Since sum of linear pair angles is 180° ∴ 90° + x = 180° ⇒ x+180°-90° = 90° and z+30°= 180° ⇒ z=180°-30°=150° also y=90°+30°=120° [Exterior angle property] ∴ x+y+x = 90°+120°+150°=360° (b) Using angle sum property of a quadrilateral, 60°+80°+120°+n=360° ⇒ 260°+n=360° ⇒ n=360°-260° ⇒ n=100° Since sumRead more
Since sum of linear pair angles is 180°
∴ 90° + x = 180°
⇒ x+180°-90° = 90°
and z+30°= 180°
⇒ z=180°-30°=150°
also y=90°+30°=120°
[Exterior angle property]
∴ x+y+x = 90°+120°+150°=360°
(b) Using angle sum property of a quadrilateral,
60°+80°+120°+n=360°
⇒ 260°+n=360°
⇒ n=360°-260°
⇒ n=100°
Since sum of linear pair angles is 180°
∴ w+100=180° …………(i)
x+120°=180° …………(ii)
y+80°=180° …………(iii)
z+60°=180° …………(iv)
Adding eq. (i), (ii), (iii) and (iv),
⇒ x + y + z + w + 100 °+ 120° + 80° + 60° = 180° + 180° + 180° + 180°
⇒ x + y + z + w +360°=720°
⇒ x + y + z + w = 720°-360°
⇒ x + y + z + w = 360°
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
(a) Using angle sum property of a quadrilateral, 50°+130°+120°+x=360° ⇒ 300°+x=360° ⇒ x=360°-300° ⇒x=360° (b) Using angle sum property of a quadrilateral, 90°+60°+70°+x=360° ⇒ 220°+x=360° ⇒x=360°-220° ⇒x=140° (c) First base interior angle = 180°-70°=110° Second base interior angle = 180°-60°=120° ThRead more
(a) Using angle sum property of a quadrilateral,
50°+130°+120°+x=360°
⇒ 300°+x=360°
⇒ x=360°-300° ⇒x=360°
(b) Using angle sum property of a quadrilateral,
90°+60°+70°+x=360°
⇒ 220°+x=360°
⇒x=360°-220° ⇒x=140°
(c) First base interior angle = 180°-70°=110°
Second base interior angle = 180°-60°=120°
There are 5 sides, = 5 n
∴ Angle sum of a polygon = (n-2)x180°
= (5-2)x180°=3×180 =540°
∴ 30°+x+110°+120°+x=540°
⇒ 260°+2x+540° ⇒2x=540°-260°
⇒ 2x=280° ⇒x=140°
(d) Angle sum of a polygon = (n-2)x180°
= (5-2)x180°= 3×180° = 540°
∴ x+x+x+x+x =540°
⇒ 5x=540° ⇒ x=108°
Hence each interior angle is 108°
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
A regular polygon: A polygon having all sides of equal length and the interior angles of equal size is known as regular polygon. (i) 3 sides. Polygon having three sides is called a triangle. (ii) 4 sides. Polygon having four sides is called a quadrilateral. (iii) 6 sides. Polygon having six sides isRead more
A regular polygon: A polygon having all sides of equal length and the interior angles of
equal size is known as regular polygon.
(i) 3 sides. Polygon having three sides is called a triangle.
(ii) 4 sides. Polygon having four sides is called a quadrilateral.
(iii) 6 sides. Polygon having six sides is called a hexagon.
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
(a) When = 7, then n Angle sum of a polygon = (n-2) x 180° = (7-2) x 180°=5x180° = 900° (b) When n = 8, then Angle sum of a polygon = (n-2) x 180°=(8-2) x 180° = 6 x 180°=1080° (b) When n = 8, then Angle sum of a polygon = (n-2) x 180°=(10-2) x 180° = 8 x 180°=1440° (d) When n = n, then, angle sum oRead more
(a) When = 7, then n
Angle sum of a polygon = (n-2) x 180° = (7-2) x 180°=5×180° = 900°
(b) When n = 8, then
Angle sum of a polygon = (n-2) x 180°=(8-2) x 180° = 6 x 180°=1080°
(b) When n = 8, then
Angle sum of a polygon = (n-2) x 180°=(10-2) x 180° = 8 x 180°=1440°
(d) When n = n, then, angle sum of a polygon = (n-2) x 180°
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
Let ABCD is a convex quadrilateral, then we draw a diagonal AC which divides the quadrilateral in two triangles. ∠A +∠B+ ∠C + ∠D = ∠1+ ∠6+ ∠5+ ∠4+ ∠3+ ∠2 = ( ∠1+ ∠2+ ∠3) +(∠4+ ∠5+ ∠6) = [By Angle sum property of triangle] 180°+180° = 360° Hence, the sum of measures of the triangles of a convex quadrRead more
Let ABCD is a convex quadrilateral, then we draw a diagonal AC which divides the quadrilateral in two triangles.
∠A +∠B+ ∠C + ∠D = ∠1+ ∠6+ ∠5+ ∠4+ ∠3+ ∠2
= ( ∠1+ ∠2+ ∠3) +(∠4+ ∠5+ ∠6)
= [By Angle sum property of triangle] 180°+180°
= 360°
Hence, the sum of measures of the triangles of a convex quadrilateral is 360°.
Yes, if quadrilateral is not convex then, this property will also be applied.
Let ABCD is a non-convex quadrilateral and join BD, which also divides the quadrilateral in two triangles.
Using angle sum property of triangle,
In ∆ABD, ∠1+ ∠2 + ∠3=180° …………(i)
In ∆ABD, ∠4+ ∠5 + ∠6=180° …………(ii)
Adding equation (i) and (ii),
∠1+ ∠2 + ∠3+∠4+ ∠5 + ∠6 = 360°
⟹ ∠1+ ∠2 + (∠3+ ∠4 ) + ∠5+ ∠6 = 360°
⟹ ∠A+ ∠B + ∠C+∠D = 360°
Hence proved.
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
(a) Find (a) x+y+ z (b)Find x+y+z+w
Since sum of linear pair angles is 180° ∴ 90° + x = 180° ⇒ x+180°-90° = 90° and z+30°= 180° ⇒ z=180°-30°=150° also y=90°+30°=120° [Exterior angle property] ∴ x+y+x = 90°+120°+150°=360° (b) Using angle sum property of a quadrilateral, 60°+80°+120°+n=360° ⇒ 260°+n=360° ⇒ n=360°-260° ⇒ n=100° Since sumRead more
Since sum of linear pair angles is 180°
∴ 90° + x = 180°
⇒ x+180°-90° = 90°
and z+30°= 180°
⇒ z=180°-30°=150°
also y=90°+30°=120°
[Exterior angle property]
∴ x+y+x = 90°+120°+150°=360°
(b) Using angle sum property of a quadrilateral,
60°+80°+120°+n=360°
⇒ 260°+n=360°
⇒ n=360°-260°
⇒ n=100°
Since sum of linear pair angles is 180°
∴ w+100=180° …………(i)
x+120°=180° …………(ii)
y+80°=180° …………(iii)
z+60°=180° …………(iv)
Adding eq. (i), (ii), (iii) and (iv),
⇒ x + y + z + w + 100 °+ 120° + 80° + 60° = 180° + 180° + 180° + 180°
⇒ x + y + z + w +360°=720°
⇒ x + y + z + w = 720°-360°
⇒ x + y + z + w = 360°
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
See lessFind the angle measures in the following figures: x
(a) Using angle sum property of a quadrilateral, 50°+130°+120°+x=360° ⇒ 300°+x=360° ⇒ x=360°-300° ⇒x=360° (b) Using angle sum property of a quadrilateral, 90°+60°+70°+x=360° ⇒ 220°+x=360° ⇒x=360°-220° ⇒x=140° (c) First base interior angle = 180°-70°=110° Second base interior angle = 180°-60°=120° ThRead more
(a) Using angle sum property of a quadrilateral,
50°+130°+120°+x=360°
⇒ 300°+x=360°
⇒ x=360°-300° ⇒x=360°
(b) Using angle sum property of a quadrilateral,
90°+60°+70°+x=360°
⇒ 220°+x=360°
⇒x=360°-220° ⇒x=140°
(c) First base interior angle = 180°-70°=110°
Second base interior angle = 180°-60°=120°
There are 5 sides, = 5 n
∴ Angle sum of a polygon = (n-2)x180°
= (5-2)x180°=3×180 =540°
∴ 30°+x+110°+120°+x=540°
⇒ 260°+2x+540° ⇒2x=540°-260°
⇒ 2x=280° ⇒x=140°
(d) Angle sum of a polygon = (n-2)x180°
= (5-2)x180°= 3×180° = 540°
∴ x+x+x+x+x =540°
⇒ 5x=540° ⇒ x=108°
Hence each interior angle is 108°
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
See lessWhat is a regular polygon? State the name of a regular polygon of: (a) 3 sides (b) 4 sides (c) 6 sides
A regular polygon: A polygon having all sides of equal length and the interior angles of equal size is known as regular polygon. (i) 3 sides. Polygon having three sides is called a triangle. (ii) 4 sides. Polygon having four sides is called a quadrilateral. (iii) 6 sides. Polygon having six sides isRead more
A regular polygon: A polygon having all sides of equal length and the interior angles of
equal size is known as regular polygon.
(i) 3 sides. Polygon having three sides is called a triangle.
(ii) 4 sides. Polygon having four sides is called a quadrilateral.
(iii) 6 sides. Polygon having six sides is called a hexagon.
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/
Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.) What can you say about the angle sum of a convex polygon with number of sides?
(a) When = 7, then n Angle sum of a polygon = (n-2) x 180° = (7-2) x 180°=5x180° = 900° (b) When n = 8, then Angle sum of a polygon = (n-2) x 180°=(8-2) x 180° = 6 x 180°=1080° (b) When n = 8, then Angle sum of a polygon = (n-2) x 180°=(10-2) x 180° = 8 x 180°=1440° (d) When n = n, then, angle sum oRead more
(a) When = 7, then n
Angle sum of a polygon = (n-2) x 180° = (7-2) x 180°=5×180° = 900°
(b) When n = 8, then
Angle sum of a polygon = (n-2) x 180°=(8-2) x 180° = 6 x 180°=1080°
(b) When n = 8, then
Angle sum of a polygon = (n-2) x 180°=(10-2) x 180° = 8 x 180°=1440°
(d) When n = n, then, angle sum of a polygon = (n-2) x 180°
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/
What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try)
Let ABCD is a convex quadrilateral, then we draw a diagonal AC which divides the quadrilateral in two triangles. ∠A +∠B+ ∠C + ∠D = ∠1+ ∠6+ ∠5+ ∠4+ ∠3+ ∠2 = ( ∠1+ ∠2+ ∠3) +(∠4+ ∠5+ ∠6) = [By Angle sum property of triangle] 180°+180° = 360° Hence, the sum of measures of the triangles of a convex quadrRead more
Let ABCD is a convex quadrilateral, then we draw a diagonal AC which divides the quadrilateral in two triangles.
∠A +∠B+ ∠C + ∠D = ∠1+ ∠6+ ∠5+ ∠4+ ∠3+ ∠2
= ( ∠1+ ∠2+ ∠3) +(∠4+ ∠5+ ∠6)
= [By Angle sum property of triangle] 180°+180°
= 360°
Hence, the sum of measures of the triangles of a convex quadrilateral is 360°.
Yes, if quadrilateral is not convex then, this property will also be applied.
Let ABCD is a non-convex quadrilateral and join BD, which also divides the quadrilateral in two triangles.
Using angle sum property of triangle,
In ∆ABD, ∠1+ ∠2 + ∠3=180° …………(i)
In ∆ABD, ∠4+ ∠5 + ∠6=180° …………(ii)
Adding equation (i) and (ii),
∠1+ ∠2 + ∠3+∠4+ ∠5 + ∠6 = 360°
⟹ ∠1+ ∠2 + (∠3+ ∠4 ) + ∠5+ ∠6 = 360°
⟹ ∠A+ ∠B + ∠C+∠D = 360°
Hence proved.
Class 8 Maths Chapter 3 Exercise 3.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/