To construction: ∆PQR, in which PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. Steps of construction: (a) Draw a line segment QR = 3.5 cm. (b) Taking Q as centre and radius 4 cm, draw an arc. (c) Similarly, taking R as centre and radius 4 cm, draw an another arc which intersects first arc at P. (d) Join PQ aRead more
To construction: ∆PQR, in which PQ = 4 cm, QR = 3.5 cm and PR = 4 cm.
Steps of construction:
(a) Draw a line segment QR = 3.5 cm.
(b) Taking Q as centre and radius 4 cm, draw an arc.
(c) Similarly, taking R as centre and radius 4 cm, draw an another arc which
intersects first arc at P.
(d) Join PQ and PR.
It is the required isosceles ∆PQR
To construct: A ∆ABC where AB = BC = CA = 5.5 cm Steps of construction: (a) Draw a line segment BC = 5.5 cm (b) Taking points B and C as centers and radius 5.5 cm, draw arcs which intersect at point A. (c) Join AB and AC. It is the required ∆ABC. Class 7 Maths Chapter 10 Exercise 10.2 for more answeRead more
To construct: A ∆ABC where AB = BC = CA = 5.5 cm
Steps of construction:
(a) Draw a line segment BC = 5.5 cm
(b) Taking points B and C as centers and radius 5.5 cm, draw arcs which intersect at
point A.
(c) Join AB and AC.
It is the required ∆ABC.
To construct: ∆XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm. Steps of construction: (a) Draw a line segment YZ = 5 cm. (b) Taking Z as centre and radius 6 cm, draw an arc. (c) Similarly, taking Y as centre and radius 4.5 cm, draw another arc which intersects first arc at point X. (d) Join XY andRead more
To construct: ∆XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Steps of construction:
(a) Draw a line segment YZ = 5 cm.
(b) Taking Z as centre and radius 6 cm, draw an arc.
(c) Similarly, taking Y as centre and radius 4.5 cm, draw another arc which
intersects first arc at point X.
(d) Join XY and XZ.
It is the required ∆XYZ.
To construct: A pair of parallel lines intersecting other part of parallel lines. Steps of construction: (a) Draw a line l and take a point P outside of l . (b) Take point Q on line l and join PQ. (c) Make equal angle at point P such that ∠Q = ∠P. (d) Extend line at P to get line m. (e) Similarly, tRead more
To construct: A pair of parallel lines intersecting other part of parallel lines.
Steps of construction:
(a) Draw a line l and take a point P outside of l .
(b) Take point Q on line l and join PQ.
(c) Make equal angle at point P such that ∠Q = ∠P.
(d) Extend line at P to get line m.
(e) Similarly, take a point R online m, at point R, draw angles such that ∠P = ∠R.
(f) Extended line at R which intersects at S online l. Draw line RS.
Thus, we get parallelogram PQRS
To construct: A line parallel to given line when perpendicular line is also given. Steps of construction: (a) Draw a line l and take a point P on it. (b) At point P, draw a perpendicular line n. (c) Take PX = 4 cm on line n. (d) At point X, again draw a perpendicular line m. It is the required constRead more
To construct: A line parallel to given line when perpendicular line is also given.
Steps of construction:
(a) Draw a line l and take a point P on it.
(b) At point P, draw a perpendicular line n.
(c) Take PX = 4 cm on line n.
(d) At point X, again draw a perpendicular line m.
It is the required construction
Draw ∆PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
To construction: ∆PQR, in which PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. Steps of construction: (a) Draw a line segment QR = 3.5 cm. (b) Taking Q as centre and radius 4 cm, draw an arc. (c) Similarly, taking R as centre and radius 4 cm, draw an another arc which intersects first arc at P. (d) Join PQ aRead more
To construction: ∆PQR, in which PQ = 4 cm, QR = 3.5 cm and PR = 4 cm.
Steps of construction:
(a) Draw a line segment QR = 3.5 cm.
(b) Taking Q as centre and radius 4 cm, draw an arc.
(c) Similarly, taking R as centre and radius 4 cm, draw an another arc which
intersects first arc at P.
(d) Join PQ and PR.
It is the required isosceles ∆PQR
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/
Construct an equilateral triangle of side 5.5 cm.
To construct: A ∆ABC where AB = BC = CA = 5.5 cm Steps of construction: (a) Draw a line segment BC = 5.5 cm (b) Taking points B and C as centers and radius 5.5 cm, draw arcs which intersect at point A. (c) Join AB and AC. It is the required ∆ABC. Class 7 Maths Chapter 10 Exercise 10.2 for more answeRead more
To construct: A ∆ABC where AB = BC = CA = 5.5 cm
Steps of construction:
(a) Draw a line segment BC = 5.5 cm
(b) Taking points B and C as centers and radius 5.5 cm, draw arcs which intersect at
point A.
(c) Join AB and AC.
It is the required ∆ABC.
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/
Construct ∆XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
To construct: ∆XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm. Steps of construction: (a) Draw a line segment YZ = 5 cm. (b) Taking Z as centre and radius 6 cm, draw an arc. (c) Similarly, taking Y as centre and radius 4.5 cm, draw another arc which intersects first arc at point X. (d) Join XY andRead more
To construct: ∆XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Steps of construction:
(a) Draw a line segment YZ = 5 cm.
(b) Taking Z as centre and radius 6 cm, draw an arc.
(c) Similarly, taking Y as centre and radius 4.5 cm, draw another arc which
intersects first arc at point X.
(d) Join XY and XZ.
It is the required ∆XYZ.
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/
Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?
To construct: A pair of parallel lines intersecting other part of parallel lines. Steps of construction: (a) Draw a line l and take a point P outside of l . (b) Take point Q on line l and join PQ. (c) Make equal angle at point P such that ∠Q = ∠P. (d) Extend line at P to get line m. (e) Similarly, tRead more
To construct: A pair of parallel lines intersecting other part of parallel lines.
Steps of construction:
(a) Draw a line l and take a point P outside of l .
(b) Take point Q on line l and join PQ.
(c) Make equal angle at point P such that ∠Q = ∠P.
(d) Extend line at P to get line m.
(e) Similarly, take a point R online m, at point R, draw angles such that ∠P = ∠R.
(f) Extended line at R which intersects at S online l. Draw line RS.
Thus, we get parallelogram PQRS
Class 7 Maths Chapter 10 Exercise 10.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/
Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.
To construct: A line parallel to given line when perpendicular line is also given. Steps of construction: (a) Draw a line l and take a point P on it. (b) At point P, draw a perpendicular line n. (c) Take PX = 4 cm on line n. (d) At point X, again draw a perpendicular line m. It is the required constRead more
To construct: A line parallel to given line when perpendicular line is also given.
Steps of construction:
(a) Draw a line l and take a point P on it.
(b) At point P, draw a perpendicular line n.
(c) Take PX = 4 cm on line n.
(d) At point X, again draw a perpendicular line m.
It is the required construction
Class 7 Maths Chapter 10 Exercise 10.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-10/