NCERT Solutions for Class 7 Maths Practical Geometry [Old Book]
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Book Name: Maths
Subject: Maths
Chapter Number: 10
Chapter Name: Practical Geometry
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Book Name: Maths
Subject: Maths
Chapter Number: 10
Chapter Name: Practical Geometry
You can also bookmark this page by clicking + Save to Library icon and refer to these NCERT Solutions anytime from your account. More Detail
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Page 1
NCERT
SOLUTIONS
CLASS - 7TH
aglase .co
SOLUTIONS
CLASS - 7TH
aglase .co
Page 2
Book : Mathematics Ncert Solutions | Chapter-10 Maths
Class : 7th
Subject : Maths
Chapter : 10
Chapter Name : Practical Geometry
Exercise 10.1
Q1 Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using
ruler and compasses only.
Answer. The steps of construction are as follows.
(i) Draw a line AB. Take a point P on it. Take a point C outside this line. Join C to P.
(ii) Taking P as centre and with a convenient radius, draw an arc intersecting line AB at point
Dand PC at point E.
(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at
H.
(iv) Adjust the compasses up to the length of DE. Without changing the opening of compasses
and taking H as the centre, draw an arc to intersect the previously drawn arc FG at point I.
Page 1 of 16 Aglasem Schools
Class : 7th
Subject : Maths
Chapter : 10
Chapter Name : Practical Geometry
Exercise 10.1
Q1 Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using
ruler and compasses only.
Answer. The steps of construction are as follows.
(i) Draw a line AB. Take a point P on it. Take a point C outside this line. Join C to P.
(ii) Taking P as centre and with a convenient radius, draw an arc intersecting line AB at point
Dand PC at point E.
(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at
H.
(iv) Adjust the compasses up to the length of DE. Without changing the opening of compasses
and taking H as the centre, draw an arc to intersect the previously drawn arc FG at point I.
Page 1 of 16 Aglasem Schools
Page 3
Book : Mathematics Ncert Solutions | Chapter-10 Maths
(v) Join the points C and I to draw a line 'l'.
This is the required line which is parallel to line AB.
Page : 196 , Block Name : Exercise 10.1
Q2 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.
Answer. The steps of construction are as follows.
(i) Draw a line I and take a point P on line I. Then, draw a perpendicular at point P.
(ii) Adjusting the compasses up to the length of 4 cm, draw an arc to intersect
this perpendicular at point X. Choose any point Y on line l. Join X to Y.
(iii) Taking Y as centre and with a convenient radius, draw an arc intersecting I at A and XY at
B.
(iv) Taking X as centre and with the same radius as before, draw an arc CD cutting XY at E.
Page 2 of 16 Aglasem Schools
(v) Join the points C and I to draw a line 'l'.
This is the required line which is parallel to line AB.
Page : 196 , Block Name : Exercise 10.1
Q2 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.
Answer. The steps of construction are as follows.
(i) Draw a line I and take a point P on line I. Then, draw a perpendicular at point P.
(ii) Adjusting the compasses up to the length of 4 cm, draw an arc to intersect
this perpendicular at point X. Choose any point Y on line l. Join X to Y.
(iii) Taking Y as centre and with a convenient radius, draw an arc intersecting I at A and XY at
B.
(iv) Taking X as centre and with the same radius as before, draw an arc CD cutting XY at E.
Page 2 of 16 Aglasem Schools
Page 4
Book : Mathematics Ncert Solutions | Chapter-10 Maths
(v) Adjust the compasses up to the length of AB. Without changing the opening of
compassesand taking E as the centre, draw an arc to intersect the previously drawn arc CD
at point F.
(vi) Join the points X and F to draw a line m.
Line m is the required line which is parallel to line l.
Page : 196 , Block Name : Exercise 10.1
Q3 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?
Answer. (i) Draw a line I and take a point A on it. Take a point P not on I and join A to P.
(ii) Taking A as centre and with a convenient radius, draw an arc cutting I at B and AP at C.
(iii) Taking P as centre and with the same radius as before, draw an arc DE to intersect AP at F.
Page 3 of 16 Aglasem Schools
(v) Adjust the compasses up to the length of AB. Without changing the opening of
compassesand taking E as the centre, draw an arc to intersect the previously drawn arc CD
at point F.
(vi) Join the points X and F to draw a line m.
Line m is the required line which is parallel to line l.
Page : 196 , Block Name : Exercise 10.1
Q3 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?
Answer. (i) Draw a line I and take a point A on it. Take a point P not on I and join A to P.
(ii) Taking A as centre and with a convenient radius, draw an arc cutting I at B and AP at C.
(iii) Taking P as centre and with the same radius as before, draw an arc DE to intersect AP at F.
Page 3 of 16 Aglasem Schools
Page 5
Book : Mathematics Ncert Solutions | Chapter-10 Maths
(iv) Adjust the compasses up to the length of BC. Without changing the opening of compasses
and taking F as the centre, draw an arc to intersect the previously drawn arc DE at point G.
(v) Join P to G to draw a line m. Line m will be parallel to line l.
(vi) Join P to any point Q on line l. Choose another point R on line m. Similarly, a line can
bedrawn through point R and parallel to PQ.
Let it meet line I at point S. In quadrilateral PQSR, opposite lines are parallel to each other. PQ
II RS and PR II QS Thus, &mnSq1 PQSR is a parallelogram.
Page : 196 , Block Name : Exercise 10.1
Exercise 10.2
Q1 Construct ∆XYZ in which XY= 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Answer. The rough gure of this triangle is as follows.
Page 4 of 16 Aglasem Schools
(iv) Adjust the compasses up to the length of BC. Without changing the opening of compasses
and taking F as the centre, draw an arc to intersect the previously drawn arc DE at point G.
(v) Join P to G to draw a line m. Line m will be parallel to line l.
(vi) Join P to any point Q on line l. Choose another point R on line m. Similarly, a line can
bedrawn through point R and parallel to PQ.
Let it meet line I at point S. In quadrilateral PQSR, opposite lines are parallel to each other. PQ
II RS and PR II QS Thus, &mnSq1 PQSR is a parallelogram.
Page : 196 , Block Name : Exercise 10.1
Exercise 10.2
Q1 Construct ∆XYZ in which XY= 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Answer. The rough gure of this triangle is as follows.
Page 4 of 16 Aglasem Schools
Page 6
Book : Mathematics Ncert Solutions | Chapter-10 Maths
The required triangle is constructed as follows.
(i) Draw a line segment YZ of length 5 cm.
(ii) Point X is at a distance of 4.5 cm from point Y. Therefore, taking point Y as centre, draw an
arc of 4.5 cm radius.
(iii) Point X is at a distance of 6 cm from point Z. Therefore, taking point Z as centre, draw an
arc of 6 cm radius. Mark the point of intersection of the arcs as X. Join XY and XZ.
XYZ is the required triangle.
Page : 199 , Block Name : Exercise 10.2
Q2 Construct an equilateral triangle of side 5.5 cm.
Answer. An equilateral triangle of side 5.5 cm has to be constructed. We know that all sides of
an equilateral triangle are of equal length. Therefore, a triangle ABC has to be
constructed with AB :: BC = CA 5.5 cm. The steps of construction are as follows.
(i) Draw a line segment BC of length 5.5 cm.
(ii) Taking point B as centre, draw an arc of 5.5 cm radius.
Page 5 of 16 Aglasem Schools
The required triangle is constructed as follows.
(i) Draw a line segment YZ of length 5 cm.
(ii) Point X is at a distance of 4.5 cm from point Y. Therefore, taking point Y as centre, draw an
arc of 4.5 cm radius.
(iii) Point X is at a distance of 6 cm from point Z. Therefore, taking point Z as centre, draw an
arc of 6 cm radius. Mark the point of intersection of the arcs as X. Join XY and XZ.
XYZ is the required triangle.
Page : 199 , Block Name : Exercise 10.2
Q2 Construct an equilateral triangle of side 5.5 cm.
Answer. An equilateral triangle of side 5.5 cm has to be constructed. We know that all sides of
an equilateral triangle are of equal length. Therefore, a triangle ABC has to be
constructed with AB :: BC = CA 5.5 cm. The steps of construction are as follows.
(i) Draw a line segment BC of length 5.5 cm.
(ii) Taking point B as centre, draw an arc of 5.5 cm radius.
Page 5 of 16 Aglasem Schools
Page 7
Book : Mathematics Ncert Solutions | Chapter-10 Maths
(iii) Taking point C as centre, draw an arc of 5.5 cm radius to meet the previous arc at point A.
(iv) Join A to B and C.
ABC is the required equilateral triangle.
Page : 199 , Block Name : Exercise 10.2
Q3 Draw ∆PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Answer. The steps of construction are as follows.
(i) Draw a line segment QR of length 3.5 cm.
(ii) Taking point Q as centre, draw an arc of 4 cm radius.
(iii) Taking point R as centre, draw an arc of 4 cm radius to intersect the previous arc at point
P.
Page 6 of 16 Aglasem Schools
(iii) Taking point C as centre, draw an arc of 5.5 cm radius to meet the previous arc at point A.
(iv) Join A to B and C.
ABC is the required equilateral triangle.
Page : 199 , Block Name : Exercise 10.2
Q3 Draw ∆PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?
Answer. The steps of construction are as follows.
(i) Draw a line segment QR of length 3.5 cm.
(ii) Taking point Q as centre, draw an arc of 4 cm radius.
(iii) Taking point R as centre, draw an arc of 4 cm radius to intersect the previous arc at point
P.
Page 6 of 16 Aglasem Schools
Page 8
Book : Mathematics Ncert Solutions | Chapter-10 Maths
(iv) Join P to Q and R.
PQR is the required triangle. As the two sides of this triangle are of the same length (PQ = PR),
therefore, ?PQR is an isosceles triangle.
Page : 199 , Block Name : Exercise 10.2
Q4 Construct ∆ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Answer. The steps of construction are as follows.
(i) Draw a line segment BC of length 6 cm.
(ii) Taking point C as centre, draw an arc of 6.5 cm radius.
(iii) Taking point B as centre, draw an arc of radius 2.5 cm to meet the previous arc at point A.
(iv) Join A to B and C.
Page 7 of 16 Aglasem Schools
(iv) Join P to Q and R.
PQR is the required triangle. As the two sides of this triangle are of the same length (PQ = PR),
therefore, ?PQR is an isosceles triangle.
Page : 199 , Block Name : Exercise 10.2
Q4 Construct ∆ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Answer. The steps of construction are as follows.
(i) Draw a line segment BC of length 6 cm.
(ii) Taking point C as centre, draw an arc of 6.5 cm radius.
(iii) Taking point B as centre, draw an arc of radius 2.5 cm to meet the previous arc at point A.
(iv) Join A to B and C.
Page 7 of 16 Aglasem Schools
Page 9
Book : Mathematics Ncert Solutions | Chapter-10 Maths
ABC is the required triangle. ∠B can be measured with the help of protractor. It comes to 90°.
Page : 199 , Block Name : Exercise 10.2
Exercise 10.3
Q1 Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
Answer. The rough sketch of the required ?DEF is as follows.
The steps of construction are as follows.
(i) Draw a line segment DE of length 5 cm.
(ii) At point D, draw a ray DX making an angle of 90° with DE.
(iii) Taking D as centre, draw an arc of 3 cm radius. It will intersect DX at point F.
(iv) Join F to E. ?DEF is the required triangle.
Page 8 of 16 Aglasem Schools
ABC is the required triangle. ∠B can be measured with the help of protractor. It comes to 90°.
Page : 199 , Block Name : Exercise 10.2
Exercise 10.3
Q1 Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.
Answer. The rough sketch of the required ?DEF is as follows.
The steps of construction are as follows.
(i) Draw a line segment DE of length 5 cm.
(ii) At point D, draw a ray DX making an angle of 90° with DE.
(iii) Taking D as centre, draw an arc of 3 cm radius. It will intersect DX at point F.
(iv) Join F to E. ?DEF is the required triangle.
Page 8 of 16 Aglasem Schools
Page 10
Book : Mathematics Ncert Solutions | Chapter-10 Maths
Page : 200 , Block Name : Exercise 10.3
Q2 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°.
Answer. An isosceles triangle PQR has to be constructed with PQ QR = 6.5 cm. A rough sketch
of the required triangle can be drawn as follows.
The steps of construction are as follows.
(i) Draw the line segment QR of length 6.5 cm.
(ii) At point Q, draw a ray QX making an angle 110° with QR.
(iii) Taking Q as centre, draw an arc of 6.5 cm radius. It intersects at point P.
(iv) Join P to R to obtain the required triangle PQR.
Page 9 of 16 Aglasem Schools
Page : 200 , Block Name : Exercise 10.3
Q2 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°.
Answer. An isosceles triangle PQR has to be constructed with PQ QR = 6.5 cm. A rough sketch
of the required triangle can be drawn as follows.
The steps of construction are as follows.
(i) Draw the line segment QR of length 6.5 cm.
(ii) At point Q, draw a ray QX making an angle 110° with QR.
(iii) Taking Q as centre, draw an arc of 6.5 cm radius. It intersects at point P.
(iv) Join P to R to obtain the required triangle PQR.
Page 9 of 16 Aglasem Schools
Page 11
Book : Mathematics Ncert Solutions | Chapter-10 Maths
Page : 200 , Block Name : Exercise 10.3
Q3 Construct ∆ABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
Answer. A rough sketch of the required triangle is as follows.
The steps of construction are as follows.
(i) Draw a line segment BC of length 7.5 cm.
(ii) At point C, draw a ray CX making 60° with BC.
(iii) Taking C as centre, draw an arc of 5 cm radius. It intersects CX at point A.
(iv) Join A to B to obtain triangle ABC.
Page 10 of 16 Aglasem Schools
Page : 200 , Block Name : Exercise 10.3
Q3 Construct ∆ABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.
Answer. A rough sketch of the required triangle is as follows.
The steps of construction are as follows.
(i) Draw a line segment BC of length 7.5 cm.
(ii) At point C, draw a ray CX making 60° with BC.
(iii) Taking C as centre, draw an arc of 5 cm radius. It intersects CX at point A.
(iv) Join A to B to obtain triangle ABC.
Page 10 of 16 Aglasem Schools
Page 12
Book : Mathematics Ncert Solutions | Chapter-10 Maths
Page : 200 , Block Name : Exercise 10.3
Exercise 10.4
Q1 Construct ∆ABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.
Answer. A rough sketch of the required ?ABC is as follows.
The steps of construction are as follows.
(i) Draw a line segment AB of length 5.8
cm.
(ii) At point A, draw a ray AX making 60° angle with AB.
(iii) At point B, draw a ray BY, making 30° angle with AB.
(iv) Point C has to lie on both the rays, AX and BY. Therefore, C is the point of intersection of
these two rays.
Page 11 of 16 Aglasem Schools
Page : 200 , Block Name : Exercise 10.3
Exercise 10.4
Q1 Construct ∆ABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.
Answer. A rough sketch of the required ?ABC is as follows.
The steps of construction are as follows.
(i) Draw a line segment AB of length 5.8
cm.
(ii) At point A, draw a ray AX making 60° angle with AB.
(iii) At point B, draw a ray BY, making 30° angle with AB.
(iv) Point C has to lie on both the rays, AX and BY. Therefore, C is the point of intersection of
these two rays.
Page 11 of 16 Aglasem Schools
Page 13
Book : Mathematics Ncert Solutions | Chapter-10 Maths
This is the required triangle ABC.
Page : 202 , Block Name : Exercise 10.4
Q2 Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°. (Hint: Recall angle-sum
property of a triangle).
Answer. A rough sketch of the required ?PQR is as follows.
In order to construct ?PQR, the measure of ∠RPQ has to be calculated.
According to the angle sum property of triangles,
∠PQR + ∠PRQ + ∠RPQ = 180°
105° + 40° + ∠RPQ = 180°
145° + ∠RPQ = 180°
∠RPQ = 180° - 145° = 35°
The steps of construction are as follows.
(i) Draw a line segment PQ of length 5 cm.
(ii) At P, draw a ray PX making an angle of 35° with PQ.
(iii) At point Q, draw a ray QY making an angle of 105° with PQ.
Page 12 of 16 Aglasem Schools
This is the required triangle ABC.
Page : 202 , Block Name : Exercise 10.4
Q2 Construct ∆PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°. (Hint: Recall angle-sum
property of a triangle).
Answer. A rough sketch of the required ?PQR is as follows.
In order to construct ?PQR, the measure of ∠RPQ has to be calculated.
According to the angle sum property of triangles,
∠PQR + ∠PRQ + ∠RPQ = 180°
105° + 40° + ∠RPQ = 180°
145° + ∠RPQ = 180°
∠RPQ = 180° - 145° = 35°
The steps of construction are as follows.
(i) Draw a line segment PQ of length 5 cm.
(ii) At P, draw a ray PX making an angle of 35° with PQ.
(iii) At point Q, draw a ray QY making an angle of 105° with PQ.
Page 12 of 16 Aglasem Schools
Page 14
Book : Mathematics Ncert Solutions | Chapter-10 Maths
(iv) Point R has to lie on both the rays, PX and QY. Therefore, R is the point of intersection
ofthese two rays.
This is the required triangle PQR.
Page : 202 , Block Name : Exercise 10.4
Q3 Examine whether you can construct ∆DEF such that EF = 7.2 cm, m∠E = 110° and m∠F =
80°. Justify your answer.
Answer. Given that,
m∠E = 110° and m∠F = 80°
Therefore,
m∠E + m∠F = 110° + 80° = 190°
However, according to the angle sum property of triangles, we should obtain
m∠E+ m∠F + m∠D = 180°
Therefore, the angle sum property is not followed by the given triangle. And thus, we cannot
construct ?DEF with the given measurements.
Also, it can be observed that point D Should lie on both rays, EX and FY, for constructing the
required triangle. However, both rays are not intersecting each other. Therefore, the required
triangle cannot be formed.
Page : 202 , Block Name : Exercise 10.4
Exercise 10.5
Page 13 of 16 Aglasem Schools
(iv) Point R has to lie on both the rays, PX and QY. Therefore, R is the point of intersection
ofthese two rays.
This is the required triangle PQR.
Page : 202 , Block Name : Exercise 10.4
Q3 Examine whether you can construct ∆DEF such that EF = 7.2 cm, m∠E = 110° and m∠F =
80°. Justify your answer.
Answer. Given that,
m∠E = 110° and m∠F = 80°
Therefore,
m∠E + m∠F = 110° + 80° = 190°
However, according to the angle sum property of triangles, we should obtain
m∠E+ m∠F + m∠D = 180°
Therefore, the angle sum property is not followed by the given triangle. And thus, we cannot
construct ?DEF with the given measurements.
Also, it can be observed that point D Should lie on both rays, EX and FY, for constructing the
required triangle. However, both rays are not intersecting each other. Therefore, the required
triangle cannot be formed.
Page : 202 , Block Name : Exercise 10.4
Exercise 10.5
Page 13 of 16 Aglasem Schools
Page 15
Book : Mathematics Ncert Solutions | Chapter-10 Maths
Q1 Construct the right angled ∆PQR, where m∠Q = 90°, QR = 8cm and PR = 10 cm.
Answer. A rough sketch of ?PQR is as follows.
The steps of construction are as follows.
(i) Draw a line segment QR of length 8 cm.
(ii) At point Q, draw a ray QX making 90° with QR.
(iii) Taking R as centre, draw an arc of 10 cm radius to intersect ray QX at point P.
(iv) Join P to R. ?PQR is the required right-angled triangle.
Page : 203 , Block Name : Exercise 10.5
Q2 Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4
cm long.
Answer. A right-angled triangle ABC with hypotenuse 6 cm and one of the legs as 4 cm has to
be constructed. A rough sketch of ?ABC is as follows.
Page 14 of 16 Aglasem Schools
Q1 Construct the right angled ∆PQR, where m∠Q = 90°, QR = 8cm and PR = 10 cm.
Answer. A rough sketch of ?PQR is as follows.
The steps of construction are as follows.
(i) Draw a line segment QR of length 8 cm.
(ii) At point Q, draw a ray QX making 90° with QR.
(iii) Taking R as centre, draw an arc of 10 cm radius to intersect ray QX at point P.
(iv) Join P to R. ?PQR is the required right-angled triangle.
Page : 203 , Block Name : Exercise 10.5
Q2 Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4
cm long.
Answer. A right-angled triangle ABC with hypotenuse 6 cm and one of the legs as 4 cm has to
be constructed. A rough sketch of ?ABC is as follows.
Page 14 of 16 Aglasem Schools
Page 16
Book : Mathematics Ncert Solutions | Chapter-10 Maths
The steps of construction are as follows.
(i) Draw a line segment BC Of length 4 cm.
(ii) At point B, draw a ray BX making an angle of 90° with BC.
(iii) Taking C as centre, draw an arc of 6 cm radius to intersect ray BX at point A.
(iv) Join A to C to obtain the required ?ABC.
Page : 203 , Block Name : Exercise 10.5
Q3 Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
Answer. In an isosceles triangle, the lengths of any two sides are equal.
Let in ?ABC, AC = BC = 6 cm. A rough sketch of this ?ABC is as follows.
Page 15 of 16 Aglasem Schools
The steps of construction are as follows.
(i) Draw a line segment BC Of length 4 cm.
(ii) At point B, draw a ray BX making an angle of 90° with BC.
(iii) Taking C as centre, draw an arc of 6 cm radius to intersect ray BX at point A.
(iv) Join A to C to obtain the required ?ABC.
Page : 203 , Block Name : Exercise 10.5
Q3 Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
Answer. In an isosceles triangle, the lengths of any two sides are equal.
Let in ?ABC, AC = BC = 6 cm. A rough sketch of this ?ABC is as follows.
Page 15 of 16 Aglasem Schools
Page 17
Book : Mathematics Ncert Solutions | Chapter-10 Maths
The steps of construction are as follows.
(i) Draw a line segment AC of length 6 cm.
(ii) At point C, draw a ray CX making an angle of 90° with AC.
(iii) Taking point C as centre, draw an arc of 6 cm radius to intersect CX at point B.
(iv) Join A to B to obtain the required ?ABC.
Page : 203 , Block Name : Exercise 10.5
Page 16 of 16 Aglasem Schools
The steps of construction are as follows.
(i) Draw a line segment AC of length 6 cm.
(ii) At point C, draw a ray CX making an angle of 90° with AC.
(iii) Taking point C as centre, draw an arc of 6 cm radius to intersect CX at point B.
(iv) Join A to B to obtain the required ?ABC.
Page : 203 , Block Name : Exercise 10.5
Page 16 of 16 Aglasem Schools
Document Details
| Board / Org | NCERT |
| Exam | Class 7 |
| Type | Solution |
| Pages | 17 |
| Updated | 30 Apr 2026 |