These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties InText Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties InText Questions

NCERT In-text Question Page No. 113 & 114

Question 1.

Write the six elements (i.e. the 3 sides and the 3 angles) of ΔABC.

Answer:

Six elements of ΔABC are: ∠A, ∠B, ∠C, \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\), and \(\overline{\mathrm{CA}}\).

Question 2.

Write the:

(i) Side opposite to the vertex Q of ΔPQR

(ii) Angle opposite to the side LM of ΔLMN

(iii) Vertex opposite to the side RT of ΔRST

Answer:

(i) The side opposite to the vertex Q is \(\overline{\mathrm{PR}}\).

(ii) The angle opposite to the side LM is ∠N.

(iii) Vertex opposite to the side RT is ‘S’.

Question 3.

Look at figures and classify each of the triangles according to its.

(a) Sides

(b) Angles

Answer:

(i) (a) Since \(\overline{\mathrm{AC}}=\overline{\mathrm{BC}}\) = 8 cm

∴ ΔABC is an isosceles triangle,

(b) Since all angles of ∴ABC are less than 90°.

∴ It is an acute triangle.

(ii) (a) Since PQ ≠ QR ≠ RP

∴ ΔPQR is a scalene triangle.

(b) Since ∠R = 90°

∴ ΔPRQ is a right triangle.

(iii) (a) In ΔLMN, LN = MN = 7 cm

∴ ΔLMN is an isosceles triangle,

(b) In ΔLMN, ∠N > 90°

∴ ΔLMN is an obtuse triangle.

(iv) (a) In ΔRST, RS = ST = TR = 5.2 cm

∴ It is an equilateral triangle.

(b) All the angles of ΔRST are acute.

∴ It is an acute triangle.

(v) (a) In ΔABC, \(\overline{\mathrm{AB}}=\overline{\mathrm{BC}}\) = 3 cm

∴ It is an isosceles triangle.

(b) In ΔABC, ∠B > 90°

∴ It is an obtuse triangle.

(vi) (a) In ΔPQR, latex]\overline{\mathrm{PQ}}=\overline{\mathrm{QR}}[/latex] = 6 cm

∴ It is an isosceles triangle.

(b) In ΔPQR, ∠Q = 90°

∴ It is a right triangle.

NCERT In-text Question Page No. 118

Question 1.

An exterior angle of a triangle is of measure 70° and one of its interior opposite angles is for measure 25°. Find the measure of the other interior opposite angle.

Answer:

Exterior angle = 70°

Interior opposite angles are 25° and x.

∴ x + 25° = 70°

[Using the exterior angle property of a triangle]

or x = 70°- 25°= 45°

∴ The required interior opposite angle = 45°.

Question 2.

The two interior opposite angles of an exterior angle of triangle are 60° and 80°. Find the measure of the exterior angle.

Answer:

Interior angles are 60° and 80°.

∵ [Exterior angle] = 60° + 80° = 140°

Question 3.

Is something wrong in this diagram?

Answer:

We know that an exterior angle of a triangle is equal to the sum of interior opposite angles. Here interior angles are 50° each and exterior angle is 50°.

∴ This triangle cannot be formed.

[∵ 50° ≠ 50° + 50°]

NCERT In-text Question Page No. 122

Question 1.

Two angles of a triangle are 30° and 80°. Find the third angle.

Answer:

Let the third angle be x.

∴ Using the angle sum property of a triangle,

30° + 80° + x = 180°

or x + 110° = 180°

or x = 180°- 110° = 70°

∴ The measure of the required third angle is 70°.

Question 2.

One of the angles of a triangle is 80° and the other two angles are equal. Find the measure of each of the equal angles.

Answer:

Let each of the equal angles be x

x + x + 80° = 180°

(angle sum property of a triangle)

2x + 80° = 180°

2x = 180° – 80°

2x = 100°

x = \(\frac{100^{\circ}}{2}\) = 50°

∴ The required measure of each of the equal angle is 50°.

Question 3.

The three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.

Answer:

Let the three angles of a triangle be x, 2x and x

x + 2x + x = 180°

(using the angle sum property)

2x + 2x = 180°

4x = 180°

x \(\frac{180^{\circ}}{4}\) = 45°

Thus, the three angles are 45°, 90° and 45°. It is an isosceles triangle (two angles are equal, opposite sides also equal)

As one angle is 90°.

∴ It is a right-angled triangle.

NCERT In-text Question Page No. 123 & 124

Question 1.

Find angle x in each figure:

Answer:

(i) Since the two sides in the triangle are equal, the base angle opposite to equal sides are equal.

∴ x = 40°

(ii) Since the two sides of the triangle are equal, the base angles opposite to equal sides are equal so, the other angle = 45°

Sum of three angles of a triangle = 180°

45° + 45° + x = 180°

x + 90° = 180°

x = 180° – 90°

x = 90°

(iii) Since the two sides of the triangle are equal, the base angles opposite to equal sides are equal.

x = 50°

(iv) Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles is 180°.

100° + x + x = 180°

2x = 180° – 100°

2x = 80°

x = \(\frac{80^{\circ}}{2}\) = 40°

(v) Base angles opposite to the equal sides of an isosceles triangles are equal and the sum of the measure of three angles of triangle is 180°.

x + x + 90° = 180°

2x + 90° = 180°

2x = 180° – 90°

2x = 90°

x = \(\frac{90^{\circ}}{2}\) = 45°

(vi) Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of triangle is 180°.

x + x + 40° = 180°

2x + 40° = 180°

2x = 180° – 40°

2x = 140°

x = \(\frac{140^{\circ}}{2}\)

= 70°

x = 70°

(vii) In the figure, two sides of the triangle are equal.

∴ The base angles opposite to equal sides are equal one of the base angle = x

other base angle = x

Now, x° and 120° form a linear pair = 180°

x + 120° = 180°

x = 180° -120°

= 60°

Thus, the value of x = 60°

(viii) In the figure, two sides of the triangle are equal.

since, one of the base angles = x

∴ The other base angle = x

Since, exterior angle is equal to sum of the interior opposite angles

∴ x + x = 110°

2x = 110°

x = \(\frac{110^{\circ}}{2}\) = 55°

(ix) Two sides of the triangle are equal.

∴ The base angles opposite to the equal sides are equal.

Since, one of the base angle = x

∴ The other base angle = x

Also, the vertically opposite angles 30° and x are equal.

∴ x = 30°

Question 2.

Find angles x and y in each figure.

Answer:

(i) Two sides of the triangle are equal.

∴ The base angles opposite to the equal sides are equal.

The other base angle is y.

Now y and 120° form a linear pair.

∴ y + 120° = 180°

y = 180° – 120° = 60°

∴ x + y + y = 180°

(sum of the three angles = 180°)

x + 60° + 60°= 180°

x+ 120° = 180°

x = 180° – 120°

= 60°

Thus x = 60° and y = 60°

(ii) Two sides of a triangles are equal.

∴ The base angles opposite to equal sides are equal. Since one of the base angle is ‘x’

∴ The other base angle = x

The given triangle is a right-angled triangle.

x + x + 90° = 180°

2x + 90° = 180°

2x = 180° – 90°

2x = 90°

x = \(\frac{90^{\circ}}{2}\) = 45°

Now x and y form a linear pair

x + y = 180°

45°+ y = 180°

y = 180° – 45° = 135°

Thus, x = 45° and y = 135°

(iii) In the given figure, two sides of a triangle are equal.

∴ The base angles are x and x

The third angle = 92°

(vertically opposite angles are equal)

x + x + 92° = 180°

(Sum of the three angles of a triangle is 180°)

2x + 92° = 180°

2x = 180° – 92°

2x = 88°

x = \(\frac{88^{\circ}}{2}\) = 44°

Now, x and y form a linear pair

x + y = 180°

44 + y = 180°

y = 180° – 44° = 136°

Thus, x = 44° and y = 136°

NCERT In-text Question Page No. 129 & 130

Question 1.

Find the unknown length x in the following figures.

(i) In the given right-angled triangle, the longest side (hypotenuse) is x

x^{2} = 3^{2} + 4^{2}

(using pythagoras property)

= 9 + 16

= 25

or x^{2} = 5^{2}

x = \(\sqrt{25}\)

= 5

x = 5

(ii) The given figure is a right-angled triangle.

x^{2} = 6^{2} + 8^{2}

(using pythagoras property)

= 36 + 64

x^{2} = 100

x^{2} = 10^{2}

∴ x = 10

(iii) The given figure is a right-angled triangle.

x^{2} = 8^{2} + 15^{2}

(using pythagoras property)

x^{2} = 64 + 225

x^{2} = 289

x^{2} = 17^{2}

x = 17

(iv) The given figure is a right-angled triangle

x^{2} = 7^{2} + 24^{2}

(using pythagoras property)

x^{2} = 49 + 576

x^{2} = 625

x^{2} = 25^{2}

x = 25

(v) The given figure can be labelled as ΔABC and the altitude is AD. Consider the right-angled triangle ABD

AB^{2} = AD^{2} + BD^{2}

(using pythagoras property)

37^{2} = 12^{2} + BD^{2}

1369 = 144 + BD^{2}

1369 – 144 = BD^{2}

1225 = BD^{2}

35^{2} = BD^{2}

∵ BD = 35

In the right-angled triangle ADC

AC^{2} = AD^{2} + DC^{2}

(using pythagoras property)

37^{2} = 12^{2} + DC^{2}

1369 = 144 +DC^{2}

1369 – 144 = DC^{2}

1225 = DC^{2}

35^{2} = DC^{2}

∴ DC = 35

∵ BC = BD + DC

= 35 + 35

x = 70

(vi) In the given right-angled triangle

x^{2} = 12^{2} + 5^{2}

(using the pythagoras property)

= 144 + 25

x^{2} = 169

x^{2} = 13^{2}

∴ x = 13