{"id":2020,"date":"2020-11-30T14:44:59","date_gmt":"2020-11-30T14:44:59","guid":{"rendered":"https:\/\/themindpalace.in\/?p=2020"},"modified":"2021-08-26T05:34:57","modified_gmt":"2021-08-26T05:34:57","slug":"the-triangle-and-its-properties","status":"publish","type":"post","link":"https:\/\/themindpalace.in\/index.php\/2020\/11\/30\/the-triangle-and-its-properties\/","title":{"rendered":"The Triangle and its Properties"},"content":{"rendered":"\n

Solved exercise on triangle and its properties<\/a><\/p>\n\n\n\n

<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Exercise<\/h1>\n\n\n\n

1. In \u2206PQR, D is the mid-pointof\u00a0QR
PM\u00a0is<\/span>
PD\u00a0is<\/p>\n\n\n\n

 If QM = MR?<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution:<\/span>
PM is altitude.
PD is median.
No, QM \u2260 MR.<\/p>\n\n\n\n

\n\n\n\n

2. Draw rough sketches for the following<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

(a) In \u2206ABC, BE is a median.
(b) In \u2206PQR, PQ and PR are altitudes of the triangle.
(c) In \u2206XYZ, YL is an altitude in the exterior of the triangle.<\/p>\n\n\n\n

\n\n\n\n

3. Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

\u2206ABC is an isosceles triangle in which AB = AC
 Draw AD as the median of the triangle.
Measure the angle ADC with the help of protractor we find, \u2220ADC = 90\u00b0
Thus, AD is the median as well as the altitude of the \u2206ABC. Hence Verified.<\/p>\n\n\n\n

\n\n\n\n

Exercise 6.2<\/span><\/strong><\/p>\n\n\n\n

1. Find the value of the unknown exterior angle x in the following diagrams<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution<\/span><\/strong><\/p>\n\n\n\n

(i)\u2220x = 50\u00b0 + 70\u00b0<\/p>\n\n\n\n

            = 120\u00b0<\/p>\n\n\n\n

(Exterior angle is equal to sum of its interior opposite angles)<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

(ii) \u2220x = 65\u00b0+ 45\u00b0<\/p>\n\n\n\n

          = 110\u00b0<\/p>\n\n\n\n

(Exterior angle is equal to sum of its interior opposite angles)<\/p>\n\n\n\n

\n\n\n\n

(iii) \u2220x = 30\u00b0 + 40\u00b0<\/p>\n\n\n\n

           = 70\u00b0<\/p>\n\n\n\n

(Exterior angle is equal to sum of its interior opposite angles)<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n


<\/p>\n\n\n\n

(iv) \u2220x = 60\u00b0 + 60\u00b0<\/p>\n\n\n\n

           = 120\u00b0<\/p>\n\n\n\n

(Exterior angle is equal to sum of its interior opposite angles)<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

(v) \u2220x = 50\u00b0 + 50\u00b0<\/p>\n\n\n\n

          =100\u00b0<\/p>\n\n\n\n

(Exterior angle is equal to sum of its interior opposite angles)<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

(vi) \u2220x = 30\u00b0 + 60\u00b0<\/p>\n\n\n\n

           = 90\u00b0<\/p>\n\n\n\n

 (Exterior angle is equal to sum of its interior opposite angle)<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

2. Find the value of the unknown interior angle x in the following figures:<\/span><\/strong><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

(i) \u2220x + 50\u00b0 = 115\u00b0 (Exterior angle of a triangle)
\u2234 \u2220x = 115\u00b0- 50\u00b0 = 65\u00b0

(ii) \u2220x + 70\u00b0 = 110\u00b0 (Exterior angle of a triangle)
\u2234 \u2220x = 110\u00b0 \u2013 70\u00b0 = 40\u00b0<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

(iii) \u2220 x + 90\u00b0 = 125\u00b0 (Exterior angle of a right triangle)
\u2234 \u2220x = 125\u00b0 \u2013 90\u00b0 = 35\u00b0<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

(iv) \u2220x + 60\u00b0 = 120\u00b0 (Exterior angle of a triangle)
\u2234 \u2220x = 120\u00b0 \u2013 60\u00b0 = 60\u00b0<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

(v)\u2220 X + 30\u00b0 = 80\u00b0 (Exterior angle of a triangle)
\u2234 \u2220x = 80\u00b0 \u2013 30\u00b0 = 50\u00b0<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

(vi) \u2220 x + 35\u00b0 = 75\u00b0 (Exterior angle of a triangle)
\u2234 \u2220 x = 75\u00b0 \u2013 35\u00b0 = 40\u00b0<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

Exercise 6.3<\/span><\/strong><\/p>\n\n\n\n

1. Find the value of the unknown x in the following diagrams<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution:<\/span><\/strong>i) By angle sum property of a triangle, we have
\u2220x + 50\u00b0 + 60\u00b0 = 180\u00b0
\u21d2 \u2220x + 110\u00b0 = 180\u00b0
\u2234 \u2220x = 180\u00b0 \u2013 110\u00b0 = 70\u00b0<\/p>\n\n\n\n

\n\n\n\n

(ii) By angle sum property of a triangle, we have
\u2220x + 90\u00b0 + 30 = 180\u00b0 [\u2206 is right angled triangle]
\u21d2 \u2220x + 120\u00b0 = 180\u00b0
\u2234 \u2220x \u2013 180\u00b0 \u2013 120\u00b0 = 60\u00b0<\/p>\n\n\n\n

\n\n\n\n

(iii) By angle sum property of a triangle, we have
\u2220x + 30\u00b0 + 110\u00b0 \u2013 180\u00b0
\u21d2 \u2220x + 140\u00b0 = 180\u00b0
\u2234 \u2220x = 180\u00b0 \u2013 140\u00b0 = 40\u00b0<\/p>\n\n\n\n

\n\n\n\n

Find the value of the unknown x in the following diagrams<\/span><\/strong><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n


Solution<\/span>:(iv ) By angle sum property of a triangle, we have
\u2220x + \u2220x + 50\u00b0 = 180\u00b0
\u21d2 2x + 50\u00b0 = 180\u00b0
\u21d2 2x = 180\u00b0 \u2013 50\u00b0
\u21d2 2x = 130\u00b0
\u2234 x=130\/2=65\u2218
(v) By angle sum property of a triangle, we have
\u2220x + \u2220x +\u2220x =180\u00b0
\u21d2 3 \u2220x = 180\u00b0
\u2234 \u2220x=180\/3=60\u2218<\/p>\n\n\n\n

\n\n\n\n

(vi) By angle sum property of a triangle, we have
x + 2 x + 90\u00b0 = 180\u00b0 (\u2206 is right angled triangle)
\u21d2 3x + 90\u00b0 = 180\u00b0
\u21d2 3x = 180\u00b0 \u2013 90\u00b0
\u21d2 3x = 90\u00b0
\u2234 x=90\/3=30\u2218<\/p>\n\n\n\n

\n\n\n\n

2. Find the values of the unknowns x and y in the following diagrams:<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n


(i) \u2220x + 50\u00b0 = 120\u00b0 (Exterior angle of a triangle)
\u2234 \u2220x = 120\u00b0- 50\u00b0 = 70\u00b0
\u2220x + \u2220y + 50\u00b0 = 180\u00b0 (Angle sum property of a triangle)
70\u00b0 + \u2220y + 50\u00b0 = 180\u00b0
\u2220y + 120\u00b0 = 180\u00b0
\u2220y = 180\u00b0 \u2013 120\u00b0
\u2234 \u2220y = 60\u00b0
Thus \u2220x = 70 and \u2220y \u2013 60\u00b0<\/p>\n\n\n\n

\n\n\n\n

(ii) \u2220y = 80\u00b0 (Vertically opposite angles are same)
\u2220x + \u2220y + 50\u00b0 = 180\u00b0 (Angle sum property of a triangle)
\u21d2 \u2220x + 80\u00b0 + 50\u00b0 = 180\u00b0
\u21d2 \u2220x + 130\u00b0 = 180\u00b0
\u2234 \u2220x = 180\u00b0 \u2013 130\u00b0 = 50\u00b0
Thus, \u2220x = 50\u00b0 and \u2220y = 80\u00b0<\/p>\n\n\n\n

\n\n\n\n

Find the values of the unknowns x and y in the following diagrams:<\/span><\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n


(iii) \u2220y + 50\u00b0 + 60\u00b0 = 180\u00b0 (Angle sum property of a triangle)
\u2220y + 110\u00b0 = 180\u00b0
\u2234 \u2220y = 180\u00b0- 110\u00b0 = 70\u00b0
\u2220x + \u2220y = 180\u00b0 (Linear pairs)
\u21d2 \u2220x + 70\u00b0 = 180\u00b0
\u2234 \u2220x = 180\u00b0 \u2013 70\u00b0 = 110\u00b0
Thus, \u2220x = 110\u00b0 and y = 70\u00b0<\/p>\n\n\n\n

\n\n\n\n

(iv) \u2220x = 60\u00b0 (Vertically opposite angles)
\u2220x + \u2220y + 30\u00b0 = 180\u00b0 (Angle sum property of a triangle)
\u21d2 60\u00b0 + \u2220y + 30\u00b0 = 180\u00b0
\u21d2 \u2220y + 90\u00b0 = 180\u00b0
\u21d2 \u2220y = 180\u00b0 \u2013 90\u00b0 = 90\u00b0
Thus, \u2220x = 60\u00b0 and \u2220y = 90\u00b0<\/p>\n\n\n\n

\n\n\n\n

Find the values of the unknowns x and y in the following diagrams:<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n
\n\n\n\n

(v) \u2220y = 90\u00b0 (Vertically opposite angles)
\u2220x + \u2220x + \u2220y = 180\u00b0 (Angle sum property of a triangle)
\u21d2 2 \u2220x + 90\u00b0 = 180\u00b0
\u21d2 2\u2220x = 180\u00b0 \u2013 90\u00b0
\u21d2 2\u2220x = 90\u00b0
\u2234 \u2220x=90\/2=45\u2218
Thus, \u2220x = 45\u00b0 and \u2220y = 90\u00b0<\/p>\n\n\n\n

\n\n\n\n

(vi) From the given figure, we have
 
Adding both sides, we have
\u2220y + \u22201 + \u22202 = 3\u2220x
\u21d2 180\u00b0 = 3\u2220x (Angle sum property of a triangle)
\u2234 \u2220x=180\/ 3=60\u2218
\u2220x = 60\u00b0, \u2220y = 60\u00b0<\/p>\n\n\n\n

\n\n\n\n

Exercise 6.4 & 6.5<\/strong><\/span><\/p>\n\n\n\n

1. Is it possible to have a triangle with the following sides?<\/span><\/p>\n\n\n\n

(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm<\/p>\n\n\n\n

Solution:We know that for a triangle, the sum of any two sides must be greater than the third side.<\/p>\n\n\n\n

(i) Given sides are 2 cm, 3 cm, 5 cm
Sum of the two sides = 2 cm + 3 cm = 5 cm Third side = 5 cm
We have, Sum of any two sides = the third side i.e. 5 cm = 5 cm
Hence, the triangle is not possible.<\/p>\n\n\n\n

\n\n\n\n

(ii) Given sides are 3 cm, 6 cm, 7 cm
Sum of the two sides = 3 cm + 6 cm = 9 cm Third side = 7 cm
We have sum of any two sides > the third
side. i.e. 9 cm > 7 cm
Hence, the triangle is possible.<\/p>\n\n\n\n

\n\n\n\n

(iii) Given sides are 6 cm, 3 cm, 2 cm
Sum of the two sides = 3 cm + 2 cm \u2013 5 cm Third side = 6 cm
We have, sum of any two sides < the third sid6 i.e. 5 cm < 6 cm Hence, the triangle is not possible.<\/p>\n\n\n\n

\n\n\n\n

2. Take any point O in the interior of a triangle PQR . Is<\/span><\/p>\n\n\n\n

i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution<\/span>:(i) Yes, In \u2206 OPQ, we have
OP + OQ > PQ
[Sum of any two sides of a triangle is greater than the third side]<\/p>\n\n\n\n

\n\n\n\n

(ii) Yes, In \u2206OQR, we have OQ + OR > QR
[Sum of any two sides of a triangle is greater than the third side]<\/p>\n\n\n\n

\n\n\n\n

(iii) Yes, In \u2206OPR, we have OR + OP > RP
[Sum of any two sides of a triangle is greater than the third side]<\/p>\n\n\n\n

\n\n\n\n

3. AM is a median of a triangle ABC.<\/span>
Is AB + BC + CA > 2AM ?
(Consider the sides of triangles \u2206ABM and \u2206AMC)<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n


Solution:<\/span><\/strong>Yes, In \u2206 ABM, we have<\/p>\n\n\n\n

AB + BM > AM \u2026(i)
[Sum of any two sides of a triangle is greater than the third side]
In \u2206AMC, we have
AC + CM > AM \u2026(ii)
[Sum of any two sides of a triangle is greater than the third side]
Adding eq (i) and (ii), we have
AB + AC + BM + CM > 2AM
AB + AC + BC + > 2AM
AB + BC + CA > 2AM
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

4. ABCD is a quadrilateral.
Is AB + BC + CD + DA > AC + BD?<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution:<\/span><\/strong><\/p>\n\n\n\n

In \u2206ABC, we have
AB + BC > AC \u2026(i)
[Sum of any two sides is greater than the third side]
In \u2206BDC, we have
BC + CD > BD \u2026(ii)
[Sum of any two sides is greater than the third side]
In \u2206ADC, we have
CD + DA > AC \u2026(iii)
[Sum of any two sides is greater than the third side]
In \u2206DAB, we have
DA + AB > BD \u2026(iv)
[Sum of any two sides is greater than the third side]
Adding eq. (i), (ii), (iii) and (iv), we get
2AB + 2BC + 2CD + 2DA > 2AC + 2BD or<\/p>\n\n\n\n

AB + BC + CD + DA > AC + BD    [Dividing both sides by 2]
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

5. ABCD is a quadrilateral.<\/span>
Is AB + BC + CD + DA < 2(AC + BD)?<\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution:<\/span>we have a quadrilateral ABCD.<\/p>\n\n\n\n

In \u2206AOB, we have AB < AO + BO \u2026(i)
[Any side of a triangle is less than the sum of other two sides]
In \u2206BOC, we have
BC < BO + CO \u2026(ii)
[Any side of a quadrilateral is less than the sum of other two sides]
In \u2206COD, we have
CD < CO + DO \u2026(iii)
[Any side of a triangle is less than the sum of other two sides]<\/p>\n\n\n\n

In \u2206AOD, we have
DA < DO + AO \u2026(iv)
[Any side of a triangle is less than the sum of other two sides]<\/p>\n\n\n\n

Adding eq. (i), (ii), (iii) and (iv), we have
AB + BC + CD + DA
\u22202AO + \u2220BO + \u2220CO + \u2220DO
\u22202(AO + BO + CO + DO)
\u22202 [(AO + CO) + (BO + DO)]
\u22202(AC + BD)
Thus, AB + BC + CD + DA < 2(AC + BD)
Hence, proved.<\/p>\n\n\n\n

\n\n\n\n

6. The length of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?<\/span><\/p>\n\n\n\n

Solution:<\/span>Sum of two sides
= 12 cm + 15 cm = 27 cm
Difference of the two sides
= 15 cm \u2013 12 cm = 3 cm
\u2234 The measure of third side should fall between 3 cm and 27 cm.<\/p>\n\n\n\n

\n\n\n\n


Exercise 6.5<\/span><\/strong><\/p>\n\n\n\n

1. PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution<\/span>:In right angled triangle PQR, we have
QR2<\/sup> = PQ2<\/sup> + PR2<\/sup> From Pythagoras property)
= (10)2<\/sup> + (24)2<\/sup>
= 100 + 576 = 676
\u2234 QR = \u221a676 = 26 cm
The, the required length of QR = 26 cm.<\/p>\n\n\n\n

\n\n\n\n

2. ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n

Solution:<\/span>In right angled \u2206 ABC, we have
BC2<\/sup> + (7)2<\/sup> = (25)2<\/sup> (By Pythagoras property)
\u21d2 BC2<\/sup> + 49 = 625
\u21d2 BC2<\/sup> = 625 \u2013 49
\u21d2 BC2<\/sup> = 576<\/p>\n\n\n\n

\u2234 BC = \u221a576 = 24 cm
Thus, the required length of BC = 24 cm.<\/p>\n\n\n\n

\n\n\n\n

3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.<\/span><\/p>\n\n\n\n

\"\"<\/figure><\/div>\n\n\n\n


Solution:<\/span>Here, the ladder forms a right angled triangle.
\u2234 a2<\/sup> + (12)2<\/sup> = (15)2<\/sup> (By Pythagoras property)
\u21d2 a2<\/sup>+ 144 = 225
\u21d2 a2 = 225 \u2013 144
\u21d2 a2<\/sup> = 81<\/p>\n\n\n\n

\u2234 a = \u221a 81  = 9 m
Thus, the distance of the foot from the ladder = 9m<\/p>\n\n\n\n

\n\n\n\n

4. Which of the following can be the sides of a right triangle?<\/span>
(i) 2.5 cm, 6.5 cm, 6 cm.
(ii) 2 cm, 2 cm, 5 cm.
(iii) 1.5 cm, 2 cm, 2.5 cm<\/p>\n\n\n\n

Solution:<\/span>i) Given sides are 2.5 cm, 6.5 cm, 6 cm.
Square of the longer side = (6.5)2<\/sup> = 42.25 cm.
Sum of the square of other two sides
= (2.5)2<\/sup> + (6)2<\/sup> = 6.25 + 36
= 42.25 cm.
Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.
\u2234 The given sides form a right triangle.<\/p>\n\n\n\n

\n\n\n\n

(ii) Given sides are 2 cm, 2 cm, 5 cm .
Square of the longer side = (5)2<\/sup> = 25 cm Sum of the square of other two sides
= (2)2<\/sup> + (2)2<\/sup> =4 + 4 = 8 cm
Since 25 cm \u2260 8 cm
\u2234 The given sides do not form a right triangle.<\/p>\n\n\n\n

\n\n\n\n

(iii) Given sides are 1.5 cm, 2 cm, 2.5 cm
Square of the longer side = (2.5)2<\/sup> = 6.25 cm Sum of the square of other two sides
= (1.5)2<\/sup> + (2)2<\/sup> = 2.25 + 4
Since 6.25 cm = 6.25 cm = 6.25 cm
Since the square of longer side in a triangle is equal to the sum of square of other two sides.
\u2234 The given sides form a right triangle.<\/p>\n\n\n\n

\n\n\n\n

5. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree . Find the original height of the tree.
<\/p>\n\n\n\n


<\/p>\n\n\n\n

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Solved exercise on triangle and its properties Exercise 1. In \u2206PQR, D is the mid-pointof\u00a0QRPM\u00a0isPD\u00a0is  If QM = MR? Solution:PM is altitude.PD is median.No, QM \u2260 MR. 2. Draw rough sketches for the following...<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[116,475,14],"tags":[246,173,247],"cp_meta_data":{"_edit_lock":["1629956097:2"],"_last_editor_used_jetpack":["block-editor"],"_jetpack_related_posts_cache":["a:1:{s:32:\"8f6677c9d6b0f903e98ad32ec61f8deb\";a:2:{s:7:\"expires\";i:1774994903;s:7:\"payload\";a:3:{i:0;a:1:{s:2:\"id\";i:2779;}i:1;a:1:{s:2:\"id\";i:1190;}i:2;a:1:{s:2:\"id\";i:1983;}}}}"],"_edit_last":["2"]},"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","jetpack-related-posts":[],"_links":{"self":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/2020"}],"collection":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/comments?post=2020"}],"version-history":[{"count":8,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/2020\/revisions"}],"predecessor-version":[{"id":2509,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/2020\/revisions\/2509"}],"wp:attachment":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/media?parent=2020"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/categories?post=2020"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/tags?post=2020"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}