The Karnaugh map for a Boolean function is given as Image The simplified Boolean equation for the above Karnaugh Map is (A) AB + CD + AB’ + AD (B) AB + AC + AD + BCD (C) AB + AD + BC + ACD (D) AB + AC + BC + BCD

The Karnaugh map for a Boolean function is given as

image

The simplified Boolean equation for the above Karnaugh Map is (A) AB + CD + AB’ + AD (B) AB + AC + AD + BCD (C) AB + AD + BC + ACD (D) AB + AC + BC + BCD

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1 Answer

Answer :

(B) AB + AC + AD + BCD
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The simplified form of a Boolean equation (AB’+AB’C+AC)(A’C’+B’) is : (A) AB’ (B) AB’C (C) A’B (D) ABC

Description : The simplified form of a Boolean equation (AB’+AB’C+AC)(A’C’+B’) is : (A) AB’ (B) AB’C (C) A’B (D) ABC

Last Answer : (A) AB’

Consider the following statements : (a) Boolean expressions and logic networks correspond to labelled acyclic digraphs. (b) Optimal Boolean expressions may not correspond to simplest networks. (c) Choosing essential blocks first in a Karnaugh map and then greedily choosing the largest remaining blocks to cover may not give an optimal expression. Which of these statement(s) is/ are correct? (A) (a) only (B) (b) only (C) (a) and (b) (D) (a), (b) and (c)

Description : Consider the following statements : (a) Boolean expressions and logic networks correspond to labelled acyclic digraphs. (b) Optimal Boolean expressions may not correspond to simplest networks. (c) Choosing essential blocks first in a Karnaugh map ... (b) only (C) (a) and (b) (D) (a), (b) and (c)

Last Answer : (D) (a), (b) and (c) 

Karnaugh map is used for the purpose of (A) Reducing the electronic circuits used. (B) To map the given Boolean logic function. (C) To minimize the terms in a Boolean expression. (D) To maximize the terms of a given a Boolean expression.

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Last Answer : (C) To minimize the terms in a Boolean expression.

Simplified Boolean equation for the following truth table is: Image(A) F = yz’ + y’z (B) F = xy’ + x’y (C) F = x’z + xz’ (D) F = x’z + xz’ + xyz

Description : Simplified Boolean equation for the following truth table is: (A) F = yz’ + y’z (B) F = xy’ + x’y (C) F = x’z + xz’ (D) F = x’z + xz’ + xyz 

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ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

In a trapezium ABCD, AB is parallel to CD, BD is perpendicular to AD. AC is perpendicular to BC. If AD = BC = 15 cm and AB = 25 cm, -Maths 9th

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Last Answer : answer:

In the adjoining figure, if ∠BAC = 90° and AD ⊥ BC, then (а) BD.CD = BC2 (b) AB.AC = BC2 (c) BD.CD = AD2 (d) AB.AC = AD2

Description : In the adjoining figure, if ∠BAC = 90° and AD ⊥ BC, then (а) BD.CD = BC2 (b) AB.AC = BC2 (c) BD.CD = AD2 (d) AB.AC = AD2

Last Answer : (c) BD.CD = AD2

The simplified function in product of sums of Boolean function F(W, X, Y, Z) = Σ(0, 1, 2, 5, 8, 9, 10) is (A) (W' + X') (Y' + Z') (X' + Z) (B) (W' + X') (Y' + Z') (X' + Z') (C) (W' + X') (Y' + Z) (X' + Z) (D) (W' + X') (Y + Z') (X' + Z)

Description : The simplified function in product of sums of Boolean function F(W, X, Y, Z) = Σ(0, 1, 2, 5, 8, 9, 10) is (A) (W' + X') (Y' + Z') (X' + Z) (B) (W' + X') (Y' + Z') (X' + Z') (C) (W' + X') (Y' + Z) (X' + Z) (D) (W' + X') (Y + Z') (X' + Z)

Last Answer : (A) (W' + X') (Y' + Z') (X' + Z)

If bc bisects the angle acd then b is the midpoint of ad?

Description : If bc bisects the angle acd then b is the midpoint of ad?

Last Answer : That is not necessarily true.

In given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then (a) DE || BC (b) DF || AC (c) EF || AB (d) none of

Description : In given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then (a) DE || BC (b) DF || AC (c) EF || AB (d) none of

Last Answer : (c) EF || AB

ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: -Maths 9th

Description : ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: -Maths 9th

Last Answer : Given: ABCD is a trapezium, in which AB || DC and AD = BC. To Prove: (i) ∠A = ∠B (ii) ∠C = ∠D (iii) △ABC ≅ △BAD (iv) Diagonal AC = diagonal BD. Const.: Produce AB to E, such that a line through ... △ABC ≅ △BAD [by SAS congruence axiom] (iv) ⇒ AC = BD [c.p.c.t.] Thus, diagonal AC = diagonal BD.

ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: -Maths 9th

Description : ABCD is a trapezium in which AB II CD and AD = BC (see flg). Show that: -Maths 9th

Last Answer : Given: ABCD is a trapezium, in which AB || DC and AD = BC. To Prove: (i) ∠A = ∠B (ii) ∠C = ∠D (iii) △ABC ≅ △BAD (iv) Diagonal AC = diagonal BD. Const.: Produce AB to E, such that a line through ... △ABC ≅ △BAD [by SAS congruence axiom] (iv) ⇒ AC = BD [c.p.c.t.] Thus, diagonal AC = diagonal BD.

In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

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Last Answer : Solution :-

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Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

Last Answer : Solution :-

ABCD is a square of side a cm. AB, BC, CD and AD are all chords of circles with equal radii each. -Maths 9th

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Last Answer : (b) \(\bigg[a^2+4\bigg[rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg]\bigg]\)As shown in the given figures, if a' is each side of the square, then ∠DOC = 120º ⇒ ∠ODC = ∠OCD = 30ºNow in fig. (iii), \(rac{ ... of square + Total area of 4 segments = \(a^2+4\bigg(rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg).\)

ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Description : ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Last Answer : Here is your First of all we will draw a quadrilateral ABCD with AD = BC and join AC, BD, P,Q,R,S are the mid points of AB, AC, CD and BD respectively. In the triangle ABC, P and Q are mid points of AB and AC respectively. All sides are equal so PQRS is a Rhombus.

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Description : In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Last Answer : answer:

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Description : ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Last Answer : Solution :-

In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. -Maths 9th

Description : In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. -Maths 9th

Last Answer : It is given that ∠BAD=∠EAC ∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides] ∴∠BAC=∠DAE In △BAC and △DAE AB=AD (Given) ∠BAC=∠DAE (Proved above) AC=AE (Given) ∴△BAC≅△DAE (By SAS congruence rule) ∴BC=DE (By CPCT)

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

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

Last Answer : (b) 3

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Description : Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Last Answer : Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD = 6 cm2 + 9.2 cm2 = 15.2 cm2 (approx.)

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

Show that in a quadrilateral ABCD,AB + BC + CD + DA > AC + BD. -Maths 9th

Description : Show that in a quadrilateral ABCD,AB + BC + CD + DA > AC + BD. -Maths 9th

Last Answer : Solution :-

In the given figure, AB is the diameter and `/_ADC = 2 /_BDC`. If `/_ BCD =70^(@)`, then find the angle made by AC at the centre of the circle.

Description : In the given figure, AB is the diameter and `/_ADC = 2 /_BDC`. If `/_ BCD =70^(@)`, then find the angle made by AC at the centre of the circle.

Last Answer : In the given figure, AB is the diameter and `/_ADC = 2 /_BDC`. If `/_ BCD =70^(@)`, then find the angle made by AC at the centre of the circle.

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc -Maths 10th

Description : If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc -Maths 10th

Last Answer : (c2 – ab) x2 + 2(bc - a2 ) x+ (b2 – ac) = 0 Comparing with Ax2 + Bx + C = 0 A = (c2 – ab), B = 2(bc - a2 ) and C = b2 – ac According to the question, B2 - 4AC = 0 Put the values in the above equation we get 4a(a3 + b3 + c3 -3abc) = 0 hence, a = 0 or a3 + b3 + c3 = 3ab

Let R be the rectangular window against which the lines are to be clipped using 2D Sutherland-Cohen line clipping algorithm. The rectangular window has lower left-hand corner at (-5,1) and upper righthand corner at (3,7). Consider the following three lines for clipping with the given end point co-ordinates: Line AB: A(-6,2) and B(-1,8) Line CD: C(-1,5) and D(4,8) Line EF: E(-2,3) and F(1,2) Which of the following line(s) is/are candidate for clipping? (A) AB (B) CD (C) EF (D) AB and CD

Description : Let R be the rectangular window against which the lines are to be clipped using 2D Sutherland-Cohen line clipping algorithm. The rectangular window has lower left-hand corner at (-5,1) and upper righthand corner at (3,7). ... s) is/are candidate for clipping? (A) AB (B) CD (C) EF (D) AB and CD

Last Answer : (D) AB and CD

In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Last Answer : answer:

The three outputs x1x2x3 from the 8x3 priority encoder are used to provide a vector address of the form 101x1x2x300. What is the second highest priority vector address in hexadecimal if the vector addresses are starting from the one with the highest priority? (A) BC (B) A4 (C) BD (D) AC

Description : The three outputs x1x2x3 from the 8x3 priority encoder are used to provide a vector address of the form 101x1x2x300. What is the second highest priority vector address in hexadecimal if the vector addresses are starting from the one with the highest priority? (A) BC (B) A4 (C) BD (D) AC

Last Answer : Answer: B

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF​=ABFB​=1 from (1) ⇒LF=LD so, DF=2DL

Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

Description : Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

Last Answer : answer:

ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. If F is a point on side BC such that segment -Maths 9th

Description : ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. If F is a point on side BC such that segment -Maths 9th

Last Answer : answer:

When simplified with Boolean Algebra (x + y)(x + z) simplifies to (A) x (B) x + x(y + z) (C) x(1 + yz) (D) x + yz

Description : When simplified with Boolean Algebra (x + y)(x + z) simplifies to (A) x (B) x + x(y + z) (C) x(1 + yz) (D) x + yz

Last Answer : Ans: D When simplified with Boolean Algebra (x + y)(x + z) simplifies to x + yz [(x + y) (x + z)] = xx + xz + xy + yz = x + xz + xy + yz (Qxx = x) = x(1+z) + xy + yz = x + xy + yz {Q(1+z) = 1} = x(1 + y) + yz = x + yz {Q(1+y) = 1}]

What is the output when following statement is executed ? >>>"abcd"[2:] a) a b) ab c) cd d) dc

Description : What is the output when following statement is executed ? >>>"abcd"[2:] a) a b) ab c) cd d) dc

Last Answer : c) cd

What is the output of the following? x = ['ab', 'cd'] for i in x: x.append(i.upper()) print(x) a) [‘AB’, ‘CD’]. b) [‘ab’, ‘cd’, ‘AB’, ‘CD’]. c) [‘ab’, ‘cd’]. d) none of the mentioned

Description : What is the output of the following? x = ['ab', 'cd'] for i in x: x.append(i.upper()) print(x) a) [‘AB’, ‘CD’]. b) [‘ab’, ‘cd’, ‘AB’, ‘CD’]. c) [‘ab’, ‘cd’]. d) none of the mentioned

Last Answer : d) none of the mentioned

What is the output of the following? x = ['ab', 'cd'] for i in x: i.upper() print(x) a) [‘ab’, ‘cd’]. b) [‘AB’, ‘CD’]. c) [None, None]. d) none of the mentioned

Description : What is the output of the following? x = ['ab', 'cd'] for i in x: i.upper() print(x) a) [‘ab’, ‘cd’]. b) [‘AB’, ‘CD’]. c) [None, None]. d) none of the mentioned

Last Answer : a) [‘ab’, ‘cd’].

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.