In the given figure, arcs have been drawn of radius 7cm each with vertices a, b, c and d of quadrilateral abcd as centres. find the area of the shaded region.

A circle with radius of 5 cm \greenD 1 1 cm × 1 1 cm start color #11accd, 11, start text, c, m, end text, times, 11, start text, c, m, end text, end color #11accd rectangle. • Your answer should be • an integer, like 6 6 6 6 • a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5 • a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4 • a mixed number, like 1 3 / 4 1\ 3/4 1 3 / 4 1, space, 3, slash, 4 • an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75 • a multiple of pi, like 12 pi 12\ \text^2 cm 2 start text, c, m, end text, squared

Related documents • HR offline classes undertaking • Declaration by the Candidate and Parents Filled • Annexure II- Affidavit BY THE Parent & Guardian Filled • Annexure I- Affidavit BY THE Student Filled • Unit-5 ghjhgrgu fhiiihgh fbnmk fguik • What Movies Show Realism Perception and Truth in Film Preview text Class- X Session- 2022- Subject- Mathematics (Standard) Sample Question Paper Time Allowed: 3 Hrs. Maximum Marks : 80 General Instructions: • This Question Paper has 5 Sections A-E. • Section A has 20 MCQs carrying 1 mark each • Section B has 5 questions carrying 02 marks each. • Section C has 6 questions carrying 03 marks each. • Section D has 4 questions carrying 05 marks each. 6. Section E has 3 case based integrated units of assessment (04 marks each) with sub- parts of the values of 1, 1 and 2 marks each respectively. • All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2marks questions of Section E 8. Draw neat figures wherever required. Take π =22/7 wherever required if not stated. SECTION A Section A consists of 20 questions of 1 mark each. S . MA RKS 1 Let a and b be two positive integers such that a = p 3 q 4 and b = p 2 q 3 , where p and q are prime numbers. If HCF(a,b) = pmqn and LCM(a,b) = prqs, then (m+n)(r+s)= (a) 15 (b) 30 (c) 35 (d) 72 1 2 Let p be a prime number. The quadratic equation having its roots as factors of p ...

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Let r be the radius of each sector = 21 cm Area of the shaded region = Area of the four sectors Let angles subtended at A, B, C and D be x°, y°, z° and w° respectively. Angle subtended at A, B, C, D (in radians, (θ)) be `(xpi)/180, (ypi)/180, (zpi)/180, (wpi)/180` respectively. ∴ Area of a sector with central angle at A = `1/2 r^2 theta` = `1/2 xx (21)^2 xx (xpi)1/0 = (441xpi)/360 cm^2` ∴ Area of a sector with central angle at B = `1/2 r^2 theta` = `1/2 xx (21)^2 xx (ypi)/180 = (441ypi)/360 cm^2` Area of a sector with central angle at C = `1/2 r^2 theta` = `1/2 xx (21)^2 xx (zoi)/180 = (441zpi)/360 cm^2` ∴ Area of a sector with central angle at D = `1/2 r^2 theta` = `1/2 xx (21)^2 xx (wpi)/180 = (441wpi)/360 cm^2` ∴ Area of four sectors = `((441xpi)/360 + (441ypi)/360 + (441zpi)/360 + (441wpi)/360) cm^2` Since, sum of all interior angles in any quadrilateral is 360° ∴x + y + z +w = 360° Thus, Area of four sectors = `((441pi)/360) (x + y + z + w)` = `(441pi)/360 xx 360` = 441π cm 2 = 1386 cm 2 Hence, required area of the shaded region is 1386 cm 2. • 1 more Video Tutorials For All Subjects • Areas of Sector and Segment of a Circle video tutorial 00:10:18 • Areas of Sector and Segment of a Circle video tutorial 00:08:32 • Areas of Sector and Segment of a Circle video tutorial 00:17:10 • Areas of Sector and Segment of a Circle video tutorial 00:09:23 • Areas of Sector and Segment of a Circle video tutorial 00:09:17 • Areas of Sector and Segment of a Circle video tutorial 00:0...

We're asked to find the area of the shaded region, so the area of this red-shaded region. So this is interesting. This is almost a 10 by 10 square, except we have these quarter circles that are cut out. So the area of this would be the area of what a 10 by 10 square would be minus the area of these quarter circles. And each of these quarter circles is a quarter of a circle with a radius 3. I think we can assume that all of these, if you took the distance from here to the outside of this quarter circle, have radius 3. So if you put four quarter circles together, you're going to have a complete white circle. So one way to think about this is that the area of this whole red region is going to be the area of the entire square, which is 10 by 10. So it's going to be 10 times 10, which is 100 whatever square units we have. And then we're going to subtract out the area of the four quarter circles. And that area is going to be equivalent to the area of one circle with a radius of 3. So what's the area of a circle with radius 3? Well, the formula for area of a circle is pi r squared, or r squared pi. So the radius is 3. So it's going to be 3 times 3, which is 9, times pi-- 9 pi. So we have 100 minus 9 pi is the area of the shaded region. And we got it right.

Views: 5,518 2 × 2 × 2 × 3 × 5, (2) 2 × 2 × 3 × 3 × 5, (3) 2 × 2 × 2 × 3 × 7 (4) 2 × 3 × 5 × 7. Ans.(1) 2 Given that HCF ( 306 , 657 ) = 9 ; LCM ( 306 , 657 ) will be- (1) 8667 , (2) 22338 (4) None of these. Ans.(2) If a 2 ​ a 1 ​ ​ = b 2 ​ b 1 ​ ​  = c 2 ​ c 1 ​ ​ th and a 2 ​ x + b 2 ​ y + c (3) 22 , 3 The HCF of 15 and 25 is- (1) 5 , (2) 10 (3) 15 , (4) 2. Ans.(1) (1) a unique sol (2) many solutio (3) no solution, (4) exactly two The prime factorisation of a natural number is 9 The straight lines except for the order of its factors. 2 x − 3 y = 0 and (1) Coincident, (3) Parallel, (1) Many, (2) Unique, 5 The sum and product of the zeroes of a quadratic 10) For which value a x − y = 2 and 6 x (1) c = 3, polynomial are 2 ​ and 3 1 ​ respectively then the (3) c = − 12, quadratic polynomial is- (1) 3 x 2 + 3 2 ​ x + 1 (2) 3 x 2 − 3 2 ​ x + 1 (3) 3 x 2 − 3 2 ​ x − 1 (4) None of these. Ans.(2) (11) What is the soluti 3 x + 2 y = 4 ? (1) x = 2 , y = − 1, (3) x = 3 , y = 0, A pair of linearequi (1) Consistent, (3) Caincide, In the given figure, arcs have been drawn of radius 7 cm each with vertices A , B , C and D of quadrilateral ABCD as centres. Find the area of the shaded region. Updated On Jan 3, 2023 Topic All topics Subject Mathematics Class Class 10 Answer Type Video solution: 1 Upvotes 80 Avg. Video Duration 6 min