Find the best tutors and institutes for Class 9 Tuition
Search in
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.
As the parallelogram and the rectangle have the same base and equal area, therefore, these will also lie between the same parallels.
Consider the parallelogram ABCD and rectangle ABEF as follows.
Here, it can be observed that parallelogram ABCD and rectangle ABEF are between the same parallels AB and CF.
We know that opposite sides of a parallelogram or a rectangle are of equal lengths. Therefore,
AB = EF (For rectangle)
AB = CD (For parallelogram)
∴ CD = EF
⇒ AB + CD = AB + EF ... (1)
Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.
∴ AF < AD
And similarly, BE < BC
∴ AF + BE < AD + BC ... (2)
From equations (1) and (2), we obtain
AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD
In Figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Let us draw a line segment AM ⊥ BC.
We know that,
Area of a triangle 
× Base × Altitude
It is given that DE = BD = EC
⇒ 
⇒ Area (ΔADE) = Area (ΔABD) = Area (ΔAEC)
It can be observed that Budhia has divided her field into 3 equal parts.
In Figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal.
∴ AD = BC ... (1)
Similarly, for parallelograms DCEF and ABFE, it can be proved that
DE = CF ... (2)
And, EA = FB ... (3)
In ΔADE and ΔBCF,
AD = BC [Using equation (1)]
DE = CF [Using equation (2)]
EA = FB [Using equation (3)]
∴ ΔADE ≅ BCF (SSS congruence rule)
⇒ Area (ΔADE) = Area (ΔBCF)
In Figure, 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.]
(i) Let G and H be the mid-points of side AB and AC respectively.
Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and also its length will be half of the length of BC (mid-point theorem).
⇒ GH = 
BC and GH || BD
⇒ GH = BD = DC and GH || BD (D is the mid-point of BC)
Consider quadrilateral GHDB.
GH ||BD and GH = BD
Two line segments joining two parallel line segments of equal length will also be equal and parallel to each other.
Therefore, BG = DH and BG || DH
Hence, quadrilateral GHDB is a parallelogram.
We know that in a parallelogram, the diagonal bisects it into two triangles of equal area.
Hence, Area (ΔBDG) = Area (ΔHGD)
Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and their respective diagonals are dividing them into two triangles of equal area.
ar (ΔGDH) = ar (ΔCHD) (For parallelogram DCHG)
ar (ΔGDH) = ar (ΔHAG) (For parallelogram GDHA)
ar (ΔBDE) = ar (ΔDBG) (For parallelogram BEDG)
ar (ΔABC) = ar(ΔBDG) + ar(ΔGDH) + ar(ΔDCH) + ar(ΔAGH)
ar (ΔABC) = 4 × ar(ΔBDE)
Hence, 
(ii)Area (ΔBDE) = Area (ΔAED) (Common base DE and DE||AB)
Area (ΔBDE) − Area (ΔFED) = Area (ΔAED) − Area (ΔFED)
Area (ΔBEF) = Area (ΔAFD) (1)
Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)
Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (1)]
Area (ΔABD) = Area (ΔABE) (2)
AD is the median in ΔABC.
From (2) and (3), we obtain
2 ar (ΔBDE) = ar (ΔABE)
Or, 
(iii)
ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)
ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)
Using equation (1), we obtain
ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC)
ar (ΔABD) = ar (ΔBEC)
ar (ΔABC) = 2 ar (ΔBEC)
(iv)It is seen that ΔBDE and ar ΔAED lie on the same base (DE) and between the parallels DE and AB.
∴ar (ΔBDE) = ar (ΔAED)
⇒ ar (ΔBDE) − ar (ΔFED) = ar (ΔAED) − ar (ΔFED)
∴ar (ΔBFE) = ar (ΔAFD)
(v)Let h be the height of vertex E, corresponding to the side BD in ΔBDE.
Let H be the height of vertex A, corresponding to the side BC in ΔABC.
In (i), it was shown that
In (iv), it was shown that ar (ΔBFE) = ar (ΔAFD).
∴ ar (ΔBFE) = ar (ΔAFD)
= 
= 2 ar (ΔFED)
Hence, 
(vi) Area (AFC) = area (AFD) + area (ADC)
Now, by (v), … (6)
Therefore, from equations (5), (6), and (7), we get:
In Figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE
intersects BC at F, show that
(i) ar (BDE) = ar (ABC)
(ii) ar (BDE) = ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
(i) Let G and H be the mid-points of side AB and AC respectively.
Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and also its length will be half of the length of BC (mid-point theorem).
⇒ GH = BC and GH || BD
⇒ GH = BD = DC and GH || BD (D is the mid-point of BC)
Consider quadrilateral GHDB.
GH ||BD and GH = BD
Two line segments joining two parallel line segments of equal length will also be equal and parallel to each other.
Therefore, BG = DH and BG || DH
Hence, quadrilateral GHDB is a parallelogram.
We know that in a parallelogram, the diagonal bisects it into two triangles of equal area.
Hence, Area (ΔBDG) = Area (ΔHGD)
Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and their respective diagonals are dividing them into two triangles of equal area.
ar (ΔGDH) = ar (ΔCHD) (For parallelogram DCHG)
ar (ΔGDH) = ar (ΔHAG) (For parallelogram GDHA)
ar (ΔBDE) = ar (ΔDBG) (For parallelogram BEDG)
ar (ΔABC) = ar(ΔBDG) + ar(ΔGDH) + ar(ΔDCH) + ar(ΔAGH)
ar (ΔABC) = 4 × ar(ΔBDE)
Hence,
(ii)Area (ΔBDE) = Area (ΔAED) (Common base DE and DE||AB)
Area (ΔBDE) − Area (ΔFED) = Area (ΔAED) − Area (ΔFED)
Area (ΔBEF) = Area (ΔAFD) (1)
Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)
Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (1)]
Area (ΔABD) = Area (ΔABE) (2)
AD is the median in ΔABC.
From (2) and (3), we obtain
2 ar (ΔBDE) = ar (ΔABE)
Or,
(iii)
ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)
ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)
Using equation (1), we obtain
ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC)
ar (ΔABD) = ar (ΔBEC)
ar (ΔABC) = 2 ar (ΔBEC)
(iv)It is seen that ΔBDE and ar ΔAED lie on the same base (DE) and between the parallels DE and AB.
∴ar (ΔBDE) = ar (ΔAED)
⇒ ar (ΔBDE) − ar (ΔFED) = ar (ΔAED) − ar (ΔFED)
∴ar (ΔBFE) = ar (ΔAFD)
(v)Let h be the height of vertex E, corresponding to the side BD in ΔBDE.
Let H be the height of vertex A, corresponding to the side BC in ΔABC.
In (i), it was shown that
In (iv), it was shown that ar (ΔBFE) = ar (ΔAFD).
∴ ar (ΔBFE) = ar (ΔAFD)
=
= 2 ar (ΔFED)
Hence,
(vi) Area (AFC) = area (AFD) + area (ADC)
Now, by (v), … (6)
Therefore, from equations (5), (6), and (7), we get:
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.]
Let us draw AM ⊥ BD and CN ⊥ BD
Area of a triangle 
∴ ar (APB) × ar (CPD) = ar (APD) × ar (BPC)
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) = ar (ARC)
(ii) ar (RQC) = ar (ABC)
(iii) ar (PBQ) = ar (ARC)
Take a point S on AC such that S is the mid-point of AC.
Extend PQ to T such that PQ = QT.
Join TC, QS, PS, and AQ.
In ΔABC, P and Q are the mid-points of AB and BC respectively. Hence, by using mid-point theorem, we obtain
PQ || AC and PQ 
⇒ PQ || AS and PQ = AS (As S is the mid-point of AC)
∴ PQSA is a parallelogram. We know that diagonals of a parallelogram bisect it into equal areas of triangles.
∴ ar (ΔPAS) = ar (ΔSQP) = ar (ΔPAQ) = ar (ΔSQA)
Similarly, it can also be proved that quadrilaterals PSCQ, QSCT, and PSQB are also parallelograms and therefore,
ar (ΔPSQ) = ar (ΔCQS) (For parallelogram PSCQ)
ar (ΔQSC) = ar (ΔCTQ) (For parallelogram QSCT)
ar (ΔPSQ) = ar (ΔQBP) (For parallelogram PSQB)
Thus,
ar (ΔPAS) = ar (ΔSQP) = ar (ΔPAQ) = ar (ΔSQA) = ar (ΔQSC) = ar (ΔCTQ) = ar (ΔQBP) ... (1)
Also, ar (ΔABC) = ar (ΔPBQ) + ar (ΔPAS) + ar (ΔPQS) + ar (ΔQSC)
ar (ΔABC) = ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ)
= ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ)
= 4 ar (ΔPBQ)
⇒ ar (ΔPBQ) = 
ar (ΔABC) ... (2)
(i)Join point P to C.
In ΔPAQ, QR is the median.
... (3)
In ΔABC, P and Q are the mid-points of AB and BC respectively. Hence, by using mid-point theorem, we obtain
PQ
Also, PQ || AC 
PT || AC
Hence, PACT is a parallelogram.
ar (PACT) = ar (PACQ) + ar (ΔQTC)
= ar (PACQ) + ar (ΔPBQ [Using equation (1)]
∴ ar (PACT) = ar (ΔABC) ... (4)
(ii)
(iii)In parallelogram PACT,
In Figure, 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:
(i)Triangle MBC ≅ Triangle ABD
(ii) ar (BYXD) = 2 ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) Triangle FCB ≅ Triangle ACE
(v) ar (CYXE) = 2 ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler
proof of this theorem in Class X.
(i) We know that each angle of a square is 90°.
Hence, ∠ABM = ∠DBC = 90º
⇒ ∠ABM + ∠ABC = ∠DBC + ∠ABC
⇒ ∠MBC = ∠ABD
In ΔMBC and ΔABD,
∠MBC = ∠ABD (Proved above)
MB = AB (Sides of square ABMN)
BC = BD (Sides of square BCED)
∴ ΔMBC ≅ ΔABD (SAS congruence rule)
(ii) We have
ΔMBC ≅ ΔABD
⇒ ar (ΔMBC) = ar (ΔABD) ... (1)
It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square
BDEC)
⇒ BD || AX (Two lines perpendicular to same line are parallel to each other)
ΔABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.
Area (BYXD) = 2 area (ΔMBC) [Using equation (1)] ... (2)
(iii) ΔMBC and parallelogram ABMN are lying on the same base MB and between same parallels MB and NC.
2 ar (ΔMBC) = ar (ABMN)
ar (BYXD) = ar (ABMN) [Using equation (2)] ... (3)
(iv) We know that each angle of a square is 90°.
∴ ∠FCA = ∠BCE = 90º
⇒ ∠FCA + ∠ACB = ∠BCE + ∠ACB
⇒ ∠FCB = ∠ACE
In ΔFCB and ΔACE,
∠FCB = ∠ACE
FC = AC (Sides of square ACFG)
CB = CE (Sides of square BCED)
ΔFCB ≅ ΔACE (SAS congruence rule)
(v) It is given that 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 ΔACE and parallelogram CYXE
ΔACE and parallelogram CYXE are on the same base CE and between the same parallels CE and AX.
⇒ ar (CYXE) = 2 ar (ΔACE) ... (4)
We had proved that
∴ ΔFCB ≅ ΔACE
ar (ΔFCB) ≅ ar (ΔACE) ... (5)
On comparing equations (4) and (5), we obtain
ar (CYXE) = 2 ar (ΔFCB) ... (6)
(vi) Consider ΔFCB and parallelogram ACFG
ΔFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.
⇒ ar (ACFG) = 2 ar (ΔFCB)
⇒ ar (ACFG) = ar (CYXE) [Using equation (6)] ... (7)
(vii) From the figure, it is evident that
ar (BCED) = ar (BYXD) + ar (CYXE)
⇒ ar (BCED) = ar (ABMN) + ar (ACFG) [Using equations (3) and (7)]
How helpful was it?
How can we Improve it?
Please tell us how it changed your life *
Please enter your feedback
UrbanPro.com helps you to connect with the best Class 9 Tuition in India. Post Your Requirement today and get connected.
Find best tutors for Class 9 Tuition Classes by posting a requirement.
Get started now, by booking a Free Demo Class
