{"id":22784,"date":"2022-01-18T15:41:08","date_gmt":"2022-01-18T10:11:08","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=22784"},"modified":"2022-01-18T15:41:08","modified_gmt":"2022-01-18T10:11:08","slug":"ap-board-7th-class-maths-solutions-chapter-11-intext-questions-new-syllabus","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-7th-class-maths-solutions-chapter-11-intext-questions-new-syllabus\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 11 Area of Plane Figures InText Questions"},"content":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf<\/a> Chapter 11 Area of Plane Figures InText Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 11th Lesson Area of Plane Figures InText Questions<\/h2>\n

Check Your Progress [Page No: 84]<\/span><\/p>\n

Question 1.
\nFill the missing values in the following table.
\n\"AP
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\nAnu has 4 right angled triangles with same size. Using those triangles, she makes a star like toy given below. Calculate the area of this toy star.
\n\"AP
\nAnswer:
\nGiven base of the triangle b = 5 cm
\nHeight of the triangle h = 12 cm
\nArea of the triangle = \\(\\frac{1}{2}\\) \u00d7 b \u00d7 h
\n\"AP
\n= 30 sq.cm
\nArea of star = Area of 4 triangles
\n\u2234 Area of the toy star = 4 \u00d7 30
\n= 120 sq.cm<\/p>\n

Let’s Explore [Page No. 85]<\/span><\/p>\n

Question 1.
\nThe areas of triangular field ABC and rectangular field EFGH are equal. The length and breadth of EFGH are 15 m., 10 m. respectively. The base of \u2206ABC 25 m. then find it’s height.
\n\"AP
\nAnswer:
\nGiven length of rectangle l = 15 m
\nbreadth of rectangle b = 10 m
\nArea of rectangle = l \u00d7 b
\n= 15 \u00d7 10
\n= 150 sq.m.
\nGiven base of the triangle b = 25 m
\nheight of the triangle h = ? m
\n\u2234 Area of the triangle = \\(\\frac{1}{2}\\) \u00d7 b \u00d7 h
\n\u21d2 \\(\\frac{1}{2}\\) \u00d7 25 \u00d7 h = 150
\n\u21d2 25h = 150 \u00d7 2
\n\"AP
\n\u2234 Height of triangle h = 12 m.<\/p>\n

\"AP<\/p>\n

Question 2.
\nAll the triangles in the given figure are on the base AB =12 cm. Find the height of each of the triangles corresponding to the base AB, by counting the grids and find the area of each triangle. What do you observe?
\n\"AP
\nAnswer:
\nGiven in the figure,
\nBase of \u2206 APB = 12 cm
\nHeight = 8 cm
\nArea of \u2206 APB = \\(\\frac{1}{2}\\) \u00d7 b \u00d7 h
\n\"AP
\n= 48 sq.cm
\nBase of \u2206 AQB = 12 cm
\nHeight = 8 cm
\nArea of \u2206 AQB
\n\"AP
\n= 48 sq.cm
\nBase of \u2206 ARB = 12 cm
\nHeight = 8 cm
\nArea of \u2206 ARB
\n\"AP
\n= 48 sq.cm
\nBase of \u2206 ASB = 12 cm
\nHeight = 8 cm
\nArea of \u2206 ASB
\n\"AP
\n= 48 sq.cm
\nBy observing area of each triangle is same.<\/p>\n

\"AP<\/p>\n

Let’s Explore [Page No: 88]<\/span><\/p>\n

Question 1.
\n5 cm width white tiles arranged between square shaped blue tiles along sides as shown in the figure. If the side of the total arrangement is 150 cm, find the area of the arranged white tiles.
\n\"AP
\nAnswer:
\nIn the given figure side of total arrangement ABCD = 150 cm.
\nWidth of white tile = 5 cm
\n\"AP
\nArea of white tile path EFGH = 150 \u00d7 5 = 750 sq.cm
\nArea of white tile path MNOP = 150 \u00d7 5 = 750 sq.cm
\nArea of common path IJKL = 5 \u00d7 5 = 25 sq.cm
\nArea of total white tile path
\n= Area of EFGH + Area of MNOP – Area of IJKL
\n= 750 + 750 – 25
\n\u2234 The area of the arranged white tiles
\n= 1475 sq.cm.<\/p>\n

\"AP<\/p>\n

Question 2.
\n2 m wide path is present the outer side of the square grass land of side 80 m. Find the area of path and total expenditure of the path flooring with bricks, if the cost of flooring with bricks per sq.m is \u20b9 200.
\n\"AP
\nAnswer:
\nGiven side of the grass land = 80 m
\nWidth of the path = 2 m
\nPath laid out side the grass land.
\nSo, outer side of grassland
\n= side + 2 \u00d7 width
\n= 80 + 2 \u00d7 2 = 84 m
\n\"AP
\nABCD = 80 \u00d7 80 – 6400 sq.m
\nArea of the grass land with path
\nEFGH = 84 \u00d7 84 = 7056 sq.m
\nArea of the path = Area of EFGH – Area of ABCD
\n= 7056 – 6400 = 656 sq.m.
\nCost of flooring per sq.m = \u20b9 200
\n\u2234 Cost of flooring per 656 sq.m
\n= 656 \u00d7 200 = \u20b9 1,31.200<\/p>\n

\"AP<\/p>\n

Let’s Do Activity [Page No: 88]<\/span><\/p>\n

Prepare two coloured rectangles one is red colour of length 25 cm, breadth 20 cm. Another one is green of length 20 cm, breadth 15 cm and place smaller rectangle middle to the bigger rectangle, so, that 2.5 cm red colour formed outside to green colour rectangle. Find the area of red colour path.
\n\"AP
\nAnswer:
\nArea of red rectangle = 25 \u00d7 20 = 500 cm2<\/sup>
\nArea of green rectangle = 20 \u00d7 15 = 300 cm2<\/sup>
\n\u2234 Area of the coloured path = (Area of red rectangle) – (Area of green rectangle)
\n= 500 – 300
\n= 200 cm2<\/sup><\/p>\n

[Page No: 89]<\/span><\/p>\n

In the part of ‘Nadu-Nedu’ programme, Head master decided to arrange cir\u00accular shape flower bed with radius 7 m in the premises of his school. How many flower plants needed if it takes 4 plants per sq. m. ?
\nAnswer:
\nHere, length of the rectangle (l)
\n= Half of the perimeter of circle
\n= \u03c0r
\nbreadth of the rectangle (b)
\n= radius of circle
\n= r
\nWe know that the area of the rectangle = l \u00d7 b = \u03c0r \u00d7 r (\u2235 l = \u03c0r, b = r)
\n= \u03c0r2<\/sup> – Area of the circle.
\nSo, the area of the circle A = \u03c0r2<\/sup>
\nNow we solve above problem,
\nwhere \u2018r\u2019 = 7m
\nArea of flower bed = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 7
\n= 22 \u00d7 7
\n= 154 sq.m
\nNumber of plants per sq.m = 4
\nNumber of plants for 154 sq.m
\n= 154 \u00d7 4 = 616 plants.
\nSo, Headmaster needs 616 plants for the flower bed.<\/p>\n

\"AP<\/p>\n

Check Your Progress [Page No. 91]<\/span><\/p>\n

Question 1.
\nThe circumference of the circle shaped rangoli sheet is 88 cm. Find the radius of the circle and the area of the circle.
\nAnswer:
\nGiven circumference of the rangoli
\n2\u03c0r = 88 cm
\n\u21d2 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 r = 88
\n\u21d2 44 r = 88 \u00d7 7
\n\"AP
\n\u21d2 r = 14 cm
\n\u2234 Radius of rangoli r = 14 cm
\nArea of the rangoli = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 142<\/sup>
\n\"AP
\n\u2234 Area of the circle = 616 sq.cm<\/p>\n

\"AP<\/p>\n

Question 2.
\nCalculate the areas of circles shown in the figure.
\n(i)
\n\"AP
\nAnswer:
\nFrom the figure radius r = 7 cm
\nArea of the circle = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 72<\/sup>
\n\"AP
\n= 154 sq.cm<\/p>\n

(ii)
\n\"AP
\nAnswer:
\nFrom the figure diameter of circle d = 28 cm
\nRadius of circle r = \\(\\frac{\\mathrm{d}}{2}\\) = \\(\\frac{28}{2}\\) = 14 cm.
\nArea of the circle = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 142<\/sup>
\n\"AP
\n\u2234 Area of the circle = 616 sq.cm<\/p>\n

\"AP<\/p>\n

(iii)
\n\"AP
\nAnswer:
\nFrom the figure radius of circle (r) = 21 cm
\nArea of the circle = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 212<\/sup>
\n\"AP
\n\u2234 Area of the circle = 1386 sq.cm<\/p>\n

Let’s Explore [Page No: 93]<\/span><\/p>\n

Radius of circular shaped grass land is 11 m. A goat is tied with a rope of length 4 m at the centre, then find the area of grass land that the goat cannot graze.
\nAnswer:
\nGiven the radius of grass land = 11 m
\nArea of grass land = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 11 \u00d7 11
\n= \\(\\frac{2662}{7}\\)
\n\u2234 Area of grass land = 380.29 sq.m.
\nRadius of the small circle = length of rope to which goat tied = 4m
\nArea of land that the goat can graze = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 4 \u00d7 4
\n= \\(\\frac{352}{7}\\) = 50.29 sq.m
\n\u2234 Area of land that the goat cannot graze
\n= Area of grass land – Area of grazed
\n= 380.29 – 50.29
\n= 330 sq.m.<\/p>\n

\"AP<\/p>\n

Reasoning Corner (Non-Verbal) [Page No: 96 ]<\/span><\/p>\n

Question 1.
\nEmbedded Figures :
\nThe problem figure (X) is given, answer figures as (a), (b), (c) & (d) given besides. The problem figures as a hidden figure of answer figure and one should identify that figure.
\n\"AP
\nAnswer:
\n\"AP<\/p>\n

Examples<\/span><\/p>\n

Question 1.
\nFind the area of the given triangles.
\n\"AP
\nAnswer:
\n(i) In \u2206PQR, Base (QR) = 6 cm,
\nHeight (PS) = 4 cm<\/p>\n

Area of \u2206PQR = \\(\\frac{1}{2}\\) x base x height
\n= \\(\\frac{1}{2}\\) \u00d7 QR \u00d7 PS
\n= \\(\\frac{1}{2}\\) \u00d7 6 cm \u00d7 4 cm
\n= 12 sq.cm.<\/p>\n

(ii) In \u2206LMN, Base (MN) = 3 cm,
\nHeight (LO) = 2 cm
\nArea of \u2206LMN = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height
\n= \\(\\frac{1}{2}\\) \u00d7 MN \u00d7 LO
\n= \\(\\frac{1}{2}\\) \u00d7 3 cm \u00d7 2 cm
\n= 3 sq.cm<\/p>\n

\"AP<\/p>\n

Question 2.
\nThe area of the \u2206 XYZ is 12 sq.cm and the height XL is 3 cm, then find base YZ.
\nAnswer:
\n\"AP
\nIn \u2206 XYZ, Base = YZ,
\nHeight (XL) = 3 cm,
\nArea of the XYZ = 12 sq.cm.
\nArea of the \u2206XYZ = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height
\n= \\(\\frac{1}{2}\\) \u00d7 YZ \u00d7 XL
\n\u21d2 12 = \\(\\frac{1}{2}\\) \u00d7 YZ \u00d7 3
\n\u21d2 YZ = 12 \u00d7 \\(\\frac{2}{3}\\)
\nSo, YZ = 8 cm<\/p>\n

Question 3.
\nIn \u2206 ABC, AC = 8 cm, BC = 4 cm and AE = 5 cm.
\nFind (i) the area of the AABC (ii) BD.
\nAnswer:
\n\"AP
\n(i) In \u2206 ABC, Base (BC) = 4 cm,
\nHeight (AE) = 5 cm.
\nArea of the \u2206 ABC = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height
\n= = \\(\\frac{1}{2}\\) \u00d7 4 \u00d7 5
\n10 sq.cm.<\/p>\n

(ii) In ABAC, Base (AC) = 8 cm,
\nheight (BD) = ?,
\nArea of \u2206 BAC = 10 sq.cm
\nArea of the \u2206 BAC = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height
\ni. e. 10 = \\(\\frac{1}{2}\\) \u00d7 8 \u00d7 BD
\nBD = 10 \u00d7 \\(\\frac{2}{8}\\) = \\(\\frac{10}{4}\\) = \\(\\frac{5}{2}\\) = 2.5
\nSo, height (BD) = 2.5 cm<\/p>\n

\"AP<\/p>\n

Question 4.
\nCalculate the area of the given right-angled triangle with sides having right angle are 6 cm, 6 cm.
\nAnswer:
\n\"AP
\nMethod – 1: Sides having right angles are 6 cm, 6 cm.
\nSides forming the right angle a = 6 cm, b = 6 cm
\nArea of a right-angled triangle
\n= \\(\\frac{1}{2}\\) \u00d7 Product of sides forming the right angle
\n= \\(\\frac{1}{2}\\) \u00d7 a \u00d7 b
\n= \\(\\frac{1}{2}\\) \u00d7 6 \u00d7 6
\n= 6 \u00d7 3 = 18 sq.cm.<\/p>\n

(or)<\/p>\n

\"AP
\nMethod – 2: Observe the grid carefully, the right angled triangle covering half of the area of Square.
\nArea of a right-angled triangle
\n= \\(\\frac{1}{2}\\) \u00d7 Area of square
\n= \\(\\frac{1}{2}\\) \u00d7 6 \u00d7 6 = 18 sq.cm.<\/p>\n

Question 5.
\nFind the area of the triangle shaped lawn whose base and heights are 12m.,7m. respectively. Find the total cost of laying lawn, if cost of grass is \u20b9 300 per Sq. m.
\n\"AP
\nAnswer:
\nBase of the triangle shaped lawn = 12 m.
\nHeight = 7 m.
\nArea of triangle shaped lawn 1
\n= \\(\\frac{1}{2}\\) \u00d7 b \u00d7 h
\n= \\(\\frac{1}{2}\\) \u00d7 12 \u00d7 7
\n= 6 \u00d7 7 = 42 Sq.m
\nCost of grass for laying in lawn per 1 Sq.m = \u20b9 300
\nCost of grass for laying in lawn for 42 Sq.m = \u20b9 300 \u00d7 42 = \u20b9 12,600<\/p>\n

\"AP<\/p>\n

Question 6.
\nThe length and breadth of a rectangular field is 65 m, 30 m respectively. A path of width 2.5 m is made around the park outside. Find the area of the path.
\n\"AP
\nAnswer:
\nIn the fig, ABCD is a rectangular field and shaded area shows 2.5m. wide path. EFGH is a rectangle (field with path)
\nLength of ABCD (AB) = 65m.,
\nBreadth of ABCD (AD) = 30m.
\nWidth of path = 2.5 m.
\nArea of path = Area of outer rectangular- field EFGH – Area of inner rectangular field ABCD
\nLength of rectangle EFGH = EF
\n= Length of field (AB) + (2 \u00d7 Width of path)
\n= 65m + (2 \u00d7 2.5 m)
\n= 65m + 5m = 70m
\nBreadth of rectangle EFGH = EH
\n= Breadth of field (AD) + (2 \u00d7 Width of path)
\n= 30 m + (2 \u00d7 2.5m)
\n= 30 m + 5m = 35m
\nArea of outer rectangle
\nEFGH = length \u00d7 breadth
\n70m \u00d7 35m = 2450 Sq.m
\nArea of inner rectangle
\nABCD = length \u00d7 breadth
\n= 65 m \u00d7 30 m
\n= 1950 Sq.m
\nArea of path = Area of outer rectangle EFGH – Area of inner rectangle ABCD
\n= 2450 Sq.m – 1950 Sq.m
\n= 500 Sq.m.<\/p>\n

Question 7.
\nA square shaped swimming pool of side 70m. It has 5m width path is present the outer side of the boundary. Find the area of this path. Find the expenditure of covering that path with tiles at the rate of \u20b9 150 per sq.m.
\nAnswer:
\nWXYZ shows a square shaped swimming pool of side 70m. 5m wide path to the outer side of the swimming pool.,
\nArea of the path = Area of swimming pool PQRS with path – Area of swimming pool WXYZ
\nPS = Side of swimming pool (WZ) + (2 \u00d7 breadth of path)
\n= 70m + (2 \u00d7 5m)
\n= 70m + 10m = 80m
\n\"AP
\nArea of swimming pool with path
\nPQRS = (Side)2<\/sup> = (80 m)2<\/sup>
\n= 6400 Sq.m.
\nArea of swimming pool
\nWXYZ = (Side)2<\/sup> = (70 m)2<\/sup>
\n= 4900 Sq.m.
\nArea of path = Area of swimming pool PQRS with path – Area of swimming pool WXYZ
\n= 6400 – 4900 = 1500 Sq.m.
\nIf cost of covering tiles per 1 Sq.m.
\n= \u20b9 150
\nCost of covering tiles per 1500 Sq.m.
\n= \u20b9 150 \u00d7 1500
\n= \u20b9 2,25,000<\/p>\n

\"AP<\/p>\n

Question 8.
\nThe length of rectangular grass land is 55 m and breadth is 45 m in the centre of the grass land two paths of 3 m wide one parallel to the length and another parallel to breadth are situated in such a way that they intersect each other. Find the area of the path.
\nAnswer:
\nIn fig. ABCD is rectangular grass land.
\nLength of ABCD = 55 m
\nBreadth of ABCD = 45 m
\nWidth of path = 3 m
\nArea of path EFGH = Length \u00d7 Width
\n= 55 \u00d7 3
\n= 165 Sq.m
\n\"AP
\nArea of path MNOP
\n= Breadth \u00d7 Width
\n= 45 \u00d7 3 = 135 Sq.m.
\nArea of common path IJKL (situated on both paths)
\n= Width \u00d7 Width = 3 m \u00d7 3 m = 9 Sq.m.
\nArea of square IJKL i.e., 9 Sq.m is included in both the paths. So we subtract one time.
\nArea of total path = Area of path EFGH + Area of path MNOP – Area of IJKL
\n= (165 Sq.m. + 135 Sq.m. – 9 Sq.m.)
\n= (300 9) Sq.m.
\n= 291 Sq.m.<\/p>\n

Question 9.
\nFind the area of the rangoli of it’s radius is 21 cm.
\n(Use \u03c0 = \\(\\frac{22}{7}\\))
\n\"AP
\nAnswer:
\nRadius of rangoli (r) = 21 cm
\n\"AP
\nArea of circular shaped rangoli = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 21 cm \u00d7 21 cm
\n= 1386 sq. cm
\n\u2234 Area of the circle = 1386 sq. cm<\/p>\n

\"AP<\/p>\n

Question 10.
\nFind the surface area of a circular shaped pool whose diameter is 28 m
\n(Use \u03c0 = \\(\\frac{22}{7}\\)).
\n\"AP
\nAnswer:
\nThe diameter of circular shaped pool (d) = 28 m
\nRadius (r) = \\(\\frac{28}{2}\\) m = 14 m
\nArea of a circular shaped pool = \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 (14m)2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 14 m \u00d7 14 m
\n= 22 \u00d7 2 \u00d7 14 sq.m.
\n= 616 sq.m.<\/p>\n

Question 11.
\nIn a circular shaped park inner portion is given for kids to play and outer portion is given for walking to elders. If outer radius is 35 m, width of walking track is 14 m, then find the area of walking path.
\n\"AP
\nAnswer:
\nOuter radius of park (R) = 35 m,
\nWidth of walking track = 14 m.
\nRadius of playground (r) = R – w = 35 – 14 = 21m.
\nArea of walking track = Area of the park – Area of playing ground
\n= \u03c0R2<\/sup> – \u03c0r2<\/sup>
\n= \\(\\frac{22}{7}\\) \u00d7 352<\/sup> – \\(\\frac{22}{7}\\) \u00d7 212<\/sup>
\n= \\(\\frac{22}{7}\\) (352<\/sup> – 212<\/sup>)
\n= \\(\\frac{22}{7}\\) (1225 – 441)
\n= \\(\\frac{22}{7}\\) \u00d7 784
\n= 22 \u00d7 112 = 2464 sq.m<\/p>\n

\"AP<\/p>\n

Question 12.
\nA water fountain is in circular shaped whose radius is 10 m. Its inner portion of radius 3m is arranged for fountain and remaining part is cemented. Find the area of that cemented part and total cost of cementing if cost of cementing is \u20b9 200 per sq.m.
\nAnswer:
\nRadius for total water fountain (R) = 10m
\nRadius of fountain arranged portion (r) = 3 m
\nArea of cemented part = Area of total water fountain – Area of fountain arranged portion
\n= \u03c0R2<\/sup> – \u03c0r2<\/sup>
\n\"AP
\n= \\(\\frac{22}{7}\\) \u00d7 (10)2<\/sup> – \\(\\frac{22}{7}\\) \u00d7 (3)2<\/sup>
\n= \\(\\frac{22}{7}\\) [(10)2<\/sup> – (3)2<\/sup>)
\n= \\(\\frac{22}{7}\\) (100 – 9) sq.m.
\n= \\(\\frac{22}{7}\\) \u00d7 91 sq.m.
\n= 22 \u00d7 13
\n= 286 sq.m
\nGiven the cost of cementing per sq.m = \u20b9 200
\n\u2234 Total cost of cementing = 286 \u00d7 200 = \u20b9 57,200<\/p>\n

Practice Questions [Page No: 96]<\/span><\/p>\n

In each question below, you are given a figure (X) followed by four figures (a), (b), (c) and (d) such that (X) is embedded in one of them. Trace out the correct alternative.<\/p>\n

Question 1.
\n\"AP
\nAnswer:
\na<\/p>\n

Question 2.
\n\"AP
\nAnswer:
\nc<\/p>\n

Question 3.
\n\"AP
\nAnswer:
\nb<\/p>\n

\"AP<\/p>\n

Question 4.
\n\"AP
\nAnswer:
\na<\/p>\n

Question 5.
\n\"AP
\nAnswer:
\nc<\/p>\n

Question 6.
\n\"AP
\nAnswer:
\na<\/p>\n

Question 7.
\n\"AP
\nAnswer:
\nb<\/p>\n

Question 8.
\n\"AP
\nAnswer:
\nb<\/p>\n

\"AP<\/p>\n

Question 9.
\n\"AP
\nAnswer:
\nd<\/p>\n

Question 10.
\n\"AP
\nAnswer:
\nb<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf Chapter 11 Area of Plane Figures InText Questions and Answers. AP State Syllabus 7th Class Maths Solutions 11th Lesson Area of Plane Figures InText Questions Check Your Progress [Page No: 84] Question 1. Fill the missing values in the following table. Answer: Question 2. Anu has 4 right angled …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22784"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=22784"}],"version-history":[{"count":3,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22784\/revisions"}],"predecessor-version":[{"id":22838,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22784\/revisions\/22838"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=22784"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=22784"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=22784"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}