Exercise 9.1
1. Which of the following figures lie on the same base and in-between the same parallels? In such a case, write the common base and the two parallels
Exercise 9.1
1. Which of the following figures lie on the same base and in-between the same parallels? In such a case, write the common base and the two parallels
Explanation:
Figures ΔPDC & Trapezium ABCD lie on the same base DC and in-between the same parallel lines AB and DC.
Parallelogram PQRS and trapezium SMNR lie on the same base SR but not in- between the same parallels (PQ≠MN).
Parallelogram PQRS and ΔRTQ are on the same base QR and in-between the same parallels PS and QR.
Parallelogram ABCD and ΔPQR do not lie on the same base (CB≠PQ) but they are in between the same parallel lines BC and AD.
Trapeziums ABQD and APCD lie are on the same base AD and in-between the same parallel lines AD and BQ.
Parallelograms PQRS and ABCD do not lie on the same base (RS≠CD) but in-between the same parallel lines SR and PQ.
Exercise 9.2
2.In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Exercise 9.2
2.In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Explanation:
Given, AB = CD = 16 cm (Opposite sides of a parallelogram are equal)
Heights: CF = 10 cm and AE = 8 cm
Now, Area of parallelogram = Base × Height
⇒ CD×AE = AD×CF
⇒ 16×8 = AD×10
⇒ AD = 128/10 cm ⇒ AD = 12.8 cm
3. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar(ABCD).
3. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar(ABCD).
Explanation:
Given, E, F, G and H are the mid-points of the sides of a ||gm ABCD respectively.
To Prove: ar (EFGH) = ½ ar (ABCD)
Construction: Join HF.
Proof: AD || BC and AD = BC (Opposite sides of a parallelogram are equal and parallel)
⇒ ½ AD = ½ BC
∴ AH=BF and HD=FC (H and F are midpoints of AD and BC)
Also, AH || BF and DH || CF (AD || BC)
Hence, ABFH and HFCD are parallelograms.
Now,
ΔEFH and parallelogram ABFH, both lie on the same FH the common base and in-between the same parallel lines AB and HF.
∴ area of EFH = ½ area of ABFH --- (1)
Similarly, area of GHF = ½ area of HFCD --- (2)
On Adding (1) and (2),
area of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD
⇒ area of EFGH = area of ABFH
∴ ar (EFGH) = ½ ar (ABCD)
(ABFH is half of the ||gm ABCD, F and H being the midpoints)
4.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).
4.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).
Explanation:
We know that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of triangle is equal to half the area of the parallelogram.
Here,
ΔAPB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.
∴ ar(ΔAPB) = ½ ar (parallelogram ABCD) --- (1)
ΔBQC and parallelogram ABCD lie on the same base BC and in-between same parallel AD and BC.
∴ar(ΔBQC) = ½ ar (parallelogram ABCD) --- (2)
From (1) and (2), ar(ΔAPB) = ar(ΔBQC).
5.In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
ar (APB) + ar (PCD) = ½ ar (ABCD)
ar (APD) + ar (PBC) = ar(APB) + ar(PCD)
[Hint: Through P, draw a line parallel to AB.]
5.In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that
ar (APB) + ar (PCD) = ½ ar (ABCD)
ar (APD) + ar (PBC) = ar(APB) + ar(PCD)
[Hint: Through P, draw a line parallel to AB.]
Explanation:
(i) Cons: Draw a line GH || to AB passing through P ---
(i).In ||gm ABCD, AD || BC ⇒ AG || BH --- (ii)
From equations (i) and (ii), ABHG is a parallelogram.
(Both the opposite sides are parallel)
Now, ΔAPB and parallelogram ABHG are lying on the same base AB and are in between the same parallels AB and GH.
∴ ar(ΔAPB) = ½ ar (ABHG) --- (iii)
Similarly, ΔPCD and parallelogram CDGH are lying on the same base CD and in-between the same parallel lines CD and GH.
∴ ar(ΔPCD) = ½ ar (CDGH) --- (iv)
Adding (iii) and (iv) we have,
ar(ΔAPB) + ar(ΔPCD) = ½ [ar (ABHG)+ar(CDGH)]
⇒ ar (APB)+ ar (PCD) = ½ ar (ABCD)
(ii)Cons: Draw a line EF || to AD passing through P ---(i).
In the ||gm ABCD, AB || CD ⇒ AE || DF --- (ii)
From equations (i) and (ii), AEDF is a parallelogram.
(Both the opposite sides are parallel)
Again, ΔAPD and parallelogram AEFD are lying on the same base AD and in-between the same parallel lines AD and EF.
∴ar(ΔAPD) = ½ ar (AEFD) --- (iii)
ΔPBC and parallelogram BCFE are lying on the same base BC and in-between the same parallel lines BC and EF.
∴ar(ΔPBC) = ½ ar (BCFE) --- (iv)
Adding equations (iii) and (iv) we have,
ar(ΔAPD) + ar(ΔPBC) = ½ [ar (AEFD)+ar(BCFE)]
⇒ar (APD)+ar (PBC) = ar (APB)+ar(PCD)
6.In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that ar (PQRS) = ar (ABRS) ar (AXS) = ½ ar (PQRS)
6.In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that ar (PQRS) = ar (ABRS) ar (AXS) = ½ ar (PQRS)
Explanation:
(i) From the given fig. ||gms PQRS and ABRS lie on the same base SR and in-between the same parallels SR and PB.
∴ ar (PQRS) = ar (ABRS) --- (i)
Also, ΔAXS and ||gm ABRS are lying on the same base AS and in-between the same parallel lines AS and BR.
∴ ar(ΔAXS) = ½ ar(ABRS) --- (ii)
From (i) and (ii), ar(ΔAXS) = ½ ar(PQRS).
7.A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields are divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
7.A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields are divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Explanation:
The ||gm field is divided into three triangles ΔPSA, ΔPAQ and ΔQAR.
So, Area of ΔPSA + ΔPAQ + ΔQAR = Area of ||gm PQRS --- (i)
Also, the ΔPAQ and ||gm PQRS are on the same base PQ and in-between the same parallels PQ&SR.
Area of ΔPAQ = ½ area of PQRS --- (ii)
From (i) and (ii),
Area of ΔPSA +Area of ΔQAR = ½ area of PQRS
Hence, the farmer must sow wheat or pulses in ΔPAQ alone or in both ΔPSA and ΔQAR.
8. D and E are points on sides AB and AC respectively of ΔABC such that ar (DBC) = ar (EBC). Prove that DE || BC.
8. D and E are points on sides AB and AC respectively of ΔABC such that ar (DBC) = ar (EBC). Prove that DE || BC.
Explanation:
ΔDBC and ΔEBC lie on the same base BC and have equal areas.
So, they will lie between the same parallel lines. (Converse of theorem)
∴ DE || BC.
9. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar(ΔABE) = ar(ΔACF)
9. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar(ΔABE) = ar(ΔACF)
Explanation:
Given, XY || BC, BE || AC and CF || AB.
To show: ar(ΔABE) = ar(ΔACF)
Proof: BE || AC ⇒ BE || CY and XY|| BC ⇒ EY|| BC. So, BCYE is a || gm.
Now ΔABE and ||gm BCYE are on the same base BE and between the same parallel lines BE and AC.
∴ ar (ABE) = ½ ar (BCYE) --- (1)
Now, CF || AB and XY || BC ⇒ CF || AB and XF || BC ⇒ BCFX is a || gm
Again, ΔACF and || gm BCFX are on the same base CF and in-between the same parallels AB and FC .
∴ar (ΔACF) = ½ ar (BCFX) ---(2)
Now, ||gms BCFX and BCYE are on the same base BC and between the same parallels BC and EF.
∴ar (BCFX) = ar (BCYE) ---(3)
From (1), (2) and (3), ar (ΔABE) = ar(ΔACF) ⇒ ar(BEYC) = ar(BXFC)
As the parallelograms are on the same base BC and in-between the same parallels EF and BC--(iii)
Also, △AEB and ||gm BEYC are on the same base BE and in-between the same parallels BE and AC.
∴ar(△AEB) = ½ ar (BEYC) --- (iv)
Similarly, △ACF and || gm BXFC on the same base CF and between the same parallels CF and AB.
∴ar (△ ACF) = ½ ar (BXFC) --- (v)
From (iii), (iv) and (v), ar(△ABE) = ar(△ACF)
10. 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 (see Fig. 9.26). Show that ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
10. 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 (see Fig. 9.26). Show that ar (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]
Explanation:
Cons: Join AC and PQ.
Now, ar(△ACQ) = ar(△APQ) (lying on common base AQ and between the same parallel lines AQ and CP)
Subtracting common area △ABQ, we have
⇒ ar(△ACQ)-ar(△ABQ) = ar(△APQ)-ar(△ABQ)
⇒ ar(△ABC) = ar(△QBP) --- (i)
AC and QP are diagonals of ||gms ABCD and PBQR.
∴ar (ABC) = ½ ar(ABCD) --- (ii)
ar (QBP) = ½ ar (PBQR) --- (iii)
From (ii) and (ii), ½ ar (ABCD) = ½ ar (PBQR)
⇒ ar (ABCD) = ar (PBQR).
11. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
11. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
Explanation:
From the fig. △DAC and △DBC lie on the same base DC and between the same parallels AB and CD.
∴ar(△DAC) = ar(△DBC)
Subtracting common area of △DOC,
⇒ ar(△DAC) – ar(△DOC) = ar(△DBC) – ar(△DOC)
⇒ ar(△AOD) = ar(△BOC).
Hence proved.
12. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar(△ACB) = ar(△ACF)
(ii) ar (AEDF) = ar (ABCDE)
12. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar(△ACB) = ar(△ACF)
(ii) ar (AEDF) = ar (ABCDE)
Explanation:
△ACB and △ACF lie on the same base AC and between the same parallels AC and BF.
∴ar(△ACB) = ar (△ ACF)
ar(△ACB) = ar(△ACF) (from (i))
Adding ar (ACDE) on both sides,
⇒ ar(△ACB) +ar (ACDE) = ar(△ACF) +ar (ACDE)
⇒ ar (ABCDE) = ar (AEDF).
13. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
13. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
Explanation:
Let ABCD be the quadrilateral shaped land plot.
Cons: Join the diagonal BD. Draw AE || to BD.
Join BE, intersecting AD at O.
So, △BCE is the shape of the original field & △AOB is the area for Health centre construction.
△DEO is the land joined to the plot.
To prove: ar(△DEO) = ar(△AOB)
Proof: From fig., △DEB and △DAB lie on the same base BD and in-between the same parallels BD and AE.
∴ ar(△DEB) = ar(△DAB)
⇒ar(△DEB) – ar(△DOB) = ar(△DAB) – ar(△DOB)
(Subtracting common area DOB)
⇒ar(△DEO) = ar(△AOB).
Hence proved.
14. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (△ADX) = ar (△ACY). [Hint: Join CX.]
14. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (△ADX) = ar (△ACY). [Hint: Join CX.]
Explanation:
Given, ABCD is a trapezium with AB || DC and XY || AC.
Cons: Join CX
To Prove: ar (ADX) = ar (ACY)
Proof: ar(△ADX) = ar(△AXC) --- (1)
(∵On the same base AX and in-between the same parallels AB and CD)
Similarly, ar (△AXC) = ar (△ ACY) --- (2)
(∵On the same base AC and in-between the same parallels XY and AC)
From (1) and (2), ar(△ADX) = ar(△ACY).
15. In Fig.9.28, AP || BQ || CR. Prove that ar(△AQC) = ar(△PBR).
15. In Fig.9.28, AP || BQ || CR. Prove that ar(△AQC) = ar(△PBR).
Explanation:
Given, AP || BQ || CR.
To Prove: ar (AQC) = ar(PBR)
Proof: ar(△AQB) = ar(△PBQ) --- (1) (Since they are on the same base BQ and between the same parallels AP and BQ.)
Similarly, ar(△BQC) = ar(△BQR) --- (2) (Since they are on the same base BQ and between the same parallels BQ and CR.)
Adding (1) and (2),
ar(△AQB) +ar(△BQC) = ar(△PBQ) +ar(△BQR)
∴ ar (△ AQC) = ar (△ PBR)
Hence proved.
16.
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.
16.
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.
Explanation:
Given, ar(△AOD) = ar(△BOC)
To Prove: ABCD is a trapezium.
Proof: ar(△AOD) = ar(△BOC)
⇒ ar(△AOD) + ar(△AOB) = ar(△BOC) +ar(△AOB)
(Adding common area AOB)
⇒ ar(△ADB) = ar(△ACB)
Areas of △ADB and △ACB are equal. So, they must lie between the same parallel lines. ∴ AB ∥ CD.
Hence, ABCD is a trapezium.
17. In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
17. In Fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
Explanation:
Given, ar(△DRC) = ar(△DPC) & ar(△BDP) = ar(△ARC)
To Prove: ABCD and DCPR are trapeziums.
Proof: ar(△BDP) = ar(△ARC)
⇒ ar(△BDP) - ar(△DPC) = ar(△ARC)- ar(△DRC)
⇒ ar(△BDC) = ar(△ADC)
So, ar(△BDC) and ar(△ADC) will lie between the parallel lines (converse).
∴ AB ∥ CD. Hence, ABCD is a trapezium.
Similarly, ar(△DRC) = ar(△DPC).
So, ar(△DRC) and ar(△DPC) are lying in-between the same parallel lines.
∴ DC ∥ PR. Hence proved that DCPR is also a trapezium.
Exercise 9.4
18.Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle
Exercise 9.4
18.Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle
Explanation:
Given, || gm ABCD and a rectangle ABEF have the same base AB and equal areas. To prove: Perimeter of || gm ABCD is greater than the perimeter of rectangle ABEF.
Proof: We know that, the opposite sides of || gm and rectangle are equal.
∴AB = DC & AB = EF
∴DC = EF ---(i)
Adding AB on both sides, we have
⇒AB + DC = AB + EF --- (ii)
We know, of all the segments drawn from a point not lying on it to a given line, the perpendicular line segment is the shortest.
∴BE < BC and AF < AD
⇒ BC > BE and AD > AF
⇒ BC+AD > BE+AF --- (iii)
Adding (ii) and (iii),
AB+DC+BC+AD > AB+EF+BE+AF
⇒ AB+BC+CD+DA > AB+ BE+EF+FA
⇒ perimeter of || gm ABCD > perimeter of rectangle ABEF.
∴The perimeter of the parallelogram is greater than that of the rectangle.
Hence Proved.
19. In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area? [Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into an equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide DABC into n triangles of equal areas.]
19. In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area? [Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into an equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide DABC into n triangles of equal areas.]
Explanation:
Given, BD = DE = EC
To prove: ar (△ABD) = ar (△ADE) = ar (△AEC)
Proof: In (△ABE), AD is median [BD = DE]
Also, the median of a triangle divides it into two parts of equal areas
∴ar(△ABD) = ar(△AED) ---(i)
Similarly, In (△ADC), AE is median [DE = EC]
∴ ar(△ADE) = ar(△AEC) ---(ii)
From the equation (i) and (ii), ar (ABD) = ar(ADE) = ar(AEC).
20. In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
20. In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
Explanation:
Given, ABCD, DCFE and ABFE are ||gms.
To prove: ar (△ADE) = ar (△BCF)
Proof: In △ADE and △BCF,
AD = BC
DE = CF
AE = BF
The opposite sides of || gms are equal in ABCD, DCFE and ABFE.
∴, △ADE ≅ △BCF [By SSS Congruence theorem]
∴ ar(△ADE) = ar(△BCF) [ By cpct]
21. In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).
[Hint: Join AC.]
21. In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).
[Hint: Join AC.]
Explanation:
Given: ABCD is a parallelogram and AD = CQ
To prove: ar (△BPC) = ar (△DPQ)
Proof: In △ADP and △QCP,
∠APD = ∠QPC [Vertically Opposite Angles are same]
∠ADP = ∠QCP [Alternate interior Angles AD|| BC or BQ and AQ is transversal]
AD = CQ
∴, △ADP ≅ △QCP [By AAS congruency]
∴, DP = CP [by cpct]
In △CDQ, QP is median. [DP = CP]
We know, median of a triangle divides it into two parts of equal areas.
∴, ar(△DPQ) = ar(△QPC) ---(i)
In △PBQ, PC is median. [AD = CQ and AD = BC ⇒ BC = QC]
Since, median of a triangle divides it into two parts of equal areas.
∴, ar(△QPC) = ar(△BPC) ---(ii)
From the equation (i) and (ii), we have ar(△BPC) = ar(△DPQ)
22. In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:
ar (BDE) =1/4 ar (ABC)
ar (BDE) = ½ ar (BAE)
ar (ABC) = 2 ar (BEC)
ar (BFE) = ar (AFD)
ar (BFE) = 2 ar (FED)
ar (FED) = 1/8 ar (AFC)
22. In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:
ar (BDE) =1/4 ar (ABC)
ar (BDE) = ½ ar (BAE)
ar (ABC) = 2 ar (BEC)
ar (BFE) = ar (AFD)
ar (BFE) = 2 ar (FED)
ar (FED) = 1/8 ar (AFC)
Explanation:
i. Let G and H are the mid-points of the sides AB and AC respectively. Join the mid-points with line-segment GH.
Here, GH is parallel to BC.
∴By midpoint theorem, GH =1/2 BC.
∴GH || BD
∴ GH = BD = DC and GH || BD (Since, D is the mid-point of BC)
Similarly, GD = HC = HA HD = AG = BG
∴, ΔABC is divided into 4 equal equilateral triangles ΔBGD, ΔAGH, ΔDHC and ΔGHD
We can say that, ΔBGD = ¼ ΔABC
Considering, ΔBDG and ΔBDE BD = BD (Common base) Since both triangles are equilateral triangle, we can say that, BG = BE and DG = DE
∴, ΔBDG ≅ ΔBDE [By SSS congruency]
∴, area (ΔBDG) = area (ΔBDE) ar (ΔBDE) = ¼ ar (ΔABC)
Hence proved
ar(ΔBDE) = ar(ΔAED) (Common base DE and DE||AB)
ar(ΔBDE) −ar(ΔFED) = ar(ΔAED)−ar (ΔFED)
ar(ΔBEF) = ar(ΔAFD) …(i)
Now, ar(ΔABD) = ar(ΔABF) +ar(ΔAFD)
ar(ΔABD) = ar(ΔABF)+ar(ΔBEF) [From (i)]
ar(ΔABD) = ar(ΔABE) …(ii)
AD is the median of ΔABC.
ar(ΔABD) = ½ ar (ΔABC) = (4/2) ar (ΔBDE) = 2 ar (ΔBDE) …(iii)
From (ii) and (iii), we have,2 ar (ΔBDE) = ar (ΔABE)
ar (BDE) = ½ ar (BAE)
Hence proved
iv.
ar(ΔABE) = ar(ΔBEC) [Common base BE and BE || AC]
ar(ΔABF) + ar(ΔBEF) = ar(ΔBEC)
From eqn (i), we get,
ar(ΔABF) + ar(ΔAFD) = ar(ΔBEC) ar(ΔABD)
= ar(ΔBEC) ½ ar(ΔABC)
= ar(ΔBEC) ar(ΔABC) = 2 ar(ΔBEC)
Hence proved
v.ar(ΔABE) = ar(ΔBEC) [Common base BE and BE || AC]
ar(ΔABF) + ar(ΔBEF) = ar(ΔBEC)
From eqn (i), we get, ar(ΔABF) + ar(ΔAFD) = ar(ΔBEC)
ar(ΔABD) = ar(ΔBEC) ½ ar(ΔABC) = ar(ΔBEC)
ar(ΔABC) = 2 ar(ΔBEC)
Hence proved.
= 2 ar (ΔFED) + (1/2) ar(ΔABC) [using (v) = 2 ar (ΔFED) + ½ [4ar(ΔBDE)] [Using result of Question (i)]
= 2 ar (ΔFED) +2 ar(ΔBDE)
Since, ΔBDE and ΔAED are on the same base and between same parallels = 2 ar (ΔFED) +2 ar (ΔAED)
= 2 ar (ΔFED) +2 [ar (ΔAFD) +ar (ΔFED)]
= 2 ar (ΔFED) +2 ar (ΔAFD) +2 ar (ΔFED) [From question (viii)]
= 4 ar (ΔFED) +4 ar (ΔFED)
⇒ar (ΔAFC) = 8 ar (ΔFED)
⇒ar (ΔFED) = (1/8) ar (ΔAFC)
Hence proved
23. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar (APB)×ar (CPD) = ar (APD)×ar (BPC). [Hint: From A and C, draw perpendiculars to BD.]
ar
23. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar (APB)×ar (CPD) = ar (APD)×ar (BPC). [Hint: From A and C, draw perpendiculars to BD.]
ar
Explanation:
Given: The diagonals AC and BD of the quadrilateral ABCD, intersect at E.
Cons: From A, Draw AM ⊥BD and from C, draw CN ⊥BD.
To Prove: ar(ΔAED) × ar(ΔBEC) = ar (ΔABE) ×ar (ΔCDE)
Proof: ar(ΔABE) = ½ ×BE×AM ---(i)
ar(ΔAED) = ½ ×DE×AM ---(ii)
Dividing (i) by (ii),
ar(ΔAED) / ar(ΔABE) = DE / BE --- (iii)
Similarly, ar(ΔCDE) / ar(ΔBEC) = DE/BE ---(iv)
From (iii) and (iv),
We have, ar(ΔAED) / ar(ΔABE) = ar(ΔCDE) / ar(ΔBEC)
∴, ar(ΔAED) × ar(ΔBEC) = ar(ΔABE) ×ar (ΔCDE)
Hence proved
24. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that:
(i) ar (PRQ) = ½ ar (ARC)
(ii) ar (RQC) = (3/8) ar (ABC)
(iii) ar (PBQ) = ar (ARC)
24. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that:
(i) ar (PRQ) = ½ ar (ARC)
(ii) ar (RQC) = (3/8) ar (ABC)
(iii) ar (PBQ) = ar (ARC)
Explanation:
(i) P,Q,R are the midpoints so PC,RC,PQ are the medians in triangles ABC,APC and BPC.
We know that median divides the triangle into two triangles of equal area,
ar (ΔBPC) = ar (ΔAPC) ---(1)
ar (ΔARC) = ½ ar (ΔAPC) ---(2)
ar (ΔPQC) = ½ ar (ΔBPC) ---(3)
From eq. (1) and (3), ar (ΔPQC) = ½ ar (ΔAPC) ---(4)
From eq. (2) and (3), we get, ar (ΔPQC) = ar (ΔARC) ---(5)
P and Q are the mid-points of AB and BC respectively
∴PQ||AC and, PA = ½ AC
We know, triangles between same parallel are equal in area,
So, ar (ΔAPQ) = ar (ΔPQC) ---(6)
From eq. (5) and (6), we obtain, ar (ΔAPQ) = ar (ΔARC) ---(7)
Also, R is the mid-point of AP.
∴RQ is the median of APQ.
ar (ΔPRQ) = ½ ar (ΔAPQ) ---(8)
From (7) and (8), ar (ΔPRQ) = ½ ar (ΔARC)
Hence Proved.
(ii)
PQ is the median of ΔBPC
ar (ΔPQC) = ½ ar (ΔBPC) = (½) ×(1/2 )ar (ΔABC) = ¼ ar (ΔABC) ---(9)
Also, ar (ΔPRC) = ½ ar (ΔAPC) [From (4)]
ar (ΔPRC) = (1/2) ×(1/2)ar ( ABC) = ¼ ar(ΔABC) ---(10)
Add eq. (9) and (10),
ar (ΔPQC) + ar (ΔPRC) = (1/4) × (1/4) ar (ΔABC)
ar (quad. PQCR) = ¼ ar (ΔABC) ---(11)
Subtracting ar (ΔPRQ) from both sides of eq (11) we have,
ar (quad. PQCR)–ar (ΔPRQ) = ½ ar (ΔABC)–ar (ΔPRQ)
ar (ΔRQC) = ½ ar (ΔABC) – ½ ar (ΔARC) [From (1)]
ar (ΔARC) = ½ ar (ΔABC) –(1/2)×(1/2)ar (ΔAPC)
ar (ΔRQC) = ½ ar (ΔABC) – (1/4) ar (ΔAPC)
ar (ΔRQC) = ½ ar (ΔABC) – (1/4) × (1/2)ar (ΔABC) (PC is median of ΔABC)
ar (ΔRQC) = ½ ar (ΔABC)–(1/8)ar (ΔABC)
ar (ΔRQC) = [(1/2)-(1/8)]ar (ΔABC)
ar (ΔRQC) = (3/8) ar(ΔABC)
2ar (ΔPRQ) = ar (ΔARC) ---(12)
ar (ΔPRQ) = ½ ar (ΔAPQ) [RQ is the median of APQ] ---(13)
But, we have, (ΔAPQ) = ar (ΔPQC) [From eq. (6)] ---(14)
From eq. (13) and (14),
ar (ΔPRQ) = ½ ar (ΔPQC) ---(15)
Also, ar (ΔBPQ) = ar (ΔPQC) [PQ is the median of ΔBPC] ---(16)
From eq. (15) and (16),
ar (ΔPRQ) = ½ ar (ΔBPQ) ---(17)
From eq. (12) and (17),
2×(1/2) ar(ΔBPQ)= ar (ΔARC)
⟹ ar (ΔBPQ) = ar (ΔARC)
Hence Proved.
25. In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that:
∆MBC = ∆ABD
ar (BYXD) = 2 ar (MBC)
ar (BYXD) = ar (ABMN)
∆FCB ≅ ∆ACE
ar (CYXE) = 2 ar (FCB)
ar (CYXE) = ar (ACFG)
(vii)ar (BCED) = ar (ABMN) + ar(ACFG)
25. In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that:
∆MBC = ∆ABD
ar (BYXD) = 2 ar (MBC)
ar (BYXD) = ar (ABMN)
∆FCB ≅ ∆ACE
ar (CYXE) = 2 ar (FCB)
ar (CYXE) = ar (ACFG)
(vii)ar (BCED) = ar (ABMN) + ar(ACFG)
Explanation:
Each angle of a square is 90°. Hence, ∠ABM = ∠DBC = 90º
∴∠ABM+∠ABC = ∠DBC+∠ABC
∴∠MBC = ∠ABD
In ∆MBC and ∆ABD,
∠MBC = ∠ABD
MB = AB (in square ABMN)
BC = BD (in square BCED)
∴ ∆MBC ≅ ∆ABD (SAS congruency)
From (i), ∆MBC≅ ∆ABD
∴ar (∆MBC) = ar (∆ABD) ---(i)
Given, AX ⊥ DE and BD ⊥ DE (Adjacent sides of square BDEC)
∴ BD || AX (Two lines perpendicular to same line are parallel to each other) ∆ABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.
Area (∆YXD) = 2 Area (∆MBC) (From (i)) --- (ii)
∆MBC and ||gm ABMN are lying on the same base MB and between same parallels MB and NC.
2 ar (∆MBC) = ar (ABMN)
ar (∆YXD) = ar (ABMN) (From (ii)) --- (iii)
Again, angle of a square is 90°.
∴∠FCA = ∠BCE = 90º
∴∠FCA+∠ACB = ∠BCE+∠ACB
∴∠FCB = ∠ACE
In ∆FCB and ∆ACE,
∠FCB = ∠ACE
FC = AC (Sides of square ACFG)
CB = CE (Sides of square BCED)
∆FCB ≅ ∆ACE (By SAS congruency)
AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC)
Hence, CE || AX.
(Two lines perpendicular to the same line are parallel to each other)
Consider BACE and parallelogram CYXE
They are on the same base CE and between the same parallels CE and AX.
∴ar (∆YXE) = ar (CYXE) = 2ar (∆ACE) --- (iv)
Since, ∆FCB ≅ ∆ACE (from above)
So, they have same areas, ar (∆FCB) ≅ ar (∆ACE) --- (v)
From (iv) and (v), we get ar (CYXE) = 2 ar (∆FCB) --- (vi)
(vi) Consider BFCB and parallelogram ACFG BFCB and ||gm ACFG are lying on the same base CF and between the same parallels CF and BG.
∴ar (ACFG) = 2 ar (∆FCB)
∴ar (ACFG) = ar (CYXE) [From (vi)] --- (vii)
(vii) From the figure, we see that ar (BCED) = ar (BYXD)+ar (CYXE)
∴ar (BCED) = ar (ABMN)+ar (ACFG) [From (iii) and (vii)].
