Chapter 6 The Triangle and its Properties

**NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1**

**Question 1.**

In ∆PQR, D is the mid-point of QR¯¯¯¯¯¯¯¯.

**Solution:**

PM¯¯¯¯¯¯¯¯¯ is the altitude.

PD is the median.

No! QM ≠ MR.

**Question 2.**

Draw rough sketches for the following:

**(a)** In ∆ ABC, BE is a median.

**(b)** In ∆ PQR, PQ and PR are altitudes of the triangle.

**(c)** In ∆ XYZ, YL is an altitude in the exterior of the triangle.

**Solution:
**

**Question 3.**

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

**Solution:**

AD is the median.

AL is the altitude.

Draw a line segment BC. By paper folding, locate the perpendicular bisector of BC. The folded crease meets BC at D, its mid-point.

Take any point A on this perpendicular bisector. Join AB and AC. The triangle thus obtained is an isosceles ∆ABC in which AB = AC.

Since D is the mid-point of BC, so AD is its median. Also, AD is the perpendicular bisector of BC. So, AD is the altitude of ∆ABC.

Thus, it is verified that the median and altitude of an isosceles triangle are the same.

**NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2**

**Question 1.**

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

**Solution:
**

**Question 2.**

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

**Solution:
**

**NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3**

**Question 1.**

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

**Solution:
**

**Question 2.**

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

**Solution:
**

**NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4**

**Question 1.**

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

- 2 cm, 3 cm, 5 cm
- 3 cm, 6 cm, 7 cm
- 6 cm, 3 cm, 2 cm.

**Solution:**

- Since, 2 + 3 > 5

So the given side lengths cannot form a triangle. - We have, 3 + 6 > 7, 3 + 7 > 6 and 6 + 7 > 3

i. e., the sum of any two sides is greater than the third side.

So, these side lengths form a triangle. - We have, 6 + 3 > 2, 3 + 2 Undefined control sequence \ngtr 6

So, the given side lengths cannot form a triangle.

**Question 2.**

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

- OP + OQ > PQ ?
- OQ + OR > QR?
- OR + OP > RP ?

**Solution:**

- Yes ! OP + OQ > PQ …(1)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side - Yes! OQ + OR > QR …(2)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side - Yes! OR + OP > RP …(3)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side

**Question 3.**

AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM?

(Consider the sides of triangles ∆ ABM and ∆ AMC.)

**Solution:**

Using triangle inequality property in triangles ABM and AMC, we have

AB + BM > AM …(1) and, AC + MC > AM …(2)

Adding (1) and (2) on both sides, we get

AB + (BM + MC) + AC > AM + AM

⇒ AB + BC + AC > 2AM

**Question 4.**

ABCD is a quadrilateral.

Is AB + BC + CD + DA > AC + BD ?

**Solution:**

In ∆ ABC, AB + BC > AC …(1)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side

In ∆ ACD, CD + DA > AC …(2)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side

Adding (1) and (2),

AB + BC + CD + DA > 2AC …(3)

In ∆ ABD, AB + DA > BD …(4)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side

In ∆ BCD, BC + CD > BD …(5)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side

Adding (4) and (5),

AB + BC + CD + DA > 2BD …(6)

Adding (3) and (6),

2 [AB + BC + CD + DA] > 2 (AC + BD)

⇒ AB + BC + CD + DA > AC + BD.

**Question 5.**

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

**Solution:**

In ∆ OAB, OA + OB > AB ….(1)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side.

In ∆ OBC, OB + OC > BC ….(2)

Sum of the lengths of any two sides of a triangle la greater than the length of the third side.

In ∆ OCA,OC + OA > CA ….(3)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side

In ∆ OAD, OA + OD > AD ….(4)

Sum of the lengths of any two sides of a triangle is greater than the length of the third side

Adding (1), (2), (3) and (4),

2(OA + OB + OC + OD) > AB + BC + CD + DA

⇒ AB + BC + CD + DA < 2 (OA + OB + OC + OD)

⇒ AB + BC + CD + DA < 2(OA + OC + OB + OD)

⇒ AB + BC + CD + DA < 2 (AC + BD).

**Question 6.**

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

**Solution:**

Let x cm be the length of the third side.

∴ Sum of the lengths of any two sides of a triangle is greater than the length of the third side.

∴ We should have

∴ The length of the third side should be any length between 3 cm and 27 cm.

**NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5**

**Question 1.**

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

**Solution:**

QR^{2} = 10^{2} + 24^{2} By Pythagoras Property

⇒ = 100 + 576 = 676

⇒ QR = 26 cm.

**Question 2.**

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

**Solution:**

AC^{2} + BC^{2} = AB^{2} By Pythagoras Property

⇒ 7^{2} + BC^{2} = 25^{2}

⇒ 49 + BC^{2} = 625

⇒ BC^{2} = 625 – 49

⇒ BC^{2} = 576

⇒ BC = 24 cm.

**Question 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.

**Solution:**

Let the distance of the foot of the ladder from the wall be a m. Then,

Hence, the distance of the foot of the ladder from the wall is 9 m.

**Question 4.**

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

- 2.5 cm, 6.5 cm, 6 cm.
- 2 cm, 2 cm, 5 cm.
- 1.5 cm, 2 cm, 2.5 cm.

In the case of right-angled triangles, identify the right angles.

**Solution:**

1. 2.5 cm, 6.5 cm, 6 cm We see that

(2.5)^{2} + 6^{2} = 6.25 + 36 = 42.25 = (6.5)^{2}

Therefore, the given lengths can be the sides of a right triangle. Also, the angle between the lengths, 2.5 cm and 6 cm is a right angle.

2. 2 cm, 2 cm, 5 cm

∵ 2 + 2 = 4 Undefined control sequence \ngtr 5

∴ The given lengths cannot be the sides of a triangle

The sum of the lengths of any two sides of a triangle is greater than the third side

3. 1.5 cm, 2 cm, 2.5 cm We find that

1.5^{2} + 2^{2} = 2.25 + 4 = 6.25 = 2.5^{2}

Therefore, the given lengths can be the sides of a right triangle.

Also, the angle between the lengths 1.5 cm and 2 cm is a right angle.

**Question 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.

**Solution:**

AC = CD Given

In right angled triangle DBC, DC^{2} = BC^{2} + BD^{2}

by Pythagoras Property = 5^{2} + 12^{2} = 25 + 144 = 169

⇒ DC = 13 ⇒ AC = 13

⇒ AB = AC + BC = 13 + 5 = 18

Therefore, the original height of the tree = 18 m.

**Question 6.**

Angles Q and R of a ∆ PQR are 25° and 65°. Write which of the following is true:

**(i)** PQ^{2} + QR^{2} = RP^{2}

**(ii)** PQ^{2} + RP^{2} = QR^{2}

**(iii)** RP^{2} + QR^{2} = PQ^{2}

**Solution:**

**(ii)** PQ^{2} + RP^{2} = QR^{2} is true.

**Question 7.**

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

**Solution:**

In right-angled triangle DAB, AB^{2} + AD^{2} = BD^{2}

⇒ 40^{2} + AD^{2} = 41^{2} ⇒ AD^{2} = 41^{2} – 40^{2}

⇒ AD^{2} = 1681 – 1600

⇒ AD^{2} = 81 ⇒ AD = 9

∴ Perimeter of the rectangle = 2(AB + AD) = 2(40 + 9) = 2(49) = 98 cm

Hence, the perimeter of the rectangle is 98 cm.

**Question 8.**

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

**Solution:**

Let ABCD be a rhombus whose diagonals BD and AC are of lengths 16 cm and 30 cm respectively.

Let the diagonals BD and AC intersect each other at O.

Since the diagonals of a rhombus bisect each other at right angles. Therefore

BO = OD = 8 cm,

AO = OC = 15 cm,

∠AOB = ∠BOC

= ∠COD = ∠DOA = 90°

In right-angled triangle AOB.

AB^{2} = OA^{2} + OB^{2}

By Pythagoras Property

⇒ AB^{2} = 15^{2} + 8^{2}

⇒ AB^{2} = 225 + 64

⇒ AB^{2} = 289

⇒ AB = 17cm

Therefore, perimeter of the rhombus ABCD = 4 side = 4 AB = 4 × 17 cm = 68 cm

Hence, the perimeter of the rhombus is 68 cm.