(i) In ΔABD and ΔACD AB = AC [∵ Given] BD = CD [∵ Given] AD = AD [∵ Common] Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule] (ii) In ΔABD ≅ ΔACD [∵ Proved above] ∠ BAD = ∠ CAD [∵ CPCT] In ABP and ACP, AB = AC [∵ Given] ∠ BAP = ∠ CAP [∵ Proved above] AP = AP [∵ Common] Hence, ΔABP ≅ ΔACP [∵ SAS CongruencyRead more
(i) In ΔABD and ΔACD
AB = AC [∵ Given]
BD = CD [∵ Given]
AD = AD [∵ Common]
Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule]
(ii) In ΔABD ≅ ΔACD [∵ Proved above]
∠ BAD = ∠ CAD [∵ CPCT]
In ABP and ACP,
AB = AC [∵ Given]
∠ BAP = ∠ CAP [∵ Proved above]
AP = AP [∵ Common]
Hence, ΔABP ≅ ΔACP [∵ SAS Congruency rule]
(iii) In ΔABD ≅ ΔACD [∵ Proved above]
∠ BAD = ∠ CAD [∵ CPCT]
∠ BDA = ∠ CDA [∵ CPCT]
Hence, AP bisects both the angles A and D
(iv) In ΔABP ≅ ΔACP [∵ Prove above]
BP = CP [∵ CPCT]
∠BPA = ∠CPA [∵ CPCT]
∠BPA + ∠CPA = 180° [∵ Linear Pair]
⇒ ∠CPA + ∠CPA = 180° [∵ ∠BPA = ∠CPA]
⇒ 2∠ CPA = 180° ⇒ ∠CPA = (180°/2) = 90°
⇒ AP is perpendicular to BC. ⇒ AP is perpendicular bisector of BC. [∵ BP = CP].
In ΔABD, E is mid-point of AD [∵ Given] and EG ∥ AB [∵ Given] Hence, G is mid-point of BD [∵ Converse of Mid-point Theorem] Similarly, In ΔBCD, G is mid-point of BD [∵ Prove above] and FG ∥ DC [∵ Given] Hence, F is mid-point of BC [∵ Converse of Mid-point Theorem]
In ΔABD,
E is mid-point of AD [∵ Given]
and EG ∥ AB [∵ Given]
Hence, G is mid-point of BD [∵ Converse of Mid-point Theorem]
Similarly,
In ΔBCD,
G is mid-point of BD [∵ Prove above]
and FG ∥ DC [∵ Given]
Hence, F is mid-point of BC [∵ Converse of Mid-point Theorem]
In AECD, AE ∥ DC [∵ Given] AD ∥ CE [∵ By construction] Hence, AECD is a parallelogram. AD = CE ...(1) [∵ Opposite sides of a parallelogram are equal] AD = BC ...(2) [∵ Given] Hence, CE = BC [∵ From the equation (1) and (2)] Therefore, in ΔBCE, ∠3 = ∠4 ...(3) [∵ In a triangle, the angles opposite toRead more
In AECD,
AE ∥ DC [∵ Given]
AD ∥ CE [∵ By construction]
Hence, AECD is a parallelogram.
AD = CE …(1) [∵ Opposite sides of a parallelogram are equal]
AD = BC …(2) [∵ Given]
Hence, CE = BC [∵ From the equation (1) and (2)]
Therefore, in ΔBCE,
∠3 = ∠4 …(3) [∵ In a triangle, the angles opposite to equal sides are equal]
Here, ∠2 + ∠3 = 180° …(4) [∵ Linear Pair]
∠1 + ∠4 = 180° …(5) [∵ Co-interior angles]
Therefore, ∠2 + ∠3 = ∠1 + ∠4 [∵ From the equation (4) and (5)]
⇒ ∠2 = ∠1 ⇒ ∠B = ∠A [∵ ∠3 = ∠4]
(ii) ABCD ia a trapezium in which AB ∥ DC, hence,
∠1 + ∠D = 180° …(6) [∵ Co-interior angles]
∠2 + ∠C = 180° …(7) [∵ Co-interior angles]
Therefore, ∠1 + ∠D = ∠2 + C [∵ From the equation (6) and (7)]
⇒ ∠D = ∠C [∵ ∠2 = ∠1]
(iii) In ΔABC and ΔBAD,
BC = AD [∵ Given]
∠ABC = ∠BAD [∵ Prove above]
AB = AB [∵ Common]
Hence, ΔABC ≅ ΔBAD [∵ SAS Congruency rule]
(iv) ΔABC ≅ ΔBAD [∵ Prove above]
Diagonal AC = diagonal BD [∵ CPCT]
Consider a circle with centre O. Let AB is the given line. Now draw a perpendicular from O to line AB, which intersect AB at P. Now take points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.
Consider a circle with centre O. Let AB is the given line.
Now draw a perpendicular from O to line AB, which intersect AB at P.
Now take points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.
Join CB. (i) In ΔAPC and ΔDPB, ∠APC = ∠DPB [Vertically Opposite Angles] ∠CAP = ∠BDP [Angles in the same segment] ΔAPC ∼ ΔDPB [AA similarity] (ii) We have already proved that ΔAPC ∼ ΔDPB. We know that the corresponding sides of similar triangles are proportional. So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PRead more
Join CB.
(i) In ΔAPC and ΔDPB,
∠APC = ∠DPB [Vertically Opposite Angles]
∠CAP = ∠BDP [Angles in the same segment]
ΔAPC ∼ ΔDPB [AA similarity]
(ii) We have already proved that ΔAPC ∼ ΔDPB.
We know that the corresponding sides of similar triangles are proportional. So,
AP/DP = PC/PB = CA/BD ⇒ AP/DP = PC/PB ⇒ AP . PB = PC . DP
Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip of the rod. Then, the lenght of the string is AC. In ΔABC, by pythagoras theorem AC² = AB² + BC² ⇒ AB² = (1.8m)² + (2.4m)² ⇒ AB² = (3.24 + 5.76) m² ⇒ AB² = 9.00 m² ⇒ AB = √9 = 3 m Hence, thRead more
Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip of the rod.
Then, the lenght of the string is AC.
In ΔABC, by pythagoras theorem
AC² = AB² + BC²
⇒ AB² = (1.8m)² + (2.4m)²
⇒ AB² = (3.24 + 5.76) m²
⇒ AB² = 9.00 m²
⇒ AB = √9 = 3 m
Hence, the lenght of string, which is out, is 3 m.
If she pulls in the string, at the rate of 5 cm/s, then the distance travelled by fly in 12 seconds
= 12 × 5 = 60 cm = 0.6 m
Let, Dbe the position of string that is out after 12 seconds.
The lenght of the string pulls in by Nazima = AD = AC – 12
= (3.00 – 0.6) m
= 2.4 m
In ΔADB,
AB² + BD² = AD²
⇒ (1.8 m)² + BD² = (2.4 m)²
⇒ BD² = (5.76 – 3.24) m² = 2.52m²
⇒ BD = 1.587 m
Horizontal distance travelled by fly
= BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m.
A line RT is drawn parallel to SP, which intersects QP produced at T. Given that, SP bisects angle QPR, therefore ∠QPS = ∠SPR ...(1) By construction, ∠SPR = ∠PRT(As PS II TR) ...(2) ∠QPS = ∠QTR(As PS II TR) ...(3) From the above equations, we have ∠PRT = ∠QTR ∴ PT = PR By construction, PS II TR In ΔRead more
A line RT is drawn parallel to SP, which intersects QP produced at T.
Given that, SP bisects angle QPR, therefore
∠QPS = ∠SPR …(1)
By construction,
∠SPR = ∠PRT(As PS II TR) …(2)
∠QPS = ∠QTR(As PS II TR) …(3)
From the above equations, we have
∠PRT = ∠QTR
∴ PT = PR
By construction, PS II TR
In ΔQTR, by Thales theorem
QS/SR = QP/PT ⇒ QS/SR = PQ/PR [∵ PT = TR]
Join CB. (i) In ΔAPC and DPB, ∠APC = ∠DPB [Vertically Opposite Angles] ∠CAP = ∠BDP [Angles in the some segment] ΔAPC ∼ Δ DPB [ AA similarity] (ii) We have already proved that ΔAPC ∼ ΔDPB. We know that the corresponding sides of similar triangle are proportional. So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PRead more
Join CB.
(i) In ΔAPC and DPB,
∠APC = ∠DPB [Vertically Opposite Angles]
∠CAP = ∠BDP [Angles in the some segment]
ΔAPC ∼ Δ DPB [ AA similarity]
(ii) We have already proved that ΔAPC ∼ ΔDPB.
We know that the corresponding sides of similar triangle are proportional.
So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PC/PC ⇒ AP. PB = PC . DP
Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Figure). If AD is extended to intersect BC at P, show that (i) Δ ABD ≅ Δ ACD (ii) Δ ABP ≅ Δ ACP (iii) AP bisects ∠ A as well as ∠ D. (iv) AP is the perpendicular bisector of BC.
(i) In ΔABD and ΔACD AB = AC [∵ Given] BD = CD [∵ Given] AD = AD [∵ Common] Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule] (ii) In ΔABD ≅ ΔACD [∵ Proved above] ∠ BAD = ∠ CAD [∵ CPCT] In ABP and ACP, AB = AC [∵ Given] ∠ BAP = ∠ CAP [∵ Proved above] AP = AP [∵ Common] Hence, ΔABP ≅ ΔACP [∵ SAS CongruencyRead more
(i) In ΔABD and ΔACD
See lessAB = AC [∵ Given]
BD = CD [∵ Given]
AD = AD [∵ Common]
Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule]
(ii) In ΔABD ≅ ΔACD [∵ Proved above]
∠ BAD = ∠ CAD [∵ CPCT]
In ABP and ACP,
AB = AC [∵ Given]
∠ BAP = ∠ CAP [∵ Proved above]
AP = AP [∵ Common]
Hence, ΔABP ≅ ΔACP [∵ SAS Congruency rule]
(iii) In ΔABD ≅ ΔACD [∵ Proved above]
∠ BAD = ∠ CAD [∵ CPCT]
∠ BDA = ∠ CDA [∵ CPCT]
Hence, AP bisects both the angles A and D
(iv) In ΔABP ≅ ΔACP [∵ Prove above]
BP = CP [∵ CPCT]
∠BPA = ∠CPA [∵ CPCT]
∠BPA + ∠CPA = 180° [∵ Linear Pair]
⇒ ∠CPA + ∠CPA = 180° [∵ ∠BPA = ∠CPA]
⇒ 2∠ CPA = 180° ⇒ ∠CPA = (180°/2) = 90°
⇒ AP is perpendicular to BC. ⇒ AP is perpendicular bisector of BC. [∵ BP = CP].
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F see figure. Show that F is the mid-point of BC.
In ΔABD, E is mid-point of AD [∵ Given] and EG ∥ AB [∵ Given] Hence, G is mid-point of BD [∵ Converse of Mid-point Theorem] Similarly, In ΔBCD, G is mid-point of BD [∵ Prove above] and FG ∥ DC [∵ Given] Hence, F is mid-point of BC [∵ Converse of Mid-point Theorem]
In ΔABD,
See lessE is mid-point of AD [∵ Given]
and EG ∥ AB [∵ Given]
Hence, G is mid-point of BD [∵ Converse of Mid-point Theorem]
Similarly,
In ΔBCD,
G is mid-point of BD [∵ Prove above]
and FG ∥ DC [∵ Given]
Hence, F is mid-point of BC [∵ Converse of Mid-point Theorem]
ABCD is a trapezium in which AB || CD and AD = BC see Figure. Show that (i) ∠ A = ∠ B (ii) ∠ C = ∠ D (iii) ∆ ABC ≅ ∆ BAD (iv) diagonal AC = diagonal BD
In AECD, AE ∥ DC [∵ Given] AD ∥ CE [∵ By construction] Hence, AECD is a parallelogram. AD = CE ...(1) [∵ Opposite sides of a parallelogram are equal] AD = BC ...(2) [∵ Given] Hence, CE = BC [∵ From the equation (1) and (2)] Therefore, in ΔBCE, ∠3 = ∠4 ...(3) [∵ In a triangle, the angles opposite toRead more
In AECD,
See lessAE ∥ DC [∵ Given]
AD ∥ CE [∵ By construction]
Hence, AECD is a parallelogram.
AD = CE …(1) [∵ Opposite sides of a parallelogram are equal]
AD = BC …(2) [∵ Given]
Hence, CE = BC [∵ From the equation (1) and (2)]
Therefore, in ΔBCE,
∠3 = ∠4 …(3) [∵ In a triangle, the angles opposite to equal sides are equal]
Here, ∠2 + ∠3 = 180° …(4) [∵ Linear Pair]
∠1 + ∠4 = 180° …(5) [∵ Co-interior angles]
Therefore, ∠2 + ∠3 = ∠1 + ∠4 [∵ From the equation (4) and (5)]
⇒ ∠2 = ∠1 ⇒ ∠B = ∠A [∵ ∠3 = ∠4]
(ii) ABCD ia a trapezium in which AB ∥ DC, hence,
∠1 + ∠D = 180° …(6) [∵ Co-interior angles]
∠2 + ∠C = 180° …(7) [∵ Co-interior angles]
Therefore, ∠1 + ∠D = ∠2 + C [∵ From the equation (6) and (7)]
⇒ ∠D = ∠C [∵ ∠2 = ∠1]
(iii) In ΔABC and ΔBAD,
BC = AD [∵ Given]
∠ABC = ∠BAD [∵ Prove above]
AB = AB [∵ Common]
Hence, ΔABC ≅ ΔBAD [∵ SAS Congruency rule]
(iv) ΔABC ≅ ΔBAD [∵ Prove above]
Diagonal AC = diagonal BD [∵ CPCT]
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 12 cm. Length PQ is :
In ΔOPQ, angle P is right angle. [Since radius is perpendicular to tangent] Using Pythagoras theorem, OQ² = PQ² + OP² ⇒ 12² = PQ² + 5² ⇒ 144 = PQ² + 25 ⇒ PQ² = 144 - 25 = 119 ⇒ PQ = √119 Hence, the option (D) is correct.
In ΔOPQ, angle P is right angle. [Since radius is perpendicular to tangent]
See lessUsing Pythagoras theorem, OQ² = PQ² + OP²
⇒ 12² = PQ² + 5²
⇒ 144 = PQ² + 25
⇒ PQ² = 144 – 25 = 119
⇒ PQ = √119
Hence, the option (D) is correct.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Consider a circle with centre O. Let AB is the given line. Now draw a perpendicular from O to line AB, which intersect AB at P. Now take points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.
Consider a circle with centre O. Let AB is the given line.
See lessNow draw a perpendicular from O to line AB, which intersect AB at P.
Now take points on the line PO, one at circle X and another Y inside the circle. Draw lines parallel to AB and passing through X and Y. CD and EF are the required lines.
Two chords AB and CD intersect each other at the point P. Prove that : (i) ∆APC ~ ∆ DPB (ii) AP . PB = CP . DP
Join CB. (i) In ΔAPC and ΔDPB, ∠APC = ∠DPB [Vertically Opposite Angles] ∠CAP = ∠BDP [Angles in the same segment] ΔAPC ∼ ΔDPB [AA similarity] (ii) We have already proved that ΔAPC ∼ ΔDPB. We know that the corresponding sides of similar triangles are proportional. So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PRead more
Join CB.
See less(i) In ΔAPC and ΔDPB,
∠APC = ∠DPB [Vertically Opposite Angles]
∠CAP = ∠BDP [Angles in the same segment]
ΔAPC ∼ ΔDPB [AA similarity]
(ii) We have already proved that ΔAPC ∼ ΔDPB.
We know that the corresponding sides of similar triangles are proportional. So,
AP/DP = PC/PB = CA/BD ⇒ AP/DP = PC/PB ⇒ AP . PB = PC . DP
How many tangents can a circle have?
A Circle can have infinite number of tangents because a circle have infinite number of points on it and at every point a tangent can be drawn.
A Circle can have infinite number of tangents because a circle have infinite number of points on it and at every point a tangent can be drawn.
See lessNazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip of the rod. Then, the lenght of the string is AC. In ΔABC, by pythagoras theorem AC² = AB² + BC² ⇒ AB² = (1.8m)² + (2.4m)² ⇒ AB² = (3.24 + 5.76) m² ⇒ AB² = 9.00 m² ⇒ AB = √9 = 3 m Hence, thRead more
Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip of the rod.
See lessThen, the lenght of the string is AC.
In ΔABC, by pythagoras theorem
AC² = AB² + BC²
⇒ AB² = (1.8m)² + (2.4m)²
⇒ AB² = (3.24 + 5.76) m²
⇒ AB² = 9.00 m²
⇒ AB = √9 = 3 m
Hence, the lenght of string, which is out, is 3 m.
If she pulls in the string, at the rate of 5 cm/s, then the distance travelled by fly in 12 seconds
= 12 × 5 = 60 cm = 0.6 m
Let, Dbe the position of string that is out after 12 seconds.
The lenght of the string pulls in by Nazima = AD = AC – 12
= (3.00 – 0.6) m
= 2.4 m
In ΔADB,
AB² + BD² = AD²
⇒ (1.8 m)² + BD² = (2.4 m)²
⇒ BD² = (5.76 – 3.24) m² = 2.52m²
⇒ BD = 1.587 m
Horizontal distance travelled by fly
= BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m.
PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/SR = PQ/PR.
A line RT is drawn parallel to SP, which intersects QP produced at T. Given that, SP bisects angle QPR, therefore ∠QPS = ∠SPR ...(1) By construction, ∠SPR = ∠PRT(As PS II TR) ...(2) ∠QPS = ∠QTR(As PS II TR) ...(3) From the above equations, we have ∠PRT = ∠QTR ∴ PT = PR By construction, PS II TR In ΔRead more
A line RT is drawn parallel to SP, which intersects QP produced at T.
See lessGiven that, SP bisects angle QPR, therefore
∠QPS = ∠SPR …(1)
By construction,
∠SPR = ∠PRT(As PS II TR) …(2)
∠QPS = ∠QTR(As PS II TR) …(3)
From the above equations, we have
∠PRT = ∠QTR
∴ PT = PR
By construction, PS II TR
In ΔQTR, by Thales theorem
QS/SR = QP/PT ⇒ QS/SR = PQ/PR [∵ PT = TR]
Two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) ∆ PAC ~ ∆ PDB (ii) PA . PB = PC . PD
Join CB. (i) In ΔAPC and DPB, ∠APC = ∠DPB [Vertically Opposite Angles] ∠CAP = ∠BDP [Angles in the some segment] ΔAPC ∼ Δ DPB [ AA similarity] (ii) We have already proved that ΔAPC ∼ ΔDPB. We know that the corresponding sides of similar triangle are proportional. So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PRead more
Join CB.
(i) In ΔAPC and DPB,
∠APC = ∠DPB [Vertically Opposite Angles]
∠CAP = ∠BDP [Angles in the some segment]
ΔAPC ∼ Δ DPB [ AA similarity]
(ii) We have already proved that ΔAPC ∼ ΔDPB.
See lessWe know that the corresponding sides of similar triangle are proportional.
So, AP/DP = PC/PB = CA/BD ⇒ AP/DP = PC/PC ⇒ AP. PB = PC . DP