In Fig. 8.12, ABCD is a square. Find x. -Maths 9th

In Fig. 8.12, ABCD is a square. Find x. -Maths 9th
by

1 Answer

Answer :

Solution   :-
by

Related questions

In the Fig.8.7 ABCD is a | |gm.Find x and y. -Maths 9th

Description : In the Fig.8.7 ABCD is a | |gm.Find x and y. -Maths 9th

Last Answer : Solution :-

In the Fig.8.11. ABCD is a parallelogram.what is the sum of angles x,y and z? -Maths 9th

Description : In the Fig.8.11. ABCD is a parallelogram.what is the sum of angles x,y and z? -Maths 9th

Last Answer : Solution :-

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 :-

The sides BA and DC of quad. ABCD are produced as shown in Fig. 8.23. Prove that x + y = a + b -Maths 9th

Description : The sides BA and DC of quad. ABCD are produced as shown in Fig. 8.23. Prove that x + y = a + b -Maths 9th

Last Answer : Solution :-

In Fig. 8.28, ABCD is a parallelogram. Find the value of x, y and z. -Maths 9th

Description : In Fig. 8.28, ABCD is a parallelogram. Find the value of x, y and z. -Maths 9th

Last Answer : Solution :-

In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Description : In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Last Answer : Solution :-

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.

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)

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.

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Last Answer : Q Solution: (i) In ΔAPB and ΔCQD, ∠ABP = ∠CDQ (Alternate interior angles) ∠APB = ∠CQD (= 90o as AP and CQ are perpendiculars) AB = CD (ABCD is a parallelogram) , ΔAPB ≅ ΔCQD [AAS congruency] (ii) As ΔAPB ≅ ΔCQD. , AP = CQ [CPCT]

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.

Find all the angles of the | |gm ABCD given in the Fig. 8.8. -Maths 9th

Description : Find all the angles of the | |gm ABCD given in the Fig. 8.8. -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. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Description : In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Last Answer : Solution :-

In Fig.8.38, AM and CN are perpendiculars to the diagonal BD of a paralelogram ABCD.Prove that AM = CN. -Maths 9th

Description : In Fig.8.38, AM and CN are perpendiculars to the diagonal BD of a paralelogram ABCD.Prove that AM = CN. -Maths 9th

Last Answer : Solution :-

In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Last Answer : Solution :-

ABCD is a parallelogram in which P and Q are the mid-points of opposite sides AB and CD (Fig. 8.48). If AQ intersects DP at S and BQ intersects CP at R, show that -Maths 9th

Description : ABCD is a parallelogram in which P and Q are the mid-points of opposite sides AB and CD (Fig. 8.48). If AQ intersects DP at S and BQ intersects CP at R, show that -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

In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th

Description : In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th

Last Answer : Solution :-

Let X be any point within a square ABCD. On AX a square AXYZ is described such that D is within it. Which one of the following is correct? -Maths 9th

Description : Let X be any point within a square ABCD. On AX a square AXYZ is described such that D is within it. Which one of the following is correct? -Maths 9th

Last Answer : answer:

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC. We have, ABC is a right angled triangle at B. Now, AC2=AB2+BC2 [by Pythagoras theorem]Now, AC2=AB2+BC2 ... =24(1+6-√)cm2=24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6-√−−−−−−√cm2241+6cm2 .

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC.

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

In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Description : In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Last Answer : In the given figure: Area of triangle ADC = Area of triangle BCD (Triangles on the same and parallel) Now subtract the area of triangle DOC from both of them so... (Area of triangle ADC - Area of ... => Area of triangle AOD = Area of triangle BOC Hence the area of triangle BOC is 30 cm square.

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 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

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Last Answer : The graph obtained by plotting the points A, B and C and D is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal. So, abscissa of C should be ... of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-2, - 4).

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.

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,

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

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

Last Answer : According to question prove that ar (ΔAER) = ar (ΔAFR).

In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Description : In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Last Answer : In the given figure: Area of triangle ADC = Area of triangle BCD (Triangles on the same and parallel) Now subtract the area of triangle DOC from both of them so... (Area of triangle ADC - Area of ... => Area of triangle AOD = Area of triangle BOC Hence the area of triangle BOC is 30 cm square.

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 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

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Last Answer : The graph obtained by plotting the points A, B and C and D is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal. So, abscissa of C should be ... of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-2, - 4).

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.

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,

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

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

Last Answer : According to question prove that ar (ΔAER) = ar (ΔAFR).

How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Description : How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Last Answer : NEED ANSWER

How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Description : How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Last Answer : According to question ABCD is a square with diagonal 44 cm.

Plot the point A(2,0), B(5,0) and C(5,3). Find the coordinates of the point D such that ABCD is a square. -Maths 9th

Description : Plot the point A(2,0), B(5,0) and C(5,3). Find the coordinates of the point D such that ABCD is a square. -Maths 9th

Last Answer : Solution :-

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.Show that △OCD is an isosceles triangle. -Maths 9th

Description : O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.Show that △OCD is an isosceles triangle. -Maths 9th

Last Answer : Solution :-

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 square. Another square EFGH with the same area is placed on the square ABCD such that the point of intersection of diagonals of square -Maths 9th

Description : ABCD is a square. Another square EFGH with the same area is placed on the square ABCD such that the point of intersection of diagonals of square -Maths 9th

Last Answer : (a) 32 (2 - √2)As is seen in the given figure, the sides of one square are parallel to the diagonals of another square. Also, square ABCD and EFGH have same area.⇒ Sides of square ABCD and square EFGH are 4 cm each. Let ... four Δs outside ABCD= 16 +16 (3 -2√2)= 16 + (4 - 2√2) = 32 (2 - √2) cm2.

A rigid square plate ABCD of unit side rotates in its own plane about the middle-point of CD until the new position of A coincides with -Maths 9th

Description : A rigid square plate ABCD of unit side rotates in its own plane about the middle-point of CD until the new position of A coincides with -Maths 9th

Last Answer : answer:

A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD. The crease will divide BC in the ratio -Maths 9th

Description : A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD. The crease will divide BC in the ratio -Maths 9th

Last Answer : answer:

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.

ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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

ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]