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In fig the vertices of square DEFG are on the sides of triangle ABC. angle A = 90°. Then prove that
In figure 1.81, the vertices of square DEFG are on the sides of triangle ABC., angle A = 90°., Then prove that DE2 = BD ´ EC
In figure 1.81, the vertices of square DEFG are on the sides of triangle ABC., angle A = 90°., Then prove that DE square = BD ´ EC,
(Hint : Show that D GBD is similar to D CFE. Use GD = FE = DE.)
#Kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
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In fig XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC3:17In fig XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC
In fig XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC
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In fig XY || seg AC., If 2AX = 3BX and XY = 9., Complete the activity to find the value of AC, In fig 1.80, XY parallel seg AC., If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC #Kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
In Quadrilateral ABCD, seg AD parallel seg BC. Diagonal AC and diagonal BD intersect each other in4:20In Quadrilateral ABCD, seg AD parallel seg BC. Diagonal AC and diagonal BD intersect each other in
In Quadrilateral ABCD, seg AD parallel seg BC. Diagonal AC and diagonal BD intersect each other in
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In Quadrilateral ABCD, seg AD || seg BC., Diagonal AC and diagonal BD intersect each other in point P., Then show that (AP / PD) = (PC / PB) In Quadrilateral ABCD, seg AD parallel seg BC., Diagonal AC and diagonal BD intersect each other in point P., Then show that (AP / PD) = (PC / PB) #Kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
bisectors of angle B and angle C of triangle ABC intersect each other inpoint X. Line AX intersects8:37bisectors of angle B and angle C of triangle ABC intersect each other inpoint X. Line AX intersects
bisectors of angle B and angle C of triangle ABC intersect each other inpoint X. Line AX intersects
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In fig 1.78, bisectors of angle B and angle C of triangle ABC intersect each other inpoint X., Line AX intersects side BC in point Y., AB = 5, AC = 4, BC = 6, then find AX upon XY, In fig 1.78, bisectors of angle B and angle C of triangle ABC intersect each other inpoint X., Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6, then find AX /XY #Kanojesir #maharashtrastateboard #studen...
In triangle PQR seg PM is a median. Angle bisectors of angle PMQ and angle PMR intersect side PQ2:53In triangle PQR seg PM is a median. Angle bisectors of angle PMQ and angle PMR intersect side PQ
In triangle PQR seg PM is a median. Angle bisectors of angle PMQ and angle PMR intersect side PQ
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In triangle PQR seg PM is a median. Angle bisectors of angle PMQ and angle PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. #Kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
In the figure seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70,10:44In the figure seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70,
In the figure seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70,
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In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR and RS. #kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
Chhatrapati Shivaji Maharaj Bhashan in Marathi | शिव जयंती भाषण | shivaji jayanti Bhashan in Marathi5:27Chhatrapati Shivaji Maharaj Bhashan in Marathi | शिव जयंती भाषण | shivaji jayanti Bhashan in Marathi
Chhatrapati Shivaji Maharaj Bhashan in Marathi | शिव जयंती भाषण | shivaji jayanti Bhashan in Marathi
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Chhatrapati Shivaji Maharaj Bhashan in Marathi | शिव जयंती भाषण | shivaji jayanti Bhashan in Marathi shivaji maharaj speech in marathi, shivaji maharaj bhashan in marathi, shivaji maharaj bhashan, chhatrapati shivaji maharaj bhashan marathi, chhatrapati shivaji maharaj bhashan, shivaji maharaj,shivaji maharaj bhashan marathi, shivaji maharaj jayanti bhashan, shiv jayanti bhashan marathi, shivaj...
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.4:44In 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.
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.
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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. #kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
triangle MNT ~ triangle QRS. Length of altitude drawn from point T is 5 and length of altitude drawn3:42triangle MNT ~ triangle QRS. Length of altitude drawn from point T is 5 and length of altitude drawn
triangle MNT ~ triangle QRS. Length of altitude drawn from point T is 5 and length of altitude drawn
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triangle MNT ~ triangle QRS., Length of altitude drawn from point T is 5, and length of altitude drawn from point S is 9., Find the ratio area of triangle MNT upon area of triangle QRS, #Kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths D MNT ~ D QRS., Length of altitude drawn from point T is 5, and length of altitude drawn from point S is 9., Find the ratio A( D MNT)...
In figure PM = 10 cm area of triangle PQS = 100 sq.cm area of triangle QRS = 110 sq.cm then find NR.3:47In figure PM = 10 cm area of triangle PQS = 100 sq.cm area of triangle QRS = 110 sq.cm then find NR.
In figure PM = 10 cm area of triangle PQS = 100 sq.cm area of triangle QRS = 110 sq.cm then find NR.
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In figure PM = 10 cm area of triangle PQS = 100 sq.cm, area of triangle QRS = 110 sq.cm, then find NR., In figure 1.74, PM = 10 cm A(DPQS) = 100 sq.cm, A(DQRS) = 110 sq.cm then find NR., #kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
In figure angle ABC = angle DCB = 90° AB = 6, DC = 8 then area of triangle ABC upon area of triangle2:33In figure angle ABC = angle DCB = 90° AB = 6, DC = 8 then area of triangle ABC upon area of triangle
In figure angle ABC = angle DCB = 90° AB = 6, DC = 8 then area of triangle ABC upon area of triangle
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In figure angle ABC = angle DCB = 90° AB = 6, DC = 8 then area of triangle ABC upon area of triangle DCB, In figure 1.73, angle ABC = angle DCB = 90° AB = 6, DC = 8 then area triangle ABC) / A( triangle DCB) = ?, #class10thmaths #Kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm3:08Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm
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Ratio of areas of two triangles with equal heights is 2 : 3., If base of the smaller triangle is 6 cm, then what is the corresponding base of the bigger triangle ?, #Kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
In D ABC, B - D - C and BD = 7, BC = 20 then find following ratios.5:27In D ABC, B - D - C and BD = 7, BC = 20 then find following ratios.
In D ABC, B - D - C and BD = 7, BC = 20 then find following ratios.
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In D ABC, B - D - C and BD = 7, BC = 20 then find following ratios. #kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths
A A test for similarity of triangles | SAS test of similarity of triangles | SSS test for similarity6:35A A test for similarity of triangles | SAS test of similarity of triangles | SSS test for similarity
A A test for similarity of triangles | SAS test of similarity of triangles | SSS test for similarity
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#Kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths SSS test for similarity of triangles A A test for similarity of triangles SAS test of similarity of triangles
Proof of AAA test | AAA test for similarity of triangles | Theorem AAA test15:10Proof of AAA test | AAA test for similarity of triangles | Theorem AAA test
Proof of AAA test | AAA test for similarity of triangles | Theorem AAA test
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Proof of AAA test | AAA test for similarity of triangles | Theorem AAA test #Kanojesir #maharashtrastateboard #studentpointacademy #maths #class10thmaths similarity of triangles, aaa test of similarity, triangles,aa test of similarity, similar triangles,ncert aaa test for similarity of triangles, tests for similarity of triangles, aa test of similarity between two triangles proof, test of simil...
In figure seg PQ || seg DE, area of triangle PQF = 20 units, PF = 2 DP, then find area of DPQE4:27In figure seg PQ || seg DE, area of triangle PQF = 20 units, PF = 2 DP, then find area of DPQE
In figure seg PQ || seg DE, area of triangle PQF = 20 units, PF = 2 DP, then find area of DPQE
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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.2:03ABC and DEF are equilateral triangles. If area of triangles ABC : DEF = 1:2 and AB = 4, find DE.
ABC and DEF are equilateral triangles. If area of triangles ABC : DEF = 1:2 and AB = 4, find DE.
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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 cm3:51Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm
Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm
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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.2:56triangle LMN ~ triangle PQR, 9 x area of triangle PQR = 16 area of LMN. If QR = 20 then find MN.
triangle LMN ~ triangle PQR, 9 x area of triangle PQR = 16 area of LMN. If QR = 20 then find MN.
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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.2:24If triangle ABC ~ triangle PQR, area of triangle ABC = 80, A (D PQR) = 125, then fill in the blanks.
If triangle ABC ~ triangle PQR, area of triangle ABC = 80, A (D PQR) = 125, then fill in the blanks.
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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 .1:54The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas .
The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas .
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The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas .
Theorem of areas of similar triangles Theorem11:50Theorem of areas of similar triangles Theorem
Theorem of areas of similar triangles Theorem
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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 that6:12In the figure, in triangle ABC, point D on side BC is such that, angle BAC = angle ADC. Prove that
In the figure, in triangle ABC, point D on side BC is such that, angle BAC = angle ADC. Prove that
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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 that2:39In the figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP. Prove that
In the figure, seg AC and seg BD intersect each other in point P and AP/CP = BP/DP. Prove that
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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 that5:30ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that
ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that
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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 = 64:27In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6
In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6
Переглядів 414 місяці тому
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 = 5PQ7:18In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ
In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ
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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 that5:53In triangle ABC, AP perpendicular BC, BQ perpendicular AC, B- P-C, A-Q - C then prove that
In triangle ABC, AP perpendicular BC, BQ perpendicular AC, B- P-C, A-Q - C then prove that
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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 ground3:50As shown in figure , two poles of height 8 m and 4 m are perpendicular to the ground
As shown in figure , two poles of height 8 m and 4 m are perpendicular to the ground
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As shown in figure , two poles of height 8 m and 4 m are perpendicular to the ground
Are the triangles in figure similar? If yes, by which test ?3:01Are the triangles in figure similar? If yes, by which test ?
Are the triangles in figure similar? If yes, by which test ?
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Are the triangles in figure similar? If yes, by which test ?

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