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0:42:08
Lesson 1 What is History ? | EVS 2 | Chapter 1 What is History ? | EVS part 2
0:11:51
In fig the vertices of square DEFG are on the sides of triangle ABC. angle A = 90°. Then prove that
0:03:17
In fig XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC
0:08:37
bisectors of angle B and angle C of triangle ABC intersect each other inpoint X. Line AX intersects
0:02:53
In triangle PQR seg PM is a median. Angle bisectors of angle PMQ and angle PMR intersect side PQ
0:10:44
In the figure seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70,
0:04:44
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.
0:03:42
triangle MNT ~ triangle QRS. Length of altitude drawn from point T is 5 and length of altitude drawn
0:03:47
In figure PM = 10 cm area of triangle PQS = 100 sq.cm area of triangle QRS = 110 sq.cm then find NR.
0:02:33
In figure angle ABC = angle DCB = 90° AB = 6, DC = 8 then area of triangle ABC upon area of triangle
0:03:08
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm
0:05:27
In D ABC, B - D – C and BD = 7, BC = 20 then find following ratios.
0:06:35
A A test for similarity of triangles | SAS test of similarity of triangles | SSS test for similarity
0:15:10
Proof of AAA test | AAA test for similarity of triangles | Theorem AAA test
0:04:27
In figure seg PQ || seg DE, area of triangle PQF = 20 units, PF = 2 DP, then find area of DPQE
0:02:56
triangle LMN ~ triangle PQR, 9 x area of triangle PQR = 16 area of LMN. If QR = 20 then find MN.
0:02:24
If triangle ABC ~ triangle PQR, area of triangle ABC = 80, A (D PQR) = 125, then fill in the blanks.
0:01:54
The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas .
0:11:50
Theorem of areas of similar triangles Theorem
0:06:12
In the figure, in triangle ABC, point D on side BC is such that, angle BAC = angle ADC. Prove that
0:02:39
In the figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP. Prove that
0:05:30
ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that
0:04:27
In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6
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