Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 13 Constructions are provided here with simple step-by-step explanations. These solutions for Constructions are extremely popular among Class 10 students for Math Constructions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Question 1:

Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{5}$.                   [CBSE 2011]

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 $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{5}$.

#### Question 2:

Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

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)

#### Question 3:

Construct a ∆PQR, in which PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then, construct another triangle whose sides are $\frac{4}{5}$ times the corresponding sides of ∆PQR.                                                                                                                                                    [CBSE 2013, 14]

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.

#### Question 4:

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of first triangle.

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.

#### Question 5:

Construct a ∆ABC, with BC = 7 cm, $\angle$B = 60º and AB = 6 cm. Construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of ∆ABC.                                                                                                                                                                          [CBSE 2008, '09, '15]

Steps of Construction

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

Step 2. At B, draw $\angle$XBC = 60º.

Step 3. With B as centre and radius 6 cm, draw an arc cutting the ray BX at A.

Step 4. Join AC. Thus, ∆ABC is the required triangle.

Step 5. Below BC, draw an acute angle $\angle$YBC.

Step 6. Along BY, mark four points B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.

Step 7. Join CB4.

Step 8. From B3, draw B3C' || CB4 meeting BC at C'.

Step 9. From C', draw A'C' || AC meeting AB in A'. Here, ∆A'BC' is the required triangle whose sides are $\frac{3}{4}$ times the corresponding sides of ∆ABC.

#### Question 6:

Construct a ∆ABC in which AB = 6 cm, $\angle$A = 30º and $\angle$B = 60º. Construct another ∆AB'C' similar to ∆ABC with base
AB' = 8 cm.                                                                                                                                                                                   [CBSE 2015]

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.

#### Question 7:

Construct a ∆ABC in which BC = 8 cm, $\angle$B = 45º and $\angle$C = 60º. Construct another triangle similar to ∆ABC such that its sides are $\frac{3}{5}$ of the corresponding sides of ∆ABC.                                                                                                                                             [CBSE 2010, '12, '14]

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.

#### Question 8:

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

(a) 3 : 4                             (b) 4 : 7                             (c) 3 : 10                             (d) 3 : 7                                                 [CBSE 2012]

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.

#### Question 9:

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are $1\frac{1}{2}$ times the corresponding sides of the isosceles triangle.

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.

#### Question 10:

Draw a right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then, construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle.                                                                                                      [CBSE 2011]

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.

#### Question 1:

Draw a circle of radius 3 cm. From a point P, 7 cm away from the centre of the circle, draw two tangents to the circle. Also, measure the lengths of the tangents.                                                                                                                                                                [CBSE 2010]

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)

#### Question 2:

Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6.2 cm from its centre.

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.

#### Question 3:

Draw a circle of radius 3.5 cm. Take two point A and B on one of its extended diameter, each at a distance of 5 cm from its centre. Draw tangents to the circle from each of these points A and B.

Steps of Construction

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

Step 2. Extends its diameter on both sides and mark two points A and B on it such that OA = OB = 5 cm.

Step 3. Draw the perpendicular bisectors of OA and OB. Let C and D be the mid-points of OA and OB, respectively.

Step 4. Draw a circle with C as centre and radius OC (or AC), to intersect the circle with centre O at the points P and Q.

Step 5. Draw another circle with D as centre and radius OD (or BD), to intersect the circle with centre O at the points R and S.

Step 6. Join AP and AQ. Also, join BR and BS. Here, AP and AQ are the tangents to the circle from A. Also, BR and BS are the tangents to the circle from B.

#### Question 4:

Draw a circle with centre O and radius 4 cm. Draw any diameter AB of this circle. Construct tangents to the circle at each of the two end points of the diameter AB.

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.

#### Question 5:

Draw a circle with the help of a bangle. Take any point P outside the circle. Construct the pair of tangents from the point P to the circle.

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.

#### Question 6:

Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.                                                                                    [CBSE 2014]

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.

#### Question 7:

Draw a circle of radius 4.2 cm. Draw a pair of tangents to this circle inclined to each other at an angle of 45°.

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.

#### Question 8:

Write the steps of construction for drawing a pair of tangents to a circle of radius 3 cm, which are inclined to each other at an angle of 60º.
[CBSE 2011, '12, '14]

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.

#### Question 9:

Draw a circle of radius 3 cm. Draw a tangent to the circle making an angle of 30° with a line passing through the centre.

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 . #### Question 10:

Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.

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,

#### Question 1:

Draw a line segment AB of length 5.4 cm. Divide it into six equal parts. Write the steps of construction.

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.

#### Question 2:

Draw a line segment AB of length 6.5 cm and divide it in the ratio 4 : 7. Measure each of the two parts.

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

#### Question 3:

Construct a ∆ABC, in which BC = 6.5 cm, AB = 4.5 cm and ∠ABC = 60°. Construct a triangle similar to this triangle whose sides are $\frac{3}{4}$ the corresponding sides of ∆ABC.

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.

#### Question 4:

Construct a ∆ABC, in which BC = 5 cm, ∠C = 60° and altitude from A is equal to 3 cm. Construct a ∆ADE similar to ∆ABC, such that each side of ∆ADE is $\frac{3}{2}$ times the corresponding side of ∆ABC. Write the steps of construction.

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.

#### Question 5:

Construct an isosceles triangle whose base is 9 cm and altitude 5 cm. Construct another triangle whose sides are $\frac{3}{4}$  the corresponding sides of the first isosceles triangle.

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.

#### Question 6:

Draw ∆ABC, right-angled at B, such that AB = 3 cm and BC = 4 cm. Now, construct a triangle similar to ∆ABC, each of whose sides is $\frac{7}{5}$ times the corresponding sides of ∆ABC.

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.

#### Question 7:

Draw a circle of radius 4.8 cm. Take a point P on it. Without using the centre of the circle, construct a tangent at the point P. Write the steps of construction.

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.

#### Question 8:

Draw a circle of radius 3.5 cm. Draw a pair of tangents to this circle, which are inclined to each other at an angle of 60°. Write the steps of construction.

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.

#### Question 9:

Draw a circle of radius 4 cm. Draw a tangent to the circle, making an angle of 60° with a line passing through the centre. 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°

#### Question 10:

Draw two concentric circles of radii 4 cm and 6 cm. Construct a tangent to the smaller circle from a point on the larger circle. Measure the length of this tangent.

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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