Important Questions & Solutions For Class 9 Maths Chapter 9 (Areas of Parallelograms & Triangles)
Q.1. In Figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Solution:
Given,
AB = CD = 16 cm (Opposite sides of a ||gm are equal)
CF = 10 cm and AE = 8 cm
Now,
Area of parallelogram = Base × Altitude
= CD × AE = AD × CF
⇒ 16 × 8 = AD × 10
⇒ AD = 128/10
⇒ AD = 12.8 cm
Q.2. 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).
Solution:
Given,
E, F, G and H are the mid-points of the sides of a parallelogram ABCD, respectively.
To Prove: ar (EFGH) = ½ ar(ABCD)
Construction: H and F are joined.
Proof:
AD || BC and AD = BC (Opposite sides of a ||gm)
⇒ ½ AD = ½ BC
Also,
AH || BF and and DH || CF
⇒ AH = BF and DH = CF (H and F are mid points)
Therefore, ABFH and HFCD are parallelograms.
Now,
As we know,, ΔEFH and ||gm 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 — (i)
And,
area of GHF = ½ area of HFCD — (ii)
Adding (i) and (ii),
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)
Q.3. In Figure, P is a point in the interior of a parallelogram ABCD. Show that
(i)ar(APB) + ar(PCD) = ½ ar(ABCD)
(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
[Hint : Through P, draw a line parallel to AB.]
(i) A line GH is drawn parallel to AB passing through P.
In a parallelogram,
AB || GH (by construction) — (i)
∴,
AD || BC ⇒ AG || BH — (ii)
From equations (i) and (ii),
ABHG is a parallelogram.
Now,
ΔAPB and ||gm ABHG are lying on the same base AB and in-between the same parallel lines AB and GH.
∴ ar(ΔAPB) = ½ ar(ABHG) — (iii)
also,
ΔPCD and ||gm 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 equations (iii) and (iv),
ar(ΔAPB) + ar(ΔPCD) = ½ {ar(ABHG) + ar(CDGH)}
⇒ ar(APB) + ar(PCD) = ½ ar(ABCD)
(ii)A line EF is drawn parallel to AD passing through P.
In the parallelogram,
AD || EF (by construction) — (i)
∴,
AB || CD ⇒ AE || DF — (ii)
From equations (i) and (ii),
AEDF is a parallelogram.
Now,
ΔAPD and ||gm AEFD are lying on the same base AD and in-between the same parallel lines AD and EF.
∴ ar(ΔAPD) = ½ ar(AEFD) — (iii)
also,
ΔPBC and ||gm BCFE is 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),
ar(ΔAPD) + ar(ΔPBC) = ½ {ar(AEFD) + ar(BCFE)}
⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)
Q.4. In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = 1/4 ar(ABC).
Solution:
ar(BED) = ½ × BD × DE
Since, E is the mid-point of AD,
AE = DE
Since, AD is the median on side BC of triangle ABC,
BD = DC
DE = ½ AD — (i)
BD = ½ BC — (ii)
From (i) and (ii), we get,
ar(BED) = (1/2) × (½) BC × (1/2)AD
⇒ ar(BED) = (½) × (½) ar(ABC)
⇒ ar(BED) = ¼ ar(ABC)
Q.5. In Figure, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that: ar(ABC) = ar(ABD).
Solution:
In ΔABC, AO is the median. (CD is bisected by AB at O)
∴ ar(AOC) = ar(AOD) — (i)
also,
ΔBCD, BO is the median. (CD is bisected by AB at O)
∴ ar(BOC) = ar(BOD) — (ii)
Adding (i) and (ii),
We get,
ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD)
⇒ ar(ABC) = ar(ABD)
6. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.
Show that
(i) BDEF is a parallelogram.
(ii) ar(DEF) = ¼ ar(ABC)
(iii) ar (BDEF) = ½ ar(ABC)
Solution:
(i)In ΔABC,
EF || BC and EF = ½ BC (by midpoint theorem)
also,
BD = ½ BC (D is the midpoint)
So, BD = EF
also,
BF and DE are parallel and equal to each other.
∴, the pair opposite sides are equal in length and parallel to each other.
∴ BDEF is a parallelogram.
(ii)Proceeding from the result of (i),
BDEF, DCEF, AFDE are parallelograms.
Diagonal of a parallelogram divides it into two triangles of equal area.
∴ ar(ΔBFD) = ar(ΔDEF) (For ||gm BDEF) — (i)
also,
ar(ΔAFE) = ar(ΔDEF) (For ||gm DCEF) — (ii)
ar(ΔCDE) = ar(ΔDEF) (For ||gm AFDE) — (iii)
From (i), (ii) and (iii)
ar(ΔBFD) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF)
⇒ ar(ΔBFD) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = arar(ΔABC)
⇒ 4 ar(ΔDEF) = ar(ΔABC)
⇒ ar(DEF) = ¼ ar(ABC)
(iii)Area (||gm BDEF) = ar(ΔDEF) + ar(ΔBDE)
⇒ ar(BDEF) = ar(ΔDEF) + ar(ΔDEF)
⇒ ar(BDEF) = 2× ar(ΔDEF)
⇒ ar(BDEF) = 2× ¼ ar(ΔABC)
⇒ ar(BDEF) = (1/2)ar(ΔABC)
Q. 7. In Figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.
If AB = CD, then show that:
ar (DOC) = ar (AOB)
ar (DCB) = ar (ACB)
DA || CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]
Given: OB = OD and AB = CD
Construction: DE ⊥ AC and BF ⊥ AC are drawn.
Proof:
(i)In ΔDOE and ΔBOF,
∠DEO = ∠BFO (Perpendiculars)
∠DOE = ∠BOF (Vertically opposite angles)
OD = OB (Given)
∴, ΔDOE ≅ ΔBOF by AAS congruence condition.
∴, DE = BF (By CPCT) — (1)
also, ar(ΔDOE) = ar(ΔBOF) (Congruent triangles) — (2)
Now,
In ΔDEC and ΔBFA,
∠DEC = ∠BFA (Perpendiculars)
CD = AB (Given)
DE = BF (From eq.1)
∴, ΔDEC ≅ ΔBFA by RHS congruence condition.
∴, ar(ΔDEC) = ar(ΔBFA) (Congruent triangles) — (3)
Adding (2) and (3),
ar(ΔDOE) + ar(ΔDEC) = ar(ΔBOF) + ar(ΔBFA)
⇒ ar (DOC) = ar (AOB)
(ii)ar(ΔDOC) = ar(ΔAOB)
Adding ar(ΔOCB) in LHS and RHS, we get,
⇒ar(ΔDOC)+ar(ΔOCB)=ar(ΔAOB)+ar(ΔOCB)
⇒ ar(ΔDCB) = ar(ΔACB)
(iii)When two triangles have the same base and equal areas, the triangles will be in between the same parallel lines
ar(ΔDCB) = ar(ΔACB)
DA || BC — (4)
For quadrilateral ABCD, one pair of opposite sides are equal (AB = CD) and other pair of opposite sides are parallel.
∴, ABCD is a parallelogram.
Q. 8. 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(ΔAC)
Solution:
Given: XY || BC, BE || AC and CF || AB
To show: ar(ΔABE) = ar(ΔAC)
Proof:
BCYE is a ||gm as 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 parallelogram
As ΔACF and ||gm BCFX are on the same base CF and in-between the same parallel AB and FC.
∴, ar (ΔACF)= ½ ar (BCFX) … (2)
But,
||gm BCFX and ||gm 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) , we get
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…..(4)
Also,
△AEB and ||gm BEYC is on the same base BE and in-between the same parallels BE and AC.
⇒ ar(△AEB) = ½ ar(BEYC) ……(5)
Similarly,
△ACF and ||gm BXFC on the same base CF and between the same parallels CF and AB.
⇒ ar(△ ACF) = ½ ar(BXFC) ……..(6)
From (4), (5) and (6),
ar(△AEB) = ar(△ACF)
Q.9. 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.
Solution:
Given,
||gm ABCD and a rectangle ABEF have the same base AB and equal areas.
To prove,
The perimeter of ||gm ABCD is greater than the perimeter of rectangle ABEF.
Proof,
As we know, the opposite sides of a||gm and rectangle are equal.
AB = DC [As ABCD is a ||gm]
and, AB = EF [As ABEF is a rectangle]
DC = EF … (i)
Adding AB on both sides, we get,
⇒ AB + DC = AB + EF … (ii)
As we know, the perpendicular segment is the shortest of all the segments that can be drawn to a given line from a point not lying on it.
BE < BC and AF < AD
⇒ BC > BE and AD > AF
⇒ BC + AD > BE + AF … (iii)
Adding (ii) and (iii), we get
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.
Q.10. 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.]
Solution:
Given: The diagonal AC and BD of the quadrilateral ABCD, intersect each other at point E.
Construction:
From A, draw AM perpendicular to BD
From C, draw CN perpendicular to 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 eq. (ii) by (i) , we get,
ar(ΔAED)/ar(ΔABE) = [1/2×DE×AM]/1/2×BE×Am]
= DE/BE ………..(iii)
Similarly,
ar(ΔCDE)/ar(ΔBEC) = DE/BE …….(iv)
From eq. (iii) and (iv), we get;
ar(ΔAED)/ar(ΔABE) = ar(ΔCDE)/ar(ΔBEC)
= ar(ΔAED) × ar(ΔBEC) = ar (ΔABE) × ar (ΔCDE)
Hence proved.
