{"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

\n","protected":false},"excerpt":{"rendered":"

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_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[116,475,14],"tags":[246,173,247],"class_list":["post-2020","post","type-post","status-publish","format-standard","hentry","category-7th-class-ncert-syllabus","category-posts-in-english","category-mathematics","tag-7th-std-properties-of-triangeles","tag-7th-std-science","tag-7th-std-science-properties-of-triangel"],"cp_meta_data":{"_edit_lock":["1629956097:2"],"_last_editor_used_jetpack":["block-editor"],"_jetpack_related_posts_cache":["a:1:{s:32:\"37550b67d263a3ce789993dc25046c5f\";a:2:{s:7:\"expires\";i:1782924476;s:7:\"payload\";a:6:{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;}i:3;a:1:{s:2:\"id\";i:1899;}i:4;a:1:{s:2:\"id\";i:1490;}i:5;a:1:{s:2:\"id\";i:1130;}}}}"],"_edit_last":["2"]},"jetpack_featured_media_url":"","jetpack-related-posts":[{"id":2779,"url":"https:\/\/themindpalace.in\/index.php\/2021\/01\/27\/allotropy-of-oxygen\/","url_meta":{"origin":2020,"position":0},"title":"ALLOTROPY of OXYGEN","author":"Nancy Diana","date":"January 27, 2021","format":false,"excerpt":"Oxygen is critical to human life and that going without it for more than just a couple of minutes is incompatible with maintaining that life.","rel":"","context":"In "7th Class"","block_context":{"text":"7th Class","link":"https:\/\/themindpalace.in\/index.php\/category\/ncert-school-syllabus\/7th-class-ncert-syllabus\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/01\/Slide1-15.jpg?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/01\/Slide1-15.jpg?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/01\/Slide1-15.jpg?resize=525%2C300&ssl=1 1.5x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2021\/01\/Slide1-15.jpg?resize=700%2C400&ssl=1 2x"},"classes":[]},{"id":1190,"url":"https:\/\/themindpalace.in\/index.php\/2020\/10\/13\/fibre-to-fabric\/","url_meta":{"origin":2020,"position":1},"title":"FIBRE TO FABRIC","author":"Nancy Diana","date":"October 13, 2020","format":false,"excerpt":"Summary of fibre to fabric Summary Clothing is very important to human beings. The importance of clothes are They protect from heatThey protect from coldThey make us look nice and decent. FIBRE A material which is available in the form of thin and continuous strand is called fibre. There are\u2026","rel":"","context":"In "7th Class"","block_context":{"text":"7th Class","link":"https:\/\/themindpalace.in\/index.php\/category\/ncert-school-syllabus\/7th-class-ncert-syllabus\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide1-9.jpg?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide1-9.jpg?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide1-9.jpg?resize=525%2C300&ssl=1 1.5x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide1-9.jpg?resize=700%2C400&ssl=1 2x"},"classes":[]},{"id":1983,"url":"https:\/\/themindpalace.in\/index.php\/2020\/11\/19\/acidsbases-and-salts\/","url_meta":{"origin":2020,"position":2},"title":"ACIDS,BASES AND SALTS","author":"Nancy Diana","date":"November 19, 2020","format":false,"excerpt":"In our daily life we make use of variety of substances, they are differ in taste and physical state. The substances may be either acids ,bases or salts. ACIDS\u00a0 The word acid comes from the latin word acere which means sour .Examples of acids: vinegar, lemon juice,orange juice, curd,tamarind,amla,black tea,grape\u2026","rel":"","context":"In "7th Class"","block_context":{"text":"7th Class","link":"https:\/\/themindpalace.in\/index.php\/category\/ncert-school-syllabus\/7th-class-ncert-syllabus\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/11\/Slide2-3.jpg?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/11\/Slide2-3.jpg?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/11\/Slide2-3.jpg?resize=525%2C300&ssl=1 1.5x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/11\/Slide2-3.jpg?resize=700%2C400&ssl=1 2x"},"classes":[]},{"id":1899,"url":"https:\/\/themindpalace.in\/index.php\/2020\/11\/16\/pollution\/","url_meta":{"origin":2020,"position":3},"title":"Pollution","author":"Nancy Diana","date":"November 16, 2020","format":false,"excerpt":"Summary of pollution Solved exercise of pollution Summary These human as well as natural activities have been continuously bringing about degradation of the environment. Such changes have spoil the quality of the environment thereby affecting the life of organisms. Any such undesirable change in the air, water or soil which\u2026","rel":"","context":"In "7th Class"","block_context":{"text":"7th Class","link":"https:\/\/themindpalace.in\/index.php\/category\/ncert-school-syllabus\/7th-class-ncert-syllabus\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/11\/Slide1-6.jpg?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/11\/Slide1-6.jpg?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/11\/Slide1-6.jpg?resize=525%2C300&ssl=1 1.5x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/11\/Slide1-6.jpg?resize=700%2C400&ssl=1 2x"},"classes":[]},{"id":1490,"url":"https:\/\/themindpalace.in\/index.php\/2020\/10\/29\/energy\/","url_meta":{"origin":2020,"position":4},"title":"Energy","author":"Nancy Diana","date":"October 29, 2020","format":false,"excerpt":"Summary of energy Summary Is the capacity of a body to do work. Energy is measured by the work that the body can do. Energy exists in many forms such as heat, light, sound, wind, electricity, chemical energy, solar energy, magnetic energy and nuclear energy etc. Energy exists prominently in\u2026","rel":"","context":"In "7th Class"","block_context":{"text":"7th Class","link":"https:\/\/themindpalace.in\/index.php\/category\/ncert-school-syllabus\/7th-class-ncert-syllabus\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide2-13.jpg?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide2-13.jpg?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide2-13.jpg?resize=525%2C300&ssl=1 1.5x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide2-13.jpg?resize=700%2C400&ssl=1 2x"},"classes":[]},{"id":1130,"url":"https:\/\/themindpalace.in\/index.php\/2020\/10\/08\/sorting-materials-into-groups\/","url_meta":{"origin":2020,"position":5},"title":"Sorting Materials into Groups","author":"Nancy Diana","date":"October 8, 2020","format":false,"excerpt":"Summary of sorting materials Solved exercise of sorting materials OBJECTS AROUND US Summary There is such a vast variety of objects everywhere. We see around us, a chair, a bullock cart, a cycle, cooking utensils, books, clothes, toys, water, stones and many other objects. All objects around us are made\u2026","rel":"","context":"In "6th Class"","block_context":{"text":"6th Class","link":"https:\/\/themindpalace.in\/index.php\/category\/ncert-school-syllabus\/6th-class-ncert-syllabus\/"},"img":{"alt_text":"","src":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide1-4.jpg?resize=350%2C200&ssl=1","width":350,"height":200,"srcset":"https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide1-4.jpg?resize=350%2C200&ssl=1 1x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide1-4.jpg?resize=525%2C300&ssl=1 1.5x, https:\/\/i0.wp.com\/themindpalace.in\/wp-content\/uploads\/2020\/10\/Slide1-4.jpg?resize=700%2C400&ssl=1 2x"},"classes":[]}],"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/2020","targetHints":{"allow":["GET"]}}],"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}]}}