = âˆš(56.25)

c = 7.5 cm

âˆ´The length of hypotenuse of a triangle is 7.5 cm.

**7. The length of the sides of some triangles are given below. Which of them are right-angled?**

**(i). a = 15 cm, b = 20 cm and c = 25 cm**

**Solution:-**

Let us assume the largest value is the hypotenuse side i.e. c = 25 cm.

Then, by Pythagoras theorem,

= c^{2} = a^{2} + b^{2}

= 25^{2} = 15^{2} + 20^{2}

= 625 = 225 + 400

= 625 = 625

The sum of square of two side of triangle is equal to the square of third side,

âˆ´The given triangle is right-angled triangle.

**(ii). a = 9 cm, b = 12 cm and c = 16 cm**

**Solution:-**

Let us assume the largest value is the hypotenuse side i.e. c = 16 cm.

Then, by Pythagoras theorem,

= c^{2} = a^{2} + b^{2}

= 16^{2} = 9^{2} + 12^{2}

= 256 = 81 + 144

= 256 â‰ 225

The sum of square of two side of triangle is not equal to the square of third side,

âˆ´The given triangle is not right â€“angled triangle.

**(iii). a = 10 cm, b = 24 cm and c = 26 cm**

**Solution:-**

Let us assume the largest value is the hypotenuse side i.e. c = 26 cm.

Then, by Pythagoras theorem,

= c^{2} = a^{2} + b^{2}

= 26^{2} = 10^{2} + 24^{2}

= 676 = 100 + 576

= 676 = 676

The sum of square of two side of triangle is equal to the square of third side,

âˆ´The given triangle is right-angled triangle.

**8. In a Î”ABC, âˆ B = 35 ^{o} and âˆ C = 55^{o}. Write which of the following is true:**

**(i). AC ^{2} = AB^{2} + BC^{2}**

**(ii). AB ^{2} = BC^{2} + AC^{2}**

**(iii). BC ^{2} = AB^{2} + AC^{2}**

**Solution:-**

Given that âˆ B = 35^{o}, âˆ C = 55^{o}

Then, âˆ A =?

We know that sum of the angle of three sides of triangle is equal to 180^{o}.

= âˆ A + âˆ B + âˆ C = 180^{o}

= âˆ A + 35^{o} + 55^{o} = 180^{o}

= âˆ A + 90^{o} = 180^{o}

= âˆ A = 180 â€“ 90

= âˆ A = 90^{o}

Also, we know that side opposite to the right angle is the hypotenuse.

âˆ´ BC^{2} = AB^{2} + AC^{2}

Hence, (iii) is true

**9. A 15-m-long ladder is placed against a wall to reach a window 12 m high. Find the distance of the foot of the ladder from the wall.**

**Solution:-**

Let BC be the wall and AB be the ladder.

Then, AB = 15 m and BC = 12 m.

Now, Î”ABC being right-angled at C, we have:

AB^{2} = BC^{2} + AC^{2}

AC^{2 }= (AB^{2} – BC^{2})

AC^{2 }= (15^{2} â€“ 12^{2})

AC^{2} = (225 â€“ 144)

AC^{2} = (81)

Sending power2 from LHS to RHS it becomes square root

AC = âˆš(81)

AC = 9 cm

âˆ´The distance of the foot of the ladder from the wall is 9 cm.

## RS Aggarwal Solutions for Class 7 Maths Chapter 15 – Properties of Triangles

Chapter 15 – Properties of Triangles contains 4 exercises and the RS Aggarwal Solutions available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.

- Various Types of Triangles
- Naming Triangles by Considering their Angles
- Angle Sum Property of a Triangle
- Exterior and Interior Opposite Angles
- Exterior Angles Property of a Triangle
- Triangle Inequality
- Pythagoras Theorem

### Chapter Brief of RS Aggarwal Solutions for Class 7 Maths Chapter 15 – Properties of Triangles

RS Aggarwal Solutions for Class 7 Maths Chapter 15 – Properties of Triangles. In Class 6, students learnt the definition of a triangle and various types of triangles based on angles and sides. Now, in this chapter review these results and study some more properties of triangles.