Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RS Aggarwal Solutions: Constructions
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Page 1
Question:1
Draw a line AB and take a point P outside it. Draw a line CD parallel to AB and passing through the point P.
Solution:
Steps of construction:
1. Draw a line AB.
2. Take a point Q on AB and a point P outside AB, and join PQ.
3. With Q as the centre and any radius, draw on arc to cut AB at X and PQ at Z.
4. With P as the centre and the same radius, draw an arc cutting QP at Y .
5. With Y as the centre and the radius equal to XZ, draw an arc to cut the previous arc at E.
6. Join PE and produce it on both the sides to get the required line.
Question:2
Draw a line AB and draw another line CD parallel to AB at a distance of 3.5 cm from it.
Solution:
Steps for construction:
1. Let AB be the given line.
2. Take any two points P and Q on AB.
3. Construct ?BPE = 90° and ?BQF = 90°
4. With P as the centre and the radius equal to 3.5 cm, cut PE at R.
5. With Q as the centre and the radius equal to 3.5cm, cut QF at S.
6. Join RS and produce it on both the sides to get the required line, parallel to AB and at a distance of 3.5 cm from it.
Question:3
Draw a line l and draw another line m parllel to l at a distance of 4.3 cm from it.
Solution:
Steps of construction:
1. Let l be the given line.
2. Take any two points A and B on line l.
3. Construct ?BAE = 90° and ?ABF=90°
4. With A as the centre and the radius equal to 4.3 cm, cut AE at C.
5. With B as the centre and the radius equal to 4.3 cm, cut BF at D.
6. Join CD and produce it on either side to get the required line m, parallel to l and at a distance of 4.3 cm from it.
Question:4
Construct a ?ABC in which BC = 3.6 cm, AB = 5 cm and AC = 5.4 cm. Draw the perpendicular bisector of the side BC.
Solution:
Steps of construction:
1. Draw a line segment (AB) of length 5 cm.
2. Draw an arc of radius 5.4 cm from the centre (A).
3. With B as the centre, draw another arc of radius 3.6 cm, cutting the previous arc at C.
4. Join AC and BC.
5. Taking B as the centre and the radius more than half of BC, draw two arcs on both the sides of BC.
6. Similarly, taking C as the centre and the same radius, draw arcs on both the sides of BC, cutting the previous arcs at P and Q.
7. Join PQ.
Then, PQ is the required perpendicuar bisector of BC, meeting BC at D.
Question:5
Construct a ?PQR in which QR = 6 cm, PQ = 4.4 cm and PR = 5.3 cm. Draw the bisector of ?P.
Page 2
Question:1
Draw a line AB and take a point P outside it. Draw a line CD parallel to AB and passing through the point P.
Solution:
Steps of construction:
1. Draw a line AB.
2. Take a point Q on AB and a point P outside AB, and join PQ.
3. With Q as the centre and any radius, draw on arc to cut AB at X and PQ at Z.
4. With P as the centre and the same radius, draw an arc cutting QP at Y .
5. With Y as the centre and the radius equal to XZ, draw an arc to cut the previous arc at E.
6. Join PE and produce it on both the sides to get the required line.
Question:2
Draw a line AB and draw another line CD parallel to AB at a distance of 3.5 cm from it.
Solution:
Steps for construction:
1. Let AB be the given line.
2. Take any two points P and Q on AB.
3. Construct ?BPE = 90° and ?BQF = 90°
4. With P as the centre and the radius equal to 3.5 cm, cut PE at R.
5. With Q as the centre and the radius equal to 3.5cm, cut QF at S.
6. Join RS and produce it on both the sides to get the required line, parallel to AB and at a distance of 3.5 cm from it.
Question:3
Draw a line l and draw another line m parllel to l at a distance of 4.3 cm from it.
Solution:
Steps of construction:
1. Let l be the given line.
2. Take any two points A and B on line l.
3. Construct ?BAE = 90° and ?ABF=90°
4. With A as the centre and the radius equal to 4.3 cm, cut AE at C.
5. With B as the centre and the radius equal to 4.3 cm, cut BF at D.
6. Join CD and produce it on either side to get the required line m, parallel to l and at a distance of 4.3 cm from it.
Question:4
Construct a ?ABC in which BC = 3.6 cm, AB = 5 cm and AC = 5.4 cm. Draw the perpendicular bisector of the side BC.
Solution:
Steps of construction:
1. Draw a line segment (AB) of length 5 cm.
2. Draw an arc of radius 5.4 cm from the centre (A).
3. With B as the centre, draw another arc of radius 3.6 cm, cutting the previous arc at C.
4. Join AC and BC.
5. Taking B as the centre and the radius more than half of BC, draw two arcs on both the sides of BC.
6. Similarly, taking C as the centre and the same radius, draw arcs on both the sides of BC, cutting the previous arcs at P and Q.
7. Join PQ.
Then, PQ is the required perpendicuar bisector of BC, meeting BC at D.
Question:5
Construct a ?PQR in which QR = 6 cm, PQ = 4.4 cm and PR = 5.3 cm. Draw the bisector of ?P.
Solution:
Steps of construction:
1. Draw a line segment QR of length 6 cm.
2. Draw arcs of 4.4 cm and 5.3 cm from Q and R, respectively. They intersect at P.
3. Draw an arc of any radius from the centre (P), cutting PQ and PR at S and T, respectively.
4. With S as the centre and the radius more than half of ST, draw an arc .
5. With T as the centre and the same radius, draw another arc cutting the previously drawn arc at X.
6. Join P and X.
Then, PX is the bisector of ?P.
Question:6
Construct an equilateral triangle each of whose sides measures 6.2 cm. Measure each of its angles.
Solution:
Steps of construction:
1. Draw AB of length 6.2 cm.
2. By taking the centres as A and B, draw equal arcs of 6.2 cm on the same side of AB, cutting each other at C.
3. Join AC and BC.
When we will measure angles of triangle using protractor then we find that all angles are equal to 60°
Question:7
Construct a ?ABC in which AB = AC = 4.8 cm and BC = 5.3 cm. Measure ?B and ?C. Draw AD ? BC.
Solution:
Steps of construction:
1. Draw BC=5.3 cm
2. Draw an arc of radius 4.8 cm from the centre, B.
3. Draw another arc of radius 4.8 cm from the centre, C.
4. Both of these arcs intersect at A.
5. Join AB and AC.
6. With A as the centre and any radius, draw an arc cutting BC at M and N.
7. With M as the centre and the radius more than half of MN, draw an arc.
8. With N as the centre and the same radius, draw another arc cutting the previously drawn arc at P.
9. Join AP, cutting BC at D.
Then, AD ? BC
Question:8
Construct a ?ABC in which AB = 3.8 cm, ?A = 60° and AC = 5 cm.
Solution:
Steps of construction:
1. Draw AB of length 3.8 cm.
2. Draw ?BAZ=60°
3. With the centre as A, cut ray AZ at 5 cm at C.
4 Join BC.
Then, ABC is the required triangle.
Question:9
Construct a ?ABC in which BC = 4.3 cm, ?C = 45° and AC = 6 cm.
Solution:
Page 3
Question:1
Draw a line AB and take a point P outside it. Draw a line CD parallel to AB and passing through the point P.
Solution:
Steps of construction:
1. Draw a line AB.
2. Take a point Q on AB and a point P outside AB, and join PQ.
3. With Q as the centre and any radius, draw on arc to cut AB at X and PQ at Z.
4. With P as the centre and the same radius, draw an arc cutting QP at Y .
5. With Y as the centre and the radius equal to XZ, draw an arc to cut the previous arc at E.
6. Join PE and produce it on both the sides to get the required line.
Question:2
Draw a line AB and draw another line CD parallel to AB at a distance of 3.5 cm from it.
Solution:
Steps for construction:
1. Let AB be the given line.
2. Take any two points P and Q on AB.
3. Construct ?BPE = 90° and ?BQF = 90°
4. With P as the centre and the radius equal to 3.5 cm, cut PE at R.
5. With Q as the centre and the radius equal to 3.5cm, cut QF at S.
6. Join RS and produce it on both the sides to get the required line, parallel to AB and at a distance of 3.5 cm from it.
Question:3
Draw a line l and draw another line m parllel to l at a distance of 4.3 cm from it.
Solution:
Steps of construction:
1. Let l be the given line.
2. Take any two points A and B on line l.
3. Construct ?BAE = 90° and ?ABF=90°
4. With A as the centre and the radius equal to 4.3 cm, cut AE at C.
5. With B as the centre and the radius equal to 4.3 cm, cut BF at D.
6. Join CD and produce it on either side to get the required line m, parallel to l and at a distance of 4.3 cm from it.
Question:4
Construct a ?ABC in which BC = 3.6 cm, AB = 5 cm and AC = 5.4 cm. Draw the perpendicular bisector of the side BC.
Solution:
Steps of construction:
1. Draw a line segment (AB) of length 5 cm.
2. Draw an arc of radius 5.4 cm from the centre (A).
3. With B as the centre, draw another arc of radius 3.6 cm, cutting the previous arc at C.
4. Join AC and BC.
5. Taking B as the centre and the radius more than half of BC, draw two arcs on both the sides of BC.
6. Similarly, taking C as the centre and the same radius, draw arcs on both the sides of BC, cutting the previous arcs at P and Q.
7. Join PQ.
Then, PQ is the required perpendicuar bisector of BC, meeting BC at D.
Question:5
Construct a ?PQR in which QR = 6 cm, PQ = 4.4 cm and PR = 5.3 cm. Draw the bisector of ?P.
Solution:
Steps of construction:
1. Draw a line segment QR of length 6 cm.
2. Draw arcs of 4.4 cm and 5.3 cm from Q and R, respectively. They intersect at P.
3. Draw an arc of any radius from the centre (P), cutting PQ and PR at S and T, respectively.
4. With S as the centre and the radius more than half of ST, draw an arc .
5. With T as the centre and the same radius, draw another arc cutting the previously drawn arc at X.
6. Join P and X.
Then, PX is the bisector of ?P.
Question:6
Construct an equilateral triangle each of whose sides measures 6.2 cm. Measure each of its angles.
Solution:
Steps of construction:
1. Draw AB of length 6.2 cm.
2. By taking the centres as A and B, draw equal arcs of 6.2 cm on the same side of AB, cutting each other at C.
3. Join AC and BC.
When we will measure angles of triangle using protractor then we find that all angles are equal to 60°
Question:7
Construct a ?ABC in which AB = AC = 4.8 cm and BC = 5.3 cm. Measure ?B and ?C. Draw AD ? BC.
Solution:
Steps of construction:
1. Draw BC=5.3 cm
2. Draw an arc of radius 4.8 cm from the centre, B.
3. Draw another arc of radius 4.8 cm from the centre, C.
4. Both of these arcs intersect at A.
5. Join AB and AC.
6. With A as the centre and any radius, draw an arc cutting BC at M and N.
7. With M as the centre and the radius more than half of MN, draw an arc.
8. With N as the centre and the same radius, draw another arc cutting the previously drawn arc at P.
9. Join AP, cutting BC at D.
Then, AD ? BC
Question:8
Construct a ?ABC in which AB = 3.8 cm, ?A = 60° and AC = 5 cm.
Solution:
Steps of construction:
1. Draw AB of length 3.8 cm.
2. Draw ?BAZ=60°
3. With the centre as A, cut ray AZ at 5 cm at C.
4 Join BC.
Then, ABC is the required triangle.
Question:9
Construct a ?ABC in which BC = 4.3 cm, ?C = 45° and AC = 6 cm.
Solution:
Steps of construction:
1. Draw AC= 6 cm
2. Draw ?ACZ = 45°
3. With C as the centre, cut ray BZ at 4. 3 cm at point B.
4. Join AB.
Then, ABC is the required triangle.
Question:10
Construct a ?ABC in which AB = AC = 5.2 cm and ?A = 120°. Draw AD ? BC.
Solution:
Steps of construction:
1. Draw AB=5.2 cm
2. Draw ?BAX=120°
3. With A as the centre, cut the ray AX at 5.3 cm at point C.
4. Join BC.
5. With A as the centre and any radius, draw an arc cutting BC at M and N.
6. With M as the centre and the radius more than half of MN, draw an arc.
7. With N as the centre and the same radius as before, draw another arc cutting the previously drawn arc at P.
8. Join AP meeting BC at D.
? AD ? BC
Question:11
Construct a ?ABC in which BC = 6.2 cm, ?B = 60° and ?C = 45°.
Solution:
Steps of construction:
1. Draw BC=6.2 cm
2. Draw ?BCX=45°
3. Draw ?CBY=60°
4. The ray CX and BY intersect at A. Then, ABC is the required triangle.
Question:12
Construct a ?ABC in which BC = 5.8 cm, ?B = ?C = 30°. Measure AB and AC. What do you observe?
Solution:
Steps of construction:
1. Draw BC=5.8 cm
2. Draw ?BCY = 30°3. Draw ?CBX = 30°4. The ray BX and CY intersect at A. Then, ABC is the required triangle. On measuring AB and AC: AB = AC = 3. 4 cm
Question:13
Construct a ?ABC in which AB = 7 cm, ?A = 45° and ?C = 75°.
Solution:
By angle sum property: ?B = 180°- ?A - ?C = 180°-45°-75° = 60°
Steps of construction:
1. Draw AB=7cm
2 Draw ?BAX= 45°
3. Draw ?ABY= 60°
4.The ray AX and BY intersect at C.
Then, ABC is the required triangle.
Page 4
Question:1
Draw a line AB and take a point P outside it. Draw a line CD parallel to AB and passing through the point P.
Solution:
Steps of construction:
1. Draw a line AB.
2. Take a point Q on AB and a point P outside AB, and join PQ.
3. With Q as the centre and any radius, draw on arc to cut AB at X and PQ at Z.
4. With P as the centre and the same radius, draw an arc cutting QP at Y .
5. With Y as the centre and the radius equal to XZ, draw an arc to cut the previous arc at E.
6. Join PE and produce it on both the sides to get the required line.
Question:2
Draw a line AB and draw another line CD parallel to AB at a distance of 3.5 cm from it.
Solution:
Steps for construction:
1. Let AB be the given line.
2. Take any two points P and Q on AB.
3. Construct ?BPE = 90° and ?BQF = 90°
4. With P as the centre and the radius equal to 3.5 cm, cut PE at R.
5. With Q as the centre and the radius equal to 3.5cm, cut QF at S.
6. Join RS and produce it on both the sides to get the required line, parallel to AB and at a distance of 3.5 cm from it.
Question:3
Draw a line l and draw another line m parllel to l at a distance of 4.3 cm from it.
Solution:
Steps of construction:
1. Let l be the given line.
2. Take any two points A and B on line l.
3. Construct ?BAE = 90° and ?ABF=90°
4. With A as the centre and the radius equal to 4.3 cm, cut AE at C.
5. With B as the centre and the radius equal to 4.3 cm, cut BF at D.
6. Join CD and produce it on either side to get the required line m, parallel to l and at a distance of 4.3 cm from it.
Question:4
Construct a ?ABC in which BC = 3.6 cm, AB = 5 cm and AC = 5.4 cm. Draw the perpendicular bisector of the side BC.
Solution:
Steps of construction:
1. Draw a line segment (AB) of length 5 cm.
2. Draw an arc of radius 5.4 cm from the centre (A).
3. With B as the centre, draw another arc of radius 3.6 cm, cutting the previous arc at C.
4. Join AC and BC.
5. Taking B as the centre and the radius more than half of BC, draw two arcs on both the sides of BC.
6. Similarly, taking C as the centre and the same radius, draw arcs on both the sides of BC, cutting the previous arcs at P and Q.
7. Join PQ.
Then, PQ is the required perpendicuar bisector of BC, meeting BC at D.
Question:5
Construct a ?PQR in which QR = 6 cm, PQ = 4.4 cm and PR = 5.3 cm. Draw the bisector of ?P.
Solution:
Steps of construction:
1. Draw a line segment QR of length 6 cm.
2. Draw arcs of 4.4 cm and 5.3 cm from Q and R, respectively. They intersect at P.
3. Draw an arc of any radius from the centre (P), cutting PQ and PR at S and T, respectively.
4. With S as the centre and the radius more than half of ST, draw an arc .
5. With T as the centre and the same radius, draw another arc cutting the previously drawn arc at X.
6. Join P and X.
Then, PX is the bisector of ?P.
Question:6
Construct an equilateral triangle each of whose sides measures 6.2 cm. Measure each of its angles.
Solution:
Steps of construction:
1. Draw AB of length 6.2 cm.
2. By taking the centres as A and B, draw equal arcs of 6.2 cm on the same side of AB, cutting each other at C.
3. Join AC and BC.
When we will measure angles of triangle using protractor then we find that all angles are equal to 60°
Question:7
Construct a ?ABC in which AB = AC = 4.8 cm and BC = 5.3 cm. Measure ?B and ?C. Draw AD ? BC.
Solution:
Steps of construction:
1. Draw BC=5.3 cm
2. Draw an arc of radius 4.8 cm from the centre, B.
3. Draw another arc of radius 4.8 cm from the centre, C.
4. Both of these arcs intersect at A.
5. Join AB and AC.
6. With A as the centre and any radius, draw an arc cutting BC at M and N.
7. With M as the centre and the radius more than half of MN, draw an arc.
8. With N as the centre and the same radius, draw another arc cutting the previously drawn arc at P.
9. Join AP, cutting BC at D.
Then, AD ? BC
Question:8
Construct a ?ABC in which AB = 3.8 cm, ?A = 60° and AC = 5 cm.
Solution:
Steps of construction:
1. Draw AB of length 3.8 cm.
2. Draw ?BAZ=60°
3. With the centre as A, cut ray AZ at 5 cm at C.
4 Join BC.
Then, ABC is the required triangle.
Question:9
Construct a ?ABC in which BC = 4.3 cm, ?C = 45° and AC = 6 cm.
Solution:
Steps of construction:
1. Draw AC= 6 cm
2. Draw ?ACZ = 45°
3. With C as the centre, cut ray BZ at 4. 3 cm at point B.
4. Join AB.
Then, ABC is the required triangle.
Question:10
Construct a ?ABC in which AB = AC = 5.2 cm and ?A = 120°. Draw AD ? BC.
Solution:
Steps of construction:
1. Draw AB=5.2 cm
2. Draw ?BAX=120°
3. With A as the centre, cut the ray AX at 5.3 cm at point C.
4. Join BC.
5. With A as the centre and any radius, draw an arc cutting BC at M and N.
6. With M as the centre and the radius more than half of MN, draw an arc.
7. With N as the centre and the same radius as before, draw another arc cutting the previously drawn arc at P.
8. Join AP meeting BC at D.
? AD ? BC
Question:11
Construct a ?ABC in which BC = 6.2 cm, ?B = 60° and ?C = 45°.
Solution:
Steps of construction:
1. Draw BC=6.2 cm
2. Draw ?BCX=45°
3. Draw ?CBY=60°
4. The ray CX and BY intersect at A. Then, ABC is the required triangle.
Question:12
Construct a ?ABC in which BC = 5.8 cm, ?B = ?C = 30°. Measure AB and AC. What do you observe?
Solution:
Steps of construction:
1. Draw BC=5.8 cm
2. Draw ?BCY = 30°3. Draw ?CBX = 30°4. The ray BX and CY intersect at A. Then, ABC is the required triangle. On measuring AB and AC: AB = AC = 3. 4 cm
Question:13
Construct a ?ABC in which AB = 7 cm, ?A = 45° and ?C = 75°.
Solution:
By angle sum property: ?B = 180°- ?A - ?C = 180°-45°-75° = 60°
Steps of construction:
1. Draw AB=7cm
2 Draw ?BAX= 45°
3. Draw ?ABY= 60°
4.The ray AX and BY intersect at C.
Then, ABC is the required triangle.
Question:14
Construct a ?ABC in which BC = 4.8 cm, ?C = 90° and AB = 6.3 cm.
Solution:
Steps of construction:
1.Draw BC=4.8 cm
2.Draw a perpendicular on C such that ?C is equal to 90°.
3.Draw an arc of radius 6.3 cm from the centre B.
4. Join AB.
Question:15
Construct a right-angled triangle one side of which measures 3.5 cm and the length of whose hypotenuse is 6 cm.
Solution:
Steps of construction:
1. Draw AB=3.5 cm
2. Construct ?ABX = 90°
3. With centre A, draw an arc of radius 6 cm cutting BX at C.
4. Join AC.
Then, ABC is the required triangle.
Question:16
Construct a right triangle having hypotenuse of length 5.6 cm and one of whose acute angles measures 30°.
Solution:
Here, ?A=30° and ?C=90°
By angle sum property:
?B=60°
1. Draw the hypotenuse AB of length 5.6 cm. 2. Draw ?BAX=30° and ?ABY=60° 3. The ray AX and BY intersect at C. Then, ABC is the required triangle.
Question:17
Mark
? against the correct answer
The supplement of 45° is
a
45°
b
75°
c
135°
d
155°
Solution:
c 135°
Supplement of 45
°
=180
°
-45
°
=135
°
Question:18
Mark
? against the correct answer
The complement of 80° is
a
100°
b
10°
( )
Page 5
Question:1
Draw a line AB and take a point P outside it. Draw a line CD parallel to AB and passing through the point P.
Solution:
Steps of construction:
1. Draw a line AB.
2. Take a point Q on AB and a point P outside AB, and join PQ.
3. With Q as the centre and any radius, draw on arc to cut AB at X and PQ at Z.
4. With P as the centre and the same radius, draw an arc cutting QP at Y .
5. With Y as the centre and the radius equal to XZ, draw an arc to cut the previous arc at E.
6. Join PE and produce it on both the sides to get the required line.
Question:2
Draw a line AB and draw another line CD parallel to AB at a distance of 3.5 cm from it.
Solution:
Steps for construction:
1. Let AB be the given line.
2. Take any two points P and Q on AB.
3. Construct ?BPE = 90° and ?BQF = 90°
4. With P as the centre and the radius equal to 3.5 cm, cut PE at R.
5. With Q as the centre and the radius equal to 3.5cm, cut QF at S.
6. Join RS and produce it on both the sides to get the required line, parallel to AB and at a distance of 3.5 cm from it.
Question:3
Draw a line l and draw another line m parllel to l at a distance of 4.3 cm from it.
Solution:
Steps of construction:
1. Let l be the given line.
2. Take any two points A and B on line l.
3. Construct ?BAE = 90° and ?ABF=90°
4. With A as the centre and the radius equal to 4.3 cm, cut AE at C.
5. With B as the centre and the radius equal to 4.3 cm, cut BF at D.
6. Join CD and produce it on either side to get the required line m, parallel to l and at a distance of 4.3 cm from it.
Question:4
Construct a ?ABC in which BC = 3.6 cm, AB = 5 cm and AC = 5.4 cm. Draw the perpendicular bisector of the side BC.
Solution:
Steps of construction:
1. Draw a line segment (AB) of length 5 cm.
2. Draw an arc of radius 5.4 cm from the centre (A).
3. With B as the centre, draw another arc of radius 3.6 cm, cutting the previous arc at C.
4. Join AC and BC.
5. Taking B as the centre and the radius more than half of BC, draw two arcs on both the sides of BC.
6. Similarly, taking C as the centre and the same radius, draw arcs on both the sides of BC, cutting the previous arcs at P and Q.
7. Join PQ.
Then, PQ is the required perpendicuar bisector of BC, meeting BC at D.
Question:5
Construct a ?PQR in which QR = 6 cm, PQ = 4.4 cm and PR = 5.3 cm. Draw the bisector of ?P.
Solution:
Steps of construction:
1. Draw a line segment QR of length 6 cm.
2. Draw arcs of 4.4 cm and 5.3 cm from Q and R, respectively. They intersect at P.
3. Draw an arc of any radius from the centre (P), cutting PQ and PR at S and T, respectively.
4. With S as the centre and the radius more than half of ST, draw an arc .
5. With T as the centre and the same radius, draw another arc cutting the previously drawn arc at X.
6. Join P and X.
Then, PX is the bisector of ?P.
Question:6
Construct an equilateral triangle each of whose sides measures 6.2 cm. Measure each of its angles.
Solution:
Steps of construction:
1. Draw AB of length 6.2 cm.
2. By taking the centres as A and B, draw equal arcs of 6.2 cm on the same side of AB, cutting each other at C.
3. Join AC and BC.
When we will measure angles of triangle using protractor then we find that all angles are equal to 60°
Question:7
Construct a ?ABC in which AB = AC = 4.8 cm and BC = 5.3 cm. Measure ?B and ?C. Draw AD ? BC.
Solution:
Steps of construction:
1. Draw BC=5.3 cm
2. Draw an arc of radius 4.8 cm from the centre, B.
3. Draw another arc of radius 4.8 cm from the centre, C.
4. Both of these arcs intersect at A.
5. Join AB and AC.
6. With A as the centre and any radius, draw an arc cutting BC at M and N.
7. With M as the centre and the radius more than half of MN, draw an arc.
8. With N as the centre and the same radius, draw another arc cutting the previously drawn arc at P.
9. Join AP, cutting BC at D.
Then, AD ? BC
Question:8
Construct a ?ABC in which AB = 3.8 cm, ?A = 60° and AC = 5 cm.
Solution:
Steps of construction:
1. Draw AB of length 3.8 cm.
2. Draw ?BAZ=60°
3. With the centre as A, cut ray AZ at 5 cm at C.
4 Join BC.
Then, ABC is the required triangle.
Question:9
Construct a ?ABC in which BC = 4.3 cm, ?C = 45° and AC = 6 cm.
Solution:
Steps of construction:
1. Draw AC= 6 cm
2. Draw ?ACZ = 45°
3. With C as the centre, cut ray BZ at 4. 3 cm at point B.
4. Join AB.
Then, ABC is the required triangle.
Question:10
Construct a ?ABC in which AB = AC = 5.2 cm and ?A = 120°. Draw AD ? BC.
Solution:
Steps of construction:
1. Draw AB=5.2 cm
2. Draw ?BAX=120°
3. With A as the centre, cut the ray AX at 5.3 cm at point C.
4. Join BC.
5. With A as the centre and any radius, draw an arc cutting BC at M and N.
6. With M as the centre and the radius more than half of MN, draw an arc.
7. With N as the centre and the same radius as before, draw another arc cutting the previously drawn arc at P.
8. Join AP meeting BC at D.
? AD ? BC
Question:11
Construct a ?ABC in which BC = 6.2 cm, ?B = 60° and ?C = 45°.
Solution:
Steps of construction:
1. Draw BC=6.2 cm
2. Draw ?BCX=45°
3. Draw ?CBY=60°
4. The ray CX and BY intersect at A. Then, ABC is the required triangle.
Question:12
Construct a ?ABC in which BC = 5.8 cm, ?B = ?C = 30°. Measure AB and AC. What do you observe?
Solution:
Steps of construction:
1. Draw BC=5.8 cm
2. Draw ?BCY = 30°3. Draw ?CBX = 30°4. The ray BX and CY intersect at A. Then, ABC is the required triangle. On measuring AB and AC: AB = AC = 3. 4 cm
Question:13
Construct a ?ABC in which AB = 7 cm, ?A = 45° and ?C = 75°.
Solution:
By angle sum property: ?B = 180°- ?A - ?C = 180°-45°-75° = 60°
Steps of construction:
1. Draw AB=7cm
2 Draw ?BAX= 45°
3. Draw ?ABY= 60°
4.The ray AX and BY intersect at C.
Then, ABC is the required triangle.
Question:14
Construct a ?ABC in which BC = 4.8 cm, ?C = 90° and AB = 6.3 cm.
Solution:
Steps of construction:
1.Draw BC=4.8 cm
2.Draw a perpendicular on C such that ?C is equal to 90°.
3.Draw an arc of radius 6.3 cm from the centre B.
4. Join AB.
Question:15
Construct a right-angled triangle one side of which measures 3.5 cm and the length of whose hypotenuse is 6 cm.
Solution:
Steps of construction:
1. Draw AB=3.5 cm
2. Construct ?ABX = 90°
3. With centre A, draw an arc of radius 6 cm cutting BX at C.
4. Join AC.
Then, ABC is the required triangle.
Question:16
Construct a right triangle having hypotenuse of length 5.6 cm and one of whose acute angles measures 30°.
Solution:
Here, ?A=30° and ?C=90°
By angle sum property:
?B=60°
1. Draw the hypotenuse AB of length 5.6 cm. 2. Draw ?BAX=30° and ?ABY=60° 3. The ray AX and BY intersect at C. Then, ABC is the required triangle.
Question:17
Mark
? against the correct answer
The supplement of 45° is
a
45°
b
75°
c
135°
d
155°
Solution:
c 135°
Supplement of 45
°
=180
°
-45
°
=135
°
Question:18
Mark
? against the correct answer
The complement of 80° is
a
100°
b
10°
( )
c
20°
d
280°
Solution:
b 10°
Complement of 80
°
= 90
°
-80
°
=10
°
Question:19
Mark
? against the correct answer
An angle is its own complement. The measure of the angle is
a
30°
b
45°
c
90°
d
60°
Solution:
(b)45°Suppose the angle is x°. Then, the complement is also x°. Complement of x° = 90°-x° ? x° = 90°-x° ? x°+x° = 90° ? 2x° = 90° ? x =
90
2
? x = 45
Question:20
Mark
? against the correct answer
An angle is one-fifth of its supplement. The measure of the angle is
a
30°
b
15°
c
75°
d
150°
Solution:
a 30°Suppose the angle is x. x =
(180-x)
5
? 5x = 180 -x ? 5x +x = 180 ? x =
180
6
? x = 30°
Question:21
Mark
? against the correct answer
An angle is 24° more than its complement. The measure of the angle is
a
47°
b
57°
c
53°
d
66°
Solution:
b 57°Suppose the angle is x. x = 90 -x +24 ? x +x = 114 ? 2x = 114 ? x =
114
2
? x = 57°
Question:22
Mark
? against the correct answer
An angle is 32° less than its supplement. The measure of the angle is
a
37°
b
74°
c
148°
d
none of these
Solution:
b 74°Suppose the angle is x. x = 180 -x -32 ? x +x = 148 ? 2x = 148 ? x =
148
2
? x = 74°
Question:23
Mark
? against the correct answer
Two supplementary angles are in the ratio 3 : 2. The smaller angle measures
a
108°
b
81°
c
( )
( )
( )
( )
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