# NCERT solutions for class 7 Maths Chapter 10 Practical Geometry

#### Solution for Exercise 10.4

Q.1 Construct $$\triangle$$ABC, given m $$\angle$$A =60°, m $$\angle$$B = 30° and AB = 5.8 cm.

1. Draw a line segment AB = 5.8 cm.
2. At point A, draw an angle 60° i.e. $$\angle$$ A = 60°.
3. At point B, draw an angle of 30° i.e. $$\angle$$B= 30°.
4. Now the two rays intersect at the point C.
Then, $$\triangle$$ ABC is the required triangle.

Q.2 Construct $$\triangle$$PQR if PQ = 5 cm, m?PQR = 105° and $$\angle$$QRP = 40°.
(Hint: Recall angle-sum property of a triangle)

Given: $$\angle PQR = 40°, \angle QRP = 105°$$
$$\therefore \angle PQR + \angle QRP + \angle QPR = 180°$$ (Angle sum property of a triangle)
105° + 40° +$$\angle$$QPR = 180°
145° + $$\angle$$QPR = 180°
$$\therefore \angle QPR = 180° – 145° = 35°$$
Step 1. Draw a line segment PQ = 5 cm.
Step 2. At point P, draw an angle $$\angle$$ P = 35°
Step 3 At point Q draw an angle $$\angle$$ Q = 105°
Step 4. Now the two rays intersect at the point R.
Then, $$\triangle$$PQR is the required triangle.

Q.3 Examine whether you can construct ?DEF such that EF = 7.2 cm, m$$\angle$$E = 110° and m$$\angle$$F = 80°. Justify your answer.

To construct $$\triangle$$DEF with the given measurement, is not possible.
$$\therefore m\angle E + m\angle F = 110° + 80°$$
= 190° > 180° [Sum of the angles of a triangle = 180°]
$$\therefore \triangle$$DEF is not possible to construct.

#### Solution for Exercise 10.1

Q.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.

(i) Draw a line AB.
(ii) Take any point R on it.
(iii) Join the given point C to R and mark $$\angle$$CRA.
(iv) Mark $$\angle CRA = \angle QCB$$ at C and produce to both side.
(v) PQ is the required line.

Q.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.

(i) Draw a given line l and take any point P on it.
(ii) Draw a perpendicular line at P to the line l such that PX = 4 cm.
(iii) Draw $$\angle$$2 = $$\angle$$1 i.e. 90° at PX and produce the line both sides.
(iv) m is the required line parallel to l through X. Using Properties of Alternate Angles

Q.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?

1. Draw a line l.
2. Take any point Q on L and a point P outside L and join PQ.
3. Make sure that angles at point P and point Q are equal i.e. $$\angle Q = \angle P$$
4. At point P extend line to get line M which is parallel L.
5. Then take any point R on line M.
6. At point R draw angle such that $$\angle P = \angle R$$
7. At point R extend line which intersects line L at S and draw a line RS.

#### Solution for Exercise 10.2

Q.1 Construct $$\triangle XYZ$$in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm

1. Draw a line segment YZ = 5 cm.
2. With Z as a center and radius 6 cm, draw an arc.
3. With Y as a center and radius 4.5 cm, draw another arc, cutting the previous arc at X.<
4. Join XY and XZ.
Then, $$\triangle$$ XYZ is the required triangle.

Q.2 Construct an equilateral triangle of side 5.5 cm.

1. Draw a line segment AB = 5.5 cm.
2. With A as a center and radius 5.5 cm, draw an arc.
3. With B as a center and radius 5.5 cm, draw another arc, cutting the previous arc at C.
4. Join CA and CB.
Then, $$\triangle$$ABC is the required equilateral triangle.

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

1. Draw a line segment QR = 3.5 cm.
2. With Q as a center and radius 4 cm, draw an arc.
3. With R as a center and radius 4 cm, draw another arc, cutting the previous arc at P.
4. Join PQ and PR.
Then, ?PQR is the required isosceles triangle.

Q.4 Construct $$\triangle$$ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure $$\angle$$B.

1. Draw a line segment BC = 6 cm.
2. With B as a center and radius 2.5 cm, draw an arc.
3. With C as a center and radius 6.5 cm, draw another arc, cutting the previous arc at A.
4. Join AB and AC.
Then, ?ABC is the required triangle.
5. When we will measure the angle B of triangle by protractor, then angle is equal to ?B = 90°

#### Solution for Exercise 10.3

Q.1 Construct $$\triangle$$DEF such that DE = 5 cm, DF = 3 cm and $$\angle$$ EDF = 90°.

1. Draw a line segment DE = 5 cm
2. Draw an angle $$\angle$$ EDF = 90 °
3. Cut an arc from D on DF of 3 cm
Then, $$\triangle$$ EDF is the required right angled triangle.

Q.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°.

Step 1.Draw a line segment QR = 6.5 cm.
Step 2. At point A, draw a line QP to making an angle of 110° i.e. $$\angle PQR$$ = 110°. And QP = 6.5 cm
4. Join PR.
Then, $$\triangle$$ PQRis the required isosceles triangle.

Q.3 Construct $$\triangle$$ABC with BC = 7.5 cm, AC = 5 cm and $$\angle$$C = 60°.

1. Draw a line segment BC = 7.5 cm.
2. At point C, draw an angle $$\angle$$ACB = 60°. And AC= 5cm
3. Join AB.
Then, $$\triangle$$ ABC is the required triangle.

#### Solution for Exercise 10.5

Q.1 Construct the right angled $$\triangle$$PQR, where m$$\angle$$ Q = 90°, QR = 8 cm and PR = 10 cm.

Step 1 Draw QR = 8 cm.
Step 2 Draw $$\angle$$Q = 90°.
Step 3 Draw an arc with centre R and radius 10 cm to cut the perpendicular line at P.
Step 4 $$\triangle$$PQR is the required triangle.
[Using RHS criterion]

Q.2 Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

Step 1 Draw BC = 4 cm.
Step 2 Draw $$\angle$$ B= 90°.
Step 3 Cut AC = 6 cm.
Step 4 join AB And AC
Step 4 $$\triangle$$ ABC is the required right-angled triangle.

Q.3 Construct an isosceles right-angled triangle ABC, where m$$\angle ACB$$= 90° and AC = 6 cm.

Step 2 Draw $$\angle$$C = 90°.
Step 5 $$\triangle$$ ACB is the required right-angled triangle.