RS Aggarwal 2015 Solutions for Class 10 Math Chapter 13 Construction are provided here with simple step-by-step explanations. These solutions for Construction are extremely popular among class 10 students for Math Construction Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2015 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 2015 Solutions. All RS Aggarwal 2015 Solutions for class 10 Math are prepared by experts and are 100% accurate.

Page No 547:

Question 1:

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

Answer:

Steps of Construction :

Step 1 . Draw a line segment AB = 6.5 cm.
Step 2. Draw a ray AX, making an acute 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 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 547:

Question 2:

Draw a line segment PQ of length 5.8 cm and divide it in the ratio 5 : 3. Measure each part.

Answer:

Steps of Construction :

Step 1 . Draw a line segment PQ = 5.8 cm.
Step 2. Draw a ray PX, making an acute angle QPX .
Step 3. Along PX, mark (5+3) =8 points A1, A2, A3, A4, A5, A6, A7 and A8 , such that
            PA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = A7A8
Step 4. Join A8Q.
Step 5. From A5, draw A5C A8Q, meeting PQ at C.
            Thus, C is the point on PQ, which divides it in the ratio 5:3.



Thus, PC : CQ = 5:3
From the figure, PC = 3.6 cm
                           CQ = 2.2  cm

Page No 547:

Question 3:

Construct a ∆ABC in which AB = 5 cm, BC = 6 cm and CA = 7 cm. Construct a triangle similar to ∆ABC, such that each of its side is 57 the corresponding sides of ∆ABC.

Answer:

Steps of Construction :

Step 1. Draw a line segment BC = 6 cm.
Step 2. With B as centre and radius equal to 5 cm, draw an arc.
Step 3. With C as centre and radius equal to 7 cm, draw another arc, cutting the previous arc at A.
Step 4. Join AB and AC.
            Thus, ABC is obtained.
Step 5. Below BC, draw an acute CBX.
Step 6. Along BX, mark off seven points B1, B2, B3, B4, B5, B6, B7, such that B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
Step 7. Join B7C.
Step 8. From B5, draw B5 B7C, meeting BC at D.
Step 9. From D, draw DECA , meeting AB at E.


Thus, EBD is the required triangle, each of whose sides is  57 the corresponding sides of ∆ABC.

Page No 547:

Question 4:

Construct a ∆PQR in which QR = 6 cm, PQ = 5 cm and ∠PQR = 60°. Now , construct a triangle similar to ∆PQR such that each of its sides is 35 the corresponding sides of ∆PQR.

Answer:

Steps of Construction :

Step 1. Draw a line segment QR = 6 cm.
Step 2. With Q as centre, draw an angle of 60o .
Step 3. With Q as centre and radius equal to 5 cm, draw another arc, cutting the previous arc at P.
Step 4. Join PQ and PR.
            Thus, △ PQR  is obtained .
Step 5. Below QR, draw an acute RQX.
Step 6. Along QX, mark off five points Q1, Q2, Q3, Q4, Q5, such that Q1Q2 = Q2Q3 = Q3Q4 = Q4Q5 .
Step 7. Join Q5R.
Step 8. From Q3, draw Q3S  ∥ Q5R, meeting QR at S.
Step 9. From S, draw ST ∥ PR, meeting PQ at T.


Thus, △TQS is the required triangle, each of whose sides is  35 the corresponding sides of ∆PQR.

Page No 547:

Question 5:

Construct an isosceles ∆ABC with base BC = 6 cm and altitude = 4 cm. Now, construct a triangle similar to ∆ABC, each of whose sides is 32 times the corresponding sides of ∆ABC.

Answer:

 Steps of Construction :

Step 1. Draw a line segment BC = 6 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 and thus, we obtain the perpendicular bisector of BC. Let it intersect BC at M.
Step 5. From D, cut an arc of radius 4 cm and mark the point as A.
Step 6. Join AB and AC.
            Thus, △ABC is obtained .
Step 7. Extend BC to D, such that BD = 32BC = 32(6) cm =  9 cm.
Step 8. Draw DE AC , cutting AB produced to E.


Thus, △EBD is the required triangle, each of whose sides is  32 the corresponding sides of ∆ABC.

Page No 547:

Question 6:

Draw ∆ABC in which BC = 5.4 cm, ∠B = 45° and ∠A = 105°. Now, construct a triangle similar to ∆ABC, each of whose sides is 43 the corresponding side of ∆ABC.

Answer:

Steps of Construction :

Step 1. Draw a line segment BC = 5.4 cm.
Step 2. With B as centre, draw an angle of 45o .
Step 3. With C as centre, draw an angle measuring 180°-105°+45° = 30°.
Step 4. Name the point of intersection A.
Step 5. Join AB and AC.
            Thus, △ABC is obtained .
Step 6. Extend BC to D, such that BD =  43 BC =  43(5.4) cm =  7.2 cm.
Step 7. Draw DE ∥ AC, cutting AB produced to E.


Thus, △EBD is the required triangle, each of whose sides is 43 the corresponding sides of ∆ABC.

Page No 547:

Question 7:

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 75 times the corresponding sides of ∆ABC.

Answer:

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 = 75 BC =   75 (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  75 the corresponding sides of ∆ABC.

Page No 547:

Question 8:

Construct a ∆ABC in which BC = 5 cm, CA = 6 cm and AB = 7 cm. Construct a ∆A' BC' similar to ∆ABC, each of whose sides is 75 times the corresponding sides of ∆ABC.

Answer:

Steps of Construction :

Step 1. Draw a line segment BC = 5 cm.
Step 2. With B as centre and radius of 7 cm, cut an arc.
Step 3. With C as centre and radius of 6 cm, cut an arc. Name the point of intersection A.
Step 4. Join AB and AC.
            Thus, △ABC is obtained .
Step 5. Extend BC to C', such that BC' = 75 BC = 75(5) cm =  7 cm.
Step 6. Draw A'C' ∥ AC, cutting BA produced at A'.

Thus, △A'BC'is the required triangle, each of whose sides is  75 the corresponding sides of ∆ABC.

Page No 547:

Question 9:

Construct a ∆ABC, in which AB = 6.5 cm, ∠B = 60° and BC = 5.5 cm. Also, construct a ∆AB'C' similar to ∆ABC,whose each side is 32 times the corresponding sides of ∆ABC.

Answer:

Steps of Construction :

Step 1. Draw a line segment AB = 6.5 cm.
Step 2. With B as centre, draw an angle of 60o .
Step 3. With B as centre and radius 5.5 cm, draw an arc cutting the angle at C.
Step 6. Join BC and AC.
            Thus, △ABC is obtained .
Step 7. Extend AB to B', such that AB' =32 AB =  32 (6.5) cm =  9.75 cm.
Step 8. Draw B'C' ∥ BC, cutting AC produced at C'.


Thus, △AB'C' is the required triangle, each of whose sides is 32 the corresponding sides of ∆ABC.

Page No 547:

Question 10:

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 34 the corresponding sides of ∆ABC.

Answer:

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 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 is34 the corresponding sides of ∆ABC.



Page No 548:

Question 11:

Construct an isosceles triangle whose base is 9 cm and altitude 5 cm. Construct another triangle whose sides are 34  the corresponding sides of the first isosceles triangle.

Answer:

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 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  34 the corresponding sides of the first triangle.



Page No 550:

Question 1:

Drawn a circle of radius 5 cm. From a point P, 8 cm away from it centre, construct a pair of tangents of the circle. Measure the length of each tangent.

Answer:

Steps of Construction :

Step 1. Draw a circle with O as centre and radius 5 cm.
Step 2 . Mark a point P outside the circle, such that OP = 8 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 at points T and T'.
Step 5. Join PT and P T'.



Thus, PT and P T'are the required tangents, measuring 6.2 cm each.

Page No 550:

Question 2:

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.

Answer:

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.

Page No 550:

Question 3:

Draw a circle of radius 3.5 cm. Take two points A and B on one of its extended diameter, each at a distance of 7 cm from its centre. Draw tangents to the circle form each of these points A and B.

Answer:

Steps of Construction :

Step 1. Draw a circle with O as centre and radius 3.5 cm.
Step 2 . Mark two points A and B outside the circle, such that OA = OB = 7 cm.
Step 3. Join OA and OB and bisect OA and OB at M and M' respectively.
Step 4. Draw a circle with M' as centre and radius equal to M'B to intersect the given circle at points T1 and T2.
Step 5. Draw a circle with M as centre and radius equal to MA to intersect the given circle at points T3 and T4.
Step 6. Join BT1, B T2 , AT3, AT4.



Thus, BT1, B T2 , AT3 and AT4 are the required tangents.

Page No 550:

Question 4:

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°.

Answer:

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 BOC  =45°, such that the radius OC meets the circle at C.
Step 4. Draw AM AB and CNOC.
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 550:

Question 5:

Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2.5 cm. Construct tangents to each circle from the centre of the other circle.

Answer:

Steps of Construction :
Step 1. Draw a line segment AB = 8.5 cm.
Step 1. Draw circles with A and B as centres and radii 4 cm and 3 cm, respectively.
Step 2  Bisect AB at M.
Step 4. Draw a circle with M as centre and radius equal to MB and MA to intersect the given circles at points T1, T2, T3 and T4.
Step 5. Join AT1 , AT2 ,BT3 and BT4.


Thus, AT1 , AT2 ,BT3 and BT4 are the required tangents.

Page No 550:

Question 6:

Draw a line segment AB of length 8.5 cm. With A as centre, draw a circle of radius 4 cm. With B as centre, draw another circle of radius 3 cm. From the centre of each circle, draw a tangent to the other circle.

Answer:

Steps of Construction :
Step 1. Draw a line segment AB = 7 cm.
Step 1. Draw circles with A and B  as centres and radius 3 cm and 2.5 cm respectively.
Step 2  Bisect it at M.
Step 4. Draw a circle with M as centre and radius equal to MA and MB to intersect the given circles at points T1, T2, T3, T4.
Step 5. Join AT1 , AT2 ,BT3, BT4.


Then AT1 , AT2 ,BT3, BT4 are the required tangents.

Page No 550:

Question 7:

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.

Answer:

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 AOP = 60°.
Step 4. Draw PQOP, meeting OB at Q.
Step 5. Then, PQ is the desired tangent, such that OQP = 30°.

Page No 550:

Question 8:

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.

Answer:

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 = 30°AOP = 60°.
Step 4. Draw PQ OP , meeting OB at Q.
Step 5. Then, PQ is the desired tangent, such that OQP = 60°.

OQP = 30°

Page No 550:

Question 1:

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

Answer:

Steps of Construction :

Step 1 . Draw a line segment AB = 5.4 cm.
Step 2. Draw a ray AX, making an acute 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 550:

Question 2:

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.

Answer:

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 BOC = 60° , such that the radius OC meets the circle at C.
Step 4. Draw MA AB and NCOC.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 551:

Question 3:

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.

Answer:

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 QPT = PRQ
Step 7. Produce TP to T', as shown in the figure.



T'PT is the required tangent.

Page No 551:

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 32 times the corresponding side of ∆ABC. Write the steps of construction.

Answer:

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 ACB = 60°.
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 =12 BC.
Step 8. Draw DE BC, cutting AC produced to E.



Then, △ADE is the required triangle, each of whose sides is  32of the corresponding sides of △ABC.



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