The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th
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According to question the mid-points of the sides of a rhombus, taken in order.
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The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Description : The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Last Answer : According to question the mid-points of the sides of a rhombus, taken in order.

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

Description : The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

Last Answer : According to question mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only.

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

Description : The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

Last Answer : According to question mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only.

The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Description : The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Last Answer : (d) Given, the quadrilateral ABCD is a rhombus. So, sides AB, BC, CD and AD are equal.

The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Description : The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Last Answer : (d) Given, the quadrilateral ABCD is a rhombus. So, sides AB, BC, CD and AD are equal.

A rectangle is formed by joining the mid-points of the sides of a rhombus. Show that the area of rectangle is half the area of rhombus. -Maths 9th

Description : A rectangle is formed by joining the mid-points of the sides of a rhombus. Show that the area of rectangle is half the area of rhombus. -Maths 9th

Last Answer : hope its clear

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if -Maths 9th

Description : The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if -Maths 9th

Last Answer : According to question the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle,

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if -Maths 9th

Description : The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if -Maths 9th

Last Answer : According to question the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle,

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 :-

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.

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.

Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Description : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Description : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

Description : 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Now, In ΔACD, R and S are the mid points of CD and DA respectively. , ... , PQRS is parallelogram. PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th

Description : Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th

Last Answer : Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th

Description : Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th

Last Answer : Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

Name the quadrilateral formed by joining the mid - points of the sides of any quadrilateral ABCD. -Maths 9th

Description : Name the quadrilateral formed by joining the mid - points of the sides of any quadrilateral ABCD. -Maths 9th

Last Answer : Solution :- Parallelogram.

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.

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

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

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

Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Description : Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Last Answer : Please see Exercise 8.2 - question 1 here in Quadrilaterals.

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Description : Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Last Answer : Please see Exercise 8.2 - question 1 here in Quadrilaterals.

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

Find the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points -Maths 9th

Description : Find the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points -Maths 9th

Last Answer : (d) \(rac{3\sqrt3a^2}{32}\)Let AB = a be the side of the outermost square.Then AG = AH = \(rac{a}{2}\)⇒ GH = \(\sqrt{rac{a^2}{4}+rac{a^2}{4}}\) = \(rac{a}{\sqrt2}\)∴ Diameter of circle = \(rac{a} ... rac{\sqrt3}{2}\) = \(rac{\sqrt3a^2}{32}\)∴ Area of ΔPQR = 3 (Area of ΔPOQ) = \(rac{\sqrt3a^2}{32}\)

ABCD is a square. P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. By joining AR, BS, CP, DQ, we get a quadrilateral which is a -Maths 9th

Description : ABCD is a square. P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. By joining AR, BS, CP, DQ, we get a quadrilateral which is a -Maths 9th

Last Answer : According to the given statement, the figure will be a shown alongside; using mid-point theorem: In △ABC,PQ∥AC and PQ=21 AC .......(1) In △ADC,SR∥AC and SR=21 AC .... ... are perpendicular to each other) ∴PQ⊥QR(angle between two lines = angle between their parallels) Hence PQRS is a rectangle.

The mid-point of the line joining the points (–10, 8) and (–6, 12) divides the line joining the points (4, –2) and (– 2, 4) in the ratio -Maths 9th

Description : The mid-point of the line joining the points (–10, 8) and (–6, 12) divides the line joining the points (4, –2) and (– 2, 4) in the ratio -Maths 9th

Last Answer : (d) 2 : 1 externallyThe mid-point of the line joining the points (-10, 8) and (- 6, 12) is\(\bigg(rac{-10+(-6)}{2},rac{8+12}{2}\bigg)\), i.e., (-8, 10).Let (-8, 10) divide the join of (4 ... 6k = 12 ⇒ k = -2 Since the value of k is negative, it is a case of external division and the ratio is 2 : 1.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Description : ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Last Answer : Clearly, EF = AB + DC / 2 = a + b / 2 Let h be the height , then ar(Trap. ABFE) : ar(Trap. EFCD) ⇒ 1/2 [a + (a+b / 2)] × h : 1/2 [b + (a+b / 2)] × h ⇒ 2a + a + b / 2 : 2b + a + b / 2 ⇒ 3a + b : 3b + a

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

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

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

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

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 and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Description : P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Last Answer : Given In a parallelogram ABCD, P and Q are the mid-points of AS and CD, respectively. To show PRQS is a parallelogram. Proof Since, ABCD is a parallelogram. AB||CD ⇒ AP || QC

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .