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 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 $\angle BAX$.

Step 3. Along AX, mark (4+7) =11 points A_{1}, A_{2}_{,} A_{3}_{, }A_{4}_{,} A_{5}, A_{6}, A_{7}, A_{8}, A_{9}, A_{10}, A_{11}_{, }such that

AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}_{ }= A_{5}A_{6} = A_{6}A_{7} = A_{7}A_{8} = A_{8}A_{9} = A_{9}A_{10} = A_{10}A_{11}

Step 4. Join A_{11}B.

Step 5. From A_{4}_{, }draw A_{4}C $\parallel $A_{11}B, 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 $\angle QPX$ .

Step 3. Along* PX*, mark (5+3) =8 points A_{1}, A_{2}_{,} A_{3}_{, }A_{4}_{,} A_{5}, A_{6}, A_{7} and A_{8}_{ , }such that

*PA*_{1}* = **A*_{1}*A*_{2}* = **A*_{2}*A*_{3}* = **A*_{3}*A*_{4}* = **A*_{4}*A*_{5}_{ }= *A*_{5}*A*_{6}* = **A*_{6}*A*_{7}* = **A*_{7}*A*_{8}

Step 4. Join *A*_{8}*Q*.

Step 5. From *A*_{5}_{, }draw *A*_{5}*C** *∥* **A*_{8}*Q*, 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 $\frac{5}{7}$ 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, $\u25b3ABC\mathrm{is}\mathrm{obtained}.$

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

Step 6. Along BX, mark off seven points B_{1}, B_{2}, B_{3}, B_{4}, B_{5}_{,} B_{6}_{,} B_{7}_{,} such that B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}_{ = }B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7}.

Step 7. Join B_{7}C.

Step 8. From B_{5}, draw B_{5}D $\parallel $ B_{7}C, meeting BC at D.

Step 9. From D, draw $DE\parallel CA$, meeting AB at E.

Thus, $\u25b3EBD$ is the required triangle, each of whose sides is $\frac{5}{7}$ the corresponding sides of ∆*ABC*.

#### Page No 547:

#### Question 4:

Construct a ∆*P**QR* 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 $\frac{3}{5}$ 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 60^{o} .

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 $\angle RQX$.

Step 6. Along QX, mark off five points Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}, such that Q_{1}Q_{2} = Q_{2}Q_{3} = Q_{3}Q_{4}_{ = }Q_{4}Q_{5} .

Step 7. Join Q_{5}R.

Step 8. From Q_{3}, draw Q_{3}S ∥ Q_{5}R, 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 $\frac{3}{5}$ 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 $\frac{3}{2}$ 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 = $\frac{3}{2}$BC = $\frac{3}{2}$(6) cm = 9 cm.

Step 8. Draw DE $\parallel AC$, cutting AB produced to E.

Thus, △EBD is the required triangle, each of whose sides is $\frac{3}{2}$ 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 $\frac{4}{3}$ 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 45^{o} .

Step 3. With C as centre, draw an angle measuring $180\xb0-\left(105\xb0+45\xb0\right)=30\xb0$.

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 = $\frac{4}{3}$ BC = $\frac{4}{3}$(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 $\frac{4}{3}$ 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 $\frac{7}{5}$ 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 90^{o}.

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 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 $\frac{7}{5}$ 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' = $\frac{7}{5}$ BC = $\frac{7}{5}$(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 $\frac{7}{5}$ 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 $\frac{3}{2}$ 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 60^{o} .

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' =$\frac{3}{2}$ AB = $\frac{3}{2}$ (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 $\frac{3}{2}$ 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 $\frac{3}{4}$ 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 60^{o}.

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 B_{1}, B_{2}, B_{3}, B_{4}, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}_{ .}

Step 7. Join B_{4}C.

Step 8. From B_{3}, draw B_{3}D ∥ B_{4}C, 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 548:

#### Question 11:

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.

#### 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 $\angle CBX$.

Step 6. Along BX, mark off four points B_{1}, B_{2}, B_{3}, B_{4}, such that BB_{1}=B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}_{.}

Step 7. Join B_{4}C.

Step 8. From B_{3}, draw B_{3}E ∥ B_{4}C, 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 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 T_{1} and T_{2}.

Step 5. Draw a circle with M as centre and radius equal to MA to intersect the given circle at points T_{3} and T_{4}.

Step 6. Join BT_{1}, B T_{2} , AT_{3}_{,} AT_{4}.

Thus, BT_{1}, B T_{2} , AT_{3}_{ and} AT_{4} 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 $\angle BOC=45\xb0$, such that the radius OC meets the circle at C.

Step 4. Draw AM $\perp AB\mathrm{and}CN\perp OC.$

AM and CN intersect at P.

Thus, PA and PC are the required tangents to the given circle inclined at an angle of 45^{o}.

#### 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 T_{1}_{, }T_{2}, T_{3} and T_{4}.

Step 5. Join AT_{1} , AT_{2}_{ ,}BT_{3} and BT_{4}_{.}

Thus, AT_{1} , AT_{2}_{ ,}BT_{3} and BT_{4}_{ }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 T_{1}_{, }T_{2}, T_{3}, T_{4}.

Step 5. Join AT_{1} , AT_{2}_{ ,}BT_{3}, BT_{4}_{.}

Then AT_{1} , AT_{2}_{ ,}BT_{3}, BT_{4}_{ }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 $\angle AOP=60\xb0$.

Step 4. Draw PQ$\perp OP$, meeting OB at Q.

Step 5. Then, PQ is the desired tangent, such that $\angle OQP=30\xb0$.

#### 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 $\angle AOP=30\xb0$

Step 4. Draw PQ $\perp OP$ , meeting OB at Q.

Step 5. Then, PQ is the desired tangent, such that

∠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, $\angle BAX$.

Step 3. Along AX, mark 6 points A_{1}, A_{2}_{,} A_{3}_{, }A_{4}_{,} A_{5}, A_{6}_{ }such that,

AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}_{ }= A_{5}A_{6} .

Step 4. Join A_{6}B.

Step 5. Draw A_{1}C _{2}D, A_{3}D, A_{4}F and A_{5}G .

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

Step 4. Draw MA

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 60^{o}.

#### 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 $\angle QPT=\angle PRQ$

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 $\frac{3}{2}$ 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 90^{o} 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 30^{o}^{ }and let it cut *l *at C. We get $\angle ACB=60\xb0$.

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 =

Step 8. Draw DE

Then, △ADE is the required triangle, each of whose sides is $\frac{3}{2}$of the corresponding sides of △ABC.

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