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#### Page No 522:

Steps of Construction:

Step 1. Draw a line segment AB = 7 cm.

Step 2. Draw a ray AX, making an acute angle $\angle$BAX.

Step 3. Along AX, mark 5 points (greater of 3 and 5)  A1, A2, A3, A4 and A5 such that

AA1 = A1A2 = A2A3 = A3A4 = A4A5

Step 4. Join A5B.

Step 5. From A3, draw A3P parallel to A5B (draw an angle equal to $\angle$AA5B), meeting AB in P. Here, P is the point on AB such that $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{3}{2}$ or $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{5}$.

#### Page No 522:

(i) Steps of Construction:

Step 1. Draw a line segment AB = 8 cm

Step 2. Draw a ray AX, making an acute angle $\angle$BAX.

Step 3. Along AX, mark (4 + 5 =) 9 points A1, A2, A3, A4, A5, A6, A7, A8 and A9 such that

AA= A1A2 = A2A= A3A4 = A4A5 = A5A6 = A6A7 = A7A= A8A9

Step 4. Join A9B.

Step 5. From A4, draw A4D parallel to A9B (draw an angle equal to $\angle$AA9B), meeting AB in D. Here, D is the point on AB which divides it in the ratio 4 : 5.

(ii) Steps of Construction:

Step 1. Draw a line segment AB = 7.6 cm

Step 2. Draw a ray AX, making an acute angle $\angle$BAX.

Step 3. Along AX, mark (5 + 8 =) 13 points A1, A2, A3, A4, A5, A6, A7, A8, A9, A10, A11, A12 and A13 such that

AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = A7A8 = A8A9 = A9A10 = A10A11 = A11A12 = A12A13

Step 4. Join A13B.

Step 5. From A5, draw A5P parallel to A13B (draw an angle equal to $\angle$AA13B), meeting AB in P. Here, P is the point on AB which divides it in the ratio 5 : 8.

∴ Length of AP = 2.9 cm (Approx)

Length of BP = 4.7 cm (Approx)

#### Page No 522:

Steps of Construction

Step 1. Draw a line segment QR = 7 cm.

Step 2. With Q as centre and radius 6 cm, draw an arc.

Step 3. With R as centre and radius 8 cm, draw an arc cutting the previous arc at P.

Step 4. Join PQ and PR. Thus, ∆PQR is the required triangle.

Step 5. Below QR, draw an acute angle $\angle$RQX.

Step 6. Along QX, mark five points R1, R2, R3, R4 and R5 such that QR1 = R1R2 = R2R3 = R3R4 = R4R5.

Step 7. Join RR5.

Step 8. From R4, draw R4R' || RR5 meeting QR at R'.

Step 9. From R', draw P'R' || PR meeting PQ in P'. Here, ∆P'QR' is the required triangle, each of whose sides are $\frac{4}{5}$ times the corresponding sides of ∆PQR.

#### Page No 522:

Steps of Construction :

Step 1. Draw a line segment BC = 4 cm.
Step 2. With B as centre, draw an angle of 90o.
Step 3. With B as centre and radius equal to 3 cm, cut an arc at the right angle and name it A.
Step 4. Join AB and AC.
Thus, △ ABC is obtained .
Step 5. Extend BC to D, such that BD = $\frac{7}{5}$ BC =   $\frac{7}{5}$ (4) cm =  5.6 cm.
Step 6. Draw DE ∥ CA, cutting AB produced to E. Thus, △EBD is the required triangle, each of whose sides is  $\frac{7}{5}$ the corresponding sides of ∆ABC.

#### Page No 523:

Given: In ΔABC,
BC = 6 cm,
B  = 60°
AB = 5 cm

Steps of construction:
(1) Draw a line segment AB = 5 cm.
(2) From the point B, draw an ∠ABY = 60°
(3) Taking B as center, 6 cm radius, draw an arc on the ray BY,
Let the point where the arc intersects the ray named as C.
(4) Join AC. Hence,  ΔABC is the required triangle.

Now, we construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of ΔABC.

Steps of construction:
(1) Draw any ray BX making an acute angle with BA on the side opposite to the vertex C.
(2) Mark four points B1B2B3 and B4 on BX, so that BB1 = B1B2B2B3B3B4.
(3) Join B4A and draw a line through Bparallel to B4to intersect AB at A'.
(4) Draw a line through A' parallel to AC to intersect BC at C'. Hence,  Δ​A'BC' is the required triangle.

#### Page No 523:

Steps of Construction

Step 1. Draw a line segment AB = 6 cm.

Step 2. At A, draw $\angle$XAB = 30º.

Step 3. At B, draw $\angle$YBA = 60º. Suppose AX and BY intersect at C.

Thus, ∆ABC is the required triangle.

Step 4. Produce AB to B' such that AB' = 8 cm.

Step 5. From B', draw B'C' || BC meeting AX at C'. Here, ∆AB'C' is the required triangle similar to ∆ABC.

#### Page No 523:

Steps of Construction

Step 1. Draw a line segment BC = 8 cm.

Step 2. At B, draw $\angle$XBC = 45º.

Step 3. At C, draw $\angle$YCB = 60º. Suppose BX and CY intersect at A.

Thus, ∆ABC is the required triangle.

Step 4. Below BC, draw an acute angle $\angle$ZBC.

Step 5. Along BZ, mark five points Z1, Z2, Z3, Z4 and Z5 such that BZ1 = Z1Z2 = Z2Z3 = Z3Z4 = Z4Z5.

Step 6. Join CZ5.

Step 7. From Z3, draw Z3C' || CZ5 meeting BC at C'.

Step 8. From C', draw A'C' || AC meeting AB in A'. Here, ∆A'BC' is the required triangle whose sides are $\frac{3}{5}$ of the corresponding sides of ∆ABC.

#### Page No 523:

To construct a triangle similar to ∆ABC in which BC = 4.5 cm, $\angle$B = 45º and $\angle$C = 60º, using a scale factor of $\frac{3}{7}$, BC will be divided in the ratio 3 : 4. Here, ∆ABC ∼ ∆A'BC'

BC' : C'C = 3 : 4

or BC' : BC = 3 : 7

Hence, the correct answer is option A.

#### Page No 523:

Steps of Construction

Step 1. Draw a line segment BC = 8 cm.

Step 2. Draw the perpendicular bisector XY of BC, cutting BC at D.

Step 3. With D as centre and radius 4 cm, draw an arc cutting XY at A.

Step 4. Join AB and AC. Thus, an isosceles ∆ABC whose base is 8 cm and altitude 4 cm is obtained.

Step 5. Extend BC to E such that BE = $\frac{3}{2}$BC = $\frac{3}{2}×8$ cm = 12 cm.

Step 6. Draw EF || CA, cutting BA produced in F. Here, ∆BEF is the required triangle similar to ∆ABC such that each side of ∆BEF is times the corresponding side of ∆ABC.

#### Page No 523:

Steps of Construction

Step 1. Draw a line segment BC = 3 cm.

Step 2. At B, draw $\angle$XBC = 90º.

Step 3. With B as centre and radius 4 cm, draw an arc cutting BX at A.

Step 4. Join AC. Thus, a right ∆ABC is obtained.

Step 5. Extend BC to D such that BD = $\frac{5}{3}$BC = $\frac{5}{3}×3$ cm = 5 cm.

Step 6. Draw DE || CA, cutting BX in E. Here, ∆BDE is the required triangle similar to ∆ABC such that each side of ∆BDE is $\frac{5}{3}$ times the corresponding side of ∆ABC.

#### Page No 529:

Steps of Construction

Step 1. Draw a circle with O as centre and radius 3 cm.

Step 2. Mark a point P outside the circle such that OP = 7 cm.

Step 3. Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q.

Step 4. Draw a circle with Q as centre and radius PQ (or OQ), to intersect the given circle at the points T and T'.

Step 5. Join PT and PT'. Here, PT and PT' are the required tangents.

PT = PT' = 6.3 cm (Approx)

#### Page No 529:

Steps of Construction

Step 1. Draw a circle with O as centre and radius 3.5 cm.

Step 2. Mark a point P outside the circle such that OP = 6.2 cm.

Step 3. Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q.

Step 4. Draw a circle with Q as centre and radius PQ (or OQ), to intersect the given circle at the points T and T'.

Step 5. Join PT and PT'. Here, PT and PT' are the required tangents.

#### Page No 529:

Steps of Construction

Step 1. Draw a circle with O as centre and radius 3 cm.

Step 2. Mark a point P and Q on one of its diameters extended on both sides outside the circle such that OP = OQ = 7 cm.

Step 3. Join OP and OQ. Draw the perpendicular bisector XY of OP and X'Y' of OQ, cutting OP at L and OQ at M.

Step 4. Draw a circle with L as centre and radius PL (or OL), to intersect the given circle at the points A and B. Draw another circle with M as centre and radius MQ (or OM), to intersect the given circle at the points C and D.

Step 5. Join PA and PB. Join QC and QD. Here, PA, PB and QC, QD are the required tangents.

#### Page No 529:

Steps of Construction

Step 1. Draw a circle with centre O and radius 4 cm.

Step 2. Draw any diameter AOB of the circle.

Step 3. At A, draw $\angle$OAX = 90º. Produce XA to Y.

Step 4. At B, draw $\angle$OBX' = 90º. Produce X'B to Y'. Here, XAY and X'BY' are the tangents to the circle at the end points of the diameter AB.

#### Page No 530:

Steps of Construction

Step 1. Draw a circle with the help of a bangle.

Step 2. Mark a point P outside the circle.

Step 3. Through P, draw a secant PAB to intersect the circle at A and B.

Step 4. Produce AP to C such that PA = PC.

Step 5. Draw a semicircle with CB as diameter.

Step 6. Draw PD ⊥ BC, intersecting the semicircle at D.

Step 7. With P as centre and PD as radius, draw arcs to intersect the circle at T and T'.

Step 8. Join PT and PT'. Here, PT and PT' are the required pair of tangents.

#### Page No 530:

Steps of Construction

Step 1. Draw a line segment AB = 8 cm.

Step 2. With A as centre and radius 4 cm, draw a circle.

Step 3. With B as centre and radius 3 cm, draw another circle.

Step 4. Draw the perpendicular bisector XY of AB, cutting AB at C.

Step 5. With C as centre and radius AC (or BC), draw a circle intersecting the circle with centre A at P and P'; and the circle with centre B at Q and Q'.

Step 6. Join BP and BP'. Also, join AQ and AQ'. Here, AQ and AQ' are the tangents from A to the circle with centre B. Also, BP and BP' are the tangents from B to the circle with centre A.

#### Page No 530:

Steps of Construction:

Step 1. Draw a circle with centre O and radius = 4.2 cm.
Step 2. Draw any diameter AOB of this circle.
Step 3. Construct , such that the radius OC meets the circle at C.
Step 4. Draw AM
AM and CN intersect at P. Thus, PA and PC are the required tangents to the given circle inclined at an angle of 45o.

#### Page No 530:

Steps of Construction

Step 1. Draw a circle with centre O and radius 3 cm.

Step 2. Draw any diameter AOB of the circle.

Step 3. Construct $\angle$BOC = 60º such that radius OC cuts the circle at C.

Step 4. Draw AM ⊥ AB and CN ⊥ OC. Suppose AM and CN intersect each other at P. Here, AP and CP are the pair of tangents to the circle inclined to each other at an angle of 60º.

#### Page No 530:

Steps Of construction:

Step 1. Draw a circle with centre O and radius 3 cm.
Step 2. Draw radius OA and produce it to B.
Step 3. Make .
Step 4. Draw PQ$\perp OP$, meeting OB at Q.
Step 5. Then, PQ is the desired tangent, such that . #### Page No 530:

Steps of Construction

Step 1. Mark a point O on the paper.

Step 2. With O as centre and radii 4 cm and 6 cm, draw two concentric circles.

Step 3. Mark a point P on the outer circle.

Step 4. Join OP.

Step 5. Draw the perpendicular bisector XY of OP, cutting OP at Q.

Step 6. Draw a circle with Q as centre and radius OQ (or PQ), to intersect the inner circle in points T and T'.

Step 7. Join PT and PT'. Here, PT and PT' are the required tangents.

PT = PT' = 4.5 cm (Approx)

Verification by actual calculation

Join OT to form a right ∆OTP.     (Radius is perpendicular to the tangent at the point of contact)

In right ∆OTP,

#### Page No 530:

Steps of Construction

Step 1. Mark a point O on paper.

Step 2. Taking O as a centre, draw two concentric circles of radius 3 cm and 5 cm. Mark any random point P on outer circle.

Step 3. Join OP and draw its perpendicular bisector which meets OP at M.

Step 4. Draw a circle with M as centre and radius PM (or OM), to intersect the inner circle at the points A and B.

Step 5. Join PA and PB. Here, PA and PB are the required tangents.

#### Page No 530:

Steps of Construction

Step 1. Draw a circle with O as centre and some radius.

Step 2. Mark a point P outside the circle. Join OP.

Step 3. Draw the perpendicular bisector of OP cutting it at M.

Step 4. Draw another circle with M as centre and radius MP (or OM), to intersect the given circle at the points A and B.

Step 5. Join PA and PB. Here, PA and  PB are the required tangents.

#### Page No 530:

Steps of Construction :

Step 1 . Draw a line segment AB = 5.4 cm.
Step 2. Draw a ray AX, making an acute angle, $\angle BAX$.BAX.
Step 3. Along AX, mark 6 points A1, A2, A3, A4, A5, A6 such that,
AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 .
Step 4. Join A6B.
Step 5. Draw A1C A2D, A3D, A4F and A5G . Thus, AB is divided into six equal parts.

#### Page No 530:

Steps of Construction :

Step 1 . Draw a line segment AB = 6.5 cm.
Step 2. Draw a ray AX, making an acute angle $\angle BAX$.
Step 3. Along AX, mark (4+7) =11 points A1, A2, A3, A4, A5, A6, A7, A8, A9, A10, A11such that
AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = A7A8 = A8A9 = A9A10 = A10A11
Step 4. Join A11B.
Step 5. From A4, draw A4C $\parallel$A11B, meeting AB at C.
Thus, C is the point on AB, which divides it in the ratio 4:7. Thus, AC : CB = 4:7
From the figure, AC = 2.36 cm
CB = 4.14 cm

#### Page No 531:

Steps of Construction :

Step 1. Draw a line segment BC = 6.5 cm.
Step 2. With B as centre, draw an angle of 60o.
Step 3. With B as centre and radius equal to 4.5 cm, draw an arc, cutting the angle at A.
Step 4. Join AB and AC.
Thus, △ ABC is   obtained .
Step 5. Below BC, draw an acute $\angle CBX$.
Step 6. Along BX, mark off four points B1, B2, B3, B4, such that BB1 = B1B2 = B2B3 = B3B4 .
Step 7. Join B4C.
Step 8. From B3, draw B3D  ∥ B4C, meeting BC at D.
Step 9. From D, draw DE ∥ CA, meeting AB at E. Thus, △ EBD  is the required triangle, each of whose sides is$\frac{3}{4}$ the corresponding sides of ∆ABC.

#### Page No 531:

Steps of Construction :

Step 1. Draw a line l .
Step 2. Draw an angle of 90o at M on l .
Step 3. Cut an arc of radius 3 cm on the perpendicular. Mark the point as A.
Step 4. With A as centre, make an angle of 30o and let it cut l at C. We get .
Step 5. Cut an arc of 5 cm from C on l and mark the point as B.
Step 6. Join AB.
Thus, △ABC is obtained .
Step 7. Extend AB to D, such that BD =$\frac{1}{2}$ BC.
Step 8. Draw DE $\parallel BC$, cutting AC produced to E. Then, △ADE is the required triangle, each of whose sides is  $\frac{3}{2}$of the corresponding sides of △ABC.

#### Page No 531:

Steps of Construction :

Step 1. Draw a line segment BC = 9 cm.
Step 2. With B as centre, draw an arc each above and below BC.
Step 3. With C as centre, draw an arc each above and below BC.
Step 4. Join their points of intersection to obtain the perpendicular bisector of BC. Let it intersect BC at D.
Step 5. From D, cut an arc of radius 5 cm and mark the point as A.
Step 6. Join AB and AC.
Thus, △ABC is obtained .
Step 5. Below BC, make an acute $\angle CBX$.
Step 6. Along BX, mark off four points B1, B2, B3, B4, such that BB1=B1B2 = B2B3 = B3B4.
Step 7. Join B4C.
Step 8. From B3, draw B3E ∥ B4C, meeting BC at E.
Step 9. From E, draw EF ∥ CA, meeting AB at F. Thus, △FBE is the required triangle, each of whose sides is  $\frac{3}{4}$ the corresponding sides of the first triangle.

#### Page No 531:

Steps of Construction :

Step 1. Draw a line segment BC = 4 cm.
Step 2. With B as centre, draw an angle of 90o.
Step 3. With B as centre and radius equal to 3 cm, cut an arc at the right angle and name it A.
Step 4. Join AB and AC.
Thus, △ ABC is obtained .
Step 5. Extend BC to D, such that BD = $\frac{7}{5}$ BC =   $\frac{7}{5}$ (4) cm =  5.6 cm.
Step 6. Draw DE ∥ CA, cutting AB produced to E. Thus, △EBD is the required triangle, each of whose sides is  $\frac{7}{5}$ the corresponding sides of ∆ABC.

#### Page No 531:

Steps of Construction  :

Step 1. Draw a circle of radius 4.8 cm.
Step 2. Mark a point P on it.
Step 3. Draw any chord PQ.
Step 4. Take a point R on the major arc QP.
Step 5. Join PR and RQ.
Step 6. Draw
Step 7. Produce TP to T', as shown in the figure. T'PT is the required tangent.

#### Page No 531:

Steps of Construction:

Step 1. Draw a circle with centre O and radius = 3.5 cm.
Step 2. Draw any diameter AOB of this circle.
Step 3. Construct , such that the radius OC meets the circle at C.
Step 4. Draw MA AB and CNOC.
Let AM and CN intersect at P. Then, PA and PC are the required tangents to the given circle that are inclined at an angle of 60o.

#### Page No 531:

Steps Of construction:

Step 1. Draw a circle with centre O and radius 4 cm.
Step 2. Draw radius OA and produce it to B.
Step 3. Make AOP = 60°.
Step 4. Draw PQ $\perp OP$ , meeting OB at Q.
Step 5. Then, PQ is the desired tangent, such that . OQP = 30°

#### Page No 531:

Steps of Construction :

Step 1. Draw a circle with O as centre and radius 6 cm.
Step 2. Draw another circle with O as centre and radius 4 cm.
Step 2 . Mark a point P on the circle with radius 6 cm.
Step 3. Join OP and bisect it at M.
Step 4. Draw a circle with M as centre and radius equal to MP to intersect the given circle with radius 4 cm at points T and T'.
Step 5. Join PT and P T'. Thus, PT or P T'are the required tangents and measure 4.4 cm each.

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