Rd Sharma 2020 2021 for Class 7 Maths Chapter 17 - Constructions

Answer:

Steps of construction:

1. Draw angle BAC =  50$°$ such that AB = 5 cm and AC = 7 cm.
2. Cut an arc through C at an angle of 50$°$.
3. Draw a straight line passing through C and the arc. This line will be parallel to AB since $\angle CAB=\angle RCA=50°$.
4. Alternate angles are equal; therefore the line is parallel to AB.
5. Again through B, cut an arc at an angle of 50$°$ and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.
6. $\angle SBA\hspace{0.17em}=\angle BAC=50°$, since they are alternate angles. Therefore BD$\parallel$AC.
7. Also we can measure BD = 7 cm and CD = 5 cm.

Answer:

1. Draw a line PQ.
2. Take any two points A and B on the line.
3. Construct $\angle PBF=90° and \angle QAE=90°$.
4. With A as centre and radius 3 cm cut AE at C.
5. With B as centre and radius 3 cm cut BF  at D.
6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.

                                     

Answer:

Steps of construction:

4. Join AS and produce in both directions to obtain the line parallel to BC.

                 

Answer:

Steps of construction:

2. Take any two points A and B on the line.
3. Construct $\angle PBF=90° and \angle QAE=90°$.
4. With A as centre and radius 5 cm cut AE at C.
5. With B as centre and radius 5 cm cut BF  at D.
6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.
   
 

Answer:

Steps of construction:

1. Draw a line segment AB of length 5.5 cm.
2. From B, cut an arc of radius 6 cm.
3. With centre A, draw an arc of radius 7 cm intersecting the previously drawn arc at say, C.
4. Join AC and BC to obtain the desired triangle.
5. With centre B and radius more than $\frac{1}{2}BC$, draw two arcs on both sides of BC.
6. With centre C and the same radius as in the previous step, draw two arcs intersecting the arcs drawn in the previous step at X and Y.
7. Join XY to get the perpendicular bisector of BC.

Answer:

Steps of construction:

1. Draw a line segment PQ of length 3 cm.
2. With Q as centre and radius 4 cm, draw an arc.
3. With P as centre and radius 5 cm, draw an arc intersecting the previously drawn arc at R.
4. Join PR and QR to obtain the required triangle.
5. From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.
6. From M and N, cut arcs of equal radius intersecting at point S.

8. Verify that angle PQS and angle SQR are equal to 45$°$ each.

Answer:

Steps of construction:

1. Draw a line segment AB of length 7 cm.
2. With centre A, draw an arc of radius 7 cm.
3. With centre B, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
4. Join AC and BC to get the required triangle.

Answer:

Steps of construction:

1. Draw a line segment PR of length 7 cm.
2. With centre P, draw an arc of radius 5 cm.
3. With centre R, draw an arc of radius 4 cm intersecting the previously drawn arc at Q.
4. Join PQ and QR to obtain the required triangle.
5. From P, draw arcs with radius more than half of PR on either sides.
6. With the same radius as in the previous step, draw arcs from R on either sides of PR intersecting the arcs drawn in the previous step at M and N.
7. MN is the required perpendicular bisector of the largest side.

 

Answer:

Steps of construction:

1. Draw a line segment BC of length 7 cm.
2. With centre B, draw an arc of radius 6 cm.
3. With centre C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.
4. Join AC and BC to get the required triangle.

5. Angle LAD is 15^o.

Answer:

Steps of construction:

1. Draw a line segment EF of length 6 cm.
2. With E as centre, draw an arc of radius 4 cm.
3. With F as centre, draw an arc of radius 4 cm intersecting the previous arc at D.
4. Join DE and DF to get the desired triangle.DF, .

Answer:

We first draw a triangle ABC with each side = 6 cm.
 

4. Join XD and extend it to intersect AC at E.
5. DE is the required parallel line.

Answer:

Steps of construction:

1. Draw a line segment AB of length 3 cm.
2. Draw $\angle XBA = 70°$.
3. Cut an arc on BX at a distance of 5 cm at C.
4. Join AC to get the required triangle.

Answer:

Steps of construction:

1. Draw a line segment AC of length 6 cm.
2. Draw $\angle$XAC = 70$°$.
3. Cut an arc on AX at a distance of 4 cm at B.
4. Join BC to get the desired triangle.
5. We see that BC = 6 cm.

Answer:

Steps of construction:

1. Draw a line segment PQ of length 3 cm.
2. Draw $\angle QPX = 45°$.
3. Cut an arc on PX at a distance of 3 cm at R.
4. Join QR to get the required triangle.

Answer:

Steps of construction:

1. Draw a line segment AC of length 3 cm.
2. Draw $\angle XAC = 120°$.
3. Cut an arc on AX at a distance of 3 cm at B.
4. Join BC to get the required triangle.                                             

Answer:

Steps of construction:

1. Draw a line segment BC of length 4 cm.
2. AT C, draw $\angle BCY = 90°$.
3. Cut an arc on CY at a distance of 4 cm at A.
4. Join AB.
5. ABC is the required triangle.  

Answer:

Steps of construction:

1. Draw a line segment AB of length 3 cm.
2. Draw an angle of 45$°$ and cut an arc at this angle at a radius of 4 cm at C.
3. Join AC to get the required triangle.
4. With A as centre, draw intersecting arcs at M and N.
5. With centre M and radius more that $\frac{1}{2}MN$, cut an arc on the opposite side of A.
6. With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
7. Join AE,  it meets BC at D, then AE is the required perpendicular.

Answer:

Steps of construction:

1. Draw a line segment BC = 4 cm.
2. Draw $\angle CBX\hspace{0.17em}= 60°$.
3. Draw an arc on BX at a radius of 3 cm cutting BX at A.
4. Join AC to get the required triangle.

Angle $\angle$COA = 120^o

Answer:

Steps of construction:

1. Draw a line segment BC of length 4 cm.
2. Draw $\angle CBX \mathrm{such} \mathrm{that} \angle CBX\hspace{0.17em}=50°$.
3. Draw $\angle BCY$ with Y on the same side of BC as X such that $\angle$BCY = 70$°$.
4. Let CY and BX intersect at A.
5. ABC is the required triangle.

Answer:

$$\begin{gathered} \angle ABC+\angle BCA+\angle CAB=180° \\ \angle BCA=180°-\angle ABC-\angle CAB \\ \angle BCA=180°-100°=80° \end{gathered}$$

Steps of construction:

1. Draw a line segment BC of length 8 cm.
2. Draw $\angle CBX \mathrm{such} \mathrm{that} \angle CBX = 50°$.
3. Draw $\angle$BCY with Y on the same side of BC as X such that $\angle$BCY = 80$°$.
4. Let CY and BX intersect at A.

Answer:

Steps of construction:

1. Draw a line segment QR = 4.5 cm.
2. Draw $\angle RQX = 80° and \angle QRY =55°$.
3. Let QX and RY intersect at P so that PQR is the required triangle.
4. With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
5. With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
6. Join MN; MN is the required perpendicular bisector of QR.

Answer:

Steps of construction:

1. Draw a line segment AB = 6.4 cm.
2. Draw $\angle BAX = 45°$.
3. Draw $\angle$ABY with Y on the same side of AB as X such that $\angle$ABY = 60$°$.

Answer:

We can see that ∠A+∠B+∠C = 180°. Therefore ∠C = 180$°$ − 60$°$ − 90$°$ = 30°.

Steps of construction:

1. Draw a line segment AC = 6 cm.
2. Draw $\angle ACX=30°$.
3. Draw $\angle$CAY with Y on the same side of AC as X such that $\angle$CAY = 90$°$.
4. Join CX and AY. Let these intersect at B.
5. ABC is the required triangle where angle $\angle$ABC = 60$°$.

Answer:

Steps of construction:

1. Draw a line segment QR = 4 cm.
2. Draw $\angle$QRX of measure 90$°$.
3. With centre Q and radius PQ = 5 cm, draw an arc of the circle to intersect ray RX at P.
4. Join PQ to obtain the desired triangle PQR.
5. PQR is the required triangle.

Answer:

Steps of construction:

1. Draw a line segment QR = 2.5 cm.
2. Draw $\angle$QRX of measure 90$°$.
3. With centre Q and radius PQ = 4 cm, draw an arc of the circle to intersect ray RX at P.
4. Join PQ to obtain the desired triangle PQR.
5. PQR is the required triangle.

Answer:

Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let $\angle C = 30°$.
Therefore $$\begin{gathered} \angle A+\angle B+\angle C =180° \\ \angle B =180-30-90 = 60° \end{gathered}$$
Steps of construction:

1. Draw a line segment BC = 5.4 cm.
2. Draw angle CBY = 60^o.
3. Draw angle BCX of measure 30^o with X on the same side of BC as Y.
4. Let BY and CX intersect at A.

Answer:

Steps of construction:

1. Draw a line segment BC = 4.5 cm.
2. Draw $\angle$BCX of measure 90$°$.
3. With centre B and radius AB =5.8 cm, draw an arc of the circle to intersect ray BX at A.
4. Join AB to obtain the desired triangle ABC.
5. ABC is the required triangle.                                                

Answer:

Steps of construction:

1. Draw a line segment BC = 4.6 cm.
2. Draw $\angle$BCX of measure 90$°$.
3. With centre B and radius AB = 5.2 cm, draw an arc of the circle to intersect ray CX at A.
4. Join AB to obtain the desired triangle ABC.
5. ABC is the required triangle.