SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles InText Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles InText Questions

Check Your Progress [Page No. 92]

Question 1.

Classify the following triangles according to the length of their sides:

Answer:

[Page No. 93]

Question 2.

Classify the following triangles according to the measure of their angles.

Answer:

Let’s Think [Page No. 97]

Question 1.

Can you find a triangle in which each angle is less than 60°?

Answer:

Sum of angles in a triangle is 180°.

If each angle is less than 60°, then the sum of angles of a triangle is < 180°. So, triangle cannot form. So, we cannot find a triangle in which each angle is less than 60°.

Check Your Progress [Page No. 99]

Question 1.

Find the value of ‘x’ in the given figure.

Answer:

In the given figure,

∠HAC = ∠TAE = a (∵ Vertically opposite angles)

In ∆AHC,

we know, ∠H + ∠C + ∠HAC 180°

⇒ 60° + 80° + a = 180°

⇒ 140° + a – 140° = 180°- 140° … a = 40°.

∴ ∠HAC = ∠TAE = a = 40°

In ∆ATE, we know,

∠T + ∠TAE + ∠E = 180°

⇒ x + a + 70° = 180°

⇒ x + 40° + 70° = 180°

⇒ x + 110°- 110° = 180°- 110°

∴ x = 70°

Let’s Do Activity [Page No. 101]

Question 1.

Measure the length of the sides of ∆ABC and fill the following table.

Answer:

From the above table, we can conclude that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Question 2.

Take the same measurements of the previous activity and note the results as given below.

Answer:

From the above table we can conclude that the difference of the lengths of any two sides of a triangle is less than the length of the third side.

Question 3.

Fill the following table as per measurements given in the above triangles:

What do you observe from the last four columns in above table ?

- In any triangle the opposite side to the biggest angle is bigger than the other two sides.
- In any triangle the opposite side to the smallest angle is smaller than the other two sides.

Answer:

Let’s Think [Page No. 102]

Question 1.

What are the measurements of angles of an equilateral triangle?

Answer:

In ∆ADI

AD = DI – AI = 6 cm (All angles are equal)

∠A = ∠D = ∠I = 60° (All angles are equal)

So, in an equilateral triangle,

All Sides are equal in length. Each angle is equal to 60°.

Puzzle Time [Page No. 104]

Question 1.

Find the interior angles of all triangles by using given dues.

Answer:

Hints:

Each angle in an equilateral triangle is 60°.

Straight angle = 180°

In an isosceles triangle, two angles are equal.

In a scalene triangle all angles are unequal.

Right angle = 90°.

Obtuse angle > 90°.

Examples:

Question 1.

In a triangle, two angles are 43° and 57°. Find the third angle.

Answer:

Given two angles of a triangle are 43° and 57°.

Sum of these two angles = 43° + 57° = 100°

In a triangle the sum of the interior angles is 180°.

∴ Third angle = 180° – 100° = 80°

Question 2.

Find the value of ‘x’ in the following triangles :

Answer:

From the figure

∠A + ∠B + ∠C = 180° (v Sum of three angles in triangle is 180°)

⇒ 65° + 60° + x = 180°

⇒ 125° + x = 180°

∴ x =180°- 125° = 55°

Answer:

From the figure

∠P + ∠Q + ∠R = 180°

x + x + 80° = 180°

⇒ 2x + 80° = 180°

⇒ 2x = 180°-80°

⇒ 2x = 100°

⇒ x = \(\frac{100^{\circ}}{2}\) = 50°

∴ x = 50°

Answer:

From the figure

∠X + ∠Y + ∠Z – 180°

⇒ x + 90° + 42° = 180°

⇒ x + 132° = 180°

⇒ x = 180°- 132°

∴ x = 48°

Question 3.

Find the values of x and y in the given triangle.

Answer:

In the ∆ABC

∠ACD + ∠ACB = 180° (linear pair of angles)

110° + y = 180°

⇒ y = 180°- 110°

∴ y = 70° ……………….(i)

∠BAC + ∠ACB + ∠CBA = 180°

⇒ x + y + 60° = 180°

⇒ x + 70° + 60° = 180° (From (i))

⇒ x + 130° = 180°

∴ x = 50°

Question 4.

In a right angled triangle one acute angle is 44°, then find the other acute angle.

Answer:

We know that, sum of the two acute angles in a right angled triangle is 90°.

Given that, in the right angled tri-angle one of the acute angles = 44°

Other acute angle in right angled triangle = 90° – 44° = 46°

Question 5.

The angles of a triangle are (x + 10)°, (x – 20)° and (x + 40)°. Find the value of x and the measure of the angles.

Answer:

Given that the angles of the triangle are

(x + 10)°, (x – 20)° and (x + 40)°

(x + 10)° + (x – 20)° + (x + 40)° = 180°

⇒ x + 10° + x – 20° + x + 40° = 180°

⇒ 3x + 30° = 180°

⇒ 3x – 180° – 30°

⇒ 3x = 150°

⇒ x = \(\frac{150^{\circ}}{3}\) = 50°

The angles are,

x + 10° = 50° + 10° = 60°

x – 20° = 50° – 20° = 30°

x + 40° = 50° + 40° = 90°

∴ Measure of the angles are 60°, 30° and 90°.

Question 6.

In the ∆ABC, exterior angle at ∠C = 105° and ∠A=65°. Find the other interior opposite angle.

Answer:

Given that, exterior angle at ∠C = 105°.

One of the interior opposite angle ∠A = 65°.

The other interior opposite angle is ∠B .

∠A + ∠B =105°(∵ Exterior angle property of a triangle)

65° + ∠B = 105°

⇒ ∠B = 105° – 65° = 40°

Question 7.

Find the exterior angle of the given triangle.

Answer:

From the figure, ∠P = 30°, ∠Q = 35°

Exterior angle at R = ∠P + ∠Q (∵ Exterior angle property of a triangle)

= 30° + 35° = 65°

Question 8.

Find the values of ‘x’ and ‘y’ in the following figure.

Answer:

∠LMN + 110° = 180° (Linear pair angles)

∠LNM = 180° -110° = 70°

∠LMN = ∠LNM (Angles opposite to equal sides)

y = 70°

⇒ x + y – 110° (Exterior angle property)

⇒ x + 70° = 110°

∴ x = 110°-70° – 40°