# Practical Geometry| NCERT Solutions| Class 7

NCERT Solutions of Chapter: Practical Geometry. NCERT Solutions along with worksheets and notes for Class 7.

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.

Ans-

Construction:-

(i)Draw a line AB. Take a point C outside it and take a point P on it.

(ii)Join point C to point P, with P as centre and a convenient radius; draw an arc cutting AB at D and PC at E.

(iii)Now with same radius and C as centre, draw an arc FG cutting PC at H.

(iv)Place the tip of the compasses at D and adjust the opening so that the pencil tip is at E.

It does not need any drawings.

(v)With the same opening and with H as centre, draw an arc cutting the arc FG at I.

(vi) Join C and I to draw a parallel 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.

Ans-

Construction:-

(i)Draw a line l. Take two points P and Q on it.

(ii) Take P as centre and a radius of more than half of PQ. Draw an arc upside the line l.

(iii) Now, with same radius and Q as center draw an arc upside the line l, cutting previous arc at R.

(iv)Join R to line l at S. RS is a perpendicular to line l.

(v)Take point X on RS at 4 cm away from line l. Take another point Y on PQ.

(vi)Join X and Y.

(vii) With Y as center and a convenient radius, draw an arc cutting XY at F and PQ at E.

(viii) With X as center and the same radius, draw an arc AB, cutting RS at C.

(ix)Place the pointed tip of the compasses at E and adjust the opening so that the pencil tip is at F.

It does not need drawing.

(x)With the same opening and with C as enter, draw an arc cutting the arc AB at D.

(xi) Now, join D and X to draw a line m.

(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?

Ans-

(i) Draw a line l. Take a point P outside it. Take a point T on line l. Join P and T.

(ii) With T as center and a convenient radius, draw an arc cutting PT at A and Line l at B.

(iii) Now with P as center and the same radius, draw an arc CD cutting PT at E.

(iv) Place the pointed tip of the compasses at B and adjust the opening so that the pencil tip is at A.

It does not need drawing.

(v) With the same opening and E as center, draw an arc cutting the arc CD at Z.

(vi) Now join P and Z to draw a line m.

(vii) Take a point R on line m. Join P to a point Q on line l.  With P as center and a convenient radius, draw an arc cutting PQ at G and line m at F.

(viii) Now with R as center and the same radius, draw an arc HI.

(ix) Place the pointed tip of the compasses at G and adjust the opening so that the pencil tip is at F.

It does not need any drawing.

(x) With the same opening and R as center, draw an arc cutting the arc HI at J.

(xi) Now join R and J to draw a line RS.

Exercise 10.2

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

Ans-

Construction of ΔXYZ-

(i) Draw a line XY.

(ii) With center X, draw an arc of 4.5 cm.

(iii) With center Y, draw an arc of 6 cm.

(iv) Mark the point of intersection of arcs as Z. Join ZY and ZX. XYZ is required triangle.

(2) Construct an equilateral triangle of side 5.5 cm.

Ans-

Construction of ΔXYZ

(i) Draw a line XY=5.5cm.

(ii) With center X, draw an arc of 5.5 cm.

(iii) With center Y, draw an arc of 5.5 cm.

(iv) Mark the point of intersection of arcs as Z. Join ZY and ZX. XYZ is required triangle.

(3)Draw PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

Ans-

Construction of ΔPQR

(i) Draw a line QR=3.5 cm.

(ii) With center Q, draw an arc of 4 cm.

(iii) With center R, draw an arc of 4 cm.

(iv) Mark the point of intersection of arcs as P. Join PQ and PR.

PQR is required isosceles triangle.

(4) Construct ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

Ans-

Construction of ΔABC

(i)Draw a line BC=6 cm.

(ii) With center B, draw an arc of 2 cm.

(iii) With center C, draw an arc of 6.5 cm.

(iv)Mark the point of intersection of arcs as A. Join AB and AC.

ΔABC is required right angle triangle. So, B = 900

Exercise 10.3

(1)Construct ΔDEF such that DE = 5 cm, DF = 3 cm and m ∠EDF = 900.

Ans-

Construction-

(i)Draw a line segment DE of length 5 cm.

(ii) At D, draw DX making 900 with DE.

(iii) With D as center, draw an arc of radius 3 cm. It cuts DX at the point F.

(iv) Join EF, ΔDEF is now obtained.

(2) Construct an isosceles triangle in which the lengths of each of its equal sides are 6.5 cm and the angle between them is 1100.

Construction-

(i) Draw a line segment DE of length 6.5 cm.

(ii) At D, draw DX making 1100 with DE.

(iii) With D as center, draw an arc of radius 6.5 cm. It cuts DX at the point F.

(iv) Join EF, ΔDEF is now obtained.

(3) Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m ∠C = 600

Ans-

Construction-

(i)Draw a line segment BC of length 7.5 cm.

(ii) At C, draw CX making 600 with BC.

(iii) With C as center, draw an arc of radius 5 cm. It cuts CX at the point A.

(iv) Join AB, ΔABC is now obtained.

Exercise 10.4

(1) Construct ΔABC, given m ∠A = 600, m ∠B = 300 and AB = 5.8 cm

Ans-

Construction-

(i) Draw AB of length 5.8 cm.

(ii) At point A, draw a ray XA making an angle of 600 with line AB.

(iii) At point B, draw a ray BY making an angle of 300 with line AB.

(iv) C has to lie on both the rays XA and YB. So, the point of intersection of the two rays is C.

ΔABC is required triangle.

(2) Construct ΔPQR if PQ = 5 cm, m ∠PQR = 1050 and m ∠QRP = 400

Ans-

Construction-

(i)Draw PQ of length 5 cm.

(ii)

At point P, draw a ray XP.

∠PQR+ ∠QRP + ∠QPR = 1800

1050+ 400 +  ∠QPR = 1800

∠QPR = 1800 – 1450

∠QPR = 350

Therefore, make an angle of 350 with line PQ.

(iii) At point Q, draw a ray YQ making an angle of 1050 with line PQ.

(iv) R has to lie on both the rays XP and YQ. So, the point of intersection of the two rays is R.

ΔPQR is required triangle.

Angle sum property

Exercise 10.5

(1)Construct the right angled ΔPQR, where m ∠Q = 900, QR = 8 cm and PR = 10 cm.

Ans-

Construction-

(i) Draw QR of length 8 cm.

(ii) At Q draw QX ⊥ QR.

(iii) With R as center, draw an arc of radius 10 cm.

(iv) P has to be on the perpendicular line QX as well as on the arc drawn with centre R.

Therefore, P is the meeting point of these two.

ΔPQR is now obtained.

(2)Construct right angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

Ans-

Construction-

(i) Draw QR of length 4 cm.

(ii) At Q draw QX ⊥ QR.

(iii) With R as center, draw an arc of radius 6 cm.

(iv) P has to be on the perpendicular line QX as well as on the arc drawn with centre R.

Therefore, P is the meeting point of these two.

ΔPQR is now obtained.

(3)Construct an isosceles right angled triangle ABC, where m ∠ACB = 900 and AC = 6 cm.

Ans-

Construction-

(i) Draw AC of length 6 cm.

(ii) At C draw CX ⊥ AC.

(iii) With C as center, draw an arc of radius 6 cm. The arc will cut CX at point B.

(iv) B has to be on the perpendicular line CX as well as on the arc drawn with centre A.

Therefore, B is the meeting point of these two.

ΔABC is now obtained.

Helping Topics

Practical Geometry

Worksheet Class 7