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

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

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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 :-

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.

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

Description : In Fig. 8.12, ABCD is a square. Find x. -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 :-

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

Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Description : Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Last Answer : Solution :-

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

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

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

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

Last Answer : answer:

In 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 :-

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

Two opposite angles of a ||gm are (60–x) degree and (3x –4) degree. Find the measure of each angles of the IIgm. -Maths 9th

Description : Two opposite angles of a ||gm are (60–x) degree and (3x –4) degree. Find the measure of each angles of the IIgm. -Maths 9th

Last Answer : Solution :-

Two adjacent angles of a ||gm are in the ratio 2:3. Find all the four angles of the parallelogram. -Maths 9th

Description : Two adjacent angles of a ||gm are in the ratio 2:3. Find all the four angles of the parallelogram. -Maths 9th

Last Answer : Solution :-

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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.

Three Angles of a quadrilateral ABCD are equal.Is it a parallelogram? -Maths 9th

Description : Three Angles of a quadrilateral ABCD are equal.Is it a parallelogram? -Maths 9th

Last Answer : Solution :-

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

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

Last Answer : Solution :-

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

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

Last Answer : answer:

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]

From the fig. 4.8, write the following: -Maths 9th

Description : From the fig. 4.8, write the following: -Maths 9th

Last Answer : Solution:- (i) A =(3,2), B =(4,0), C =(0,4) (ii) D (iii) 2 (iv) -3

In Fig.8.6, BDEF and FDCE are parallelograms.Can you say that BD = CD? -Maths 9th

Description : In Fig.8.6, BDEF and FDCE are parallelograms.Can you say that BD = CD? -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 :-

l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.

If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Description : If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Last Answer : Solution :- As, ∠ C = 90° (Angle in the semicircle) ∴ AC2 + BC2 = AB2 (By Pythagoras Theorem)

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

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

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

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