NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Exercise 6.1

1. In Δ PQR, D is the mid-point of QR.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

PM is Altitude.

Answer: Altitude

Explanation: Altitude is the perpendicular line to the line QR. It makes a right angle. An altitude has one end point at a vertex of the triangle and the other on the line containing the opposite side.

PD is Median.

Answer: Median

Explanation: Median is the line segment that joins the mid-point of the line QR to its opposite vertex P. A median divides the line segment QR into two equal parts.

Here,

D is the midpoint of the line QR, which divides it into two equal parts.

QD = DR

Is QM = MR?

Answer: No

Explanation: D is the midpoint of the line QR, which divides it into two equal parts. M is not a mid-point. Hence, QM is not equal to MR.

Instead, we can say that:

QD = DR

2. Draw rough sketches for the following:

(a) In ΔABC, BE is a median.

Answer:

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Explanation: Median is the line segment that joins the mid-point of the line AC to its opposite vertex B

(b) In ΔPQR, PQ and PR are altitudes of the triangle.

Answer:

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Explanation: An altitude has one end point at a vertex of the triangle and the other on the line containing the opposite side.

(c) In ΔXYZ, YL is an altitude in the exterior of the triangle.

Answer:

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Explanation: An altitude makes a right angle with the intersecting line. Its one end point is connected to the vertex of the triangle (Y) and the other point connected to the opposite side.

3. Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.

Answer:

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Explanation: Yes, median and altitude of an isosceles triangle can be the same.

Isosceles triangle is the triangle with its two sides equal. Here, AD is the median as it divides the line segment BC into two equal parts. AD is also the altitude as it forms a right angle with the opposite side.

Exercise 6.2

1. Find the value of the unknown exterior angle x in the following diagrams:

(i)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 120°

Explanation: Here, x is the exterior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Exterior angle x = 50° + 70°

x = 50° + 70°

x = 120°

(ii)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 110°

Explanation: Here, x is the exterior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Exterior angle x = 65° + 45°

x = 65° + 45°

x = 110°

(iii)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 70°

Explanation: Here, x is the exterior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Exterior angle x = 30° + 40°

x = 30° + 40°

x = 70°

(iv)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 120°

Explanation: Here, x is the exterior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Exterior angle x = 60° + 60°

x = 60° + 60°

x = 120°

(v)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 100°

Explanation: Here, x is the exterior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Exterior angle x = 50° + 50°

x = 50° + 50°

x = 100°

(vi)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 90°

Explanation: Here, x is the exterior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Exterior angle x = 60° + 30°

x = 60° + 30°

x = 90°

2. Find the value of the unknown interior angle x in the following figures:

(i)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 65°

Explanation: Here, x is the interior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Sum of interior angles = Exterior angle

x + 50° = 115°

x = 115° - 50°

x = 65°

(ii)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 30°

Explanation: Here, x is the interior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Sum of interior angles = Exterior angle

x + 70° = 100°

x = 100° - 70°

x = 30°

(iii)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 35°

Explanation: Here, x is the interior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Sum of interior angles = Exterior angle

x + 90° = 125°

The other angle is the right angle.

x = 125° - 90°

x = 35°

(iv)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 60°

Explanation: Here, x is the interior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Sum of interior angles = Exterior angle

x + 60° = 120°

x = 120° - 60°

x = 60°

(v)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 50°

Explanation: Here, x is the interior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Sum of interior angles = Exterior angle

x + 30° = 80°

x = 80° - 30°

x = 50°

(vi)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 40°

Explanation: Here, x is the interior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Sum of interior angles = Exterior angle

x + 35° = 75°

x = 75° - 35°

x = 40°

Exercise 6.3

1. Find the value of the unknown x in the following diagrams:

(i)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 70°

Explanation: The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are 50°, 60°, and x.

So,

50° + 60° + x = 180°

110° + x = 180°

x = 180° - 110°

x = 70°

(ii)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 60°

Explanation: The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are 30°, 90° (right angle), and x.

So,

30° + 90° + x = 180°

120° + x = 180°

x = 180° - 120°

x = 60°

(iii)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 40°

Explanation: The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are 30°, 110°, and x.

So,

30° + 110° + x = 180°

140° + x = 180°

x = 180° - 140°

x = 40°

(iv)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 65°

Explanation: The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are 50°, x, and x.

So,

50° + x + x = 180°

50° + 2x = 180°

2x = 180° - 50°

2x = 130°

x = 130/2°

x = 65°

(v)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 60°

Explanation: The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are x, x, and x.

So,

x + x + x = 180°

3x = 180°

x = 130/3°

x = 60°

Thus, each angle of the triangle is equal to 60°.

(vi)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 30°

Explanation: The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are 2x, 90° (right angle), and x.

So,

2x + 90° + x = 180°

90° + 3x = 180°

3x = 180° - 90°

3x = 90°

x = 90/3°

x = 30°

2. Find the values of the unknowns x and y in the following diagrams:

(i)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: x = 70°, y = 60°

Explanation:

120° is the exterior angle of the triangle.

Exterior angle is equal to the sum of the opposite interior angles.

Sum of interior angles = Exterior angle

x + 50° = 120°

x = 120° - 50°

x = 70°

The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are x, 50° (right angle), and y.

So,

70° + 50° + y = 180°

120° + y = 180°

y = 180° - 120°

y = 60°

Or

Y and the exterior angle 120° form a linear pair.

So,

y + 120° = 180°

y = 180° - 120°

y = 60°

(ii)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: x = 50°, y = 80°

Explanation: Y is the vertically opposite angle to the 80°. The vertically opposite angles are equal.

So,

y = 80°

The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are x, 50° (right angle), and y.

So,

80° + 50° + x = 180°

130° + x = 180°

x = 180° - 130°

x = 50°

(iii)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: x = 110°, y = 70°

Explanation: 50° and 60° are the interior angles of the triangle. X is the exterior angle.

Exterior angle is equal to the sum of the opposite interior angles.

Exterior angle = Sum of interior angles

Exterior angle = 60° + 50°

Exterior angle = 110°

x = 110°

The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are 60°, 50° (right angle), and y.

So,

60° + 50° + y = 180°

110° + y = 180°

y = 180° - 110°

y = 70°

Or

Y and the exterior angle 110° form a linear pair.

So,

y + 110° = 180°

y = 180° - 110°

y = 70°

(iv)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: x = 60°, y = 90°

Explanation: x is the vertically opposite angle to the 60°. The vertically opposite angles are equal.

So,

x = 60°

The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are x, 30° (right angle), and y.

So,

60° + 30° + y = 180°

90° + y = 180°

y = 180° - 90°

y =90°

(v)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: x = 45°, y = 90°

Explanation: Y is the vertically opposite angle to the 90°. The vertically opposite angles are equal.

So,

y = 90°

The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are x, x (right angle), and y.

So,

x + 90° + x = 180°

90° + 2x = 180°

2x = 180° - 90°

2x = 90°

x = 90°/2

x = 45°

(vi)

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: x = 60°, y = 60°

Explanation: Y is the vertically opposite angle to the x, The vertically opposite angles are equal.

So,

y = x

The other angles opposite to the vertically opposite angle will also become x.

The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are x, x (right angle), and y.

(y = x)

So,

x + x + x = 180°

3x = 180°

x = 180°/3

x = 60°

y = x

So,

y = 60°

Exercise 6.4

1. Is it possible to have a triangle with the following sides?

(i) 2 cm, 3 cm, 5 cm

Answer: Not possible

Explanation: For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

Let' check.

2 + 3 > 5

No

3 + 5 > 2

Yes

2 + 5 > 3

Yes

Therefore, the triangle with the above dimensions is not possible. It is because the sum of the two sides (2 cm and 3 cm) is not greater than the third side (5 cm).

(ii) 3 cm, 6 cm, 7 cm

Answer: Possible

Explanation: For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

Let' check.

6 + 3 > 5

Yes

3 + 7 > 6

Yes

6 + 7 > 3

Yes

Therefore, the triangle with the above dimensions is possible.

(iii) 6 cm, 3 cm, 2 cm

Answer: Not possible

Explanation: For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

Let' check.

6 + 3 > 2

Yes

3 + 2 > 6

No

2 + 6 > 3

Yes

Therefore, the triangle with the above dimensions is not possible. It is because the sum of the two sides (2 cm and 3 cm) is not greater than the third side (6 cm).

2. Take any point O in the interior of a triangle PQR. Is

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

(i) OP + OQ > PQ?

Answer: Yes

Explanation: Let's connect the point 0 to the three vertices of the triangle.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

POQ forms a triangle.

For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

It means,

OP + OQ > PQ

Hence, the above condition is possible.

(ii) OQ + OR > QR?

Answer: Yes

Explanation: Let's connect the point 0 to the three vertices of the triangle.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

ROQ forms a triangle.

For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

It means,

OQ + OR > QR

Hence, the above condition is possible.

(iii) OR + OP > RP?

Answer: Yes

Explanation: Let's connect the point 0 to the three vertices of the triangle.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

POR forms a triangle.

For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

It means,

OR + OP > RP

Hence, the above condition is possible.

3. AM is a median of a triangle ABC.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Is AB + BC + CA > 2 AM? (Consider the sides of triangles ΔABM and ΔAMC.)

Answer: Yes

Explanation: AM divides the triangle ABC into two triangles.

For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

It means,

AB + BM > AM ... (1)

AC + CM > AM ... (2)

Adding equation 1 and 2, we get:

AB + BC + BM + CM > 2 AM

CA = BM + CM

Substituting the value of CM,

AB + BC + CA > 2 AM

Hence, the above condition is satisfied.

4. ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: Yes

Explanation: The two diagonal divide the quadrilateral into four triangles.

For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

Let the midpoint be O.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

For Δ ABC,

AB + BC > AC ... (1)

For Δ BCD,

BC + CD > BD ... (2)

For Δ CAD,

CD + AD > AC ... (3)

For Δ ABD,

AB + AD > BD ... (4)

Adding equations 1, 2, 3, and 4, we get:

2 (AB + BC + CD + AD) > 2 (AC + BD)

AB + BC + CD + AD > AC + BD

Hence, the above condition is satisfied.

5. ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?

Answer: Yes

Explanation: The two diagonal divide the quadrilateral into four triangles.

For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

Let the midpoint be O.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

For Δ AOB,

AO + OB > AB ... (1)

For Δ BOC,

BO + OC > BC ... (2)

For Δ COD,

CO + OD > CD ... (3)

For Δ AOD,

AO + OD > AD ... (4)

Adding equations 1, 2, 3, and 4, we get:

AO + OB + BO + OC + CO + OD + AO + OD > AB + BC + CD + AD

(AO + OC = AC)

(OB + OD = AD)

= 2 (AC + AD) > AB + BC + CD + AD

Or

AB + BC + CD + AD < 2 (AC + AD)

Hence, the above condition is satisfied.

6. The lengths of two sides of a triangle are 12 cm and 15 cm. between what two measures should the length of the third side fall?

Answer: 3 and 7

Explanation: For the triangle to exist, the sum of any two sides of a triangle should be greater than its third side.

The range of the third side of triangle falls between the sum of its two sides and the difference of its two sides.

Sum = 12 cm + 15 cm = 27 cm

Difference = 15 cm - 12 cm = 3 cm

Thus, the length of the third side of the triangle falls between 3 cm and 27 cm.

Exercise 6.5

1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 26 cm

Explanation: According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

Here,

QR = Hypotenuse

So,

(QR)2 = (PQ)2 + (PR)2

(QR)2 = (10)2 + (24)2

(QR)2 = 100 + 576

(QR)2 = 676

QR = 26

Thus, the length of the QR is 26 cm.

2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Answer: 24 cm

Explanation: The right angle triangle is shown below:

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

Here,

AB = Hypotenuse

So,

(AB)2 = (AC)2 + (BC)2

(25)2 = (7)2 + (BC)2

(BC)2 = (25)2 - (7)2

(BC)2 = 625 - 49

(BC)2 = 576

BC = 24

Thus, the length of the BC is 24 cm.

3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Answer: 9 m

Explanation: The given figure is a right angle triangle.

According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

Here,

15 m = Hypotenuse

So,

(15)2 = (12)2 + (a)2

(a)2 = (15)2 - (12)2

(a)2 = 225 - 144

(a)2 = 81

a = 9

Thus, the length of the side 'a' is 9 m.

4. Which of the following can be the sides of a right triangle?

(i) 2.5 cm, 6.5 cm, 6 cm.

Answer: Yes, these can be the sides of a right triangle.

Explanation: According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

Or

Square of larger side = Sum of squares of the smaller side

Here,

6.5 cm = Hypotenuse (larger side)

2.5 cm and 6 cm are the smaller sides.

So,

(6.5)2 = (6)2 + (2.5)2

42.25 = 36 + 6.25

42.25 = 42.25

LHS = RHS

Hence, the given triangle is a right angle triangle.

(ii) 2 cm, 2 cm, 5 cm.

Answer: No, these cannot be the sides of a right triangle.

Explanation: According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

Or

Square of larger side = Sum of squares of the smaller side

Here,

5 cm = Hypotenuse (larger side)

2 cm and 2 cm are the smaller sides.

So,

(5)2 = (2)2 + (2)2

25 = 4 + 4

25 = 8

LHS is not equal to RHS

Hence, the given triangle is a not right angle triangle.

(iii) 1.5 cm, 2cm, 2.5 cm.

Answer: Yes, these can be the sides of a right triangle.

Explanation: According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

Or

Square of larger side = Sum of squares of the smaller side

Here,

2.5 cm = Hypotenuse (larger side)

1.5 cm and 2 cm are the smaller sides.

So,

(2.5)2 = (2)2 + (1.5)2

6.25 = 4 + 2.25

6.25 = 6.25

LHS = RHS

Hence, the given triangle is a right angle triangle.

5. 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. Find the original height of the tree.

Answer: 18 m

Explanation: The figure for the above question can be represented as:

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Total height of the tree = 5 m + Broken tree

Broken tree = Hypotenuse of the triangle

The triangle is a right angle triangle.

According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

(Hypotenuse)2 = (5)2 + (12)2

(Hypotenuse)2 = 25 + 144

(Hypotenuse)2 = 169

Hypotenuse = 13

Broken tree length = 13 m

Total height of the tree = 5m + Broken tree

Total height of the tree = 5 m + 13 m

Total height of the tree = 18 m

6. Angles Q and R of a ΔPQR are 25º and 65º. Write which of the following is true:

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

(i) PQ2 + QR2 = RP2

Answer: False

Explanation: PQR is a triangle.

The total measure of the three angles of a triangle is 180°.

The three angles of the given triangle are 25°, 65°, and P.

So,

25° + 65° + P = 180°

90° + P = 180°

P = 180° - 90°

P = 90°

It means that the above triangle is the right angle triangle, which is right angled at P.

According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

Hypotenuse = the opposite side of the right angle P

Hypotenuse = QR

(QR)2 = (PQ)2 + (PR)2

So, the above statement is False.

(ii) PQ2 + RP2 = QR2

Answer: True

Explanation: The three angles of the given triangle are 25°, 65°, and P.

So,

25° + 65° + P = 180°

90° + P = 180°

P = 180° - 90°

P = 90°

It means that the above triangle is the right angle triangle, which is right angled at P.

According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

Hypotenuse = the opposite side of the right angle P

Hypotenuse = QR

(QR)2 = (PQ)2 + (PR)2

(PR can also be written as RP. Both are the same).

So, the above statement is true.

(iii) RP2 + QR2 = PQ2

Answer: False

Explanation: The three angles of the given triangle are 25°, 65°, and P.

So,

25° + 65° + P = 180°

90° + P = 180°

P = 180° - 90°

P = 90°

It means that the above triangle is the right angle triangle, which is right angled at P.

According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

Hypotenuse = the opposite side of the right angle P

Hypotenuse = QR

(QR)2 = (PQ)2 + (PR)2

(PR can also be written as RP. Both are the same).

So, the above statement is false.

7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Answer: 98 cm

Explanation: Let ABCD be the rectangle with BD as the diagonal.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Perimeter of a rectangle = 2 × (Length + Width)

Length of the rectangle = 40 cm

Width =?

BCD forms a right angle triangle with BD as the hypotenuse.

According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

(BD)2 = (BC)2 + (CD)2

(41)2 = (BC)2 + (40)2

1681 = (BC)2 + 1600

(BC)2 = 1681 - 1600

(BC)2 = 81

BC = 9

Thus, the width of the rectangle is 9 cm.

Perimeter of a rectangle = 2 × (Length + Width)

Perimeter of a rectangle = 2 × (40 + 9)

Perimeter of a rectangle = 2 × 49

Perimeter of a rectangle = 98 cm

8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Answer: 68 cm

Explanation: All the sides of a rhombus are equal.

Let ABCD be the rhombus.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

Perimeter of rhombus = 4 × Side

BOC forms a right angle triangle with BC as the hypotenuse.

NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties

BC = Side of the rhombus

According to the Pythagoras theorem,

Hypotenuse square = Sum of the square of the sides of the right angle triangle

(BC)2 = (BO)2 + (OC)2

O divides the diagonal into two equal parts.

BO = 16/2 = 8 cm

OC = 30/2 = 15 cm

(BC)2 = (8)2 + (15)2

(BC)2 = 64 + 225

(BC)2 = 289

BC = 17

Thus, the side of the rhombus is 17 cm.

Perimeter of rhombus = 4 × Side

Perimeter of rhombus = 4 × 17

Perimeter of rhombus = 68 cm






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