Since, Exterior angle = Sum of interior opposite angles, therefore (i) x = 50° + 70° = 120° (ii) x = 65° + 45° = 110° (iii) x = 30° + 40° = 70° (iv) x = 60° + 60° = 120° (v) x = 50° + 50° =100° (vi) x = 60° + 30° = 90° Class 7 Maths Chapter 6 Exercise 6.2 for more answers vist to: https://www.tiwariRead more
Since, Exterior angle = Sum of interior opposite angles, therefore
(i) x = 50° + 70° = 120°
(ii) x = 65° + 45° = 110°
(iii) x = 30° + 40° = 70°
(iv) x = 60° + 60° = 120°
(v) x = 50° + 50° =100°
(vi) x = 60° + 30° = 90°
Isosceles triangle means any two sides are same. Take ∆ABC and draw the median when AB = AC. AL is the median and altitude of the given triangle. Class 7 Maths Chapter 6 Exercise 6.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Isosceles triangle means any two sides are same.
Take ∆ABC and draw the median when AB = AC.
AL is the median and altitude of the given triangle.
follow the link for more answers and video: Class 7 Maths Chapter 6 Exercise 6.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Given: QD = DR ∴ PM is altitude. PD is median. No, QM ≠ MR as D is the mid-point of QR. Class 7 Maths Chapter 6 Exercise 6.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Given: QD = DR
∴ PM is altitude.
PD is median.
No, QM ≠ MR as D is the mid-point of QR.
Find the value of the unknown interior angle x in the following figures:
Since, Exterior angle = Sum of interior opposite angles, therefore (i) x + 50° = 115° ⇒ x = 115° - 50° = 65° (ii) 70° + x = 100° ⇒ x = 100° - 70° = 30° (iii) x + 90° = 125° ⇒ x = 120° - 90° = 35° (iv) 60° + x = 120° ⇒ x = 120° - 60° = 60° (v) 30° + x = 80° ⇒ x = 80° - 30° = 50° (vi) x + 35° = 75° ⇒Read more
Since, Exterior angle = Sum of interior opposite angles, therefore
(i) x + 50° = 115° ⇒ x = 115° – 50° = 65°
(ii) 70° + x = 100° ⇒ x = 100° – 70° = 30°
(iii) x + 90° = 125° ⇒ x = 120° – 90° = 35°
(iv) 60° + x = 120° ⇒ x = 120° – 60° = 60°
(v) 30° + x = 80° ⇒ x = 80° – 30° = 50°
(vi) x + 35° = 75° ⇒ x = 75° – 35° = 40°
Class 7 Maths Chapter 6 Exercise 6.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Find the value of the unknown exterior angle x in the following diagrams:
Since, Exterior angle = Sum of interior opposite angles, therefore (i) x = 50° + 70° = 120° (ii) x = 65° + 45° = 110° (iii) x = 30° + 40° = 70° (iv) x = 60° + 60° = 120° (v) x = 50° + 50° =100° (vi) x = 60° + 30° = 90° Class 7 Maths Chapter 6 Exercise 6.2 for more answers vist to: https://www.tiwariRead more
Since, Exterior angle = Sum of interior opposite angles, therefore
(i) x = 50° + 70° = 120°
(ii) x = 65° + 45° = 110°
(iii) x = 30° + 40° = 70°
(iv) x = 60° + 60° = 120°
(v) x = 50° + 50° =100°
(vi) x = 60° + 30° = 90°
Class 7 Maths Chapter 6 Exercise 6.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same.
Isosceles triangle means any two sides are same. Take ∆ABC and draw the median when AB = AC. AL is the median and altitude of the given triangle. Class 7 Maths Chapter 6 Exercise 6.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Isosceles triangle means any two sides are same.
Take ∆ABC and draw the median when AB = AC.
AL is the median and altitude of the given triangle.
Class 7 Maths Chapter 6 Exercise 6.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
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.
follow the link for more answers and video: Class 7 Maths Chapter 6 Exercise 6.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
follow the link for more answers and video:
Class 7 Maths Chapter 6 Exercise 6.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
In ∆ PQR, D is the mid-point of QR. PM is _______________. PD is ________________ Is QM = MR?
Given: QD = DR ∴ PM is altitude. PD is median. No, QM ≠ MR as D is the mid-point of QR. Class 7 Maths Chapter 6 Exercise 6.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Given: QD = DR
∴ PM is altitude.
PD is median.
No, QM ≠ MR as D is the mid-point of QR.
Class 7 Maths Chapter 6 Exercise 6.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/