icse maths book for class 9 solved

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 9 Logarithms

Exercise 9.1

Question 1.
Convert the following to logarithmic form:
(i) 5² = 25
(ii) a⁵ =64
(iii) 7^x =100
(iv) 9° = 1
(v) 6¹ = 6
(vi) 3^-2 = \(\frac { 1 }{ 9 }\)
(vii) 10^-2 = 0.01
(viii) (81)^{\(\frac { 3 }{ 4 }\)} = 27
Solution:

Question 2.
Convert the following into exponential form:
(i) log₂ 32 = 5
(ii) log₃ 81=4
(iii) log₃\(\frac { 1 }{ 3 }\)= -1
(iv) log₃ 4= \(\frac { 2 }{ 3 }\)
(v) log₈ 32= \(\frac { 5 }{ 3 }\)
(vi) log₁₀ (0.001) = -3
(Vii) log₂ 0.25 = -2
(viii) log_a (\(\frac { 1 }{ a }\)) =-1
Solution:

Question 3.
By converting to exponential form, find the values of:
(i) log₂ 16
(ii) log₅ 125
(iii) log₄ 8
(iv) log₉ 27
(v) log₁₀(.01)
(vi) log₇ \(\frac { 1 }{ 7 }\)
(vii) log₅ 256
(Viii) log₂ 0.25
Solution:

Question 4.
Solve the following equations for x.

Solution:

Question 5.
Given log₁₀a = b, express 10^2b-3 in terms of a.
Solution:

Question 6.
Given log₁₀ x= a, log₁₀ y = b and log₁₀ z =c,
(i) write down 10^2a-3 in terms of x.
(ii) write down 10^3b-1 in terms of y.
(iii) if log₁₀ P = 2a + \(\frac { b }{ 2 }\)– 3c, express P in terms of x, y and z.
Solution:

Question 7.
If log₁₀x = a and log₁₀y = b, find the value of xy.
Solution:

Question 8.
Given log₁₀ a = m and log₁₀ b = n, express \(\frac { { a }^{ 3 } }{ { b }^{ 2 } }\) in terms of m and n.
Solution:

Question 9.
Given log₁₀a= 2a and log₁₀y = –\(\frac { b }{ 2 }\)
(i) write 10^a in terms of x.
(ii) write 10^2b+1 in terms of y.
(iii) if log₁₀P= 3a -2b, express P in terms of x and y .
Solution:

Question 10.
If log₂ y = x and log₃ z = x, find 72^x in terms of y and z.
Solution:

Question 11.
If log₂ x = a and log₅y = a, write 100^2a-1 in terms of x and y.
Solution:

Exercise 9.2

Question 1.
Simplify the following :

Solution:

Question 2.
Evaluate the following:


Solution:

Question 3.
Express each of the following as a single logarithm:

Solution:

Question 4.
Prove the following :
(i) log₁₀ 4 ÷ log₁₀ 2 = l0g₃ 9
(ii) log₁₀ 25 + log₁₀ 4 = log₅ 25
Solution:

Question 5.
If x = 100)^a , y = (10000)^b and z = (10)^c, express

Solution:

Question 6.
If a = log₁₀x, find the following in terms of a :
(i) x
(ii) log₁₀\(\sqrt [ 5 ]{ { x }^{ 2 } }\)
(iii) log₁₀5x
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
If x = log₁₀ 12, y = log₄ 2 x log₁₀ 9 and z = log₁₀ 0.4, find the values of
(i)x-y-z
(ii) 7^x-y-z
Solution:

Question 10.
If log V + log3 = log π + log4 + 3 log r, find V in terns of other quantities.
Solution:

Question 11.
Given 3 (log 5 – log3) – (log 5-2 log 6) = 2 – log n , find n.
Solution:

Question 12.
Given that log₁₀y + 2 log₁₀x= 2, express y in terms of x.
Solution:

Question 13.
Express log₁₀2+1 in the from log₁₀x.
Solution:

Question 14.

Solution:

Question 15.
Given that log m = x + y and log n = x-y, express the value of log m²n in terms of x and y.
Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.
Solve for x:

Solution:

Question 19.
Given 2 log₁₀x+1= log₁₀250, find
(i) x
(ii) log₁₀2x
Solution:

Question 20.

Solution:

Question 21.
Prove the following :
(i) 3^{_{log 4}} = 4^{_{log 3}}
(ii) 27^{_{log 2}} = 8^{_{log 3}}
Solution:

Question 22.
Solve the following equations :
(i) log (2x + 3) = log 7
(ii) log (x +1) + log (x – 1) = log 24
(iii) log (10x + 5) – log (x – 4) = 2
(iv) log₁₀5 + log₁₀(5x+1) = log₁₀(x + 5) + 1
(v) log (4y – 3) = log (2y + 1) – log3
(vi) log₁₀(x + 2) + log₁₀(x – 2) = log₁₀3 + 31og₁₀4.
(vii) log(3x + 2) + log(3x – 2) = 5 log 2.
Solution:

Question 23.
Solve for x :
log₃ (x + 1) – 1 = 3 + log₃ (x – 1)
Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.
If p = log₁₀20 and q = log₁₀25, find the value of x if 2 log₁₀ (x +1) = 2p – q.
Solution:

Question 28.
Show that:

Solution:

Question 29.
Prove the following identities:

Solution:

Question 30.

Solution:

Question 31.
Solve for x :

Solution:

Multiple Choice Questions

correct Solution from the given four options (1 to 7):
Question 1.
If log√3 27 = x, then the value of x is
(a) 3
(b) 4
(c) 6
(d) 9
Solution:

Question 2.
If log₅ (0.04) = x, then the vlaue of x is
(a) 2
(b) 4
(c) -4
(d) -2
Solution:

Question 3.
If log_0.5 64 = x, then the value of x is
(a) -4
(b) -6
(c) 4
(d) 6
Solution:

Question 4.
If log_{10\(\sqrt [ 3 ]{ 5 }\)} x = -3, then the value of x is

Solution:

Question 5.
If log (3x + 1) = 2, then the value of x is

Solution:

Question 6.
The value of 2 + log₁₀ (0.01) is
(a)4
(b)3
(c)1
(d)0
Solution:

Question 7.


Solution:

Chapter Test

Question 1.

Solution:

Question 2.
Find the value of log√3 3√3 – log₅ (0.04)
Solution:

Question 3.
Prove the following:

Solution:

Question 4.
If log (m + n) = log m + log n, show that n = \(\frac { m }{ m-1 }\)
Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.
Solve the following equations for x:

Solution:

Question 8.
Solve for x and y:

Solution:

Question 9.
If a = 1 + log_xyz, 6 = 1+ log_y zx and c=1 + log_zxy, then show that ab + bc + ca = abc.
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 10 Triangles

Exercise 10.1

Question 1.
It is given that ∆ABC ≅ ∆RPQ. Is it true to say that BC = QR ? Why?
Solution:

Question 2.
“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?
Solution:
No, it is not true statement as the angles should be included angle of there two given sides.

Question 3.
In the given figure, AB=AC and AP=AQ. Prove that
(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠APC = ∠AQB.
Solution:

Question 4.
In the given figure, AB = AC, P and Q are points on BA and CA respectively such that AP = AQ. Prove that
(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠ACP = ∠ABQ.
Solution:

Question 5.
In the given figure, AD = BC and BD = AC. Prove that :
∠ADB = ∠BCA and ∠DAB = ∠CBA.

Solution:

Question 6.
In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Solution:

Question 7.
In the given figure, AB = DC and AB || DC. Prove that AD = BC.
Solution:

Question 8.
In the given figure. AC = AE, AB = AD and ∠BAD = ∠CAE. Show that BC = DE.

Solution:

Question 9.
In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that
(i) ∆ACE ≅ ∆DBF
(ii) AE = DF.
Solution:

Question 10.
In the given figure, AB = AC and D is mid-point of BC. Use SSS rule of congruency to show that
(i) ∆ABD ≅ ∆ACD
(ii) AD is bisector of ∠A
(iii) AD is perpendicular to BC.
Solution:

Question 11.
Two line segments AB and CD bisect each other at O. Prove that :
(i) AC = BD
(ii) ∠CAB = ∠ABD
(iii) AD || CB
(iv) AD = CB.

Solution:

Question 12.
In each of the following diagrams, find the values of x and y.

Solution:

Exercise 10.2

Question 1.
In triangles ABC and PQR, ∠A= ∠Q and ∠B = ∠R. Which side of APQR should be equal to side AB of AABC so that the two triangles are congruent? Give reason for your answer.
Solution:

Question 2.
In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of APQR should be equal to side BC of AABC so that the two triangles are congruent? Give reason for your answer.
Solution:

Question 3.
“If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent”. Is the statement true? Why?
Solution:
The given statement can be true only if the corresponding (included) sides are equal otherwise not.

Question 4.
In the given figure, AD is median of ∆ABC, BM and CN are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BM = CN.
Solution:

Question 5.
In the given figure, BM and DN are perpendiculars to the line segment AC. If BM = DN, prove that AC bisects BD.

Solution:

Question 6.
In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

Solution:

Question 7.
In the given figure, two lines AB and CD intersect each other at the point O such that BC || DA and BC = DA. Show that O is the mid-point of both the line segments AB and CD.
Solution:

Question 8.
In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Show that
(i) ∆ACD ≅ ∆BDC
(ii) BC = AD
(iii) ∠A = ∠B.
Solution:

Question 9.
In the given figure, ∠ABC = ∠ACB, D and E are points on the sides AC and AB respectively such that BE = CD. Prove that
(i) ∆EBC ≅ ∆DCB
(ii) ∆OEB ≅ ∆ODC
(iii) OB = OC.
Solution:

Question 10.
ABC is an isosceles triangle with AB=AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:

Question 11.
In the given figure, BA ⊥ AC, DE⊥ DF such that BA = DE and BF = EC.
Solution:

Question 12.
ABCD is a rectanige. X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
Solution:

Question 13.
(a) In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of A PQR. If XS⊥ QR and XT ⊥ PQ, prove that
(i) ∆XTQ ≅ ∆XSQ
(ii) PX bisects the angle P.
(b) In the figure (2) given below, AB || DC and ∠C = ∠D. Prove that
(i) AD = BC
(ii) AC = BD.
(c) In the figure (3) given below, BA || DF and CA II EG and BD = EC . Prove that, .
(i) BG = DF
(ii) EG = CF.


Solution:

Question 14.
In each of the following diagrams, find the values of x and y.

Solution:

Exercise 10.3

Question 1.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:

Question 2.
Show that the angles of an equilateral triangle are 60° each.
Solution:

Question 3.
Show that every equiangular triangle is equilateral.
Solution:

Question 4.
In the following diagrams, find the value of x:

Solution:

Question 5.
In the following diagrams, find the value of x:

Solution:

Question 6.
(a) In the figure (1) given below, AB = AD, BC = DC. Find ∠ ABC.
(b)In the figure (2) given below, BC = CD. Find ∠ACB.
(c) In the figure (3) given below, AB || CD and CA = CE. If ∠ACE = 74° and ∠BAE =15°, find the values of x and y.
Solution:

Question 7.
In ∆ABC, AB = AC, ∠A = (5x + 20)° and each of the base angle is \(\frac { 2 }{ 5 }\) th of ∠A. Find the measure of ∠A.
Solution:

Question 8.
(a) In the figure (1) given below, ABC is an equilateral triangle. Base BC is produced to E, such that BC’= CE. Calculate ∠ACE and ∠AEC.
(b) In the figure (2) given below, prove that ∠ BAD : ∠ ADB = 3 : 1.
(c) In the figure (3) given below, AB || CD. Find the values of x, y and ∠.

Solution:

Question 9.
In the given figure, D is mid-point of BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that ABC is an isosceles triangle.

Solution:

Question 10.
In the given figure, AD, BE and CF arc altitudes of ∆ABC. If AD = BE = CF, prove that ABC is an equilateral triangle.
Solution:

Question 11.
In a triangle ABC, AB = AC, D and E are points on the sides AB and AC respectively such that BD = CE. Show that:
(i) ∆DBC ≅ ∆ECB
(ii) ∠DCB = ∠EBC
(iii) OB = OC,where O is the point of intersection of BE and CD.
Solution:

Question 12.
ABC is an isosceles triangle in which AB = AC. P is any point in the interior of ∆ABC such that ∠ABP = ∠ACP. Prove that
(a) BP = CP
(b) AP bisects ∠BAC.
Solution:

Question 13.
In the adjoining figure, D and E are points on the side BC of ∆ABC such that BD = EC and AD = AE. Show that ∆ABD ≅ ∆ACE.
Solution:

Question 14.
(a) In the figure (i) given below, CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ∆ADE ≅ ∆BCE and hence, AEB is an isosceles triangle.

(b) In the figure (ii) given below, O is a point in the interior of a square ABCD such that OAB is an equilateral trianlge. Show that OCD is an isosceles triangle.

Question 15.
In the given figure, ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
Solution:

Exercise 10.4

Question 1.
In ∆PQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is longest? Give reason for your answer.
Solution:

Question 2.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:

Question 3.
PQR is a right angle triangle at Q and PQ : QR = 3:2. Which is the least angle.
Solution:

Question 4.
In ∆ ABC, AB = 8 cm, BC = 5.6 cm and CA = 6.5 cm. Which is (i) the greatest angle ?
(ii) the smallest angle ?
Solution:

Question 5.
In ∆ABC, ∠A = 50°, ∠B= 60°, Arrange the sides of the triangle in ascending order.
Solution:

Question 6.
In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show
(i) BD > AD
(ii) DC > AD
(iii) AC > DC
(iv) AB > BD

Solution:

Question 7.
(a) In the figure (1) given below, AD bisects ∠A. Arrange AB, BD and DC in the descending order of their lengths.
(b) In the figure (2) given below, ∠ ABD = 65°, ∠DAC = 22° and AD = BD. Calculate ∠ ACD and state (giving reasons) which is greater : BD or DC ?

Solution:

Question 8.
(a) In the figure (1) given below, prove that (i) CF> AF (ii) DC>DF.
(b) In the figure (2) given below, AB = AC.
Prove that AB > CD.
(c) In the figure (3) given below, AC = CD. Prove that BC < CD.

Solution:

Question 9.
(a) In the figure (i) given below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC. (b) In the figure (ii) given below, D is any point on the side BC of ∆ABC. If AB > AC, show that AB > AD.
Solution:

Question 10.
(i) Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer,
(ii) Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer.
(iii) Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer.
Solution:
(i) Length of sides of a triangle are 4 cm, 3 cm and 7 cm
We know that sum of any two sides of a triangle is greatar than its third side But 4 + 3 = 7 cm
Which is not possible
Hence to construction of a triangle with sides 4 cm, 3 cm and 7 cm is not possible.
(ii) Length of sides of a triangle are 9 cm, 7 cm and 17 cm
We know that sum of any two sides of a triangle is greater than its third side Now 9 + 7 = 16 < 17 ∴ It is not possible to construct a triangle with these sides.
(iii) Length of sides of a triangle are 8 cm, 7 cm and 4 cm We know that sum of any two sides of a triangle is greater than its third side Now 7 + 4 = 11 > 8
Yes, It is possible to construct a triangle with these sides.

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 18):
Question 1.
Which of the following is not a criterion for congruency of triangles?
(a) SAS
(b) ASA
(c) SSA
(d) SSS
Solution:
Criteria of congruency of two triangles ‘SSA’ is not the criterion. (c)

Question 2.
In the adjoining figure, AB = FC, EF=BD and ∠AFE = ∠CBD. Then the rule by which ∆AFE = ∆CBD is
(a) SAS
(b) ASA
(c) SSS
(d) AAS

Solution:

Question 3.
In the adjoining figure, AB ⊥ BE and FE ⊥ BE. If AB = FE and BC = DE, then
(a) ∆ABD ≅ ∆EFC
(b) ∆ABD ≅ ∆FEC
(c) ∆ABD ≅ ∆ECF
(d) ∆ABD ≅ ∆CEF
Solution:
In the figure given,

Question 4.
In the adjoining figure, AB=AC and AD is median of ∆ABC, then AADC is equal to
(a) 60°
(b) 120°
(c) 90°
(d) 75°
Solution:

Question 5.
In the adjoining figure, O is mid point of AB. If ∠ACO = ∠BDO, then ∠OAC is equal to
(a) ∠OCA
(b) ∠ODB
(c) ∠OBD
(d) ∠BOD
Solution:

Question 6.
In the adjoining figure, AC = BD. If ∠CAB = ∠DBA, then ∠ACB is equal to
(a) ∠BAD
(b) ∠ABC
(c) ∠ABD
(d) ∠BDA

Solution:

Question 7.
In the adjoining figure, ABCD is a quadrilateral in which BN and DM are drawn perpendiculars to AC such that BN = DM. If OB = 4 cm, then BD is
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Solution:

Question 8.
In ∆ABC, AB = AC and ∠B = 50°. Then ∠C is equal to
(a) 40°
(b) 50°
(c) 80°
(d) 130°
Solution:

Question 9.
In ∆ABC, BC = AB and ∠B = 80°. Then ∠A is equal to
(a) 80°
(b) 40°
(c) 50°
(d) 100°
Solution:

Question 10.
In ∆PQR, ∠R = ∠P, QR = 4 cm and PR = 5 cm. Then the length of PQ is
(a) 4 cm
(b) 5 cm
(c) 2 cm
(d) 2.5 cm
Solution:

Question 11.
In ∆ABC and APQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but isosceles
(d) neither congruent nor isosceles
Solution:

Question 12.
Two sides of a triangle are of lenghts 5 cm and 1.5 cm. The length of the third side of the triangle can not be
(a) 3.6 cm
(b) 4.1 cm
(c) 3.8 cm
(d) 3.4 cm
Solution:

Question 13.
If a, b, c are the lengths of the sides of a trianlge, then
(a) a – b > c
(b) c > a + b
(c) c = a + b
(d) c < A + B
Solution:
a, b, c are the lengths of the sides of a trianlge than a + b> c or c < a + b
(Sum of any two sides is greater than its third side) (d)

Question 14.
It is not possible to construct a triangle when the lengths of its sides are
(a) 6 cm, 7 cm, 8 cm
(b) 4 cm, 6 cm, 6 cm
(c) 5.3 cm, 2.2 cm, 3.1 cm
(d) 9.3 cm, 5.2 cm, 7.4 cm
Solution:
We know that sum of any two sides of a triangle is greater than its third side 2.2 + 3.1 = 5.3 ⇒ 5.3 = 5.3 is not possible (c)

Question 15.
In ∆PQR, if ∠R> ∠Q, then
(a) QR > PR
(b) PQ > PR
(c) PQ < PR
(d) QR < PR
Solution:
In ∆PQR, ∠R> ∠Q
∴ PQ > PR (b)

Question 16.
If triangle PQR is right angled at Q, then
(a) PR = PQ
(b) PR < PQ
(c) PR < QR
(d) PR > PQ
Solution:

Question 17.
If triangle ABC is obtuse angled and ∠C is obtuse, then
(a) AB > BC
(b) AB = BC
(c) AB < BC
(d) AC > AB
Solution:

Question P.Q.
A triangle can be constructed when the lengths of its three sides are
(a) 7 cm, 3 cm, 4 cm
(b) 3.6 cm, 11.5 cm, 6.9 cm
(c) 5.2 cm, 7.6 cm, 4.7 cm
(d) 33 mm, 8.5 cm, 49 mm
Solution:
We know that in a triangle, if sum of any two sides is greater than its third side, it is possible to construct it 5.2 cm, 7.6 cm, 4.7 cm is only possible. (c)

Question P.Q.
A unique triangle cannot be constructed if its
(a) three angles are given
(b) two angles and one side is given
(c) three sides are given
(d) two sides and the included angle is given
Solution:
A unique triangle cannot be constructed if its three angle are given, (a)

Question 18.
If the lengths of two sides of an isosceles are 4 cm and 10 cm, then the length of the third side is
(a) 4 cm
(b) 10 cm
(c) 7 cm
(d) 14 cm
Solution:
Lengths of two sides of an isosceles triangle are 4 cm and 10 cm, then length of the third side is 10 cm
(Sum of any two sides of a triangle is greater than its third side and 4 cm is not possible as 4 + 4 > 10 cm.

Chapter Test

Question 1.
In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF. Will the two triangles be congruent? Give reasons for your answer.
Solution:

Question 2.
In the given figure, ABCD is a square. P, Q and R are points on the sides AB, BC and CD respectively such that AP= BQ = CR and ∠PQR = 90°. Prove that
(a) ∆PBQ ≅ ∆QCR
(b) PQ = QR
(c) ∠PRQ = 45°

Solution:

Question 3.
In the given figure, AD = BC and BD = AC. Prove that ∠ADB = ∠BCA.
Solution:

Question 4.
In the given figure, OA ⊥ OD, OC X OB, OD = OA and OB = OC. Prove that AB = CD.

Solution:

Question 5.
In the given figure, PQ || BA and RS CA. If BP = RC, prove that:
(i) ∆BSR ≅ ∆PQC
(ii) BS = PQ
(iii) RS = CQ.

Solution:

Question 6.
In the given figure, AB = AC, D is a point in the interior of ∆ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of ∆ABC.
Solution:

Question 7.
In the adjoining figure, AB || DC. CE and DE bisects ∠BCD and ∠ADC respectively. Prove that AB = AD + BC.
Solution:

Question 8.
In ∆ABC, D is a point on BC such that AD is the bisector of ∠BAC. CE is drawn parallel to DA to meet BD produced at E. Prove that ∆CAE is isosceles
Solution:

Question 9.
In the figure (ii) given below, ABC is a right angled triangle at B, ADEC and BCFG are squares. Prove that AF = BE.

Solution:

Question 10.
In the given figure, BD = AD = AC. If ∠ABD = 36°, find the value of x .

Solution:

Question 11.
In the adjoining figure, TR = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that RB = SA.

Solution:
Given: In the figure , RST is a triangle

Question 12.
(a) In the figure (1) given below, find the value of x.
(b) In the figure (2) given below, AB = AC and DE || BC. Calculate
(i)x
(ii) y
(iii) ∠BAC
(c) In the figure (1) given below, calculate the size of each lettered angle.

Solution:

Question 13.
(a) In the figure (1) given below, AD = BD = DC and ∠ACD = 35°. Show that
(i) AC > DC (ii) AB > AD.
(b) In the figure (2) given below, prove that
(i) x + y = 90° (ii) z = 90° (iii) AB = BC

Solution:

Question 14.
In the given figure, ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Solution:

Question 15.
In the given figure, AP ⊥ l and PR > PQ. Show that AR > AQ.
Solution:

Question 16.
If O is any point in the interior of a triangle ABC, show that
OA + OB + OC > \(\frac { 1 }{ 2 }\)
(AB + BC + CA).

Solution:

Question P.Q.
Construct a triangle ABC given that base BC = 5.5 cm, ∠ B = 75° and height = 4.2 cm.
Solution:

Question P.Q.
Construct a triangle ABC in which BC = 6.5 cm, ∠ B = 75° and ∠ A = 45°. Also construct median of A ABC passing through B.
Solution:

Question P.Q.
Construct triangle ABC given that AB – AC = 2.4 cm, BC = 6.5 cm. and ∠ B = 45°.
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 12 Pythagoras Theorem

Question 1.
Lengths of sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse:
(i) 3 cm, 8 cm, 6 cm
(ii) 13 cm, .12 cm, 5 cm
(iii) 1.4 cm, 4.8 cm, 5 cm
Solution:

Question 2.
Foot of a 10 m long ladder leaning against a vertical well is 6 m away from the base of the wail. Find the height of the point on the wall where the top of the ladder reaches.
Solution:

Question 3.
A guy attached a wire 24 m long to a vertical pole of height 18 m and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taught?
Solution:

Question 4.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Solution:

Question 5.
In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is 4:3, find the sides.
Solution:

Question 6.
If the sides of a triangle are in the ratio 3:4:5, prove that it is right-angled triangle.
Solution:

Question 7.
For going to a city B from city A, there is route via city C such that AC ⊥ CB, AC = 2x km and CB=2(x+ 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of highway.
Solution:

Question 8.
The hypotenuse of right triangle is 6m more than twice the shortest side. If the third side is 2m less than the hypotenuse, find the sides of the triangle.
Solution:

Question 9.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:

Question 10.
In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = AC² + BD².
Solution:

Question 11.
In ∆PQR, PD ⊥ QR, such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a – b) = (c + d) (c – d).
Solution:

Question 12.
ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. Find the altitude on BC and Hence, calculate its area.
Solution:

Question 13.
Find the area and the perimeter of a square whose diagonal is 10 cm long.
Solution:

Question 14.
(a) In fig. (i) given below, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm, ∠ ABD = ∠BCD = 90°. Calculate the length of AB.
(b) In fig. (ii) given below, ABCD is a quadrilateral in which AB = AD, ∠A = 90° =∠C, BC = 8 cm and CD = 6 cm. Find AB and calculate the area of ∆ ABD.

Solution:

Question 15.
(a) In figure (i) given below, AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm.Calculate the length of BD.
(b) In figure (ii) given below, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.
(c) In figure (iii) given below, ∠ D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm. Find CD.

Solution:

Question 16.
(a) In figure (i) given below, BC = 5 cm,
∠B =90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.
(b) In the figure (ii) given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB2 = 4AD² – 3AC².

Solution:

Question 17.
In ∆ ABC, AB = AC = x, BC = 10 cm and the area of ∆ ABC is 60 cm². Find x.
Solution:

Question 18.
In a rhombus, If diagonals are 30 cm and 40 cm, find its perimeter.
Solution:

Question 19.
(a) In figure (i) given below, AB || DC, BC = AD = 13 cm. AB = 22 cm and DC = 12cm. Calculate the height of the trapezium ABCD.
(b) In figure (ii) given below, AB || DC, ∠ A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.
(c) In figure (iii) given below, ABCD is a square of side 7 cm. if
AE = FC = CG = HA = 3 cm,
(i) prove that EFGH is a rectangle.
(ii) find the area and perimeter of EFGH.

Solution:

Question 20.
AD is perpendicular to the side BC of an equilateral Δ ABC. Prove that 4AD² = 3AB².
Solution:

Question 21.
In figure (i) given below, D and E are mid-points of the sides BC and CA respectively of a ΔABC, right angled at C.

Solution:

Question 22.
If AD, BE and CF are medians of ΕABC, prove that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).
Solution:

Question 23.
(a) In fig. (i) given below, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that
AB² + CD² = AD² + BC².
(b) In figure (ii) given below, OD⊥BC, OE ⊥CA and OF ⊥ AB. Prove that :
(i) OA² + OB² + OC² = AF² + BD² + CE² + OD² + OE² + OF².
(ii) OAF² + BD² + CE² = FB² + DC² + EA².

Solution:

Question 24.
In a quadrilateral, ABCD∠B = 90° = ∠D. Prove that 2 AC² – BC2 = AB² + AD² + DC².
Solution:

Question 25.
In a ∆ ABC, ∠ A = 90°, CA = AB and D is a point on AB produced. Prove that :
DC² – BD² = 2AB. AD.
Solution:

Question 26.
In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD² = AC² + BD.CD.
Solution:

Question P.Q.
(a) In figure (i) given below, PQR is a right angled triangle, right angled at Q. XY is parallel to QR. PQ = 6 cm, PY = 4 cm and PX : OX = 1:2. Calculate the length of PR and QR.
(b) In figure (ii) given below, ABC is a right angled triangle, right angled at B.DE || BC.AB = 12 cm, AE = 5 cm and AD : DB = 1: 2. Calculate the perimeter of A ABC.
(c)In figure (iii) given below. ABCD is a rectangle, AB = 12 cm, BC – 8 cm and E is a point on BC such that CE = 5 cm. DE when produced meets AB produced at F.
(i) Calculate the length DE.
(ii) Prove that ∆ DEC ~ AEBF and Hence, compute EF and BF.

Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 7):
Question 1.
In a ∆ABC, if AB = 6√3 cm, BC = 6 cm and AC = 12 cm, then ∠B is
(a) 120°
(b) 90°
(c) 60°
(d) 45°
Solution:

Question 2.
If the sides of a rectangular plot are 15 m and 8 m, then the length of its diagonal is
(a) 17 m
(b) 23 m
(c) 21 m
(d) 17 cm
Solution:

Question 3.
The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of the side of the rhombus is
(a) 9 cm
(b) 10 cm
(c) 8 cm
(d) 20 cm
Solution:

Question 4.
If a side of a rhombus is 10 cm and one of the diagonals is 16 cm, then the length of the other diagonals is
(a) 6 cm
(b) 12 cm
(c) 20 cm
(d) 12 cm
Solution:

Question 5.
If a ladder 10 m long reaches a window 8 m above the ground, then the distance of the foot of the ladder from the base of the wall is
(a) 18 m
(b) 8 m
(c) 6 m
(d) 4 m
Solution:

Question 6.
A girl walks 200 m towards East and then she walks ISO m towards North. The distance of the girl from the starting point is
(a) 350 m
(b) 250 m
(c) 300 m
(d) 225 m
Solution:

Question 7.
A ladder reaches a window 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9 m high. If the length of the ladder is 15 m, then the width of the street is
(a) 30 m
(b) 24 m
(c) 21 m
(d) 18 m
Solution:

Chapter Test

Question 1.
(a) In fig. (i) given below, AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC.
(b) In figure (ii) given below, ∠BAC = 90°, ∠ADC = 90°, AD = 6 cm, CD = 8 cm and BC = 26 cm. Find :
(i) AC (ii) AB (iii) area of the shaded region.
(c) In figure (iii) given below, triangle ABC is right angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are mid-points of the sides AB and AC respectively, calculate
(i) the length of BC (ii) the area of ∆ ADE.

Solution:

Question 2.
If in ∆ ABC, AB > AC and ADI BC, prove that AB² – AC² = BD² – CD².
Solution:

Question 3.
In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2:1. Prove that
(i) 9AQ² = 9AC² + 4BC²
(ii) 9BP² = 9BC² + 4AC²
(iii) 9(AQ² + BP²) = 13AB².
Solution:
A right angled ∆ ABC in which ∠ C

Question 4.
In the given figure, ∆PQR is right angled at Q and points S and T trisect side QR. Prove that 8PT² – 3PR² + 5PS².
Solution:

Question 5.
In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², prove that ∠ACD = 90°.
Solution:

Question 6.
In the given figure, find the length of AD in terms of b and c.
Solution:

Question 7.
ABCD is a square, F is mid-point of AB and BE is one-third of BC. If area of ∆FBE is 108 cm², find the length of AC.
Solution:

Question 8.
In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC x CD, Prove that BD = BC.
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 14 Theorems on Area

Question 1.
Prove that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.
Solution:

Question 2.
Prove that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:

Question 3.
(a) In the figure (1) given below, AD is median of ∆ABC and P is any point on AD. Prove that
(i) area of ∆PBD = area of ∆PDC.
(ii) area of ∆ABP = area of ∆ACP.
(b) In the figure (2) given below, DE || BC. prove that (i) area of ∆ACD = area of ∆ ABE.
(ii) area of ∆OBD = area of ∆OCE.

Solution:

Question 4.
(a) In the figure (1) given below, ABCD is a parallelogram and P is any point in BC. Prove that: Area of ∆ABP + area of ∆DPC = Area of ∆APD.
(b) In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that:
(i) area of ∆OAB + area of ∆OCD = \(\frac { 1 }{ 2 }\) area of || gm ABCD.
(ii) area of ∆ OBC + area of ∆ OAD = \(\frac { 1 }{ 2 }\) area of ||gmABCD

Solution:

Question 5.
If E, F, G and H are mid-points of the sides AB, BC, CD and DA respectively of a parallelogram ABCD, prove that area of quad. EFGH = 1/2 area of || gm ABCD.
Solution:

Question 6.
(a) In the figure (1) given below, ABCD is a parallelogram. P, Q are any two points on the sides AB and BC respectively. Prove that, area of ∆ CPD = area of ∆ AQD.

(b) In the figure (2) given below, PQRS and ABRS are parallelograms and X is any point on the side BR. Show that area of ∆ AXS = \(\frac { 1 }{ 2 }\) area of ||gm PQRS
Solution:

Question 7.
D,EandF are mid-point of the sides BC, CA and AB respectively of a ∆ ABC. Prove that
(i) FDCE is a parallelogram
(ii) area of ADEF = \(\frac { 1 }{ 4 }\) area of A ABC
(iii) area of || gm FDCE = \(\frac { 1 }{ 2 }\) area of ∆ ABC.
Solution:

Question 8.
In the given figure, D, E and F are mid points of the sides BC, CA and AB respectively of AABC. Prove that BCEF is a trapezium and area of trap. BCEF = \(\frac { 3 }{ 4 }\) area of ∆ ABC.

Solution:

Question P.Q.
Prove that two triangles having equal areas and having one side of one of the triangles equal to one side of the other, have their corresponding altitudes equal.
Solution:

Question 9.
(a) In the figure (1) given below, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that area of ∆ ABD: area of ∆ ADC = m : n.
(b) In the figure (2) given below, P is a point on the sidoBC of ∆ABC such that PC = 2BP, and Q is a point on AP such that QA = 5 PQ, find area of ∆AQC : area of ∆ABC.
(c) In the figure (3) given below, AD is a median of ∆ABC and P is a point in AC such that area of ∆ADP : area of AABD = 2:3. Find
(i) AP : PC (ii) area of ∆PDC : area of ∆ABC.

Solution:

Question 10.
(a) In the figure (1) given below, area of parallelogram ABCD is 29 cm2. Calculate the height of parallelogram ABEF if AB = 5.8 cm
(b) In the figure (2) given below, area of ∆ABD is 24 sq. units. If AB = 8 units, find the height of ABC.
(c) In the figure (3) given below, E and F are mid points of sides AB and CD respectively of parallelogram ABCD. If the area of parallelogram ABCip is 36 cm2.
(i) State the area of ∆ APD.
(ii) Name the parallelogram whose area is equal to the area of ∆ APD.

Solution:

Question 11.
(a) In the figure (1) given below, ABCD is a parallelogram. Points P and Q on BC trisect BC into three equal parts. Prove that :
area of ∆APQ = area of ∆DPQ = \(\frac { 1 }{ 6 }\) (area of ||gm ABCD)
(b) In the figure (2) given below, DE is drawn parallel to the diagonal AC of the quadrilateral ABCD to meet BC produced at the point E. Prove that area of quad. ABCD = area of ∆ABE.
(c) In the figure (3) given below, ABCD is a parallelogram. O is any point on the diagonal AC of the parallelogram. Show that the area of ∆AOB is equal to the area of ∆AOD.

Solution:

Question P.Q.
(a) In the figure (1) given below, two parallelograms ABCD and AEFB are drawn on opposite sides of AB, prove that: area of || gm ABCD + area of || gm AEFB = area of || gm EFCD.
(b) In the figure (2) given below, D is mid-point of the side AB of ∆ABC. P is any point on BC, CQ is drawn parallel to PD to meet AB in Q. Show that area of ∆BPQ = \(\frac { 1 }{ 2 }\) area of ∆ABC.
(c) In the figure (3) given below, DE is drawn parallel to the diagonal AC of the quadrilateral ABCD to meet BC produced at the point E. Prove that area of quad. ABCD = area of ∆ABE.


Solution:

Question 12.
(a) In the figure given, ABCD and AEFG are two parallelograms.
Prove that area of || gm ABCD = area of || gm AEFG.
(b) In the fig. (2) given below, the side AB of the parallelogram ABCD is produced to E. A st. line At through A is drawn parallel to CE to meet CB produced at F and parallelogram BFGE is Completed prove that area of || gm BFGE=Area of || gmABCD.

(c) In the figure (3) given below AB || DC || EF, AD || BEandDE || AF. Prove the area ofDEFH is equal to the area of ABCD.

Solution:

Question 13.
Any point D is taken on the side BC of, a ∆ ABC and AD is produced to E such that AD=DE, prove that area of ∆ BCE = area of ∆ ABC.
Solution:

Question 14.
ABCD is a rectangle and P is mid-point of AB. DP is produced to meet CB at Q. Prove that area of rectangle ∆BCD = area of ∆ DQC.
Solution:

Question P.Q.
ABCD is a square, E and F are mid-points of the sides AB and AD respectively Prove that area of ∆CEF = \(\frac { 3 }{ 8 }\) (area of square ABCD).
Solution:

Question P.Q.
A line PQ is drawn parallel to the side BC of ∆ABC. BE is drawn parallel to CA to meet QP (produced) at E and CF is drawn parallel to BA to meet PQ (produced) at F. Prove that
area of ∆ABE=area of ∆ACF.
Solution:

Question 15.
(a) In the figure (1) given below, the perimeter of parallelogram is 42 cm. Calculate the lengths of the sides of the parallelogram.
(b) In the figure (2) given below, the perimeter of ∆ ABC is 37 cm. If the lengths of the altitudes AM, BN and CL are 5x, 6x, and 4x respectively, Calculate the lengths of the sides of ∆ABC.
(c) In the fig. (3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and area of ∆BCP = 15 cm². Find
(i) area of || gm ABCD
(ii) DP : PC.

Solution:

Question 16.
In the adjoining figure, E is mid-point of the side AB of a triangle ABC and EBCF is a parallelogram. If the area of ∆ ABC is 25 sq. units, find the area of || gm EBCF.
Solution:

Question 17.
(a) In the figure (1) given below, BC || AE and CD || BE. Prove that: area of ∆ABC= area of ∆EBD.
(b) In the llgure (2) given below, ABC is right angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that:
(i) ∆BCF ≅ ∆ ABE.
(ii)arca of square ABFG = area of rectangle BENM.


Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 8):
Question 1.
In the given figure, if l || m, AF || BE, FC ⊥ m and ED ⊥ m , then the correct statement is
(a) area of ||ABEF = area of rect. CDEF
(b) area of ||ABEF = area of quad. CBEF
(c) area of ||ABEF = 2 area of ∆ACF
(d) area of ||ABEF = 2 area of ∆EBD

Solution:
In the given figure,
l ||m, AF || BE, FC ⊥ m and ED ⊥ m
∵ ||gm ABEF and rectangle CDEF are on the same base EF and between the same parallel
∴ area ||gm ABEF = area rect. CDEF (a)

Question 2.
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 3 : 1
Solution:
A triangle and a parallelogram are on the same base and between same parallel, then
∴ They are equal in area
∴ Their ratio 1:1 (b)

Question 3.
If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of area of the triangle to the area of parallelogram is
(a) 1 : 3
(b) 1 : 2
(c) 3 : 1
(d) 1 : 4
Solution:
A triangle and a parallelogram are on the same base and between same parallel, then area of
triangle = \(\frac { 1 }{ 2 }\) area ||gm
∴ Their ratio 1 : 2 (b)

Question 4.
A median of a triangle divides it into two
(a) triangles of equal area
(b) congruent triangles
(c) right triangles
(d) isosceles triangles
Solution:
A median of a triangle divides it into two triangle equal in area. (a)

Question 5.
In the given figure, area of parallelogram ABCD is
(a) AB x BM
(b) BC x BN
(c) DC x DL
(d) AD x DL

Solution:
In the given figure,
Area of ||gm ABCD = AB x DL or DC x DL (∵ AB = DC) (c)

Question 6.
The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to
(a) \(\frac { 1 }{ 2 }\) area of ∆ABC
(b) \(\frac { 1 }{ 3 }\) area of ∆ABC
(c) \(\frac { 1 }{ 4 }\) area of ∆ABC
(d) area of ∆ABC
Solution:

Question 7.
In the given figure, ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are mid-points of the non parallel sides. The ratio of area of ABEF and area of EFCD is
(a) a : b
(b) (3a + b) : (a + 3b)
(c) (a + 3b) : (3a + b)
(d) (2a + b) : (3a + b)
Solution:

Question 8.
In the given figure, AB || DC and AB ≠ DC. If the diagonals AC and BD of the trapezium ABCD intersect at O, then which of the following statements is not true?
(a) area of ∆ABC = area of ∆ABD
(b) area of ∆ACD = area of ∆BCD
(c) area of ∆OAB = area of ∆OCD
(d) area of ∆OAD = area of ∆OBC

Solution:

Chapter test

Question 1.
(a) In the figure (1) given below, ABCD is a rectangle (not drawn to scale ) with side AB = 4 cm and AD = 6 cm. Find :
(i) the area of parallelogram DEFC
(ii) area of ∆EFG.
(b) In the figure (2) given below, PQRS is a parallelogram formed by drawing lines parallel to the diagonals of a quadrilateral ABCD through its corners. Prove that area of || gm PQRS = 2 x area of quad. ABCD.

Solution:

Question P.Q.
In the adjoining figure, ABCD and ABEF are parallelogram and P is any point on DC. If area of || gm ABCD = 90 cm2, find:
(i) area of || gm ABEF
(ii) area of ∆ABP.
(iii) area of ∆BEF.

Solution:

Question 2.
In the parallelogram ABCD, P is a point on the side AB and Q is a point on the side BC. Prove that
(i) area of ∆CPD = area of ∆AQD
(ii)area of ∆ADQ = area of ∆APD + area of ∆CPB.

Solution:

Question 3.
In the adjoining figure, X and Y are points on the side LN of triangle LMN. Through X, a line is drawn parallel to LM to meet MN at Z. Prove that area of ∆LZY = area of quad. MZYX.

Solution:

Question P.Q.
If D is a point on the base BC of a triangle ABC such that 2BD = DC, prove that area of ∆ABD= \(\frac { 1 }{ 3 }\) area of ∆ ABC.
Solution:

Question 4.
Perpendiculars are drawn from a point within an equilateral triangle to the three sides. Prove that the sum of the three perpendiculars is equal to the altitude of the triangle.
Solution:

Question 5.
If each diagonal of a quadrilateral’ divides it into two triangles of equal areas, then prove that the quadrilateral is a parallelogram.
Solution:

Question 6.
In the given figure, ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of ∆DFB = 3 cm², find the area of parallelogram ABCD.

Solution:

Question 7.
In the given figure, ABCD is a square. E and F are mid-points of sides BC and CD respectively. If R is mid-point of EF, prove that: area of ∆AER = area of ∆AFR.
Solution:

Question 8.
In the given figure, X and Y are mid-points of the sides AC and AB respectively of ∆ABC. QP || BC and CYQ and BXP are straight lines. Prove that area of ∆ABP = area of ∆ACQ.

Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 15 Circle

EXERCISE 15.1

Question 1.
Calculate the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm.
Solution:

Question 2.
A chord of length 48 cm is drawn in a circle of radius 25 cm. Calculate its distance from the centre of the circle.
Solution:

Question 3.
A chord of length 8 cm is at a distance of 3 cm from the centre of the circle. Calculate the radius of the circle.
Solution:

Question 4.
Calculate the length of the chord which is at a distance of 6 cm from the centre of a circle of diameter 20 cm.
Solution:

Question 5.
A chord of length 16 cm is at a distance of 6 cm from the centre of the circle. Find the length of the chord of the same circle which is at a distance of 8 cm from the centre.
Solution:

Question 6.
In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords if they are on :
(i) the same side of the centre.
(ii) the opposite sides of the centre.
Solution:

Question 7.
(a) In the figure given below, O is the centre of the circle. AB and CD are two chords of the circle, OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the:
(i) radius of the circle.
(ii) length of chord CD.

(b) In the figure (ii) given below, CD is the diameter which meets the chord AB in E such that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.

Solution:

Question 8.
In the adjoining figure, AB and CD ate two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.

Solution:

Question 9.
AB and CD are two parallel chords of a circle of lengths 10 cm and 4 cm respectively. If the chords lie on the same side of the centre and the distance between them is 3 cm, find the diameter of the circle.
Solution:

Question 10.
ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12√5 cm and BC = 24 cm, find the radius of the circle.
Solution:

Question 11.
An equilateral triangle of side 6 cm is inscribed in a circle. Find the radius of the circle.
Solution:

Question 12.
AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.
Solution:

Question 13.
A rectangle with one side of length 4 cm is inscribed in a circle of diameter 5 cm. Find the area of the rectangle.
Solution:

Question 14.
The length of the common chord of two intersecting circles is 30 cm. If the radii of the two circles are 25 cm and 17 cm, find the distance between their centres.
Solution:

Question 15.
The line joining the mid-points of two chords of a circle passes through its centre. Prove that the chords are parallel.
Solution:

Question 16.
If a diameter of a circle is perpendicular to one of two parallel chords of the circle, prove that it is perpendicular to the other and bisects it.
Solution:

Question 17.
In an equilateral triangle, prove that the centroid and the circumcentre of the triangle coincide.
Solution:

Question 18.
(a) In the figure (i) given below, OD is perpendicular to the chord AB of a circle whose centre is O. If BC is a diameter, show that CA = 2 OD.
(b) In the figure (ii) given below, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC and OP ⊥ AB, OQ ⊥ AC, Prove that PB = QC.

Solution:

Question 19.
(a) In the figure (i) given below, a line l intersects two concentric circles at the points A, B, C and D. Prove that AB = CD.
(b) In the figure (it) given below, chords AB and CD of a circle with centre O intersect at E. If OE bisects ∠AED, Prove that AB = CD.

Solution:

Question 20.
(a) In the figure (i) given below, AD is a diameter of a circle with centre O.
If AB || CD, prove that AB = CD.
(b) In the figure (ii) given below, AB and CD are equal chords of a circle with centre O. If AB and CD meet at E (outside the circle) Prove that :
(i) AE = CE (ii) BE = DE.

Solution:

EXERCISE 15.2

Question 1.
If arcs APB and CQD of a circle are congruent, then find the ratio of AB: CD.
Solution:

Question 2.
A and B are points on a circle with centre O. C is a point on the circle such that OC bisects ∠AOB, prove that OC bisects the arc AB.
Solution:

Question 3.
Prove that the angle subtended at the centre of a circle is bisected by the radius passing through the mid-point of the arc.
Solution:

Question 4.
In the given figure, two chords AB and CD of a circle intersect at P. If AB = CD, prove that arc AD = arc CB.
Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 6) :
Question 1.
If P and Q are any two points on a circle, then the line segment PQ is called a
(a) radius of the circle
(b) diameter of the circle
(c) chord of the circle
(d) secant of the circle
Solution:
chord of the circle (c)

Question 2.
If P is a point in the interior of a circle with centre O and radius r, then
(a) OP = r
(b) OP > r
(c) OP ≥ r
(d) OP < r
Solution:
OP > r (b)

Question 3.
The circumference of a circle must be
(a) a positive real number
(b) a whole number
(c) a natural number
(d) an integer
Solution:
a positive real number (a)

Question 4.
AD is a diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm, then the distance of AB from the centre of circle is
(a) 17 cm
(b) 15 cm
(c) 4 cm
(d) 8 cm
Solution:

Question 5.
If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Solution:

Question 6.
In the given figure, O is the centre of the circle. If OA = 5 cm, AB = 8 cm and OD ⊥ AB, then length of CD is equal to
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
Solution:

Chapter Test

Question 1.
In the given figure, a chord PQ of a circle with centre O and radius 15 cm is bisected at M by a diameter AB. If OM = 9 cm, find the lengths of :
(i) PQ
(ii) AP
(iii) BP

Solution:

Question 2.
The radii of two concentric circles are 17 cm and 10 cm ; a line PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, calculate PQ.
Solution:

Question 3.
A chord of length 48 cm is at a distance of 10 cm from the centre of a circle. If another chord of length 20 cm is drawn in the same circle, find its distance from the centre of the circle.
Solution:

Question 4.
(a) In the figure (i) given below, two circles with centres C, D intersect in points P, Q. If length of common chord is 6 cm and CP = 5 cm, DP = 4 cm, calculate the distance CD correct to two decimal places.

(b) In the figure (ii) given below, P is a point of intersection of two circles with centres C and D. If the st. line APB is parallel to CD, Prove that AB = 2 CD.
Solution:

Question 5.
(a) In the figure (i) given below, C and D are centres of two intersecting circles. The line APQB is perpendicular to the line of centres CD.Provethat:
(i) AP=QB
(ii) AQ = BP.
(b) In the figure (ii) given below, two equal chords AB and CD of a circle with centre O intersect at right angles at P. If M and N are mid-points of the chords AB and CD respectively, Prove that NOMP is a square.

Solution:

Question 6.
In the given figure, AD is diameter of a circle. If the chord AB and AC are equidistant from its centre O, prove that AD bisects ∠BAC and ∠BDC.
Solution: