To construct: ∆ABC where m∠A = 60 , m∠B = 30 and AB = 5.8 cm. Steps of construction: (a) Draw a line segment AB = 5.8 cm. (b) At point A, draw an angle ∠YAB = 60 with the help of compass. (c) At point B, draw ∠XBA = 30 with the help of compass. (d) AY and BX intersect at the point C. It is the requiRead more
To construct: ∆ABC where m∠A = 60 , m∠B = 30 and AB = 5.8 cm.
Steps of construction:
(a) Draw a line segment AB = 5.8 cm.
(b) At point A, draw an angle ∠YAB = 60 with the help of compass.
(c) At point B, draw ∠XBA = 30 with the help of compass.
(d) AY and BX intersect at the point C.
It is the required triangle ABC.
To construct: ∆ABC where BC = 7.5 cm, AC = 5 cm and m∠C = 60 . Steps of construction: (a) Draw a line segment BC = 7.5 cm. (b) At point C, draw an angle of 60 with the help of protractor, i.e., ∠XCB = 60 . (c) Taking C as centre and radius 5 cm, draw an arc, which cuts XC at the point A. (d) Join ABRead more
To construct: ∆ABC where BC = 7.5 cm, AC = 5 cm and m∠C = 60 .
Steps of construction:
(a) Draw a line segment BC = 7.5 cm.
(b) At point C, draw an angle of 60 with the help of protractor, i.e., ∠XCB = 60 .
(c) Taking C as centre and radius 5 cm, draw an arc, which cuts XC at the point A.
(d) Join AB
It is the required triangle ABC.
To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and ∠Q = 110°. Steps of construction: (a) Draw a line segment QR = 6.5 cm. (b) At point Q, draw an angle of 110° with the help of protractor, i.e., ∠YQR = 110°. (c) Taking Q as centre, draw an arc with radius 6.5 cm, which cuts QY at poiRead more
To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and ∠Q = 110°.
Steps of construction:
(a) Draw a line segment QR = 6.5 cm.
(b) At point Q, draw an angle of 110° with the
help of protractor, i.e., ∠YQR = 110°.
(c) Taking Q as centre, draw an arc with radius
6.5 cm, which cuts QY at point P.
(d) Join PR
It is the required isosceles triangle PQR.
To construct: ∆DEF where DE = 5 cm, DF = 3 cm and m∠EDF = 90°. Steps of construction: (a) Draw a line segment DF = 3 cm. (b) At point D, draw an angle 90° of with the help of compass i.e., ∠XDF = 90°. (c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E. (d) Join EF. It iRead more
To construct: ∆DEF where DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
Steps of construction: (a) Draw a line segment DF = 3 cm.
(b) At point D, draw an angle 90° of with the help of compass i.e., ∠XDF = 90°.
(c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E.
(d) Join EF.
It is the required right angled triangle DEF.
To construct: ∆ABC in which AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Steps of construction: (a) Draw a line segment BC = 6 cm. (b) Taking B as centre and radius 2.5 cm, draw an arc. (c) Similarly, taking C as centre and radius 6.5 cm, draw another arc which intersects first arc at point A. (d) Join ARead more
To construct: ∆ABC in which AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm.
Steps of construction:
(a) Draw a line segment BC = 6 cm.
(b) Taking B as centre and radius 2.5 cm, draw an arc.
(c) Similarly, taking C as centre and radius 6.5 cm, draw another arc which
intersects first arc at point A.
(d) Join AB and AC.
(e) Measure angle B with the help of protractor.
It is the required ∆ABC where ∠B = 80°.
Construct ∆ABC, given m∠A =60°,m∠B =30°, and AB = 5.8 cm.
To construct: ∆ABC where m∠A = 60 , m∠B = 30 and AB = 5.8 cm. Steps of construction: (a) Draw a line segment AB = 5.8 cm. (b) At point A, draw an angle ∠YAB = 60 with the help of compass. (c) At point B, draw ∠XBA = 30 with the help of compass. (d) AY and BX intersect at the point C. It is the requiRead more
To construct: ∆ABC where m∠A = 60 , m∠B = 30 and AB = 5.8 cm.
Steps of construction:
(a) Draw a line segment AB = 5.8 cm.
(b) At point A, draw an angle ∠YAB = 60 with the help of compass.
(c) At point B, draw ∠XBA = 30 with the help of compass.
(d) AY and BX intersect at the point C.
It is the required triangle ABC.
Class 7 Maths Chapter 10 Exercise 10.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/
Construct ∆ABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
To construct: ∆ABC where BC = 7.5 cm, AC = 5 cm and m∠C = 60 . Steps of construction: (a) Draw a line segment BC = 7.5 cm. (b) At point C, draw an angle of 60 with the help of protractor, i.e., ∠XCB = 60 . (c) Taking C as centre and radius 5 cm, draw an arc, which cuts XC at the point A. (d) Join ABRead more
To construct: ∆ABC where BC = 7.5 cm, AC = 5 cm and m∠C = 60 .
Steps of construction:
(a) Draw a line segment BC = 7.5 cm.
(b) At point C, draw an angle of 60 with the help of protractor, i.e., ∠XCB = 60 .
(c) Taking C as centre and radius 5 cm, draw an arc, which cuts XC at the point A.
(d) Join AB
It is the required triangle ABC.
Class 7 Maths Chapter 10 Exercise 10.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/
Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110°.
To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and ∠Q = 110°. Steps of construction: (a) Draw a line segment QR = 6.5 cm. (b) At point Q, draw an angle of 110° with the help of protractor, i.e., ∠YQR = 110°. (c) Taking Q as centre, draw an arc with radius 6.5 cm, which cuts QY at poiRead more
To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and ∠Q = 110°.
Steps of construction:
(a) Draw a line segment QR = 6.5 cm.
(b) At point Q, draw an angle of 110° with the
help of protractor, i.e., ∠YQR = 110°.
(c) Taking Q as centre, draw an arc with radius
6.5 cm, which cuts QY at point P.
(d) Join PR
It is the required isosceles triangle PQR.
Class 7 Maths Chapter 10 Exercise 10.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/
Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
To construct: ∆DEF where DE = 5 cm, DF = 3 cm and m∠EDF = 90°. Steps of construction: (a) Draw a line segment DF = 3 cm. (b) At point D, draw an angle 90° of with the help of compass i.e., ∠XDF = 90°. (c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E. (d) Join EF. It iRead more
To construct: ∆DEF where DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
Steps of construction: (a) Draw a line segment DF = 3 cm.
(b) At point D, draw an angle 90° of with the help of compass i.e., ∠XDF = 90°.
(c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E.
(d) Join EF.
It is the required right angled triangle DEF.
Class 7 Maths Chapter 10 Exercise 10.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/
Construct ∆ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
To construct: ∆ABC in which AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Steps of construction: (a) Draw a line segment BC = 6 cm. (b) Taking B as centre and radius 2.5 cm, draw an arc. (c) Similarly, taking C as centre and radius 6.5 cm, draw another arc which intersects first arc at point A. (d) Join ARead more
To construct: ∆ABC in which AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm.
Steps of construction:
(a) Draw a line segment BC = 6 cm.
(b) Taking B as centre and radius 2.5 cm, draw an arc.
(c) Similarly, taking C as centre and radius 6.5 cm, draw another arc which
intersects first arc at point A.
(d) Join AB and AC.
(e) Measure angle B with the help of protractor.
It is the required ∆ABC where ∠B = 80°.
Class 7 Maths Chapter 10 Exercise 10.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/