ML Aggarwal

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

Question 1.
In the given figure, ∠1 = ∠2 and ∠3 = ∠4. Show that PT × QR = PR × ST.
Solution:

Question 2.
In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM × PC = PN × PB.

Solution:

Question 3.
(a) In the figure (1) given below. ∠AED = ∠ABC. Find the values of x and y.
(b) In the fig. (2) given below, CD = \(\\ \frac { 1 }{ 2 } \) AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that:
(i) CE || AG
(ii) 3ED = GD.

Solution:

Question 4.
In the given figure, 2AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:
(i) DF || BH
(ii) AH = 3AF.

Solution:

Question 5.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.
Solution:

Question 6.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8cm. Is DE || BC? Justify your answer.
Solution:

Question 7.
In a ∆ABC, DE is parallel to the base BC, with D on AB and E on AC. If \(\frac { AD }{ DB } =\frac { 2 }{ 3 } ,\frac { BC }{ DE } \)
Solution:

Question 8.
If the area of two similar triangles are 360 cm² and 250 cm² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.
Solution:

Question 9.
In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find
(i) BC
(ii) DC
(iii) area of ∆ACD : area of ∆BCA.

Solution:

Question 10.
In the adjoining figure, the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find the area of DAOE : area of ||gm ABCD.

Solution:

Question 11.
In the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of ∆AOB and ∆COD.

Solution:

Question 12.
In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meets AB (produced) at F. Prove that

(i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4.
Solution:

Question 13.
A model of a ship is made to a scale of 1 : 250. Calculate :
(i) the length of the ship, if the length of the model is 1.6 m.
(ii) the area of the deck of the ship, if the area of the deck of the model is 2.4 m².
(iii) the volume of the model, if the volume of the ship is 1 km³.
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices MCQS

Choose the correct answer from the given four options (1 to 14):

Question 1.
If A = [a_ij]_2×2 where a_ij = i + j, then A is equal to
(a) \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
(b) \(\begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} \)
(c) \(\begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix} \)
(d) \(\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \)
Solution:

Question 2.
If \(\begin{bmatrix} x+3 & 4 \\ y-4 & x+y \end{bmatrix}=\begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix} \) then the values of x and y are
(a) x = 2, y = 7
(b) x = 7, y = 2
(c) x = 3, y = 6
(d) x = – 2, y = 7
Solution:

Question 3.
If \(\begin{bmatrix} x+2y & -y \\ 3x & 7 \end{bmatrix}=\begin{bmatrix} -4 & 3 \\ 6 & 4 \end{bmatrix} \) then the values of x and y are
(a) x = 2, y = 3
(b) x = 2, y = – 3
(c) x = – 2, y = 3
(d) x = 3, y = 2
Solution:

Question 4.
If \(\begin{bmatrix} x-2y & 5 \\ 3 & y \end{bmatrix}=\begin{bmatrix} 6 & 5 \\ 3 & -2 \end{bmatrix} \) then the value of x is
(a) – 2
(b) 0
(c) 1
(d) 2
Solution:

Question 5.
If \(\begin{bmatrix} x+2y & 3y \\ 4x & 2 \end{bmatrix}=\begin{bmatrix} 0 & -3 \\ 8 & 2 \end{bmatrix} \) then the value of x – y is
(a) – 3
(b) 1
(c) 3
(d) 5
Solution:

Question 6.
If \(x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 6 \end{matrix} \right] \) then the values of x and y are
(a) x = 2, y = 6
(b) x = 2, y = – 6
(c) x = 3, y = – 4
(d) x = 3, y = – 6
Solution:

Question 7.
If B = \(\begin{bmatrix} -1 & 5 \\ 0 & 3 \end{bmatrix} \) and A – 2B = \(\begin{bmatrix} 0 & 4 \\ -7 & 5 \end{bmatrix} \)
then the matrix A is equal to
(a) \(\begin{bmatrix} 2 & 14 \\ -7 & 11 \end{bmatrix} \)
(b) \(\begin{bmatrix} -2 & 14 \\ 7 & 11 \end{bmatrix} \)
(c) \(\begin{bmatrix} 2 & -14 \\ 7 & 11 \end{bmatrix} \)
(d) \(\begin{bmatrix} -2 & 14 \\ -7 & 11 \end{bmatrix} \)
Solution:

Question 8.
If A + B = \(\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \) and A – 2B = \(\begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix} \)
then A is equal to
(a) \(\frac { 1 }{ 3 } \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} \)
(b) \(\frac { 1 }{ 3 } \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \)
(c) \(\begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} \)
(d) \(\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \)
Solution:

Question 9.
A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) then A² =
(a) \(\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \)
(b) \(\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \)
(c) \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
(d) \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
Solution:

Question 10.
If A = \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \) , then A² =
(a) \(\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \)
(b) \(\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \)
(c) \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
(d) \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
Solution:

Question 11.
If A = \(\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \) , then A² =
(a) A
(b) O
(c) I
(d) 2A
Solution:

Question 12.
If A = \(\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \) , then A² =
(a) \(\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \)
(b) \(\begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \)
(c) \(\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)
(d) none of these
Solution:

Question 13.
If A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \) , then A² =
(a) \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \)
(b) \(\begin{bmatrix} 8 & -5 \\ 5 & 3 \end{bmatrix} \)
(c) \(\begin{bmatrix} 8 & -5 \\ -5 & -3 \end{bmatrix} \)
(d) \(\begin{bmatrix} 8 & -5 \\ -5 & 3 \end{bmatrix} \)
Solution:

Question 14.
If A = \(\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \) , then A² = pA, then the value of p is
(a) 2
(b) 4
(c) – 2
(d) – 4
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS

Choose the correct answer from the given four options (1 to 22):

Question 1.
In the given figure, ∆ABC ~ ∆QPR. Then ∠R is
(a) 60°
(b) 50°
(c) 70°
(d) 80°

Solution:

Question 2.
In the given figure, ∆ABC ~ ∆QPR. The value of x is
(a) 2.25 cm
(b) 4 cm
(c) 4.5 cm
(d) 5.2 cm

Solution:

Question 3.
In the given figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to
(a) 50°
(b) 30°
(c) 60°
(d) 100°

Solution:

Question 4.
If in two triangles ABC and PQR,
\(\frac { AB }{ QR } =\frac { BC }{ PR } =\frac { CA }{ PQ } \)
then
(a) ∆PQR ~ ∆CAB
(b) ∆PQR ~ ∆ABC
(c) ∆CBA ~ ∆PQR
(d) ∆BCA ~ ∆PQR
Solution:

Question 5.
In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE, then the two triangles are
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Solution:

Question 6.
In the given figure, if D, E and F are midpoints of the sides BC, CA and AB respectively, then the two triangles ABC and DEF are
(a) similar
(b) congruent
(c) both similar and congruent
(d) neither similar nor congruent

Solution:

Question 7.
The given figure, AB || DE. The length of CD is
(a) 2.5 cm
(b) 2.7 cm
(c) \(\\ \frac { 10 }{ 3 } \) cm
(d) 3.5 cm

Solution:

Question 8.
If in triangles ABC and DEF,\(\frac { AB }{ DE } =\frac { BC }{ FD } \) , then they will be similar when
(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠A = ∠F
Solution:

Question 9.
If ∆PQR ~ ∆ABC, PQ = 6 cm, AB = 8 cm and perimeter of ∆ABC is 36 cm, then perimeter of ∆PQR is
(a) 20.25 cm
(b) 27 cm
(c) 48 cm
(d) 64 cm
Solution:

Question 10.
In the given figure, DE || BC and all measurements are in centimetres. The length of AE is
(a) 2 cm
(b) 2.25 cm
(c) 3.5 cm
(d) 4 cm

Solution:

Question 11.
In the given figure, PQ || CA and all lengths are given in centimetres. The length of BC is
(a) 6.4 cm
(b) 7.5 cm
(c) 8 cm
(d) 9 cm

Solution:

Question 12.
In the given figure, MN || QR. If PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm, then PM is
(a) 4 cm
(b) 3.6 cm
(c) 2 cm
(d) 3 cm

Solution:

Question 13.
D and E are respectively the points on the sides AB and AC of a ∆ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then the length of DE is
(a) 2.5 cm
(b) 3 cm
(c) 5 cm
(d) 6 cm
Solution:

Question 14.
It is given that ∆ABC ~ ∆PQR with \(\frac { BC }{ QR } =\frac { 1 }{ 3 } \) then \(\frac { area\quad of\quad \Delta PQR }{ area\quad of\quad \Delta ABC } \) equal to
(a) 9
(b) 3
(c) \(\\ \frac { 1 }{ 3 } \)
(d) \(\\ \frac { 1 }{ 9 } \)
Solution:

Question 15.
If the areas of two similar triangles are in the ratio 4 : 9, then their corresponding sides are in the ratio
(a) 9 : 4
(b) 3 : 2
(c) 2 : 3
(d) 16 : 81
Solution:

Question 16.
If ∆ABC ~ ∆PQR, BC = 8 cm and QR = 6 cm, then the ratio of the areas of ∆ABC and ∆PQR is
(a) 8 : 6
(b) 3 : 4
(c) 9 : 16
(d) 16 : 9
Solution:

Question 17.
If ∆ABC ~ ∆QRP \(\frac { area\quad of\quad \Delta ABC }{ area\quad of\quad \Delta PQR } \) AB = 18 cm and BC = 15 cm, then the length of PR is equal to
(a) 10 cm
(b) 12 cm
(c) \(\\ \frac { 20 }{ 3 } \)
(d) 8 cm
Solution:

Question 18.
If ∆ABC ~ ∆PQR, area of ∆ABC = 81 cm², area of ∆PQR = 144 cm² and QR = 6 cm, then length of BC is
(a) 4 cm
(b) 4.5 cm
(c) 9 cm
(d) 12 cm
Solution:

Question 19.
In the given figure, DE || CA and D is a point on BD such that BD : DC = 2 : 1. The ratio of area of ∆ABC to area of ∆BDE is
(a) 4 : 1
(b) 9 : 1
(c) 9 : 4
(d) 3 : 2

Solution:

Question 20.
If ABC and BDE are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC and BDE is
(a) 2 : 1
(b) 1 : 2
(c) 1 : 4
(d) 4 : 1
Solution:

Question 21.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. If an altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle is
(a) 9 cm
(b) 7 cm
(c) 6 cm
(d) 4.5 cm
Solution:

Question 22.
Given ∆ABC ~ ∆PQR, area of ∆ABC = 54 cm² and area of ∆PQR = 24 cm². If AD and PM are medians of ∆’s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is
(a) \(\\ \frac { 49 }{ 3 } \) cm
(b) \(\\ \frac { 20 }{ 3 } \) cm
(c) 15 cm
(d) 22.5 cm
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3

Question 1.
Given that ∆s ABC and PQR are similar.
Find:
(i) The ratio of the area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3.
(ii) the ratio of their corresponding sides if area of ∆ABC : area of ∆PQR = 25 : 36.
Solution:

Question 2.
∆ABC ~ DEF. If area of ∆ABC = 9 sq. cm., area of ∆DEF =16 sq. cm and BC = 2.1 cm., find the length of EF.
Solution:

Question 3.
∆ABC ~ ∆DEF. If BC = 3 cm, EF = 4 cm and area of ∆ABC = 54 sq. cm. Determine the area of ∆DEF.
Solution:

Question 4.
The area of two similar triangles is 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.
Solution:

Question 5.
(a) In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm², find the area of ∆QOA. (2006)

(b) In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm.
(i) Prove that ∆OAB ~ ∆OCD.
(ii) Find CD and OB.
(iii) Find the ratio of areas of ∆OAB and ∆OCD.

Solution:

Question 6.
(a) In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm, find the area of ∆ABC.

(b) In the figure (iii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of ∆ADE and trapezium DBCE.

Solution:

Question 7.
In the given figure, DE || BC.
(i) Prove that ∆ADE and ∆ABC are similar.
(ii) Given that AD = \(\\ \frac { 1 }{ 2 } \) BD, calculate DE if BC = 4.5 cm.

(iii) If area of ∆ABC = 18 cm², find the area of trapezium DBCE
Solution:

Question 8.
In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm, calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.

Solution:

Question 9.
In the adjoining figure, ABC is a triangle.
DE is parallel to BC and \(\frac { AD }{ DB } =\frac { 3 }{ 2 } \)
(i) Determine the ratios \(\frac { AD }{ AB }, \frac { DE }{ BC } \)
(ii) Prove that ∆DEF is similar to ∆CBF.
Hence, find \(\\ \frac { EF }{ FB } \).
(iii) What is the ratio of the areas of ∆DEF and ∆CBF ? (2007)

Solution:

Question 10.
In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:
(i) area ∆APO : area ∆ABC.
(ii) area ∆APO : area ∆CQO. (2008)

Solution:

Question 11.
(a) In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of ∆AOB and ∆COD.
(b) In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find
(i) AB
(ii) BC
(iii) area of ∆ADM : area of ∆ANB.
(c) In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find
(i) EF : AD
(ii) area of ∆BEF : area of ∆ABD
(iii) area of ∆ABD : area of trap. AFED
(iv) area of ∆FEO : area of ∆OBC.

Solution:

Question 12.
In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q. If area of ∆CPQ = 20 cm², find

(i) area of ∆BPQ.
(ii) area ∆CDP.
(iii) area of || gm ABCD.
Solution:

Question 13.
(a) In the figure (i) given below, DE || BC and the ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.

(b) In the figure (ii) given below, AB || DC and AB = 2 DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find (i) ED (ii) BE (iii) area of ∆EDC : area of trapezium ABCD.

Solution:

Question 14.
(a) In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BB = 12 cm., find
(i) BP
(ii) the ratio of areas of ∆APB and ∆DPC.

(b) In the figure given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA
(ii) Find BC and CD
(iii) Find the area of ∆ACD : area of ∆ABC.

Solution:

Question 15.
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:
(i) ∆ADE ~ ∆ACB.
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of ∆ADE : area of quadrilateral BCED.

Solution:

Question 16.
Two isosceles triangles have equal vertical angles and their areas are in the ratio 7 : 16. Find the ratio of their corresponding height.
Solution:

Question 17.
On a map is drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements :
AB = 3 cm, BC = 4 cm and ∠ABC = 90°. Calculate
(i) the actual length of AB in km.
(ii) the area of the plot in sq. km:
Solution:

Question 18.
On a map is drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD has the following measurements AB = 12 cm and BG = 16 cm.
Calculate:
(i) the distance of a diagonal of the plot in km.
(ii) the area of the plot in sq. km.
Solution:

Question 19.
The model of a building is constructed with scale factor 1 : 30.
(i) If the height of the model is 80 cm, find the actual height of the building in metres.
(ii) If the actual volume of a tank at the top of the building is 27 m³, find the volume of the tank on the top of the model. (2009)
Solution:

Question 20.
A model of a ship is made to a scale of 1 : 200.
(i) If the length of the model is 4 m, find the length of the ship.
(ii) If the area of the deck of the ship is 160000 m², find the area of the deck of the model.
(iii) If the volume of the model is 200 litres, find the volume of the ship in m³. (100 litres = m³)
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions MCQS

Choose the correct answer from the given four options (1 to 33) :

Question 1.
The list of numbers – 10, – 6, – 2, 2, … is
(a) an A.P. with d = -16
(b) an A.P with d = 4
(c) an A.P with d = -4
(d) not an A.P
Solution:

Question 2.
The 10th term of the A.P. 5, 8, 11, 14, … is
(a) 32
(b) 35
(c) 38
(d) 185
Solution:

Question 3.
The 30th term of the A.P. 10, 7, 4, … is
(a) 87
(b) 77
(c) -77
(d) -87
Solution:

Question 4.
The 11th term of the A.P. -3, \(– \frac { 1 }{ 2 } \), 2, … is
(a) 28
(b) 22
(c) -38
(d) -48
Solution:

Question 5.
The 4th term from the end of the A.P. -11, -8, -5, …, 49 is
(a) 37
(b) 40
(c) 43
(d) 58
Solution:

Question 6.
The 15th term from the last of the A.P. 7, 10, 13, …,130 is
(a) 49
(b) 85
(c) 88
(d) 110
Solution:

Question 7.
If the common difference of an A.P. is 5, then a₁₈ – a₁₃ is
(a) 5
(b) 20
(c) 25
(d) 30
Solution:

Question 8.
In an A.P., if a₁₈ – a₁₄ = 32 then the common difference is
(a) 8
(b) -8
(c) -4
(d) 4
Solution:

Question 9.
In an A.P., if d = -4, n = 7, a_n = 4, then a is
(a) 6
(d) 7
(c) 20
(d) 28
Solution:

Question 10.
In an A.P., if a = 3.5, d = 0, n = 101, then a_n will be
(a) 0
(b) 3.5
(c) 103.5
(d) 104.5
Solution:

Question 11.
In an A.P., if a = -7.2, d = 3.6, a_n = 7.2, then n is
(a) 1
(b) 3
(c) 4
(d) 5
Solution:

Question 12.
Which term of the A.P. 21, 42, 63, 84,… is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th
Solution:

Question 13.
If the last term of the A.P. 5, 3, 1, -1,… is -41, then the A.P. consists of
(a) 46 terms
(b) 25 terms
(c) 24 terms
(d) 23 terms
Solution:

Question 14.
If k – 1, k + 1 and 2k + 3 are in A.P., then the value of k is
(a) – 2
(b) 0
(c) 2
(d) 4
Solution:

Question 15.
The 21st term of an A.P. whose first two terms are -3 and 4 is
(a) 17
(b) 137
(c) 143
(d) -143
Solution:

Question 16.
If the 2nd term of an A.P. is 13 and the 5th term is 25, then its 7th term is
(a) 30
(b) 33
(c) 37
(d) 38
Solution:

Question 17.
If the first term of an A.P. is -5 and the common difference is 2, then the sum of its first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15
Solution:

Question 18.
The sum of 25 terms of the A.P.\(-\frac { 2 }{ 3 } ,-\frac { 2 }{ 3 } ,-\frac { 2 }{ 3 } \) is
(a) 0
(b) \(– \frac { 2 }{ 3 } \)
(c) \(– \frac { 50 }{ 3 } \)
(d) -50
Solution:

Question 19.
In an A.P., if a = 1, a_n = 20 and S_n = 399, then n is
(a) 19
(b) 21
(c) 38
(d) 42
Solution:

Question 20.
In an A.P., if a = -5, l = 21. and S_n = 200, then n is equal to
(a) 50
(b) 40
(c) 32
(d) 25
Solution:

Question 21.
In an A.P., if a = 3 and S₈ = 192, then d is
(a) 8
(b) 7
(c) 6
(d) 4
Solution:

Question 22.
The sum of first five multiples of 3 is
(a) 45
(b) 55
(c) 65
(d) 75
Solution:

Question 23.
The number of two-digit numbers which are divisible by 3 is
(a) 33
(b) 31
(c) 30
(d) 29
Solution:

Question 24.
The number of multiples of 4 that lie between 10 and 250 is
(a) 62
(b) 60
(c) 59
(d) 55
Solution:

Question 25.
The sum of first 10 even whole numbers is
(a) 110
(b) 90
(c) 55
(d) 45
Solution:

Question 26.
The list of number \(\\ \frac { 1 }{ 9 } \) , \(\\ \frac { 1 }{ 3 } \), 1, – 3,… is a
(a) GP. with r = – 3
(b) G.P. with r = \(– \frac { 1 }{ 3 } \)
(c) GP. with r = 3
(d) not a G.P.
Solution:

Question 27.
The 11th of the G.P. \(\\ \frac { 1 }{ 8 } \) , \(– \frac { 1 }{ 4 } \) , 2, – 1, … is
(a) 64
(b) -64
(c) 128
(d) -128
Solution:

Question 28.
The 5th term from the end of the G.P. 2, 6, 18, …, 13122 is
(a) 162
(b) 486
(c) 54
(d) 1458
Solution:

Question 29.
If k, 2(k + 1), 3(k + 1) are three consecutive terms of a G.P., then the value of k is
(a) -1
(b) -4
(c) 1
(d) 4
Solution:

Question 30.
Which term of the G.P. 18, -12, 8, … is \(\\ \frac { 512 }{ 729 } \) ?
(a) 12th
(b) 11th
(c) 10th
(d) 9th
Solution:

Question 31.
The sum of the first 8 terms of the series 1 + √3 + 3 + … is
Solution:

Question 32.
The sum of first 6 terms of the G.P. 1, \(– \frac { 2 }{ 3 } \) ,\(\\ \frac { 4 }{ 9 } \) ,… is
(a) \(– \frac { 133 }{ 243 } \)
(b) \(\\ \frac { 133 }{ 243 } \)
(c) \(\\ \frac { 793 }{ 1215 } \)
(d) none of these
Solution:

Question 33.
If the sum of the GP., 1, 4, 16, … is 341, then the number of terms in the GP. is
(a) 10
(b) 8
(c) 6
(d) 5
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.5

Question 1.
Find the sum of:
(i) 20 terms of the series 2 + 6 + 18 + …
(ii) 10 terms of series 1 + √3 + 3 + …
(iii) 6 terms of the GP. 1, \(– \frac { 2 }{ 3 } \) , \(\\ \frac { 4 }{ 9 } \), …
(iv) 20 terms of the GP. 0.15, 0.015, 0.0015,…
(v) 100 terms of the series 0.7 + 0.07 + 0.007 +…
(vi) 5 terms and n terms of the series \(1+\frac { 2 }{ 3 } +\frac { 4 }{ 9 } +…\)
(vii) n terms of the G.P. √7, √21, 3√7, …
(viii)n terms of the G.P. 1, -a, a², -a³, … (a ≠ -1)
(ix) n terms of the G.P. x³, x⁵ , x⁷, … (x ≠ ±1).
Solution:

Question 2.
Find the sum of the first 10 terms of the geometric series
√2 + √6 + √18 + ….
Solution:

Question 3.
Find the sum of the series 81 – 27 + 9….\(– \frac { 1 }{ 27 } \)
Solution:

Question 4.
The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term.
Solution:

Question 5.
If the sum of the first six terms of any G.P. is equal to 9 times the sum of the first three terms, then find the common ratio of the G.P.
Solution:

Question 6.
A G.P. consists of an even number of terms. If the sum of all the terms is 3 times the sum of the odd terms, then find its common ratio.
Solution:

Question 7.
(i) How many terms of the G.P. 3, 3², 3³, … are needed to give the sum 120?
(ii) How many terms of the G.P. 1, 4, 16, … must be taken to have their sum equal to 341?
Solution:

Question 8.
How many terms of the GP. 1, √2 > 2, 2 √2,… are required to give a sum of 1023( √2 + 1)?
Solution:

Question 9.
How many terms of the \(\frac { 2 }{ 9 } -\frac { 1 }{ 3 } +\frac { 1 }{ 2 } +…\) will make the sum \(\\ \frac { 55 }{ 72 } \) ?
Solution:

Question 10.
The 2nd and 5th terms of a geometric series are \(– \frac { 1 }{ 2 } \) and sum \(\\ \frac { 1 }{ 16 } \) respectively. Find the sum of the series upto 8 terms.
Solution:

Question 11.
The first term of a G.P. is 27 and 8th term is \(\\ \frac { 1 }{ 81 } \) . Find the sum of its first 10 terms.
Solution:

Question 12.
Find the first term of the G.P. whose common ratio is 3, the last term is 486 and the sum of those terms is 728
Solution:

Question 13.
In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.
Solution:

Question 14.
Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186.
Solution:

Question 15.
If the first term of a G.P. is 5 and the sum of first three terms is \(\\ \frac { 31 }{ 5 } \), find the common ratio.
Solution:

Question 16.
The sum of first three terms of a GP. is to the sum of first six terms as 125 : 152. Find the common ratio of the GP.
Solution:

Question 17.
Find the sum of the products of the corresponding terms of the geometric progression 2, 4, 8, 16, 32 and 128, 32, 8, 2, \(\\ \frac { 1 }{ 2 } \)
Solution:

Question 18.
Evaluate \(\sum _{ n=1 }^{ 50 }{ \left( { 2 }^{ n }-1 \right) } \)
Solution:

Question 19.
Find the sum of n terms of a series whose mth term is 2^m + 2m.
Solution:

Question 20.
Sum the series
x(x + y) + x² (x² + y²) + x³ (x³ + y³) … to n terms.
Solution:

Question 21.
Find the sum of the series
1 + (1 + x) + (1 + x + x²) + … to n terms, x ≠ 1.
Solution:

Question 22.
Find the sum of the following series to n terms:
(i) 7 + 77 + 777 + …
(ii) 8 + 88 + 888 + …
(iii) 0.5 + 0.55 + 0.555 + …
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 7 Ratio and Proportion Ex 7.2

Question 1.
Find the value of x in the following proportions:
(i) 10 : 35 = x : 42
(ii) 3 : x = 24 : 2
(iii) 2.5 : 1.5 = x : 3
(iv) x : 50 :: 3 : 2
Solution:

Question 2.
Find the fourth proportional to
(i) 3, 12, 15
(ii) \(\frac { 1 }{ 3 } ,\frac { 1 }{ 4 } ,\frac { 1 }{ 5 } \)
(iii) 1.5, 2.5, 4.5
(iv) 9.6 kg, 7.2 kg, 28.8 kg
Solution:

Question 3.
Find the third proportional to
(i) 5, 10
(ii) 0.24, 0.6
(iii) Rs. 3, Rs. 12
(iv) \(5 \frac { 1 }{ 4 } \) and 7.
Solution:

Question 4.
Find the mean proportion of:
(i) 5 and 80
(ii) \(\\ \frac { 1 }{ 12 } \) and \(\\ \frac { 1 }{ 75 } \)
(iii) 8.1 and 2.5
(iv) (a – b) and (a³ – a²b), a> b
Solution:

Question 5.
If a, 12, 16 and b are in continued proportion find a and b.
Solution:

Question 6.
What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion? (2009)
Solution:

Question 7.
What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion? (2004)
Solution:

Question 8.
If 2x – 1, 5x – 6, 6x + 2 and 15x – 9 are in proportion, find the value of x.
Solution:

Question 9.
If x + 5 is the mean proportion between x + 2 and x + 9, find the value of x.
Solution:

Question 10.
What number must be added to each of the numbers 16, 26 and 40 so that the resulting numbers may be in continued proportion?
Solution:

Question 11.
Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.
Solution:

Question 12.
If b is the mean proportional between a and c, prove that a, c, a² + b², and b² + c² are proportional.
Solution:

Question 13.
If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a² + b²) and (b² + c²).
Solution:

Question 14.
If y is mean proportional between x and z, prove that
xyz (x + y + z)³ = (xy + yz + zx)³.
Solution:

Question 15.
If a + c = mb and \(\frac { 1 }{ b } +\frac { 1 }{ d } =\frac { m }{ c } \), prove that a, b, c and d are in proportion.
Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.
If ax = by = cz; prove that
\(\frac { { x }^{ 2 } }{ yz } +\frac { { y }^{ 2 } }{ zx } +\frac { { z }^{ 2 } }{ xy } \) = \(\frac { bc }{ { a }^{ 2 } } +\frac { ca }{ { b }^{ 2 } } +\frac { ab }{ { c }^{ 2 } } \)
Solution:

Question 19.

Solution:

Question 20.
If x, y, z are in continued proportion, prove that:\(\frac { { \left( x+y \right) }^{ 2 } }{ { \left( y+z \right) }^{ 2 } } =\frac { x }{ z } \). (2010)
Solution:

Question 21.
If a, b, c are in continued proportion, prove that:
\(\frac { { pa }^{ 2 }+qab+{ rb }^{ 2 } }{ { pb }^{ 2 }+qbc+{ rc }^{ 2 } } =\frac { a }{ c } \)
Solution:

Question 22.

Solution:

Question 23.

Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Ex 12.1

Question 1.
Find the slope of a line whose inclination is
(i) 45°
(ii) 30°
Solution:

Question 2.
Find the inclination of a line whose gradient is
(i) 1
(ii) √3
(iii) \(\frac { 1 }{ \sqrt { 3 } } \)
Solution:

Question 3.
Find the equation of a straight line parallel 1 to x-axis which is at a distance
(i) 2 units above it
(ii) 3 units below it.
Solution:

Question 4.
Find the equation of a straight line parallel to y-axis which is at a distance of:
(i) 3 units to the right
(ii) 2 units to the left.
Solution:

Question 5.
Find the equation of a straight line parallel to the y-axis and passing through the point (-3, 5).
Solution:

Question 6.
Find the equation of the a line whose
(i) slope = 3, y-intercept = -5
(ii) slope = \(– \frac { 2 }{ 7 } \), y-intercept = 3
(iii) gradient = √3, y-intercept = \(– \frac { 4 }{ 3 } \)
(iv) inclination = 30°, y-intercept = 2
Solution:

Question 7.
Find the slope and y-intercept of the following lines:
(i) x – 2y – 1 = 0
(ii) 4x – 5y – 9 = – 0
(iii) 3x +5y + 7 = 0
(iv) \(\frac { x }{ 3 } +\frac { y }{ 4 } =1\)
(v) y – 3 = 0
(vi) x – 3 = 0
Solution:

Question 8.
The equation of the line PQ is 3y – 3x + 7 = 0
(i) Write down the slope of the line PQ.
(ii) Calculate the angle that the line PQ makes with the positive direction of the x-axis.
Solution:

Question 9.
The given figure represents the line y = x + 1 and y = √3x – 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence determine θ.

Solution:

Question 10.
Find the value of p, given that the line \(\frac { y }{ 2 } =x-p\) passes through the point (-4, 4) (1992).
Solution:

Question 11.
Given that (a, 2a) lies on the line \(\frac { y }{ 2 } =3x-6\) find the value of a.
Solution:

Question 12.
The graph of the equation y = mx + c passes through the points (1, 4) and (-2, -5). Determine the values of m and c.
Solution:

Question 13.
Find the equation of the line passing through the point (2, -5) and making an intercept of -3 on the y-axis.
Solution:

Question 14.
Find the equation of a straight line passing through (-1, 2) and whose slope is \(\\ \frac { 2 }{ 5 } \).
Solution:

Question 15.
Find the equation of a straight line whose inclination is 60° and which passes through the point (0, -3).
Solution:

Question 16.
Find the gradient of a line passing through the following pairs of points.
(i) (0, -2), (3, 4)
(ii) (3, -7), (-1, 8)
Solution:

Question 17.
The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find:
(i) The gradient of EF
(ii) The equation of EF
(iii) The coordinates of the point where the line EF intersects the x-axis.
Solution:

Question 18.
Find the intercepts made by the line 2x – 3y + 12 = 0 on the co-ordinate axis.
Solution:

Question 19.
Find the equation of the line passing through the points P (5, 1) and Q (1, -1). Hence, show that the points P, Q and R (11, 4) are collinear.
Solution:

Question 20.
Find the value of ‘a’ for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence find the equation of the line.
Solution:

Question 21.
Use a graph paper for this question. The graph of a linear equation in x and y, passes through A (-1, -1) and B (2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and (½, k). (2005)
Solution:

Question 22.
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4). Find
(i) the coordinates of A
(ii) the equation of the diagonal BD.
Solution:

Question 23.
In ∆ABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A.
Solution:

Question 24.
Find the equation of a line passing through the point (-2, 3) and having x-intercept 4 units. (2002)
Solution:

Question 25.
Find the equation of the line whose x-intercept is 6 and y-intercept is -4.
Solution:

Question 26.
Write down the equation of the line whose gradient is \(\\ \frac { 1 }{ 2 } \) and which passes through P where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3. (2001)
Solution:

Question 27.
Find the equation of the line passing through the point (1, 4) and intersecting the line x – 2y – 11 = 0 on the y-axis.
Solution:

Question 28.
Find the equation of the straight line containing the point (3, 2) and making positive equal intercepts on axes.
Solution:

Question 29.
Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2) find:
(i) the coordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD. (2015)
Solution:

Question 30.
A and B are two points on the x-axis and y-axis respectively. P (2, -3) is the midpoint of AB. Find the

(i) the co-ordinates of A and B.
(ii) the slope of the line AB.
(iii) the equation of the line AB. (2010)
Solution:

Question 31.
Find the equations of the diagonals of a rectangle whose sides are x = -1, x = 2 , y = -2 and y = 6.

Solution:

Question 32.
Find the equation of a straight line passing through the origin and through the point of intersection of the lines 5x + 1y – 3 and 2x – 3y = 7
Solution:

Question 33.
Point A (3, -2) on reflection in the x-axis is mapped as A’ and point B on reflection in the y-axis is mapped onto B’ (-4, 3).
(i) Write down the co-ordinates of A’ and B.
(ii) Find the slope of the line A’B, hence find its inclination.
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.3

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 9 Arithmetic and Geometric Progressions Ex 9.3

Question 1.
Find the sum of the following A.P.s:
(i) 2, 7, 12, … to 10 terms
(ii) \(\frac { 1 }{ 15 } ,\frac { 1 }{ 12 } ,\frac { 1 }{ 10 } ,… \) t0 11 terms
Solution:

Question 2.
How many terms of the A.P. 27, 24, 21, …, should be taken so that their sum is zero?
Solution:

Question 3.
Find the sums given below :
(i) 34 + 32 + 30 + … + 10
(ii) -5 + ( -8) + ( -11) + … + ( -230)
Solution:

Question 4.
In an A.P. (with usual notations) :
(i) given a = 5, d = 3, a_n = 50, find n and S_n
(ii) given a = 7, a₁₃ = 35, find d and S₁₃
(iii) given d = 5, S₉ = 75, find a and a₉
(iv) given a = 8, a_n = 62, S_n = 210, find n and d
(v) given a = 3, n = 8, S = 192, find d.
Solution:

Question 5.
(i) The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
(ii) The sum of the first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.
Solution:

Question 6.
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:

Question 7.
Solve for x : 1 + 4 + 7 + 10 + … + x = 287.
Solution:

Question 8.
(i) How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116? Also, find the last term.
(ii) How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78? Explain the double answer.
Solution:

Question 9.
Find the sum of first 22 terms, of an A.P. in which d = 7 and a₂₂ is 149.
Solution:

Question 10.
(i) Find the sum of the first 51 terms of the A.P. whose second and third terms are 14 and 18 respectively.
(ii) If the third term of an A.P. is 1 and 6th term is -11, find the sum of its first 32 terms.
Solution:

Question 11.
If the sum of the first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms.
Solution:

Question 12.
Show that a₁, a₂, a₃, … form an A.P. where a_n is defined as a_n = 3 + 4n. Also, find the sum of the first 15 terms.
Solution:

Question 13.
(i) If a_n = 3 – 4n, show that a₁, a₂, a₃, … form an A.P. Also find S₂₀.
(ii) Find the common difference of an A.P. whose first term is 5 and the sum of the first four terms is half the sum of the next four terms.
Solution:

Question 14.
The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 equal to the sum of first 2n terms of another A.P. whose first term is -30 and the common difference is 8. Find n.
Solution:

Question 15.
The sum of the first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is \(\\ \frac { 1 }{ 3 } \). Calculate the first and the thirteenth term.
Solution:

Question 16.
In an A.P., the sum of its first n terms is 6n – n². Find the 25th term.
Solution:

Question 17.
If the sum of first n terms of an A.P. is 4n – n², what is the first term (i. e. S₁)? What is the sum of the first two terms? What is the second term? Also, find the 3rd term, the 10th term, and the nth terms?
Solution:

Question 18.
If S_n denotes the sum of first n terms of an A.P., prove that S₃₀ = 3(S₂₀ – S₁₀).
Solution:

Question 19.
(i) Find the sum of the first 1000 positive integers.
(ii) Find the sum of first 15 multiples of 8.
Solution:

Question 20.
(i) Find the sum of all two digit natural numbers which are divisible by 4.
(ii) Find the sum of all natural numbers between 100 and 200 which are divisible by 4.
(iii) Find the sum of all multiples of 9 lying between 300 and 700.
(iv) Find the sum of all natural numbers less than 100 which are divisible by 6.
Solution:

Question 21.
(i) Find the sum of all two digit odd positive numbers.
(ii) Find the sum of all 3-digit natural numbers which are divisible by 7.
(iii) Find the sum of all two digit numbers which when divided by 7 yields 1 as the remainder.
Solution:

Question 22.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay a penalty if he has delayed the work for 30 days?
Solution:

Question 23.
Kanika was given her pocket money on 1st Jan 2016. She puts Rs 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued on doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money and was found that at the end of the month she still has Rs 100 with her. How much money was her pocket money for the month?
Solution:

Question 24.
Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?
Solution:

Question 25.
The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she traveled carrying a flag?
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 11 Section Formula Chapter Test

Question 1.
The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, -3). If the origin is the mid-point of the base BC, find the coordinates of the points A and B
Solution:

Question 2.
A and B have co-ordinates (4, 3) and (0, 1), Find
(i) the image A’ of A under reflection in the y-axis.
(ii) the image of B’ of B under reflection in the line AA’.
(iii) the length of A’B’.
Solution:

Question 3.
Find the co-ordinates of the point that divides the line segment joining the points P (5, -2) and Q (9, 6) internally in the ratio of 3 : 1.
Solution:

Question 4.
Find the coordinates of the point P which is three-fourths of the way from A (3, 1) to B (-2, 5).
Solution:

Question 5.
P and Q are the points on the line segment joining the points A (3, -1) and B (-6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q.
Solution:

Question 6.
The centre of a circle is (α + 2, α – 5). Find the value of a given that the circle passes through the points (2, -2) and (8, -2).
Solution:

Question 7.
The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.
Solution:

Question 8.
The ends of a diameter of a circle have the co-ordinates (3, 0) and (-5, 6). PQ is another diameter where Q has the coordinates ( -1, -2). Find the co-ordinates of P and the radius of the circle.
Solution:

Question 9.
In what ratio does the point (-4, 6) divide the line segment joining the points A(-6, 10) and B (3, -8)?
Solution:

Question 10.
Find the ratio in which the point P (-3, p) divides the line segment joining the points (-5, -4) and (-2, 3). Hence find the value of p.
Solution:

Question 11.
In what ratio is the line joining the points (4, 2) and (3, -5) divided by the x-axis? Also, find the co-ordinates of the point of division.
Solution:

Question 12.
If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points (-4, -3) and (6, 3). Hence, find the co-ordinates of P.
Solution:

Question 13.
Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, -2) and B (3, 7). Also, find the co-ordinates of the point of division.
Solution:

Question 14.
Point A(2, -3) is reflected in the v-axis onto point A’. Then the point A’ is reflected in the line x = 4 onto the point A”.
(i) Write the coordinates of A’ and A”.
(ii) Find the ratio in which the line segment AA” is divided by the x-axis. Also, find the coordinates of the point of division.
Solution:

Question 15.
ABCD is a parallelogram. If the coordinates of A, B and D are (10, -6), (2, -6) and (4, -2) respectively, find the co-ordinates of C.
Solution:

Question 16.
ABCD is a parallelogram whose vertices A and B have co-ordinates (2, -3) and (-1, -1) respectively. If the diagonals of the parallelogram meet at the point M(1, -4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. find the perimeter of the parallelogram.
Solution:

Question 17.
In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.

Solution:

Question 18.
Given, O, (0, 0), P(1, 2), S(-3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram.
Find : (i) the co-ordinates of Q.
(ii)the co-ordinates of R.
(iii) the ratio in which RQ is divided by y-axis.

Solution:

Question 19.
If A (5, -1), B (-3, -2) and C (-1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.
Solution:

ML Aggarwal Class 10 Solutions for ICSE Maths