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Lesson 1 What is History ? | EVS 2 | Chapter 1 What is History ? | EVS part 2
42:08
Lesson 1 What is History ? | EVS 2 | Chapter 1 What is History ? | EVS part 2
In fig the vertices of square DEFG are on the sides of triangle ABC. angle A = 90°. Then prove that
11:51
In fig the vertices of square DEFG are on the sides of triangle ABC. angle A = 90°. Then prove that
In fig XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC
3:17
In fig XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC
In Quadrilateral ABCD, seg AD parallel seg BC. Diagonal AC and diagonal BD intersect each other in
4:20
In Quadrilateral ABCD, seg AD parallel seg BC. Diagonal AC and diagonal BD intersect each other in
bisectors of angle B and angle C of triangle ABC intersect each other inpoint X. Line AX intersects
8:37
bisectors of angle B and angle C of triangle ABC intersect each other inpoint X. Line AX intersects
In triangle PQR seg PM is a median. Angle bisectors of angle PMQ and angle PMR intersect side PQ
2:53
In triangle PQR seg PM is a median. Angle bisectors of angle PMQ and angle PMR intersect side PQ
In the figure seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70,
10:44
In the figure seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70,
Chhatrapati Shivaji Maharaj Bhashan in Marathi | शिव जयंती भाषण | shivaji jayanti Bhashan in Marathi
5:27
Chhatrapati Shivaji Maharaj Bhashan in Marathi | शिव जयंती भाषण | shivaji jayanti Bhashan in Marathi
In figure 1.75, A – D – C and B – E – C seg DE || side AB If AD = 5,DC = 3, BC = 6.4 then find BE.
4:44
In figure 1.75, A – D – C and B – E – C seg DE || side AB If AD = 5,DC = 3, BC = 6.4 then find BE.
triangle MNT ~ triangle QRS. Length of altitude drawn from point T is 5 and length of altitude drawn
3:42
triangle MNT ~ triangle QRS. Length of altitude drawn from point T is 5 and length of altitude drawn
In figure PM = 10 cm area of triangle PQS = 100 sq.cm area of triangle QRS = 110 sq.cm then find NR.
3:47
In figure PM = 10 cm area of triangle PQS = 100 sq.cm area of triangle QRS = 110 sq.cm then find NR.
In figure angle ABC = angle DCB = 90° AB = 6, DC = 8 then area of triangle ABC upon area of triangle
2:33
In figure angle ABC = angle DCB = 90° AB = 6, DC = 8 then area of triangle ABC upon area of triangle
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm
3:08
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm
In D ABC, B - D – C and BD = 7, BC = 20 then find following ratios.
5:27
In D ABC, B - D – C and BD = 7, BC = 20 then find following ratios.
A A test for similarity of triangles | SAS test of similarity of triangles | SSS test for similarity
6:35
A A test for similarity of triangles | SAS test of similarity of triangles | SSS test for similarity
Proof of AAA test | AAA test for similarity of triangles | Theorem AAA test
15:10
Proof of AAA test | AAA test for similarity of triangles | Theorem AAA test
In figure seg PQ || seg DE, area of triangle PQF = 20 units, PF = 2 DP, then find area of DPQE
4:27
In figure seg PQ || seg DE, area of triangle PQF = 20 units, PF = 2 DP, then find area of DPQE
ABC and DEF are equilateral triangles. If area of triangles ABC : DEF = 1:2 and AB = 4, find DE.
2:03
ABC and DEF are equilateral triangles. If area of triangles ABC : DEF = 1:2 and AB = 4, find DE.
Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm
3:51
Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm
triangle LMN ~ triangle PQR, 9 x area of triangle PQR = 16 area of LMN. If QR = 20 then find MN.
2:56
triangle LMN ~ triangle PQR, 9 x area of triangle PQR = 16 area of LMN. If QR = 20 then find MN.
If triangle ABC ~ triangle PQR, area of triangle ABC = 80, A (D PQR) = 125, then fill in the blanks.
2:24
If triangle ABC ~ triangle PQR, area of triangle ABC = 80, A (D PQR) = 125, then fill in the blanks.
The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas .
1:54
The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas .
Theorem of areas of similar triangles Theorem
11:50
Theorem of areas of similar triangles Theorem
In the figure, in triangle ABC, point D on side BC is such that, angle BAC = angle ADC. Prove that
6:12
In the figure, in triangle ABC, point D on side BC is such that, angle BAC = angle ADC. Prove that
In the figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP. Prove that
2:39
In the figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP. Prove that
ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that
5:30
ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that
In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6
4:27
In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6
In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ
7:18
In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ
In triangle ABC, AP perpendicular BC, BQ perpendicular AC, B- P-C, A-Q - C then prove that
5:53
In triangle ABC, AP perpendicular BC, BQ perpendicular AC, B- P-C, A-Q - C then prove that
As shown in figure , two poles of height 8 m and 4 m are perpendicular to the ground
3:50
As shown in figure , two poles of height 8 m and 4 m are perpendicular to the ground