Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RS Aggarwal Solutions: Constructions
RS Aggarwal Solutions: Constructions | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
( )
( )
( )
( )
Read More
|
76 videos|345 docs|39 tests
|
Related Exams
About this Document
|
4.73/5 Rating |
|
Nov 23, 2024 Last updated |
Document Description: RS Aggarwal Solutions: Constructions for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RS Aggarwal Solutions: Constructions have been prepared according to the Class 7 exam syllabus. Information about RS Aggarwal Solutions: Constructions covers topics
like and RS Aggarwal Solutions: Constructions Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RS Aggarwal Solutions: Constructions.
Introduction of RS Aggarwal Solutions: Constructions in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RS Aggarwal Solutions: Constructions in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: RS Aggarwal Solutions: Constructions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for RS Aggarwal Solutions: Constructions | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RS Aggarwal Solutions: Constructions
In this doc you can find the meaning of RS Aggarwal Solutions: Constructions defined & explained in the simplest way possible. Besides explaining types of
RS Aggarwal Solutions: Constructions theory, EduRev gives you an ample number of questions to practice RS Aggarwal Solutions: Constructions tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
mock tests for examination
,Previous Year Questions with Solutions
,study material
,Semester Notes
,RS Aggarwal Solutions: Constructions | Mathematics (Maths) Class 7
,past year papers
,Sample Paper
,Important questions
,RS Aggarwal Solutions: Constructions | Mathematics (Maths) Class 7
,ppt
,MCQs
,practice quizzes
,video lectures
,Extra Questions
,Exam
,Free
,Summary
,shortcuts and tricks
,Viva Questions
,RS Aggarwal Solutions: Constructions | Mathematics (Maths) Class 7
,Objective type Questions
;
Additional Information about RS Aggarwal Solutions: Constructions for Class 7 Preparation
RS Aggarwal Solutions: Constructions Free PDF Download
The RS Aggarwal Solutions: Constructions is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RS Aggarwal Solutions: Constructions now and kickstart your journey towards success in the Class 7 exam.
Importance of RS Aggarwal Solutions: Constructions
The importance of RS Aggarwal Solutions: Constructions cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RS Aggarwal Solutions: Constructions Notes
RS Aggarwal Solutions: Constructions Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RS Aggarwal Solutions: Constructions.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RS Aggarwal Solutions: Constructions Notes on EduRev are your ultimate resource for success.
RS Aggarwal Solutions: Constructions Class 7 Questions
The "RS Aggarwal Solutions: Constructions Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RS Aggarwal Solutions: Constructions on the App
Students of Class 7 can study RS Aggarwal Solutions: Constructions alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RS Aggarwal Solutions: Constructions,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RS Aggarwal Solutions: Constructions is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup