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Trigonometric Identities Class 10 ICSE ML Aggarwal Chapter Test

August 27, 2024August 27, 2024 by Raju

ML Aggarwal Class 10 Solutions Trigonometric Identities Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test.

Question 1.
(i) If θ is an acute angle and cosec θ = √5 find the value of cot θ – cos θ.
(ii) If θ is an acute angle and tan θ = \(\\ \frac { 8 }{ 15 } \), find the value of sec θ + cosec θ.
Solution:
(i) θ is an acute angle.
cosec θ = √5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q1.1

Question 2.
Evaluate the following:
(i) \(2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right) \) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
(ii) \(\frac { { sec }29^{ O } }{ { cosec }61^{ O } } \) + 2 cot 8° cot 17° cot 45° cot 73°0 cot 82° – 3(sin2 38° + sin2 52°)
(iii) \(\frac { { sin }^{ 2 }{ 22 }^{ O }+{ sin }^{ 2 }{ 68 }^{ O } }{ { cos }^{ 2 }{ 22 }^{ O }+{ cos }^{ 2 }{ 68 }^{ O } } \) + sin2 63° + cos 63° sin 27°
Solution:
(i) \(2\times \left( \frac { { cos }^{ 2 }{ 20 }^{ O }+{ cos }^{ 2 }{ 70 }^{ O } }{ { sin }^{ 2 }{ 25 }^{ O }+{ sin }^{ 2 }{ 65 }^{ O } } \right) \) – tan 45° + tan 13° tan 23° tan 30° tan 67° tan 77°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q2.1

Question 3.
If \(\\ \frac { 4 }{ 3 } \) (sec2 59° – cot2 31°) – \(\\ \frac { 2 }{ 2 } \) sin 90° + 3tan2 56° tan2 34° = \(\\ \frac { x }{ 2 } \), then find the value of x.
Solution:
Given
\(\\ \frac { 4 }{ 3 } \) (sec2 59° – cot2 31°) – \(\\ \frac { 2 }{ 2 } \) sin 90° + 3tan2 56° tan2 34° = \(\\ \frac { x }{ 2 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q3.1

Question 4.
(i) \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA \)
(ii) \(\frac { cosA }{ cosecA+1 } +\frac { cosA }{ cosecA-1 } =2tanA \)
Solution:
(i) \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } =2secA \)
L.H.S = \(\frac { cosA }{ 1-sinA } +\frac { cosA }{ 1+sinA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q4.1

Question 5.
(i) \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta \)
(ii) (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.
Solution:
(i) \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } =sec\theta \)
L.H.S = \(\frac { (cos\theta -sin\theta )(1+tan\theta ) }{ 2{ cos }^{ 2 }\theta -1 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q5.1

Question 6.
(i) sin² θ + cos⁴ θ = cos² θ + sin⁴ θ
(ii) \(\frac { cot\theta }{ cosec\theta +1 } +\frac { cosec\theta +1 }{ cot\theta } =2sec\theta \)
Solution:
L.H.S. = sin² θ + cos⁴ θ
= (1 – cos² θ + cos⁴ θ
= 1 – cos² θ + cos⁴ θ
= 1 – cos² θ (1 – cos² θ)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q6.1

Question 7.
(i) sec⁴ A (1 – sin⁴ A) – 2 tan² A = 1
(ii) \(\frac { 1 }{ sinA+cosA+1 } +\frac { 1 }{ sinA+cosA-1 } =secA+cosecA\)
Solution:
(i) sec⁴ A (1 – sin⁴ A) – 2 tan² A = 1
L.H.S = sec⁴ A (1 – sin⁴ A) – 2 tan² A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q7.1

Question 8.
(i) \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1\)
(ii) (sec A – tan A)² (1 + sin A) = 1 – sin A.
Solution:
(i) \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta =1\)
L.H.S = \(\frac { { sin }^{ 3 }\theta +{ cos }^{ 3 }\theta }{ sin\theta cos\theta } +sin\theta cos\theta\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q8.1

Question 9.
(i) \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA \)
(ii) (sec A – cosec A) (1 + tan A + cot A) = tan A sec A – cot A cosec A
(iii) \(\frac { { tan }^{ 2 }\theta }{ { tan }^{ 2 }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { sec }^{ 2 }\theta -{ cosec }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta } \)
Solution:
(i) \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } =sinA+cosA \)
L.H.S = \(\frac { cosA }{ 1-tanA } -\frac { { sin }^{ 2 }A }{ cosA-sinA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q9.1

Question 10.
\(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 } \)
Solution:
\(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } =\frac { 2 }{ 1-2{ cos }^{ 2 }A } =\frac { { 2sec }^{ 2 }A }{ { tan }^{ 2 }A-1 } \)
L.H.S = \(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q10.1

Question 11.
2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = θ
Solution:
2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = θ
L.H.S = 2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q11.1

Question 12.
If cot θ + cos θ = m, cot θ – cos θ = n, then prove that (m² – n²)² = 16 run.
Solution:
cot θ + cos θ = m…..(i)
cot θ – cos θ = n……(ii)
Adding (i)&(ii) we get
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q12.1

Question 13.
If sec θ + tan θ = p, prove that sin θ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
Solution:
sec θ + tan θ = p,
prove that sin θ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
\(\frac { 1 }{ cos\theta } +\frac { sin\theta }{ cos\theta } =p\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q13.1

Question 14.
If tan A = n tan B and sin A = m sin B, prove that cos² A = \(\frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 } \)
Solution:
m = \(\\ \frac { sinA }{ sinB } \)
n = \(\\ \frac { tanA }{ tanB } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q14.1

Question 15.
If sec A = \(x+ \frac { 1 }{ 4x } \), then prove that sec A + tan A = 2x or \(\\ \frac { 1 }{ 2x } \)
Solution:
sec A = \(x+ \frac { 1 }{ 4x } \)
To prove that sec A + tan A = 2x or \(\\ \frac { 1 }{ 2x } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q15.1

Question 16.
When 0° < θ < 90°, solve the following equations:
(i) 2 cos² θ + sin θ – 2 = 0
(ii) 3 cos θ = 2 sin² θ
(iii) sec² θ – 2 tan θ = 0
(iv) tan² θ = 3 (sec θ – 1).
Solution:
0° < θ < 90°
(i) 2 cos² θ + sin θ – 2 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 19 Trigonometric Identities Chapter Test Q16.1

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Circles Class 10 ICSE ML Aggarwal Chapter Test

August 27, 2024August 27, 2024 by Raju

ML Aggarwal Class 10 Solutions Circles Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test.

Question 1.
(a) In the figure (i) given below, triangle ABC is equilateral. Find ∠BDC and ∠BEC.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. OD is perpendicular to AB and C is a point on the arc DB. Find ∠BAD and ∠ACD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q1.1
Solution:
(a) ∆ABC is an equilateral triangle.

Question 2.
(a) In the figure given below, AB is a diameter of the circle. If AE = BE and ∠ADC = 118°, find
(i) ∠BDC (ii) ∠CAE.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q2.1
(b) In the figure given below, AB is the diameter of the semi-circle ABCDE with centre O. If AE = ED and ∠BCD = 140°, find ∠AED and ∠EBD. Also Prove that OE is parallel to BD.

Solution:
(a) Join DB, CA and CB.
∠ADC = 118° (given)
and ∠ADB = 90°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q2.3

Question 3.
(a) In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2 (∠ACB + ∠BAC).
(b) In the figure (ii) given below, O is the centre of the circle. Prove that x + y = z.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q3.1
Solution:
(a) Given : O is the centre of the circle.
To Prove : ∠AOC = 2 (∠ACB + ∠BAC).
Proof: In ∆ABC,
∠ACB + ∠BAC + ∠ABC = 180°
(Angles of a triangle)

Question 4.
(a) In the figure (i) given below, AB is diameter of a circle. If DC is parallel to AB and ∠CAB = 25°, find :
(i)∠ADC (ii) ∠DAC.
(b) In the figure (ii) given below, the centre O of the smaller circle lies on the circumference of the bigger circle. If ∠APB = 70° and ∠BCD = 60°, find :
(i) ∠AOB (ii) ∠ACB
(iii) ∠ABD (iv) ∠ADB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q4.1
Solution:
(a) AB is diameter and DC || AB,
∠CAB = 25°, join AD,BD

Question 5.
(a) In the figure (i) given below, ABCD is a cyclic quadrilateral. If AB = CD, Prove that AD = BC.
(b) In the figure (ii) given below, ABC is an isosceles triangle with AB = AC. If ∠ABC = 50°, find ∠BDC and ∠BEC.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q5.1
Solution:
(a) Given : ABDC is a cyclic quadrilateral AB = CD.
To Prove: AD = BC.

Question 6.
A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the distance of P from the nearest point of the circle.
Solution:
Join OT, OP = 13 cm and TP = 12 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q6.1

Question 7.
Two circles touch each other internally. Prove that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution:
Given : Two circles with centre O and O’ touch each other internally at P.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q7.1

Question 8.
From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that
(i) ∠AOP = ∠BOP.
(ii) OP is the perpendicular bisector of the chord AB.
Solution:
Given : From a point P, outside the circle with centre O. PA and PB are the tangents to the circle, OA, OB and OP are joined.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q8.1

Question 9.
(a) The figure given below shows two circles with centres A, B and a transverse common tangent to these circles meet the straight line AB in C. Prove that:
AP : BQ = PC : CQ.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q9.1
(b) In the figure (ii) given below, PQ is a tangent to the circle with centre O and AB is a diameter of the circle. If QA is parallel to PO, prove that PB is tangent to the circle.

Solution:
(a) Given : Two circles with centres A and B and a transverse common tangent to these circles meet AB at C.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q9.3

Question 10.
In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q10.1
Solution:
In the given figure, two chords with centre A and B touch externally. PM is a tangent to the circle with centre A and QN is tangent to the circle with centre B. PM = 15 cm, QN = 12 cm, PA = 17 cm, QB = 13 cm. We have to find AB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q10.2

Question 11.
Two chords AB, CD of a circle intersect externally at a point P. If PB = 7 cm, AB = 9 cm and PD = 6 cm, find CD.
Solution:
∵ AB and CD are two chords of a circle which intersect each other at P, outside the circle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q11.1

Question 12.
(a) In the figure (i) given below, chord AB and diameter CD of a circle with centre O meet at P. PT is tangent to the circle at T. If AP = 16 cm, AB = 12 cm and DP = 2 cm, find the length of PT and the radius of the circle
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q12.1
(b) In the figure (ii) given below, chord AB and diameter CD of a circle meet at P. If AB = 8 cm, BP = 6 cm and PD = 4 cm, find the radius of the circle. Also find the length of the tangent drawn from P to the circle. .
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q12.2
Solution:
Given : (a) AB is chord of a circle with centre O and PT is tangent and CD is the diameter of the circle which meet at P.
AP = 16 cm, AB = 12 cm, OP = 2 cm
∴PB = PA – AB = 16 – 12 = 4 cm
∵ABP is a secant and PT is tangent.
∴PT² = PA x PB .
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q12.3

Question 13.
In the figure given below, chord AB and diameter PQ of a circle with centre O meet at X. If BX = 5 cm, OX = 10 cm and.the radius of the circle is 6 cm, compute the length of AB. Also find the length of tangent drawn from X to the circle.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q13.1
Solution:
Chord AB and diameter PQ meet at X on producing outside the circle

Question 14.
(a) In the figure (i) given below, ∠CBP = 40°, ∠CPB = q° and ∠DAB = p°. Obtain an equation connecting p and q. If AC and BD meet at Q so that ∠AQD = 2 q° and the points C, P, B and Q are concyclic, find the values of p and q.
(b) In the figure (ii) given below, AC is a diameter of the circle with centre O. If CD || BE, ∠AOB = 130° and ∠ACE = 20°, find:
(i)∠BEC (ii) ∠ACB
(iii) ∠BCD (iv) ∠CED.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q14.1
Solution:
(a) (i) Given : ABCD is a cyclic quadrilateral.
Ext. ∠PBC = ∠ADC
=> 40° = ∠ADC

Question 15.
(a) In the figure (i) given below, APC, AQB and BPD are straight lines.
(i) Prove that ∠ADB + ∠ACB = 180°.
(ii) If a circle can be drawn through A, B, C and D, Prove that it has AB as diameter
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q15.1
(b) In the figure (ii) given below, AQB is a straight line. Sides AC and BC of ∆ABC cut the circles at E and D respectively. Prove that the points C, E, P and D are concyclic.

Solution:
(a) Given : In the figure, APC, AQB and BPD are straight lines.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q15.3

Question 16.
(a) In the figure (i) given below, chords AB, BC and CD of a circle with centre O are equal. If ∠BCD = 120°, find
(i) ∠BDC (ii) ∠BEC
(iii) ∠AEC (iv) ∠AOB.
Hence Prove that AOAB is equilateral.
(b) In the figure (ii) given below, AB is a diameter of a circle with centre O. The chord BC of the circle is parallel to the radius OD and the lines OC and BD meet at E. Prove that
(i) ∠CED = 3 ∠CBD (ii) CD = DA.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q16.1
Solution:
(a) In ∆BCD, BC = CD
∠CBD = ∠CDB
But ∠BCD + ∠CBD + ∠CDB = 180°
(∵ Angles of a triangle)

Question 17.
(a) In the adjoining figure, (i) given below AB and XY are diameters of a circle with centre O. If ∠APX = 30°, find
(i) ∠AOX (ii) ∠APY (iii) ∠BPY (iv) ∠OAX.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q17.1
(b) In the figure (ii) given below, AP and BP are tangents to the circle with centre O. If ∠CBP = 25° and ∠CAP = 40°, find :
(i) ∠ADB (ii) ∠AOB (iii) ∠ACB (iv) ∠APB.

Solution:
(a) AB and XY are diameters of a circle with centre O.
∠APX = 30°.
To find :
(i) ∠AOX (ii) ∠APY
(iii) ∠BPY (iv) ∠OAX
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test Q17.3

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 16 Circles Chapter Test are helpful to complete your math homework.

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Locus Class 10 ICSE ML Aggarwal Chapter Test

August 27, 2024August 27, 2024 by Raju

ML Aggarwal Class 10 Solutions Locus Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test.

Question 1.
Draw a straight line AB of length 8 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
Solution:
(i) Draw a line segment AB = 8 cm.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test Q1.1
(ii) Draw the perpendicular bisector of AB intersecting AB at D.
∴ Every point P on it will be equidistant from A and B.
(iii) Take a point P on the perpendicular bisector.
(iv) Join PA and PB.
Proof : In ∆PAD and ∆PBD
PD = PD (common)
AD = BD (D is mid-point of AB)
∠PDA = ∠PDB (each 90°)
∴ ∆ PAD ≅ ∆ PBD (SAS axiom of congruency)
∴PA = PB (c.p.c.t.)
Similarly we can prove any other point on the perpendicular bisector of AB is equidistant from A and B.
Hence Proved.

Question 2.
A point P is allowed to travel in space. State the locus of P so that it always remains at a constant distance from a fixed point C.
Solution:
The point P is moving in the space and it is at a constant distance from a fixed point C.
∴ Its locus is a sphere.

Question 3.
Draw a line segment AB of length 7 cm. Construct the locus of a point P such that area of triangle PAB is 14 cm².
Solution:
Base of ∆PAB = 7 cm
and its area = 14 cm²
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test Q3.1
Now draw a line XY parallel to AB at a distance of 4 cm.
Now take any point P on XY
Join PA and PB
area of ∆PAB = 14 cm.
Hence locus of P is the line XY which is parallel to AB at distance of 4 cm.

Question 4.
Draw a line segment AB of length 12 cm. Mark M, the mid-point of AB. Draw and describe the locus of a point which is
(i) at a distance of 3 cm from AB.
(ii) at a distance of 5 cm from the point M. Mark the points P, Q, R, S which satisfy both the above conditions. What kind of quadrilateral is PQRS ? Compute the area of the quadrilateral PQRS.
Solution:
Steps of Construction :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test Q4.1
(i) Take a line AB = 12 cm
(ii) Take M, the mid point of AB.
(iii) Draw straight lines CD and EF parallel to AB at a distance of 3 cm.
(iv) With centre M and radius 5 cm, draw areas which intersects CD at P and Q and EF at R and S.
(v) Join QR and PS.
PQRS is a rectangle where length
PQ = 8 cm.
Area of rectangle PQRS = PQ x RS = 8 x 6 = 48 cm²

Question 5.
AB and CD are two intersecting lines. Find the position of a point which is at a distance of 2 cm from AB and 1.6 cm from CD.
Solution:
(i) AB and CD are the intersecting lines which intersect each other at O.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test Q5.1
(ii) Draw a line EF parallel to AB and GH parallel to CD intersecting each other at P
P is the required point.

Question 6.
Two straight lines PQ and PK cross each other at P at an angle of 75°. S is a stone on the road PQ, 800 m from P towards Q. By drawing a figure to scale 1 cm = 100 m, locate the position of a flag staff X, which is equidistant from P and S, and is also equidistant from the road.
Solution:
1 cm = 100 cm
800 m = 8 cm.
Steps of Construction :
(i) Draw the lines PQ and PK intersecting each other at P making an angle of 75°.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test Q6.1
(ii) Take a point S on PQ such that PS = 8 cm.
(iii) Draw the perpendicular bisector of PS.
(iv) Draw the angle bisector of ∠KPS intersecting the perpendicular bisector at X. X is the required point which is equidistant from P and S and also from PQ and PK.

Question 7.
Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively. Find by construction a point X equidistant from PQ, PS and equidistant from R, S. Measure XR.
Solution:
Steps of Construction :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test Q7.1
(i) Take PR = 8 cm and draw the perpendicular bisector of PR intersecting it at O.
(ii) From O, out. off OS = OQ = 3 cm
(iii) Join PQ, QR, RS and SP.
PQRS is a rhombus. Whose diagonal are PR and QS.
(iv) PR is the bisector of ∠SPQ.
(v) Draw the perpendicular bisector of SR intersecting PR at X
∴ X is equidistant from PQ and PS and also from S and R.
On measuring length of XR = 3.2 cm (approx)

Question 8.
Without using set square or protractor, construct the parallelogram ABCD in which AB = 5.1 cm. the diagonal AC = 5.6 cm and the diagonal BD = 7 cm. Locate the point P on DC, which is equidistant from AB and BC.
Solution:
Steps of Construction :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test Q8.1
(i) Take AB = 5.1 cm
(ii) At A, with readius \(\\ \frac { 5.6 }{ 2 } \) = 2.8 cm and at B with radius \(\\ \frac { 7.0 }{ 2 } \) = 3.5 cm, draw two arcs
intersecting each other at O.
(iii) Join AO and produce it to C such that OC = AD = 2.8 cm and join BO and produce it to D such that BO = OD = 3.5 cm
(iv) Join BC, CD, DA
ABCD is a parallelogram.
(v) Draw the angle bisector of ∠ABC intersecting CD at P. P is the required point which is equidistant from AB and BC.

Question 9.
By using ruler and compass only, construct a quadrilateral ABCD in which AB = 6.5 cm, AD = 4cm and ∠DAB = 75°. C is equidistant from the sides AB and AD, also C is equidistant from the points A and B.
Solution:
Steps of Construction :
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test Q9.1
(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a ray making an angle of 75° and cut off AD = 4 cm.
(iii) Draw the bisector of ∠DAB.
(iv) Draw perpendicular bisector of AB intersecting the angle bisector at C.
(v) Join CB and CD.
ABCD is the required quadrilateral.

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Locus Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. APlusTopper try to provide online math tutoring for you.

Similarity Class 10 ICSE ML Aggarwal Chapter Test

August 27, 2024August 27, 2024 by Raju

ML Aggarwal Class 10 Solutions Similarity Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test.

Question 1.
In the given figure, ∠1 = ∠2 and ∠3 = ∠4. Show that PT x QR = PR x ST.
Solution:
Given : In the given figure,
∠1 = ∠1 and ∠3 = ∠4
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q1.1
To prove : PT x QR = PR x ST
Proof: ∠1 = ∠2
Adding ∠6 to both sides
∠1 + ∠6 = ∠2 + ∠6
∠SPT = ∠QPR
In ∆PQR and ∆PST

Question 2.
In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q2.1
Solution:
Given : In the given figure,
AB = AC, PM ⊥ AB and PN ⊥ AC
To prove : PM x PC = PN x PB
Proof: In ∆ABC, AB = AC
∠B = ∠C
Now in ∆CPN and ∆BPM,

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) 3 ED = GD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q3.1
Solution:
(a) Given : In following figure, ∠AED = ∠ABC
Required : The values of x and y.
Now, in ∆ABC and ∆ADE
∠AED = ∠ABC (given)
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE
(By A.A. axiom of similarity)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q3.2

Question 4.
In the given figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:
(i) DF || BH
(ii) AH = 3 AF.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q4.1
Solution:
Given : E is the mid-point of BD and F is mid-point of AC also 2 AD = BD and EC || BH
To Prove : (i) DF || BH
(ii) AH = 3 AF

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:
Given : In ∆ABC, D and E are the points on the sides AB and AC respectively
DE || BC
AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm, BC = 5 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q5.1

Question 6.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.
Solution:
In ∆ABC, D and E are points on the sides AB and AC respectively
AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q6.1

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:
In ∆ABC, DE || BC
D is on AB and E is on AC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q7.1

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:
Let ∆ABC and ∆DEF are similar and area of
∆ABC = 360 cm²
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q8.1

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.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q9.1
Solution:
In ∆ABC and ∆ACD
∠C = ∠C (Common)
∠ABC = ∠CAD (given)
∴ ∆ABC ~ ∆ACD

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 area of DAOE : area of ||gm ABCD.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q10.1
Solution:
(a) In the figure
Diagonals of parallelogram ABCD are AC and BD which intersect each other at O.
OE is drawn parallel to CB to meet AB in E.

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.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q11.1
Solution:
In the given figure, ABCD is trapezium in
which AB || DC, 2AB = 3DC

Question 12.
In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that .
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q12.1
(i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4.
Solution:
Given : In || gm ABCD,
E is mid point of BC and DE meets the diagonal AC at O and meet AB produced at F.
To prove : (i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4
Proof: In ∆AOD and ∆EDC
∠AOD = ∠EOC (vertically opposite angle)
∠OAD = ∠OCB (alt. angles)
∆AOD ~ ∆EOC (AA postulate)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q12.2

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 model is 1.6 m.
(ii) the area of the deck of the ship, if the area of the deck of model is 2.4 m².
(iii) the volume of the model, if the volume of the ship is 1 km³.
Solution:
Scale factor (k) of the model of the ship = \(\\ \frac { 1 }{ 250 } \)
(i) Length of model = 1.6 m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Similarity Chapter Test Q13.1

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Equation of a Straight Line Class 10 ICSE ML Aggarwal Chapter Test

August 27, 2024August 27, 2024 by Raju

ML Aggarwal Class 10 Solutions Equation of a Straight Line Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test.

Question 1.
Find the equation of a line whose inclination is 60° and y-intercept is – 4.
Solution:
Angle of inclination = 60°
Slope = tan θ = tan 60° = √3
Equation of the line will be,
y = mx + c = √3x + ( – 4)
=> y – √3x – 4 Ans.

Question 2.
Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.
Solution:
Slope of the line 3y + 2x = 12
=> 3y = 12 – 2x
=> 3y = – 2x + 12
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q2.1

Question 3.
If the equation of a line is y – √3x + 1, find its inclination.
Solution:
In the line
y = √3 x + 1
Slope = √3 => tan θ = √3
θ = 60° (∵ tan 60° = √3)

Question 4.
If the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.
Solution:
The equation of line y = mx + c
∵ it passes through (2, – 4) and ( – 3, 1)
Now substituting the value of these points – 4 = 2 m + c …(i)
and 1 = – 3 m + c …(ii)
Subtracting we get,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q4.1

Question 5.
If the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.
Solution:
Let the points to be A (1, 4), B (3, – 2) and C (p, – 5) are collinear and let B (3, – 2)
divides AC in the ratio of m1 : m2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q5.1

Question 6.
Find the inclination of the line joining the points P (4, 0) and Q (7, 3).
Solution:
Slope of the line joining the points P (4, 0) and Q (7, 3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q6.1

Question 7.
Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to \(– \frac { 3 }{ 7 } \)
Solution:
Equation of lines are
2x + y = 5 …(i)
x – 2y = 5 …(ii)
Multiply (i) by 2 and (ii) by 1, we get
4x + 2y = 10
x – 2y = 5
Adding we get,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q7.1

Question 8.
If the point A is reflected in the y-axis, the co-ordinates of its image A₁, are (4, – 3),
(i) Find the co-ordinates of A
(ii) Find the co-ordinates of A₂, A₃ the images of the points A, A₁, Respectively under reflection in the line x = – 2
Solution:
(i) ∵ A is reflected in the y-axis and its image is A₁ (4, – 3)
Co-ordinates of A will be ( – 4, – 3)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q8.1

Question 9.
If the lines \(\frac { x }{ 3 } +\frac { y }{ 4 } =7 \) and 3x + ky = 11 are perpendicular to each other, find the value of k.
Solution:
Given
Equation of lines are
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q9.1

Question 10.
Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).
Solution:
The equation of the line is x – 2y + 8 = 0
=> 2y = x + 8
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q10.1

Question 11.
Write down the equation of the line passing through ( – 3, 2) and perpendicular to the line 3y = 5 – x.
Solution:
Equations of the line is
3y = 5 – x
=> 3y = – x + 5
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q11.1

Question 12.
Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.
Solution:
Let slope of the line joining the points A (1, 2) and B (6, 7) be m₁
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q12.1

Question 13.
The points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.
Solution:
Slope of line AC (m₁)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q13.1

Question 14.
A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q14.1
(i) the co-ordinates of A and B
(ii) the equation of the line AB
Solution:
A lies on x-axis and B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y) , and P (2, 1) divides BA in the ratio 3 : 1.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q14.2

Question 15.
A straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the line passes through the point ( – 3, 8), find its equation.
Solution:
Let the line make intercept a and b with the x-axis and y-axis respectively then the line passes through
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q15.1

Question 16.
If the coordinates of the vertex A of a square ABCD are (3, – 2) and the quation of diagonal BD is 3 x – 7 y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the centre of the square.
Solution:
Co-ordinates of A are (3, – 2).
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test Q16.1