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Q1. A triangle and a rhombus are on the same base and between the same parallels. Then the ratio of area of triangle to that rhombus is:

• 1) 1 : 4
• 2) 1 : 2
• 3) 1 : 1
• 4) 1 : 3

Solution

A rhombus is also a parallelogram. We know that a triangle and a parallelogram lying on the same base and between same parallels have area in the ratio 1: 2. Thus, the ratio of area of triangle to that rhombus is 1: 2.

Q2. In ABC, E is the mid-point of median AD. Area(BED)=

• 1)
• 2)
• 3)
• 4) None of the above

Solution

We know that a median divides the triangle into two triangles of equal areas. As AD is the median of ABC. Now, E is the mid-point of median AD, so BE is the median of ABD.

Q3. In the given figure, ABCD is parallelogram and AE DC. If AB is 20 cm and the area of parallelogram ABCD is 80 cm², find AE.

Solution

AB = 20 cm AB = CD (opposite sides of a ||^gm) CD = 20 cm ar(||^gm ABCD) = base height 80 = 20 AE AE = 4 cm

Q4. Prove that parallelograms on the same base and between the same parallels are equal in area.

Solution

Let ABCD and EFCD be two parallelograms lying on the same base CD and between the same parallels AF and CD. We have to prove that area of the parallelograms ABCD and EFCD are equal. In ADE and BCF, DAE = CBF (Corresponding angles) AED = BFC (Corresponding angles) AD = BC (Opposite sides of the parallelogram ABCD) ADE BCF (By ASA congruence rule) We know that congruent figures have the same area. area (ADE) = area (BCF) … (1) Now, we have: area (ABCD) = area(ADE) + area (EDCB) = area (BCF) + area (EDCB) [Using (1)] = area (EDCF) Thus, the areas of the two parallelograms ABCD and EFCD are the same.

Q5. In given figure, if ABCD is a parallelogram then length of BE is :

• 1) 6 cm
• 2) 8 cm
• 3) 24 cm
• 4) 26 cm

Solution

Let perpendicular from D meet AB at F. In parallelogram ABCD, BC = AD = 3 cm [Opposite sides of a parallelogram are equal] We know that: Area of a parallelogram = Base Corresponding altitude Area of parallelogram ABCD = AB DF = AD BE 6 cm 4 cm = 3 cm BE BE = BE = 8 cm

Q6. D and E are the points on the sides AB and AC respectively of triangle ABC such that DE||BC. If area of DBC = 15 cm² then area of EBC is:

• 1) 30 cm²
• 2) 20 cm²
• 3) 7.5 cm²
• 4) 15 cm²

Solution

From the figure, DBC and EBC are triangles on the same base BC and between the same parallels DE and BC, therefore they are equal in area. Hence, area(EBC) = 15 cm²

Q7. A point E is taken as the midpoint of the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar(ABFC).

Solution

In triangles ABE and CFE, BE = CE (E is the mid-point of BC) (vertically opposite angles) (AB||CD and CD is produced to CF) ABE FCE (ASA criteria) AB = CF Now, AB = CD, as ABCD is a parallelogram. Therefore, AB = CF = CD C is midpoint of FD In parallelogram ABFC, AF is the diagonal, therefore, ar(ABF) = ar(ACF) ……(i) In ADF, AC is the median, since CD = CF Therefore, ar(ACF) = ar(ACD) ……(ii) From (i) and (ii) ar(ABF) = ar(ACD) Adding ar(ACF) to both sides, ar(ABF) + ar(ACF) = ar(ACD) + ar(ACF) ar(||gm ABFC) = ar(ADF)

Q8. Prove that the triangles between the same parallels and on the same base are equal in area.

Solution

ABC and PBC are on the same base BC and between the same parallels BC and AP. To Prove: ar (ABC) = ar (PBC) Construction; Through B, draw BD  CA intersecting PA produced at D, and Through C draw CQ  BP intersecting BP produced at Q. Proof: BCQP is a parallelogram with diagonal PC. ar (BCP) = ar (BCQP) Again BCAD is also a parallelogram with diagonal AB ar (ABC) = ar (ADBC) Since parallelogram BCQP and parallelogram BCAD are on the same base BC and between the same parallels BC and DQ;  ar (BCQP) = ar (BCAD) ar (BCQP) = ar (BCAD)  ar (BCP) = ar (ABC) Hence ar (ABC) = ar (BCP)

Q9. Find the area of quadrilateral ABCD, if AB || CD and AB = CD is as shown in figure :

• 1) 40 cm²
• 2) 25 cm²
• 3) 32 cm²
• 4) 16 cm²

Solution

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram. Area of parallelogram = 2 (Area of Triangle CDE)                                                                 = 32 cm²

Q10.

Solution

Q11.

Solution

Q12. If a triangle and a square are on the same base and between the same parallels, then the ratio of area of triangle to the area of square is

• 1) 1 : 3
• 2) 1 : 2
• 3) 3 : 1
• 4) 1 : 4

Solution

We know that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram. A square is a parallelogram. Therefore, required ratio = 1 : 2

Q13. In the given fig., M, N are points on sides PQ and PR respectively of PQR such that ar(QRN) = ar(QRM). Show that MN||QR.

Solution

QRN and QRM are on the same base QR and having equal areas. They lie between the same parallel lines MN||QR

Q14. PQRS and ABRS are parallelograms and X is any point on side BR. Show that: (i) area PQRS = area ABRS (ii) area AXS = area PQRS

Solution

PQRS and ABRS are parallelograms on the same base SR and between the same parallels SR and PB. area PQRS = area ABRS ... (i) AXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR. area ... (ii) From (i) and (ii), area AXS = area PQRS

Q15.

Solution

Q16. ABCD is a parallelogram of area 900 sq. cm. AP is drawn perpendicular to BC and AQ is perpendicular to DC. If AP = 24 cm and AQ = 18.75 cm. Calculate (i) AB (ii) BC (iii) area of ABC.

Solution

Area of parallelogram = base  corresponding altitude. Area of parallelogram ABCD = BC  AP 900 = BC  24 BC = 37.5 cm. Also, 900 = AB  18.75 AB = 48 cm.  

Q17. The ratio of the areas of two parallelograms on the same base and between the same parallels is:

• 1) 2 : 1
• 2) 1 : 3
• 3) 1 : 2
• 4) 1 : 1

Solution

The areas of two parallelograms on the same base and between the same parallels are equal. Hence, ratio of their areas is 1 : 1

Q18. If the base of a parallelogram is 8 cm and its altitude is 5 cm, then its area is equal to

• 1) 15 cm²
• 2) 40 cm²
• 3) 10 cm²
• 4) 20 cm²

Solution

Area of parallelogram = base height = 8 5 cm² = 40 cm²

Q19. If each diagonal of a quadrilateral separates it into two triangles of equal area, then show that the quadrilateral is a parallelogram.

Solution

ar(ABD) = ar(BDC) and ar(ACB) = ar(ADC) ar(ABD) + ar(BDC) = ar(Quad. ABCD)                    2ar(ABD) = ar(Quad. ABCD)                       ar(ABD) = ar(Quad. ABCD)       Similarly ar(ADC) = ar(Quad. ABCD)                   ar(ABD) = ar(ADC) ABD and ADC are on the same base AD Altitude from B of the ABD = altitude from C of the ADC AD  BC    Similarly, AB  DC Quadrilateral ABCD is a parallelogram     (opposite sides are parallel)

Q20. Find the figure which has two polygons having the same area:

• 1) (d)
• 2) (b)
• 3) (a)
• 4) (c)

Solution

In figure (d), PQRA and BQRS are the two polygons which lie between the same parallels and on the same base. So, they have equal areas.

Q21. In a parallelogram ABCD, P is a point in interior of parallelogram ABCD. If ar (||gm ABCD) = 18 cm² then [arAPD) + ar(CPB)] is :

• 1) 15 cm²
• 2) 9 cm²
• 3) 12 cm²
• 4) 18 cm²

Solution

Let us draw a line segment MN, passing through point P and parallel to line segment AD. In parallelogram ABCD, MN || AD (By construction) ... (1) ABCD is a parallelogram. AB || DC (Opposite sides of a parallelogram) AM || DN ... (2) From equations (1) and (2), we obtain MN || AD and AM || DN Therefore, quadrilateral AMND is a parallelogram. It can be observed that APD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN. Area (APD) =Area (AMND) ... (3) Similarly, for PCB and parallelogram MNCB, Area (PCB) =Area (MNCB) ... (4) Adding equations (3) and (4), we obtain Area (APD) + Area (PCB) = Area (AMND) + Area (MNCB) Area (APD) + Area (PCB) = Area(ABCD) = 9 cm²

Q22. ABCDE is pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) Area ACB = Area ACF (ii) area AEDF = area ABCDE

Solution

(i) We know, AC||BF ACF and ACB are triangles between the same parallels and on the same base AC. Therefore, area ACF = area ACB (ii) Now, area ACF = area ACB Add area AEDC on both the sides, ar AEDC + ar ACF = ar AEDC + ar ACB area AEDF = area ABCDE

Q23.

Solution

Q24. ABC and ABD are on the same base AB. If line segment CD is bisected by AB at O, show that ar (ABC) = ar(ABD)

Solution

ABC and ABD are on same base AB O is the mid point of CD (Given) In ACD, AO is the median ar (AOC) = ar (AOD) ...(I) [Median divides a triangle into two triangles of equal area] In BCD, BO is the median ar (BOC) = ar (DOB) ...(II) Adding I and II ar(AOC) + ar(BOC) = ar(AOD) + ar (DOB) ...(III) From III, ar (ABC) = ar(ABD)

Q25. ABCD is a parallelogram. If E and F are mid points of sides AB and CD and diagonal AC is joined then ar (FCBE) : ar (CAB) is:

• 1) 2 : 1
• 2) 1 : 4
• 3) 1 : 2
• 4) 1 : 1

Solution

Join EF. Since E and F are midpoints of AB and DC respectively, therefore EF divides parallelogram ABCD into two parallelograms (AEFD and FCBE) of equal areas. Now, ACB and parallelogram ABCD lie on the same base and between the same parallels, therefore, area ofACB is half the area of parallelogram ABCD. From (i) and (ii), ar(FCBE) = ar(CAB) Hence, the ratio = 1 : 1

Q26. In the given figure, M is a point in the interior of a parallelogram PQRS. Show that (i) ar(PMQ) + ar(MRS) = (||^gm PQRS) (ii) ar(PMS) + ar(MQR) = ar(PMQ) + ar(MRS)

Solution

Const : Draw a line parallel to SR through M and a line parallel to SP through M   Proof : (i) In parallelogram PQRS, PQ || UV(by construction) ……(i) PS || QR(opp. sides of a parallelogram) Therefore, PU || QV……(ii) From (i) and (ii), PQ || UV and PU || QR Therefore, quadrilateral PUVQ is a parallelogram. It can be observed that PMQ and parallelogram PQVU lie on the same base PQ and between the same parallels PQ and UV Similarly, for MRS and parallelogram SRVU, adding (iii) and (iv) (ii) In parallelogram PQRS, PS || XY(by construction) ……(vi) PQ || SR(opp. sides of a parallelogram) Therefore, PX || SY……(vii) From (vi) and (vii), PS || XY and PX || SY Therefore, quadrilateral PSYX is a parallelogram. It can be observed that PMS and parallelogram PSYX lie on the same base PS and between the same parallels PS and XY Similarly, for MRQ and parallelogram XYRQ, adding (viii) and (ix) From (v) and (x)

Q27. In the figure if ar(||^gm ABCD) = 30 cm², then ar(AEB) is:

• 1) 45 cm²
• 2) 30 cm²
• 3) 60 cm²
• 4) 15 cm²

Solution

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram. Therefore, ar(AEB) = 15 cm²

Q28. ABCD is a parallelogram of area 100 cm². AE is drawn perpendicular to BC and AF is drawn perpendicular to DC. If AE = 8 cm and AF is equal to 6.25 cm, then AB is

• 1) 15 cm
• 2) 20 cm
• 3) 12.5 cm
• 4) 16 cm

Solution

Area of a parallelogram = base  corresponding altitude 100 = DC  6.25 DC = 16 cm. DC=AB =16 cm................[ Opposite sides of a parallelogram are equal.] AB =16 cm

Q29.

Solution

Q30.

Solution

Q31. ABCD is a trapezium in which, AB  DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively, the mid points of AD and BC, prove that XY = 50 cm.

Solution

Join DY and produce it to meet AB produced at P. In BYP and CYD BYP =CYD                     (vertically opposite angles) DCY =PBY                     (alternate angles) BY = CY                                 (given) BYP  CYD                   (ASA) DY = YP and DC = BP   Y is the mid point of DP Also, X is the mid point of AD Hence XY = 50 cm.

Q32.

Solution

Q33. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed. Show that area ABCD = area PBQR.

Solution

Join AC and PQ. ACQ and APQ lie on the same base AQ and between the same parallels CP and AQ area ACQ = area APQ area ACQ - area ABQ = area APQ - area ABQ area ABC = area PBQ 2 area ABC = 2 area PBQ … (i) AC is the diagonal of parallelogram ABCD. area ABC = area ADC = area ABCD area ABCD = 2 area ABC ... (ii) Similarly, area PBQR = 2 area PBQ ... (iii) From (i), (ii) and (iii), area ABCD = area PBQR

Q34. If medians of a triangle ABC intersects at G show that ar (AGB) = ar (AGC) = ar(BGC) = ar (ABC).

Solution

In ABC, AD is the median Ar(ABD) = ar(ACD) ...(i) In GBC, GD is the median (GBD) =ar(GCD) ...(ii) Subtracting equation (ii) from equation (i) Ar(ABD) - ar(GBD) = ar(ACD) - ar(GCD) Ar(AGB) = ar(AGC) Similarly, arAGB = ar(BGC) arABC = arAGB + arBGC + arAGB = arAGB + arAGB + arAGB = 3 arAGB ar ABC = ar(AGB)

Q35. Show that line segment joining the mid points of a pair of opposite sides of a parallelogram, divides it into two equal parallelograms.

Solution

Since ABCD is a parallelogram AB = DC and AB  DC AE = DF and AE  DF Quadrilateral ABCD is a parallelogram. Similarly, quadrilateral EBCF is also a parallelogram. AEFD and EBCF are two parallelograms on the same base EF and between the same parallels, AD and BC. ar(AEFD) = ar(EBCF)

Q36. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that: ar(APB) = ar(BQC)

Solution

Since APB and ||^gm ABCD are on the same base AB and between the same parallels AB and DC ar(APB) = ....(i) Similarly ar(BQC) = ...(ii) From (i) and (ii) we have

Q37. In triangle ABC, D is the mid point of AB. P is any point on BC. CQ PD meets AB at Q. Show that ar (BPQ) =  ar (ABC).

Solution

Given: ABC in which D is the mid point of AB.PD CQ Construction: Join C to D and Q to P. Proof: CD is a median of the ABC      ar (BCD) = ar (DAC) PDC and PDQ are on the same base PD and between the same parallels PD and CQ.  ar (PDC) = ar (PDQ) ar (BCD) =  ar (ABC)          (median divides a triangle into two triangles of equal area)  ar (BPD) + ar (PDC) =   ar (ABC) ar (BPD) + ar (PDQ) =   ar (ABC) Hence ar (BPQ) =  ar (ABC)                                   ( BPD + PDQ =  BPQ)

Q38. In the given figure, if parallelogram ABCD and rectangle ABEF are of equal area then:

• 1) Perimeter of ABCD = (perimeter of ABEF)
• 2) Perimeter of ABCD < perimeter of ABEF
• 3) Perimeter of ABCD = perimeter of ABEF
• 4) Perimeter of ABCD > perimeter of ABEF

Solution

Consider ∆AFD, right angled at F. AD is the hypotenuse and FD and AF are the base and height of triangle. Since, hypotenuse is the longest side in a right angled triangle. AD > AF Perimeter of parallelogram ABCD = AB + BC + CD + DA = 2 AB + 2 AD = 2 (AB + AD) > 2 (AB + AF) [ As AD > AF ] Perimeter of rectangle ABEF = 2 (AB + AF) Thus, Perimeter of parallelogram ABCD > Perimeter of rectangle ABEF.

Q39. Show that the median of a triangle divides it into two triangles of equal areas.

Solution

Const.- Draw AN BC Now ar(ADB) = ar(ADC) = Since BD = DC ar(ADB) = ar(ADC)

Q40. P is a point in the interior of a parallelogram ABCD. Show that (i) Area APB + Area PCD = area ABCD (ii) Area APD + Area PBC = area APB + area PCD

Solution

(i) Through point P draw a line parallel to AB. Now, AXYB and XDCY are two parallelograms. area APB + area PCD = (area AXYB + area XDCY) area APB + area PCD = area ABCD …..(1) (ii) Similarly, we draw a line MN parallel to AD through P. Now, AMND and MNCB are two parallelograms. area APD + area BPC = (area AMND + area MNCB) area APD + area BPC = area ABCD …….(2) Comparing (1) and (2), area APB + area PCD = area APD + area BPC

Q41. In a parallelogram ABCD, AB = 20 cm. The altitude corresponding to the side AB and AD are 14 cm and 10 cm respectively. Find AD.

Solution

Area of parallelogram = base x altitude AB x BF = BC x DE 20 x 14 = BC x 10 BC = 28 cm AD = BC          (opposite sides of a parallelogram) AD = 28 cm.

Q42. If the diagonals AC, BD of a quadrilateral ABCD, intersect at O, and divide it into four triangles of equal area, Show that the quadrilateral is a parallelogram.

Solution

ar (AOD) = ar (BOC)  ar (AOD) + ar (AOB) = ar (BOC) + ar (AOB)  ar (ABD) = ar (ABC) Thus, ABD and ABC have the same base AB and have equal areas. So their corresponding altitudes must be equal. Altitude from D of ABD = Altitude from C of ABC DC  AB. Similarly, AD BC Hence, quadrilateral ABCD is a parallelogram.

Q43. If area of parallelogram ABCD is 25 cm² and on the same base CD, a triangle BCD is given such that area BCD = x cm², then value of x is

• 1) 25 cm²
• 2) 20 cm²
• 3) 15 cm²
• 4) 12.5 cm²

Solution

If a triangle and a parallelogram are on the same base and between same parallels then area of triangle is half the area of parallelogram. Therefore, area of triangle BCD = x = 25 cm² = 12.5 cm²

Q44. If E, F, G and H are respectively the mid points of the sides of a parallelogram ABCD, show that ar(EFGH) = ar(ABCD).

Solution

Join HF. Since H and f are the mid points of Ad and BC respectively, ABCD is a parallelogram AD = BC and AD  BC AH = BF and AH  BF ABFH is a parallelogram Since parallelogram FHAB and FHE are on the same base and between the same parallels HF and AB,

Q45.

Solution

Q46. In the figure AP  BQ  CR. Prove that ar (AQC) = ar (PBR).

Solution

Triangles ABQ and PBQ are on the same base BQ and between the same parallels AP and BQ.  ar (ABQ) = ar (PBQ)                      …(ii) Similarly, triangles BCQ and BRQ are on the same base BQ and between the same parallels BQ and CR.  ar (BCQ) = ar (BRQ)                      …(i) Adding (i) and (ii) ar (ABQ) + ar (BCQ) = ar (PBQ) + ar (BRQ)  ar (AQC) = ar (PBR)

Q47. In a parallelogram PQRS, Take a point A on RS and join P and Q. In how many parts, the parallelogram is divided? What are the shapes of these Parts? Divide the parallelogram into two parts of equal area.

Solution

PQRS is a parallelogram. A is a point on SR By joining A to P and Q, parallelogram is divided into three parts. Each part is triangular in shape. Three triangles are PSA, PAQ and QAR. The diagonals of a parallelogram divide it into two parts of equal area.

Q48.

Solution

Q49. Given a triangle ABC and E is mid point of median AD of ABC. If ar(BED) = 20 cm², then ar(ABC) is:

• 1) 80 cm²
• 2) 5 cm²
• 3) 60 cm²
• 4) 10 cm²

Solution

In ABC, AD is the median. ar(ABD) = ar(ADC) ... (i) [Median divides the triangle into two equal parts] Again, in ADB, BE is a median. ar(ABE) = ar(BDE) ... (ii) From (i), we have ar(ABD) = ar(ABC) ... (iii) From (ii), we have ar(BED) = ar(ABD) ... (iv) From (iii) and (iv), we have ar(BED) = ar(ABC) 20 = ar(ABC) ar (ABC) = 80 cm²

Q50. A point D is taken on the base BC of a ABC and AD is produced to E, such that DE = AD. Show that ar (BCE) = ar(ABC).

Solution

In ACE, since D is the mid-point of AE, CD is the median Hence, ar(ADC) = ar(EDC) (1) Similarly, ar(ADB) = ar(EDB) (2) Adding (1) and (2), ar(ADC) + ar(ADB) = ar(EDC) + ar(EDB) ar(ABC) = ar(BCE)

Q51. ABCD and BCFE are parallelograms. If area of triangle EBC = 480 sq. cm. Calculate (i) area of parallelogram ABCD                   (ii) area of parallelogram BCFE.

Solution

EBC and parallelogram ABCD are on the same base BC and between the same parallels AD and BC.           ar (ABCD) = 2 ar (EBC) Hence, ar (ABCD) = 2  480= 960 sq. cm Parallelogram ABCD and BCEF are on the same base BC and between the same parallels AF and BC. ar (ABCD) = ar (BCEF) = 960 sq. cm

Q52. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

Solution

Draw DN  AB and CM  AB. ar (AOD) = ar (BOC)  ar (AOD) + ar (AOB) = ar (BOC) + ar (AOB)  ar (ABD) = ar (ABC) AB DN =  AB CM DN = CM But, DN  CM as DN and CM are perpendicular to AB. DCMN is a parallelogram.  DC  AB Hence, ABCD is a trapezium.

Q53. In the given figure, ABCD is a quadrilateral. A line through D parallel to AC meets BC produced at E. Prove that ar(ABE) = ar quad. ABCD)

Solution

Proof: ACD and ACE are on the same base AC and between the same parallels as AC||DE. ...(1) On adding ar(ABC) to both sides: ar(ACE) + ar(ABC) = ar(ACD) + ar(ABC) ar(ABE) = ar(quad. ABCD)

Q54. Prove that of all the parallelograms of the given sides, a rectangle has the greatest area.

Solution

ABCD is a rectangle and ABFE is a parallelogram, such that AD = BC = AE = BF Area of rectangle = base x altitude                              = AB x AD Area of parallelogram ABFC = base x altitude           nbsp;                                      = AB x EG In AGE, AE > EG    (AE is the hypotenuse of right triangle) But AD = AE AD > EG Hence, ar(rectangle ABCD) > ar (parallelogram ABFE)

Q55. Prove that the parallelograms on the same base and between same parallels are equal in area.

Solution

Given : To ||^gm ABCD and EFCD are on the same base DC and between the same parallels AF and DC To prove: ar(||^gm ABCD) = ar(||^gm EFCD) Proof : In ADE and BCF DAE = CBF AED = BFC [Corresponding 's] ADE = BCF (Angle sum property of a triangle) AD = BC (opposite sides of the ||^gm) ar(ADE) = ar(BCF) (Congruent figure have equal areas) Now, ar(ABCD) = ar(ADE) + ar(EDCB) = ar(BCF) + ar(EDCB) = ar(EFCD) So, ||^gm ABCD and ||^gm EFCD are equal in area.

Q56. ABCD is a parallelogram in which CD = 15 cm, its corresponding altitude AM is 8 cm and CN AD. If CN = 10 cm, then find the length of AD.

Solution

ar (||gm ABCD) = base Altitude = CD AM = 15 cm 8 cm = 120 cm² Again ar(||^gm ABCD) = AD CN 120 = AD 10 AD = 12 cm

Q57. XY is a line parallel to the side BC of a ABC. BE  AC and CF  AB meets XY at E and F respectively. Show that ar (ABE) = ar (ACF).

Solution

Given: ABC, such that BE  AC and CF  AB Proof: since YE  BC and BE  CY  BCYE is a parallelogram. Similarly, BCFX is a parallelogram. Parallelograms BCFX and BCYE are on the same base BC and between same parallels BC and EF,   ar (BCYE) = ar (BCFX)                            …(i) ar (ABE) = ar (BCYE)                                 …(ii)    ( on same base and between same parallels) Similarly ar (ACF) = ar (BCFX)                   …(iii) From (i), (ii) and (iii) Hence ar (ABE) = ar (ACF)

Q58.

Solution

Q59. Diagonal AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

Solution

Given, area (AOD) = area (BOC) area (AOD) + area (AOB) = area (BOC) + area (AOB) area (ABD) = area (ABC) But, triangle ABD and ABC are on the same base AB and have equal area. They are between the same parallels AB||DC Hence, ABCD is a trapezium. [Since, a pair of opposite sides is parallel]

Q60. PQRS is a trapezium with PQ||SR. A line parallel to PR intersects PQ at L and QR at M. Prove that ar(PSL) = ar (PRM)

Solution

ar PSL) = ar (PRL) …(i) [Two triangles having same base (PL) and between same parallels (PL and SR) are equal in area] ar (PRL) = ar(PRM) ….(ii) [Two triangles having same base (PR) and between same parallels (PR and LM) are equal in area] From (i) and (ii), ar(PSL) = ar(PRM)

Q61. In the figure, ABCD is a trapezium in which AB||DC. Prove that ar(AOD) = ar(BOC).

Solution

AB||CD ar(ADC) = ar(BCD) [Triangles on the same base and between the same set of parallel lines have equal area] Subtracting ar(DOC) from both sides, we get, ar(ADC) - ar(DOC) = ar(BCD) - ar(DOC) ar(AOD) = ar(BOC)

Q62. ABCDE is a pentagon. A line through B is parallel to AC meets DC produced at F. Show that ar (ABC) = ar (ACF).

Solution

AC  BF, since ABC and ACF are on the same base AC and between the same parallels AC and BF  ar (ABC) = ar (ACF)

Q63. In the figure, PQ||DC||AB. Prove that ar(ACP) = ar(BDQ)

Solution

PQ||DC ar(PDC) = ar(QCD) ...(I) [Triangle between same parallels and on the same base] ACD and CDB are on same base CD and between same parallels, CD||AB ar(ACD) = ar(CDB) …(II) Adding (I) and (II), ar(PDC) + ar(ACD) = ar(QCD) + ar(CDB) ar(ACP) = ar(BDQ)