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Tamil Nadu Board of Secondary EducationSSLC (English Medium) इयत्ता १०

SSLC (English Medium) इयत्ता १० - Tamil Nadu Board of Secondary Education Question Bank Solutions for Mathematics

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विषय
मुख्य विषय
अध्याय
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Mathematics
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Given the LCM and GCD of the two polynomials p(x) and q(x) find the unknown polynomial in the following table

LCM GCD p(x) q(x)
(x4 –  y4)(x4 + x2y2 + y2 (x2 –  y2)   (x4 –  y4)(x2 + y2 –  xy)
[3] Algebra
Chapter: [3] Algebra
Concept: undefined >> undefined

If (x – 6) is the HCF of x2 – 2x – 24 and x2 – kx – 6 then the value of k is 

[3] Algebra
Chapter: [3] Algebra
Concept: undefined >> undefined

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Find the least common multiple of xy(k2 + 1) + k(x2 + y2) and xy(k2 – 1) + k(x2 – y2)

[3] Algebra
Chapter: [3] Algebra
Concept: undefined >> undefined

Find the GCD of the following by division algorithm

2x4 + 13x3 + 27x2 + 23x + 7, x3 + 3x2 + 3x + 1, x2 + 2x + 1

[3] Algebra
Chapter: [3] Algebra
Concept: undefined >> undefined

In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC

If `"AD"/"DB" = 3/4` and AC = 15 cm find AE

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC

If AD = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = 3x – 1, find the value of x

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

ABCD is a trapezium in which AB || DC and P, Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

In ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC

AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

In ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC

AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

If PQ || BC and PR || CD prove that `"AR"/"AD" = "AQ"/"AB"`

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

Rhombus PQRB is inscribed in ΔABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

In trapezium ABCD, AB || DC, E and F are points on non-parallel sides AD and BC respectively, such that EF || AB. Show that = `"AE"/"ED" = "BF"/"FC"`

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

DE || BC and CD || EE Prove that AD2 = AB × AF

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

Check whether AD is bisector of ∠A of ∆ABC of the following
AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

Check whether AD is bisector of ∠A of ∆ABC of the following

AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm.

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

∠QPR = 90°, PS is its bisector. If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F, respectively. Prove that EF || BD.

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

Construct a ∆PQR in which the base PQ = 4.5 cm, ∠R = 35° and the median from R to RG is 6 cm.

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined

Construct a ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.

[4] Geometry
Chapter: [4] Geometry
Concept: undefined >> undefined
< prev  101 to 120 of 418  next > 
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