The Octal equivalent of HEX and number AB.CD is

The Octal equivalent of HEX and number AB.CD is 

(A) 253.314 (B) 253.632 (C) 526.314 (D) 526.632
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The Octal equivalent of HEX and number AB.CD is 253.632

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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

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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.)

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

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.

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.

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. 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........

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ . -Maths 9th

Last Answer : Join AQ and PC . Since ABCD is a parallelogram . ⇒ AB | | DC ⇒ AP | | QC ∵ AP and QC are parts of AB and DC respectively] Also, AP = CQ [given] Thus, APCQ is a parallelogram . We know that diagonals of a parallelogram bisect each other . Hence AC and PQ bisect each other .

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.

Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Description : Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Last Answer : Given, opposite angles of a quadrilateral are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. ∴ CD = AB = 4cm

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

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 and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Description : P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Last Answer : Given In a parallelogram ABCD, P and Q are the mid-points of AS and CD, respectively. To show PRQS is a parallelogram. Proof Since, ABCD is a parallelogram. AB||CD ⇒ AP || QC

In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Description : In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Last Answer : (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = 1/2 AB = 1/2 x 8 = 4 cm given OA = 5 cm AO2 = AC2 + OC2 (5)2 = (4)2 + OC2 25 = 16 + OC2 ... length is always positive] OA = OD [same radius of a circle] OD = 5 cm CD = OD - OC = 5 - 3 = 2 cm