# NCERT Solutions for Class 7 Maths Ch 10 Practical Geometry

Chapter 10 Practical Geometry Class 7 Maths NCERT Solutions is provided here that will help you in finding the accurate and detailed answers of every questions given in the chapter, Whenever you'll be in doubt or finding difficulty while solving a specific question you can take help from the given NCERT Solutions for Class 7. These NCERT Solutions will be helpful in knowing the important points inside the chapter. Exercise 10.1

1. Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.

To construct: A line, parallel to given line by using ruler and compasses.
Steps of construction:
(a) Draw a line-segment AB and take a point C outside AB.
(b) Take any point D on AB and join C to D.
(c) With D as centre and take convenient radius, draw an arc cutting AB at E and CD at F.
(d) With C as centre and same radius as in step 3, draw an arc GH cutting CD at I.
(e) With the same arc EF, draw the equal arc cutting GH at J.
(f) Join JC to draw a line l. This the required line. 2. 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.
Given figure is the required construction. 3. 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, 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. Exercise 10.2

1. Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

To construct:  ΔXYZ, 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 ΔXYZ. 2. Construct an equilateral triangle of side 5.5 cm.

To construct:  A ΔABC 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 ΔABC. 3. Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

To construction: ΔPQR, 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 ΔPQR. 4. Construct ΔABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

To construct:  ΔABC 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 ΔABC where ∠B = 80°. Exercise 10.3

1. Construct ΔDEF such that DE = 5 cm, DF = 3 cm and  ∠EDF = 90°.

To construct: ΔDEF 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 of 90o 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. 2. 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 point P.
(d) Join PR
Given figure is the required isosceles triangle PQR. 3. Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m ∠C = 60° .

To construct: ΔABC 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. Exercise 10.4

1. Construct ΔABC, given m∠A = 60°, m∠B = 30° and AB = 5,8 cm.

To construct: ΔABC, given 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 Z YAB = 60°= with the help of a compass.
(c) At point B, draw Z XBA = 30° with the help of a compass.
(d) AY and BX intersect at the point C.
Given figure is the required triangle ABC. 2. Construct Δ PQR if PQ = 5 cm, m∠PQR = 105° and m∠.QRP = 40°.

Given:

m∠ PQR = 105° and m∠ QRP = 40°
We know that sum of angles of a triangle is 180°.
∴ m∠PQR + m∠QRP + m∠ QPR = 180°
⇒ 105° + 40° + m∠QPR= 180°
⇒ 145° + m∠QPR= 180°
⇒  m∠QPR= 180° - 145°
⇒ m∠QPR = 35°

To construct: ΔPQR where m∠P = 35°, m∠Q = 105° and PQ = 5 cm.

Steps of construction:
(a) Draw a line segment PQ = 5 cm,
(b) At point P, draw ∠ XPQ = 35° with the help of protractor.
(c) At point Q, draw ∠ YQP = 105° with the help of protractor.
(d) XP and YQ intersect at point R.
It is the required triangle PQR. 3. Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠ E= 110° and m∠F= 80°. Justify your answer.

Given: In ΔDEF, m∠E = 110° and m∠T = 80°
Using angle sum property of triangle
∠D + ∠E + ∠F = 180°
⇒∠D + 110°+ 80° = 180°
⇒ ∠D + 190° = 180°
⇒ ∠D = 180° - 190° = -10°
Which is not possible.

Exercise 10.5

1. A right angled triangle PQR where m∠Q = 9ff, QR= 8 cm and PQ = 10 cm.

To construct: Construct the right angled Δ PQR, where m∠Q = 90", QR = 8 cm and PR = 10 cm.
Steps of construction:
(a)  Draw a line segment QR = 8 cm,
(b) At point Q, draw QX ⊥ QR,
(c) Taking R as centre, draw an arc of radius 10 cm.
(d) This arc cuts QX at point P.
(e) foin PQ.
Given figure is the required right-angled triangle PQR, 2. Construct a right-angled triangle whose hypotenuse is 6 cm long and one the legs is 4 cm long.

To construct: A right-angled triangle DEF where DF = 6 cm and EF = 4 cm.
Steps of construction:
(a) Draw a line segment EF = 4 cm.
(b) At point Q, draw EX ⊥ EF.
(c) Taking F as centre and radius 6 cm, draw an arc. (Hypotenuse)
(d) This arc cuts the EX at point D.
(e) Join DF.
It is the required right-angled triangle DEF. 3. Construct an isosceles right angled triangle ABC, where m∠ACB = 90° and AC = 6 cm. 