If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th
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Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.
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If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

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.

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.

If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. -Maths 9th

Description : If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. -Maths 9th

Last Answer : Given Let ABCD be a cyclic quadrilateral and AD = BC. Join AC and BD. To prove AC = BD Proof In ΔAOD and ΔBOC, ∠OAD = ∠OBC and ∠ODA = ∠OCB [since, same segments subtends equal angle to the circle] AB = BC [ ... is DOC on both sides, we get ΔAOD+ ΔDOC ≅ ΔBOC + ΔDOC ⇒ ΔADC ≅ ΔBCD AC = BD [by CPCT]

If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. -Maths 9th

Description : If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. -Maths 9th

Last Answer : Given Let ABCD be a cyclic quadrilateral and AD = BC. Join AC and BD. To prove AC = BD Proof In ΔAOD and ΔBOC, ∠OAD = ∠OBC and ∠ODA = ∠OCB [since, same segments subtends equal angle to the circle] AB = BC [ ... is DOC on both sides, we get ΔAOD+ ΔDOC ≅ ΔBOC + ΔDOC ⇒ ΔADC ≅ ΔBCD AC = BD [by CPCT]

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]

Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Description : Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th

Last Answer : Solution :- Construction: Draw two circles having centres O and O' intersecting at points A and B. Draw a parallel line PQ to OO' ... iii) Again, OO' = MN [As OO' NM is a rectangle] ...(iv) ⇒ 2OO' = PQ Hence proved.

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Description : The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Description : The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm. -Maths 9th

Description : Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm. -Maths 9th

Last Answer : s= 2 4+4+2​ =5 Area of the triangle Δ= s(s−a)(s−b)(s−c)​ = 5(5−4)(5−4)(5−2)​ = 15​ cm 2

If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

Description : If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

Last Answer : Given ABCD is a trapezium whose non-parallel sides AD and BC are equal.

If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

Description : If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

Last Answer : Given ABCD is a trapezium whose non-parallel sides AD and BC are equal.

Prove that quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. -Maths 9th

Description : Prove that quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. -Maths 9th

Last Answer : Solution :- Given: A cyclic quadrilateral ABCD in which the angle bisectors AR, BR, CP O and DP of internal angles A, B, C and D respectively form a quadrilateral PQRS. To prove: PQRS ... the sum of a pair of opposite angles of quadrilateral PQRS is 180°. Therefore, quadrilateral PQRS is cyclic.

State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.

State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) 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........

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........

The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at P. Let O be the circumcentre of ∆APB and H be the orthocentre -Maths 9th

Description : The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at P. Let O be the circumcentre of ∆APB and H be the orthocentre -Maths 9th

Last Answer : answer:

Two congruent circles intersect each other at point A and B.Through A any line segment PAQ is drawn so that P,Q lie on the two circles.Prove that BP = BQ. -Maths 9th

Description : Two congruent circles intersect each other at point A and B.Through A any line segment PAQ is drawn so that P,Q lie on the two circles.Prove that BP = BQ. -Maths 9th

Last Answer : Solution :- Let, O and O' be the centres of two congruent circles. As, AB is the common chord of these circles. ∴ ACB = ADB As congruent arcs subtent equal angles at the centre. ∠AOB = ∠AO'B ⇒ 1/2∠AOB = 1/2∠AO'B ⇒ ∠BPA = ∠BQA ⇒ BP = BQ (Sides opposite to equal angles)

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Last Answer : Each shade of paper is divided into 3 triangles i.e., I, II, III 8 cm For triangle I: ABCD is a square [Given] ∵ Diagonals of a square are equal and bisect each other. ∴ AC = BD = 32 cm Height of AABD ... are: Area of shade I = 256 cm2 Area of shade II = 256 cm2 and area of shade III = 17.92 cm2

Prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which -Maths 9th

Description : Prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which -Maths 9th

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If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure). -Maths 9th

Description : If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure). -Maths 9th

Last Answer : According to question prove that the area of the parallelogram

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure). -Maths 9th

Description : If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure). -Maths 9th

Last Answer : According to question prove that the area of the parallelogram

If a triangle and a convex quadrilateral are drawn on the same base and no part of the quadrilateral is outside the triangle, -Maths 9th

Description : If a triangle and a convex quadrilateral are drawn on the same base and no part of the quadrilateral is outside the triangle, -Maths 9th

Last Answer : answer:

ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Description : ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Last Answer : answer:

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Last Answer : Solution of this question

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Last Answer : Solution of this question

If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Description : If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Last Answer : Solution :- Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D. To prove: D lies on BC Construction: Join A and D Proof: ∠ADB = 90° (Angle in the semi-circle ... + 90° => ∠ADB + ∠ADC = 180° => BDC is a straight line. Hence, D lies On third side BC.

Prove that the figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. -Maths 9th

Description : Prove that the figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. -Maths 9th

Last Answer : Solution :-

Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Description : Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Last Answer : We have, Three sides13cm,13cm and 20cm. By using Heron's formula We need to get the semi-perimeter s= 2 a+b+c​ = 2 13+13+20​ = 2 46​ =23 Now, put the heron's formula, s= s(s−a)(s−b)(s−c)​ = 23(23−13)(23−13)(23−20)​ = 23×10×10×3​ =10 23×3​ =83.07cm 2

In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is: -Maths 9th

Description : In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is: -Maths 9th

Last Answer : (c) 25√2 cm2.ΔABC is an isosceles triangle with AB = AC = 10 cm. ∠A = 45° ∴ Area of ΔABC= \(rac{1}{2}\) x 10 x 10 x sin 45°[Using Δ = \(rac{1}{2}\) bc sin A]= \(rac{50}{\sqrt2}\) = \(rac{50}{\sqrt2}\) x \(rac{\sqrt2}{\sqrt2}\) = 25√2 cm2.

In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th

Description : In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th

Last Answer : answer:

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral,prove that it is a rectangle. -Maths 9th

Description : If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral,prove that it is a rectangle. -Maths 9th

Last Answer : Solution :- Let, ABCD be a cyclic quadrilateral such that its diagonal AC and BD are the diameters of the circle though the vertices A,B,C and D. As angle in a semi-circle is 900 ∴ ∠ABC = 900 and ∠ADC = 900 ∠DAB = 900 ... Hence, ABCD is a rectangle.

ABCD is a cyclic quadrilateral such that AC ⊥ BD. AC meet BD at E. Prove that EA^2 + EB^2 + EC^2 + ED^2 = 4R^2, -Maths 9th

Description : ABCD is a cyclic quadrilateral such that AC ⊥ BD. AC meet BD at E. Prove that EA^2 + EB^2 + EC^2 + ED^2 = 4R^2, -Maths 9th

Last Answer : answer:

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Description : Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Last Answer : If h is the height of the triangle, then h 2 =y 2 − 4 x​ 2 ⇒h= 4 4y 2 −x 2 ​ ​ cm ∴Area= 2 1​ ×base×h = 2 x​ 4 4y 2 −x 2 ​ ​ cm 2

In the given figure, ΔABC is an equilateral triangle and ABDC is a cyclic quadrilateral, then find the measure of ∠BDC. -Maths 9th

Description : In the given figure, ΔABC is an equilateral triangle and ABDC is a cyclic quadrilateral, then find the measure of ∠BDC. -Maths 9th

Last Answer : △ABC is an equilateral triangle. ∠BAE=60 ABEC is a cyclic quadrilateral. We know, ∠BAC+∠BEC=180 =>∠BEC=180−60 =120 We know that the angles subtended by an arc on the circumference on the same side are equal. Therefore, ∠ABC=∠BDC =>∠BDC=60

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 :-

Prove that the points (1, –1) ((-1/2),(1/2)) and (1, 2) are the vertices of an isosceles triangle. -Maths 9th

Description : Prove that the points (1, –1) ((-1/2),(1/2)) and (1, 2) are the vertices of an isosceles triangle. -Maths 9th

Last Answer : (x, y) is equidistant from the points (2, 1) and (1, –2) ⇒ Distance between (x, y) and (2, 1) = Distance between (x, y) and (1, –2)⇒ \(\sqrt{(x-2)^2+(y-1)^2}\) = \(\sqrt{(x-1)^2+(y+2)^2}\)⇒ x2 – 4x + 4 + y2 – 2y + 1 = x2 – 2x + 1 + y2 + 4y + 4⇒ – 4x + 2x – 2y – 4y = 0 ⇒ –2x – 6y = 0 ⇒ x + 3y = 0

Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

Description : Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

Last Answer : answer:

The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, -Maths 9th

Description : The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, -Maths 9th

Last Answer : (b) \(rac{q^2}{4}\) (2√3 + 1).AC = q, ∠ABC = 90º ⇒ q = \(\sqrt{AB^2+BC^2}\)⇒ q = \(\sqrt{2x^2}\)⇒ q2 = 2x2 ⇒ \(x\) = \(rac{q}{\sqrt2}\)∴ Area of the re-entrant hexagon = Sum of areas of (ΔABC + ΔADC ... (rac{\sqrt3}{4}\)q2 + \(rac{\sqrt3}{8}\)q2 + \(rac{\sqrt3q^2}{8}\) = \(rac{q^2}{4}\) (2√3 + 1).

Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and equal to half of their difference. -Maths 9th

Description : Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and equal to half of their difference. -Maths 9th

Last Answer : In a parallelogram ABCD, the bisector of ∠ A also bisects BC at X.Prove that AD = 2AB.

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

Description : Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

Last Answer : According to question prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side,

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

Description : Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

Last Answer : According to question prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side,

In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Description : In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Last Answer : Step-by-step explanation: We have been provided that, Triangle ABC is an Equilateral triangle. Side of triangle is = 10 cm The arcs are drawn from each vertices of the triangle. We get three sectors ... portion is, Remaining area = Area of triangle ABC - Area of all the sectors 39.25cm square

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,