## NCERT Solutions for class 7 Maths Chapter 6: The triangles and its properties## Exercise 6.1
PM is
PD is
Here, D is the midpoint of the line QR, which divides it into two equal parts. QD = DR
Instead, we can say that: QD = DR
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
Exterior angle is equal to the sum of the opposite interior angles. Exterior angle x = 50° + 70° x = 50° + 70° x = 120°
Exterior angle is equal to the sum of the opposite interior angles. Exterior angle x = 65° + 45° x = 65° + 45° x = 110°
Exterior angle is equal to the sum of the opposite interior angles. Exterior angle x = 30° + 40° x = 30° + 40° x = 70°
Exterior angle is equal to the sum of the opposite interior angles. Exterior angle x = 60° + 60° x = 60° + 60° x = 120°
Exterior angle is equal to the sum of the opposite interior angles. Exterior angle x = 50° + 50° x = 50° + 50° x = 100°
Exterior angle is equal to the sum of the opposite interior angles. Exterior angle x = 60° + 30° x = 60° + 30° x = 90°
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°
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°
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°
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°
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°
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
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°
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°
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°
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°
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°.
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°
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°
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°
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°
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°
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°
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
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).
Let' check. 6 + 3 > 5 Yes 3 + 7 > 6 Yes 6 + 7 > 3 Yes Therefore, the triangle with the above dimensions is possible.
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).
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.
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.
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.
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.
Let the midpoint be O. For AB + BC > AC ... (1) For BC + CD > BD ... (2) For CD + AD > AC ... (3) For 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.
Let the midpoint be O. For AO + OB > AB ... (1) For BO + OC > BC ... (2) For CO + OD > CD ... (3) For 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.
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
Hypotenuse square = Sum of the square of the sides of the right angle triangle Here, QR = Hypotenuse So, (QR) (QR) (QR) (QR) QR = 26 Thus, the length of the QR is 26 cm.
According to the Pythagoras theorem, Hypotenuse square = Sum of the square of the sides of the right angle triangle Here, AB = Hypotenuse So, (AB) (25) (BC) (BC) (BC) BC = 24 Thus, the length of the BC is 24 cm.
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) (a) (a) (a) a = 9 Thus, the length of the side 'a' is 9 m.
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) 42.25 = 36 + 6.25 42.25 = 42.25 LHS = RHS Hence, the given triangle is a right angle triangle.
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) 25 = 4 + 4 25 = 8 LHS is not equal to RHS Hence, the given triangle is a not right angle triangle.
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) 6.25 = 4 + 2.25 6.25 = 6.25 LHS = RHS Hence, the given triangle is a right angle triangle.
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) (Hypotenuse) (Hypotenuse) 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
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) So, the above statement is
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) (PR can also be written as RP. Both are the same). So, the above statement is
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) (PR can also be written as RP. Both are the same). So, the above statement is
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) (41) 1681 = (BC) (BC) (BC) 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
Let ABCD be the rhombus. Perimeter of rhombus = 4 × Side BOC forms a right angle triangle with BC as the hypotenuse. 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) O divides the diagonal into two equal parts. BO = 16/2 = 8 cm OC = 30/2 = 15 cm (BC) (BC) (BC) 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 |