About triangle and use of mid point theorem in it -Maths 9th

About triangle and use of mid point theorem in it -Maths 9th
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A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted ∆ABC. A midpoint is a point on a line segment equally distant from the two endpoints. The Midpoint Theorem is used to make a bold statement regarding triangle sides and their lengths. Given a triangle, if we connect two sides with a line segment, and this line segment joins each of the two sides at the centers, or midpoints of each side, we can know two very important aspects about the triangle and the relationships between the sides. The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side. Anytime you have a line segment that connects two sides of a triangle at the midpoints, you automatically know that the sides are cut in half, and that the segment is parallel to the third side of the triangle. Parallel sides are shown using this symbol ||. You also know the line segment is one-half the length of the third side.
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About triangle and use of mid point theorem in it -Maths 9th

Description : About triangle and use of mid point theorem in it -Maths 9th

Last Answer : A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted ∆ABC. A midpoint is a point on a line ... shown using this symbol ||. You also know the line segment is one-half the length of the third side.

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th

Last Answer : Solution :-

isosceles triangle theorem -Maths 9th

Description : isosceles triangle theorem -Maths 9th

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isosceles triangle theorem -Maths 9th

Description : isosceles triangle theorem -Maths 9th

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Without using Pythagoras’ theorem, show that the points A (0, 4), B(1, 2) and C(3, 3) are the vertices of a right angle triangle. -Maths 9th

Description : Without using Pythagoras’ theorem, show that the points A (0, 4), B(1, 2) and C(3, 3) are the vertices of a right angle triangle. -Maths 9th

Last Answer : Slope (m) = \(rac{(y_2-y_1)}{(x_2-x_1)}\) = \(rac{6-2}{5-1}\) = \(rac{4}{4}\) = 1Also slope (m) = tan θ, where θ is the inclination of the line to the positive direction of the x-axis in the anticlockwise direction. tan θ = 1 ⇒ θ = tan –11 = 45º.

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 .

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

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 .

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

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 the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Description : D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Last Answer : Solution :-

One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Description : One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Last Answer : Perimeter of the largest (outermost) equilateral triangle = 3 24 = 72 cm. Now, the perimeter of the triangle formed by joining the midpoints of a given triangle will be half the perimeter of the original triangle. ∴ Required sum = 72 + ... -rac{1}{2}}\) = \(rac{72}{rac{1}{2}}\) = 72 x 2 = 144 cm.

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) = \(rac{\sqrt3a^2}{32}\)

The co-ordinates of mid-points of sides of a triangle are (1, 2), (0, –1) and (2, –1). Find its centroid. -Maths 9th

Description : The co-ordinates of mid-points of sides of a triangle are (1, 2), (0, –1) and (2, –1). Find its centroid. -Maths 9th

Last Answer : ABCD is a parallelogram, if the mid-points of diagonals AC and BD have the same co-ordinates (∵ Diagonals of a parallelogram bisect each other)Co-ordinates of mid-point of AC are \(\bigg(rac{a+2}{2},rac{-11+15}{2}\bigg)\) = \(\bigg(rac ... \(rac{a+2}{2}\) = 3 and 2 = \(rac{b+1}{2}\) ⇒ a = 4, b = 3.

If the co-ordinates of the mid-points of the sides of a triangle are (1, 1), (2, –3), (3, 4), find its incentre. -Maths 9th

Description : If the co-ordinates of the mid-points of the sides of a triangle are (1, 1), (2, –3), (3, 4), find its incentre. -Maths 9th

Last Answer : (b) 3 : 2 ; m = \(-rac{2}{5}\)Let P(m, 6) divides AB in the ratio k : 1. Then co-ordinates of P are \(\bigg(\)\(rac{2k-4}{k+1}\), \(rac{8k+3}{k+1}\)\(\bigg)\)Given, co-ordinates of P are (m, 6) ⇒\(rac{2k-4}{k+1} ... 2}-4}{rac{3}{2}+1}\) = \(rac{3-4}{rac{5}{2}}\) = \(rac{-2}{5}\)∴ m = \(rac{-2}{5}\).

In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB. -Maths 9th

Description : In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB. -Maths 9th

Last Answer : First Find the points D and E by midpoint formula. (x₂+x₁/2 , y₂+y₁/2) For DE=1/2AB In ΔsCED and CAB ∠ECD=∠ACB and the ratio of the side containing the angle is same i.e, CD=1/2BC ⇒CD/BC=1/2 EC=1/2AC ⇒EC/AC=1/2 ∴,ΔCED~ΔCAB hence the ratio of their corresponding sides will be equal, DE=1/2AB

4. 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 Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. 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 Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

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

In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Description : In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Last Answer : In △ABC, AD is the median ∴ ar(△ABD) = 1/2 ar(△ABC) Again, △ABD BL is the median ∴ ar(△ABL) = 1/2 ar(△ABD) = 1/2 × 1/2 ar((△ABC) = 1/4 ar((△ABC) Hence, value of x is 1/4.

In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Description : In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Last Answer : In ||gm ABCD, AC is the diagonal ∴ ar(△ABC) = ar(△ADC) = 1/2 ar ||gm ABCD) In△ADC, AL is the median ∴ ar(△ADL) = ar(△ACL)= 1/2 ar(△ADC) = 1/4 ar (||gm ABCD) Now, ar(quad.ABCL) = ar(△ABC) + ar(△ACL) = 3/4 ar ... ar(||gm ABCD) = 96 cm2 ∴ ar(△ADC) = 1/2 ar(||gm ABCD) = 1/2 96 = 48 cm2

ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Last Answer : Since R is the mid point of EF . ∴ AR is the median in △AEF. As, a median of a triangle divides it into two triangles of equal area . ∴ ar(△AER) = ar(△AFR)

In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Description : In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Last Answer : Solution of this question

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 the side CD of a parallelogram ABCD. -Maths 9th

Description : P is the mid-point of the side CD of a parallelogram ABCD. -Maths 9th

Last Answer : According to question prove that DA = AR and CQ = QR.

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

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

In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Description : In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Last Answer : In △ABC, AD is the median ∴ ar(△ABD) = 1/2 ar(△ABC) Again, △ABD BL is the median ∴ ar(△ABL) = 1/2 ar(△ABD) = 1/2 × 1/2 ar((△ABC) = 1/4 ar((△ABC) Hence, value of x is 1/4.

In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Description : In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Last Answer : In ||gm ABCD, AC is the diagonal ∴ ar(△ABC) = ar(△ADC) = 1/2 ar ||gm ABCD) In△ADC, AL is the median ∴ ar(△ADL) = ar(△ACL)= 1/2 ar(△ADC) = 1/4 ar (||gm ABCD) Now, ar(quad.ABCL) = ar(△ABC) + ar(△ACL) = 3/4 ar ... ar(||gm ABCD) = 96 cm2 ∴ ar(△ADC) = 1/2 ar(||gm ABCD) = 1/2 96 = 48 cm2

ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Last Answer : Since R is the mid point of EF . ∴ AR is the median in △AEF. As, a median of a triangle divides it into two triangles of equal area . ∴ ar(△AER) = ar(△AFR)

In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Description : In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Last Answer : Solution of this question

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 the side CD of a parallelogram ABCD. -Maths 9th

Description : P is the mid-point of the side CD of a parallelogram ABCD. -Maths 9th

Last Answer : According to question prove that DA = AR and CQ = QR.

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Description : If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Last Answer : NEED ANSWER

If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Description : If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Last Answer : (b) Let x and y be the lower and upper class limit of a continuous frequency distribution. Now, mid-point of a class = (x + y)/2 = m [given] ⇒ x + y = 2 m =x + l = 2m [∴ y = l = upper class limit (given)] ⇒ x = 2 m-l Hence, the lower class limit of the class is 2m – l.

In Fig.5.7, AC = XD, c is the mid-point of AB and D is the mid-point of XY. Using a Euclid's axiom,show that AB=XY. -Maths 9th

Description : In Fig.5.7, AC = XD, c is the mid-point of AB and D is the mid-point of XY. Using a Euclid's axiom,show that AB=XY. -Maths 9th

Last Answer : Solution :-

Prove that every line segment has one and only one mid-point. -Maths 9th

Description : Prove that every line segment has one and only one mid-point. -Maths 9th

Last Answer : Solution :-

In Fig. 8.31, D is the mid-point of AB and PC = 1/2AP = 3 cm. If AD = DB = 4 cm and DE||BP. Find AE. -Maths 9th

Description : In Fig. 8.31, D is the mid-point of AB and PC = 1/2AP = 3 cm. If AD = DB = 4 cm and DE||BP. Find AE. -Maths 9th

Last Answer : Solution :-

In Fig. 8.32, ABCD and PQRB are rectangles where Q is the mid-point of BD. If QR = 5 cm, find the measure of AB. -Maths 9th

Description : In Fig. 8.32, ABCD and PQRB are rectangles where Q is the mid-point of BD. If QR = 5 cm, find the measure of AB. -Maths 9th

Last Answer : Solution :-

In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF​=ABFB​=1 from (1) ⇒LF=LD so, DF=2DL

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:

ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th

Description : ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th

Last Answer : answer: