ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th
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According to question altitude from D to side AB bisects AB. Find the angles of the rhombus.
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ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Description : ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Last Answer : According to question altitude from D to side AB bisects AB. Find the angles of the rhombus.

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Description : ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Last Answer : Solution :-

Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. -Maths 9th

Description : Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. -Maths 9th

Last Answer : . Solution: (i) In ΔADC and ΔCBA, AD = CB (Opposite sides of a parallelogram) DC = BA (Opposite sides of a parallelogram) AC = CA (Common Side) , ΔADC ≅ ΔCBA [SSS congruency] Thus, ∠ACD = ∠CAB by ... are equal) Also, AB = BC = CD = DA (Opposite sides of a parallelogram) Thus, ABCD is a rhombus.

Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus -Maths 9th

Description : Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus -Maths 9th

Last Answer : (i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A. ∴ ∠DAC=∠BAC ---- ( 1 ) Now, AB∥DC and AC as traversal, ∴ ∠BAC=∠DCA [ Alternate angles ] --- ( 2 ) AD∥BC and AAC as traversal, ∴ ∠DAC= ... ---- ( 2 ) From ( 1 ) and ( 2 ), ⇒ AB=BC=CD=DA Hence, ABCD is a rhombus.

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Description : A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Last Answer : According to question parallelogram bisects one of its angles.

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Description : A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Last Answer : According to question parallelogram bisects one of its angles.

ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th

Description : ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th

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In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th

Description : In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th

Last Answer : answer:

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Description : In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Last Answer : Given, ABCD is a parallelogram. BE = AB To show, ED bisects BC Proof: AB = BE (Given) AB = CD (Opposite sides of ||gm) ∴ BE = CD Let DE intersect BC at F. Now, In ΔCDO and ΔBEO, ∠DCO = ... CD (Proved) ΔCDO ≅ ΔBEO by AAS congruence condition. Thus, BF = FC (by CPCT) Therefore, ED bisects BC. Proved

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Description : In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Last Answer : Given, ABCD is a parallelogram. BE = AB To show, ED bisects BC Proof: AB = BE (Given) AB = CD (Opposite sides of ||gm) ∴ BE = CD Let DE intersect BC at F. Now, In ΔCDO and ΔBEO, ∠DCO = ... CD (Proved) ΔCDO ≅ ΔBEO by AAS congruence condition. Thus, BF = FC (by CPCT) Therefore, ED bisects BC. Proved

ABCD is parallelogram . AB is produced to E so that BE = AB. Provethat ED bisects BC -Maths 9th

Description : ABCD is parallelogram . AB is produced to E so that BE = AB. Provethat ED bisects BC -Maths 9th

Last Answer : This answer was deleted by our moderators...

ABCD is a parallelogram. AB is produced to E such that BE = AB. Prove that ED bisects BC. -Maths 9th

Description : ABCD is a parallelogram. AB is produced to E such that BE = AB. Prove that ED bisects BC. -Maths 9th

Last Answer : answer:

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a rhombus and AB is produved to E and F such that AE=AB=BF prove that ED and FC are perpendicular to each other -Maths 9th

Description : ABCD is a rhombus and AB is produved to E and F such that AE=AB=BF prove that ED and FC are perpendicular to each other -Maths 9th

Last Answer : This answer was deleted by our moderators...

If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus. -Maths 9th

Description : If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Here, we are joining A and C. In ΔABC P is the mid point of AB Q is the mid point of BC PQ∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and ... RS=PS=RQ[All sides are equal] ∴ PQRS is a parallelogram with all sides equal ∴ So PQRS is a rhombus.

Let ABCD be a parallelogram. P is any point on the side AB. If DP and CP are joined in such a way that they bisect the angles -Maths 9th

Description : Let ABCD be a parallelogram. P is any point on the side AB. If DP and CP are joined in such a way that they bisect the angles -Maths 9th

Last Answer : answer:

Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Description : Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Last Answer : Given, opposite angles of a quadrilateral are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. ∴ CD = AB = 4cm

Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Description : Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Last Answer : Given, opposite angles of a quadrilateral are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. ∴ CD = AB = 4cm

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square (ii) Diagonal BD bisects ∠B as well as ∠D. -Maths 9th

Description : ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square (ii) Diagonal BD bisects ∠B as well as ∠D. -Maths 9th

Last Answer : Solution: (i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C) ⇒ AD = CD (Sides opposite to equal angles of a triangle are equal) also, CD = AB (Opposite sides of a rectangle) ,AB = BC = CD = AD Thus ... interior angles) ⇒ ∠CBD = ∠ABD Thus, BD bisects ∠B Now, ∠CBD = ∠ADB ⇒ ∠CDB = ∠ADB Thus, BD bisects ∠D

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

Description : In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

Last Answer : Solution :-

In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y. -Maths 9th

Description : In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y. -Maths 9th

Last Answer : Since diagonals of a rhombus bisect each other at right angle . ∴ In △AOB , we have ∠OAB + ∠x + 90° = 180° ∠x = 180° - 90° - 35° [∵ ∠ OAB = 35°] = 55° Also, ∠DAO = ∠BAO = 35° ∴ ∠y + ∠DAO + ∠BAO + ∠x ... 180° ⇒ ∠y = 180° - 125° = 55° Hence the values of x and y are x = 55°, y = 55°.

In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y. -Maths 9th

Description : In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y. -Maths 9th

Last Answer : Since diagonals of a rhombus bisect each other at right angle . ∴ In △AOB , we have ∠OAB + ∠x + 90° = 180° ∠x = 180° - 90° - 35° [∵ ∠ OAB = 35°] = 55° Also, ∠DAO = ∠BAO = 35° ∴ ∠y + ∠DAO + ∠BAO + ∠x ... 180° ⇒ ∠y = 180° - 125° = 55° Hence the values of x and y are x = 55°, y = 55°.

In Fig. 8.17, ABCD is a rhombus. Find the value of x. -Maths 9th

Description : In Fig. 8.17, ABCD is a rhombus. Find the value of x. -Maths 9th

Last Answer : Solution :-

If ABCD is a rhombus, then -Maths 9th

Description : If ABCD is a rhombus, then -Maths 9th

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The locus of a point in rhombus ABCD which is equidistant from A and C is -Maths 9th

Description : The locus of a point in rhombus ABCD which is equidistant from A and C is -Maths 9th

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(–2, –1) and (4, –5) are the co-ordinates of vertices B and D respectively of rhombus ABCD. Find the equation of the diagonal AC. -Maths 9th

Description : (–2, –1) and (4, –5) are the co-ordinates of vertices B and D respectively of rhombus ABCD. Find the equation of the diagonal AC. -Maths 9th

Last Answer : 3\(x\) - 2y + 5 = 0 ⇒ -2y = -3\(x\) - 5 ⇒ y = \(rac{3}{2}\)\(x\) + \(rac{5}{2}\)On comparing with y = m\(x\) + c, we see that slope of given line = \(rac{3}{2}\)As the required line is perpendicular to the given line, ... - 4)⇒ 3(y - 5) = - 2\(x\) + 8 ⇒ 3y - 15 = -2\(x\) + 8 ⇒ 3y + 2\(x\) - 23 = 0

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

E is the mid-point of the side AD of the trapezium ABCD with AB || DC. -Maths 9th

Description : E is the mid-point of the side AD of the trapezium ABCD with AB || DC. -Maths 9th

Last Answer : Given ABCD is a trapezium in which AB || DC and EF||AB|| CD. Construction Join, the diagonal AC which intersects EF at O. To show F is the mid-point of BC. Proof Now, in ΔADC, E is the mid-point of AD ... 0 is the mid-point of AC and OF || AB. So, by mid-point theorem, F is the mid-point of BC.

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

E is the mid-point of the side AD of the trapezium ABCD with AB || DC. -Maths 9th

Description : E is the mid-point of the side AD of the trapezium ABCD with AB || DC. -Maths 9th

Last Answer : Given ABCD is a trapezium in which AB || DC and EF||AB|| CD. Construction Join, the diagonal AC which intersects EF at O. To show F is the mid-point of BC. Proof Now, in ΔADC, E is the mid-point of AD ... 0 is the mid-point of AC and OF || AB. So, by mid-point theorem, F is the mid-point of BC.

ABCD is a square of side a cm. AB, BC, CD and AD are all chords of circles with equal radii each. -Maths 9th

Description : ABCD is a square of side a cm. AB, BC, CD and AD are all chords of circles with equal radii each. -Maths 9th

Last Answer : (b) \(\bigg[a^2+4\bigg[rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg]\bigg]\)As shown in the given figures, if a' is each side of the square, then ∠DOC = 120º ⇒ ∠ODC = ∠OCD = 30ºNow in fig. (iii), \(rac{ ... of square + Total area of 4 segments = \(a^2+4\bigg(rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg).\)

ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. If F is a point on side BC such that segment -Maths 9th

Description : ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. If F is a point on side BC such that segment -Maths 9th

Last Answer : answer:

In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Description : In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Last Answer : answer:

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

Description : 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral whose diagonals bisect each other at right angles. Given that, OA = OC OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90° To show that, if the ... a parallelogram. , ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle. Hence Proved.

Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Description : Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Last Answer : answer:

If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles. -Maths 9th

Description : If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles. -Maths 9th

Last Answer : Solution :-

ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

Description : ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

Last Answer : answer:

ABCD is a parallelogram and line segments AX, CY bisect the angles A and C, respectively. -Maths 9th

Description : ABCD is a parallelogram and line segments AX, CY bisect the angles A and C, respectively. -Maths 9th

Last Answer : Since opposite angles are equal in a parallelogram . Therefore , in parallelogram ABCD , we have ∠A = ∠C ⇒ 1 / 2 ∠A = 1 / 2 ∠C ⇒ ∠1 = ∠2 ---- i) [∵ AX and CY are bisectors of ∠A and ∠C ... intersects AX and YC at A and Y such that ∠1 = ∠3 i.e. corresponding angles are equal . ∴ AX | | CY .

If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Description : If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Last Answer : (c) Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4. Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°. 3x + 7x + 6x + 4x = 360° => 20x = 360° => x=360°/20° = 18°

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Description : If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.