Evaluate each of the following using identities: (i) (399)2 (ii) (0.98)2 (iii) 991 x 1009 (iv) 117 x 83 -Maths 9th

Evaluate each of the following using identities: (i) (399)2 (ii) (0.98)2 (iii) 991 x 1009 (iv) 117 x 83 -Maths 9th
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Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Description : Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Last Answer : (i) (2x - 1/x)2 [Use identity: (a - b)2 = a2 + b2 - 2ab ] (2x - 1/x)2 = (2x) 2 + (1/x)2 - 2 (2x)(1/x) = 4x2 + 1/x2 - 4 (ii) (2x + y) (2x - y) [Use identity: (a - b)(a + b) = a2 - b 2 ] (2x + y) (2x - ... ) = a2 - b 2 ](1.5 x 2 - 0.3y2 ) (1.5x2 + 0.3y2 ) = (1.5 x 2 ) 2 - (0.3y2 ) 2 = 2.25 x4 - 0.09y4

Evaluate each of the following: (i) (103)3 (ii) (98)3 (iii) (9.9)3 (iv) (10.4)3 (v) (598)3 (vi) (99)3 -Maths 9th

Description : Evaluate each of the following: (i) (103)3 (ii) (98)3 (iii) (9.9)3 (iv) (10.4)3 (v) (598)3 (vi) (99)3 -Maths 9th

Last Answer : answer:

Evaluate using identities -Maths 9th

Description : Evaluate using identities -Maths 9th

Last Answer : (i) 1023 = (100 + 2)3 = (100)3 + 3(100)2 (2) + 3(100) (2)2 + 23 = 10,00,000 + 60,000 + 1200 + 8 = 10,61,208 (ii) 993 = (100 - 1)3 = 1003 - 3(100)2 (1) + 3(100) (1)2 - 13 = 10,00,000 - 30,000 + 300 - 1 = 9,70,299

Evaluate using the identities -Maths 9th

Description : Evaluate using the identities -Maths 9th

Last Answer : (i) 505 × 503 = (500 + 5) (500 + 3) = 5002 + (5 + 3) (500) + 5 × 3 = 250000 + 4000 + 15 = 254015 (ii) 37 × 16 = (30 + 7) (30 - 4) = 302 + (7 - 4) (30) - 7 × 4 = 900 + 90 - 28 = 962

Evaluate using identities -Maths 9th

Description : Evaluate using identities -Maths 9th

Last Answer : (i) 1023 = (100 + 2)3 = (100)3 + 3(100)2 (2) + 3(100) (2)2 + 23 = 10,00,000 + 60,000 + 1200 + 8 = 10,61,208 (ii) 993 = (100 - 1)3 = 1003 - 3(100)2 (1) + 3(100) (1)2 - 13 = 10,00,000 - 30,000 + 300 - 1 = 9,70,299

Evaluate using the identities -Maths 9th

Description : Evaluate using the identities -Maths 9th

Last Answer : (i) 505 × 503 = (500 + 5) (500 + 3) = 5002 + (5 + 3) (500) + 5 × 3 = 250000 + 4000 + 15 = 254015 (ii) 37 × 16 = (30 + 7) (30 - 4) = 302 + (7 - 4) (30) - 7 × 4 = 900 + 90 - 28 = 962

Evaluate each of the following: (i) 1113 – 893 (ii) 463 + 343 (iii) 1043 + 963 (iv) 933 – 1073 -Maths 9th

Description : Evaluate each of the following: (i) 1113 – 893 (ii) 463 + 343 (iii) 1043 + 963 (iv) 933 – 1073 -Maths 9th

Last Answer : answer:

If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

Description : If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

Last Answer : (x+y)2 = x2 - 2xy + y2 = x2 + 2xy + y2 - 4xy = (x+y)2 - 4xy = 52 - 4 × 4 = 25 - 16 = 9 ∴ x - y = √9 = 3

If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

Description : If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

Last Answer : (x+y)2 = x2 - 2xy + y2 = x2 + 2xy + y2 - 4xy = (x+y)2 - 4xy = 52 - 4 × 4 = 25 - 16 = 9 ∴ x - y = √9 = 3

A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days, 4/7 of the work completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day? a) 79 b) 80 c) 81 d) 82 e) 83

Description : A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days, 4/7 of the work completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day? a) 79 b) 80 c) 81 d) 82 e) 83

Last Answer : work finished by 33 days = 33 117 8 = 30888 man hours 4/7th of the job is finished. remaining work = 1-(4/7) = 3/7 if 4/7th of the work is 30888 hours, then 3/7th of the work = [30888 ... 9 number of men = 198 since, 117 men are already working, additional men required = 198 - 117 = 81 Answer: c)

Using identities , find the value of -Maths 9th

Description : Using identities , find the value of -Maths 9th

Last Answer : (i) 1012 = (100 + 1)2 = 1002 + 2 100 1 + 12 = 10000 + 200 + 1 = 10201 (ii) 982 = (100 - 2)2 = 1002 - 2 100 2 + 22 = 10000 - 400 + 4 = 9604 (iii) 0.982 = (1- 0.02)2 = 12 - 2 1 0 ... 102 - 12 = 99 (v) 190 190 - 10 10 = 1902 - 102 = (190 + 10 ) (190 - 10) = 200 180 = 36000

Using identities , find the value of -Maths 9th

Description : Using identities , find the value of -Maths 9th

Last Answer : (i) 1012 = (100 + 1)2 = 1002 + 2 100 1 + 12 = 10000 + 200 + 1 = 10201 (ii) 982 = (100 - 2)2 = 1002 - 2 100 2 + 22 = 10000 - 400 + 4 = 9604 (iii) 0.982 = (1- 0.02)2 = 12 - 2 1 0 ... 102 - 12 = 99 (v) 190 190 - 10 10 = 1902 - 102 = (190 + 10 ) (190 - 10) = 200 180 = 36000

Using identities solve 103^ -Maths 9th

Description : Using identities solve 103^ -Maths 9th

Last Answer : NEED ANSWER

Using identities solve 103^ -Maths 9th

Description : Using identities solve 103^ -Maths 9th

Last Answer : FRIEND YOUR QUESTION IS NOT CLEAR

In quadratic equation ax^2 + bx + c = 0 , Identities the sum and product of Roots? -Maths 9th

Description : In quadratic equation ax^2 + bx + c = 0 , Identities the sum and product of Roots? -Maths 9th

Last Answer : If the two roots of the quadratic equation ax2 + bx + c = 0 obtained by the quadratic formula be denoted by a and b, then we have α = \(\frac{-b+\sqrt{b^2-4ac}}{2a},\) β = \(\frac{-b-\sqrt{b^2-4ac}}{2a}\) ∴ Sum of roots ... then α + β = - (- \(\frac{5}{6}\)) = \(\frac{5}{6}\), ab = \(\frac{7}{6}\).

Express the following as cm using decimals. (a) 2 mm (b) 30 mm (c) 116 mm (d) 4 cm 2 mm (e) 162 mm (f) 83 mm -Maths 9th

Description : Express the following as cm using decimals. (a) 2 mm (b) 30 mm (c) 116 mm (d) 4 cm 2 mm (e) 162 mm (f) 83 mm -Maths 9th

Last Answer : We know that 1 cm = 10 mm 1 mm = 1 / 10 cm (a) 2 mm = 2 / 10 cm = 0.2 cm (b) 30 mm = 30 / 10 cm = 3.0 cm (c) 116 mm = 116 / 10 cm = 11.6 cm (d) 4 cm 2 mm = [(4 + 2 / 10)] cm = 4.2 cm (e) 162 mm = 162 / 10 cm = 16.2 cm (f) 83 mm = 83 / 10 cm = 8.3 cm

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 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]

Find the following products: (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) (ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx) (iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) (iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz) -Maths 9th

Description : Find the following products: (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) (ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx) (iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) (iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz) -Maths 9th

Last Answer : answer:

Simplify: (i) (a + b + c)2 + (a – b + c)2 (ii) (a + b + c)2 – (a – b + c)2 (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2 (iv) (2x + p – c)2 – (2x – p + c)2 (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2 -Maths 9th

Description : Simplify: (i) (a + b + c)2 + (a – b + c)2 (ii) (a + b + c)2 – (a – b + c)2 (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2 (iv) (2x + p – c)2 – (2x – p + c)2 (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2 -Maths 9th

Last Answer : answer:

Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 and x + y + z = 15 -Maths 9th

Description : Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 and x + y + z = 15 -Maths 9th

Last Answer : Consider the equation x + y + z = 15 From algebraic identities, we know that (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) So, (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + xz) From the question, x2 + y2 + z2 ... y3 + z3 - 3xyz = 15(83 - 71) => x3 + y3 + z3 - 3xyz = 15 12 Or, x3 + y3 + z3 - 3xyz = 180

The probability thata singlebit will be inerror on a typical public telephone lineusing 4800 bps modem is 10 to the power -3. If no error detection mechanism is used, the residual error ratefor a communicationline using 9-bit frames is approximately equal to A. 0.003 B. 0.009 C. 0.991 D. 0.999 E. None of the above

Description : The probability thata singlebit will be inerror on a typical public telephone lineusing 4800 bps modem is 10 to the power -3. If no error detection mechanism is used, the residual error ratefor a communicationline using 9-bit frames is ... to A. 0.003 B. 0.009 C. 0.991 D. 0.999 E. None of the above

Last Answer : 0.009

Using suitable identity, evaluate the following. -Maths 9th

Description : Using suitable identity, evaluate the following. -Maths 9th

Last Answer : Evaluate the following

Using suitable identity, evaluate the following -Maths 9th

Description : Using suitable identity, evaluate the following -Maths 9th

Last Answer : Evaluation of following

Using suitable identity, evaluate the following. -Maths 9th

Description : Using suitable identity, evaluate the following. -Maths 9th

Last Answer : Evaluate the following

Using suitable identity, evaluate the following -Maths 9th

Description : Using suitable identity, evaluate the following -Maths 9th

Last Answer : Evaluation of following

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Description : Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Last Answer : Rationalisation of denominator

Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Description : Rationalise the denominator in each of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732 and √5 = 2.236 up to three places of decimal. -Maths 9th

Last Answer : Rationalisation of denominator

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

Evaluate 185 x 185 - 15 x 15 -Maths 9th

Description : Evaluate 185 x 185 - 15 x 15 -Maths 9th

Last Answer : Solution :-

Evaluate 105 x 108 without multiplying directly. -Maths 9th

Description : Evaluate 105 x 108 without multiplying directly. -Maths 9th

Last Answer : Solution :-

Evaluate x if log3 (3 + x) + log3 (8 – x) – log3 (9x – 8) = 2 – log39 -Maths 9th

Description : Evaluate x if log3 (3 + x) + log3 (8 – x) – log3 (9x – 8) = 2 – log39 -Maths 9th

Last Answer : (c) 4log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) = 2 - log39 ⇒ log3 (3 + x) + log3 (8 - x) - log3 (9x - 8) + log39 = 2⇒ log3 \(\bigg[rac{(3+x)(8-x)(9)}{(9x-8)}\bigg]=2\)⇒ \(rac{9(24+8x-3x-x^2)}{(9x- ... = 9x - 8 ⇒ x2 + 4x - 32 = 0 ⇒ (x + 8) (x - 4) = 0 ⇒ x = - 8, 4. Taking the positive value x = 4.

What number completes this 107 98 90 83 77 68 65?

Description : What number completes this 107 98 90 83 77 68 65?

Last Answer : 72difference decreases by 1 each time9 = 107 - 988 = 98 - 907 = 90 - 836 = 83 - 775 = 77 - 724 = 72 - 683 = 68 - 65

108,991.20 tax tables for filing jointly but at a higher single rate and married w/0 dependents?

Description : 108,991.20 tax tables for filing jointly but at a higher single rate and married w/0 dependents?

Last Answer : If my husband makes $75,000 in 2011 filing married and zero and I make $34,615 filing married but at a higher single rate and zero plus $25 will we owe taxes or get a refund?

The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : NEED ANSWER

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : Median will be a good representative of the data, because 1.each value occurs once. 2.the data is influenced by extreme values.

Evaluate: 48×52 + 61×59 + 77×83 a) 12486 b) 10996 c) 13406 d) 13206 e) None of these

Description : Evaluate: 48×52 + 61×59 + 77×83 a) 12486 b) 10996 c) 13406 d) 13206 e) None of these

Last Answer : 48×52 = (50 - 2) × (50 + 2) = 502 – 22 = 2496 61 × 59 = (60 + 1) × (60 - 1) = 602 – 12 = 3599 77 × 83 = (80 - 3) × (80 + 3) = 802 – 32 = 6391 Sum = 2496 + 3599 + 6391 = 12486 Answer: a)

Convert each of the following decimals into a percentage: (i) 0.6 (ii) 0.42 (iii) 0.07 (iv) 0.005 -Maths

Description : Convert each of the following decimals into a percentage: (i) 0.6 (ii) 0.42 (iii) 0.07 (iv) 0.005 -Maths

Last Answer : answer:

Convert each of the following into decimal form: (i) 45% (ii) 127% (iii) 3.6% (iv) 0.23% -Maths

Description : Convert each of the following into decimal form: (i) 45% (ii) 127% (iii) 3.6% (iv) 0.23% -Maths

Last Answer : answer:

Convert each of the following into a fraction: (i) 32% (ii) 614% (iii) 2623% (iv) 120% (v) 6.25% (vi) 0.8% (vii) 0.06% (viii) 22.75% -Maths

Description : Convert each of the following into a fraction: (i) 32% (ii) 614% (iii) 2623% (iv) 120% (v) 6.25% (vi) 0.8% (vii) 0.06% (viii) 22.75% -Maths

Last Answer : answer:

Evaluate: 25 to the power 1/3 × 5 to the power 1/3. -Maths 9th

Description : Evaluate: 25 to the power 1/3 × 5 to the power 1/3. -Maths 9th

Last Answer : NEED ANSWER

Evaluate: 25 to the power 1/3 × 5 to the power 1/3. -Maths 9th

Description : Evaluate: 25 to the power 1/3 × 5 to the power 1/3. -Maths 9th

Last Answer : Solution :-

How much does an object having the mass of 100kg weight in newton.  a. 981 N  b. 991 N  c. 981.6 N  d. 980.1N F = ma

Description : How much does an object having the mass of 100kg weight in newton.  a. 981 N  b. 991 N  c. 981.6 N  d. 980.1N F = ma

Last Answer : 900,000lb/hr;625 ft/s981 N

what is 0.399 written as a percent what do i do?

Description : what is 0.399 written as a percent what do i do?

Last Answer : oof

Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Description : Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Last Answer : Formula: Surface area of sphere (SA) = 4πr2 (i) Radius of sphere, r = 10.5 cm SA = 4 (22/7) 10.52 = 1386 Surface area of sphere is 1386 cm2 (ii) Radius of sphere, r = 5.6cm Using formula, SA = 4 (22 ... 75 cm Surface area of sphere = 4πr2 = 4 (22/7) 1.752 = 38.5 Surface area of a sphere is 38.5 cm2

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

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.