Ex-15.3 (Part - 1), Areas Of Parallelograms And Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q 1. In figure, compute the area of quadrilateral ABCD.

Solution:

Given:

DC = 17 cm, AD = 9 cm and BC = 8 cm

In ΔBCD we have

CD² = BD²+BC²

⇒ 17² = BD²+8²

⇒ BD² = 289−64

⇒ BD = 15

In ΔABD we have

AB²+AD² = BD²

⇒ 15² = AB²+9²

⇒ AB² = 225−81 = 144

⇒ AB = 12

ar(quadABCD) = ar(ΔABD)+ar(ΔBCD)

ar(quadABCD) = (12×9)+(8×17) = 54+68 = 122cm²

ar(quadABCD) = (12×9)+(8×15) = 54+60 = 114cm²

Q2. In figure, PQRS is a square and T and U are, respectively, the midpoints of PS and QR . Find the area of ΔOTS if PQ = 8 cm.

Solution:

From the figure,

T and U are mid points of PS and QR respectively

∴TU||PQ

⇒ TO||PQ

Thus , in ΔPQS , T   is the mid point of PS and TO||PQ

∴TO = PQ = 4cm

Also, TS =  PS = 4cm

∴ar(ΔOTS) = (TO×TS) = (4×4)cm² = 8cm²

Q3. Compute the area of trapezium PQRS in figure

Solution:

We have,

ar(trap.PQRS) = ar(rect.PSRT)+ar(ΔQRT)

⇒ ar(trap.PQRS) = PT×RT+(QT×RT)

=   = 8×RT+(8×RT) = 12×RT

In ΔQRT , we have

QR² = QT²+RT²

⇒ RT² = QR²−QT²

⇒ RT² = 17²−8² = 225

⇒ RT = 15

Hence , area of trapezium =  12×15cm²  =  180cm²

Q4. In figure, ∠AOB = 90^∘ , AC = BC , OA = 12 cm and OC = 6.5 cm .Find the area of ΔAOB.

Solution:

Since, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices

∴CA = CB = OC

⇒ CA = CB = 6.5cm

⇒ AB = 13cm

In right angled triangle OAB , we have

AB² = OB²+OA²

⇒ 13² = OB²+12²

⇒ OB² = 13²−12² = 169–144 = 25

⇒ OB = 5

∴ar(ΔAOB) = (12×5) = 30cm²

Q5. In figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm , and distance between AB and DC is 4 cm . Find the value of x and area of trapezium ABCD.

Solution:

Draw AL ⊥ DC, BM ⊥ DC then ,

AL = BM = 4 cm and LM = 7 cm.

In Δ ADL , we have

AD² = AL²+DL²

⇒ 25 = 16+DL²

⇒ DL = 3cm

Similarly, MC = 

∴x = CD = CM+ML+LD = (3+7+3)cm  = 13 cm

ar(trap.ABCD) = (AB+CD)×AL = (7+13)×4cm² = 40cm²

Q 6. In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = , find the area of the rectangle .

Solution:

Given OD = 10 cm and OE =   

By using Pythagoras theorem

∴OD² = OE²+DE²

⇒ DE =

∴ Area of rectangle OCDE =  OE×DE =  = 40cm²

Q 7. In figure, ABCD is a trapezium in which AB || DC. Prove that ar(ΔAOD) = ar(ΔBOC)

Solution:

Given:  ABCD is a trapezium in which AB || DC

To prove: ar(ΔAOD) = ar(ΔBOC)

Proof:- Since , ΔADC and  ΔBDC  are on the same base DC and between same parallels AB and DC

Then, ar(ΔADC) = ar(ΔBDC)

⇒ ar(ΔAOD)+ar(ΔDOC) = ar(ΔBOC)+ar(ΔDOC)

⇒ ar(ΔAOD) = ar(ΔBOC)

Q 8. In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(ΔADE) = ar(ΔBCF).

Solution:

Given that

ABCD is parallelogram ⇒ AD = BC

CDEF is parallelogram ⇒ DE = CF

ABFE is parallelogram ⇒ AE = BF

Thus, in Δs ADF and BCF , we have

AD = BC, DE = CF and AE = BF

So, by SSS criterion of congruence, we have

ΔADE≅ΔBCF

ar(ΔADE) = ar(ΔBCF)

Q 9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that : ar(ΔAPB)×ar(ΔCPD) = ar(ΔAPD)×ar(ΔBPC).

Solution:

Construction: – Draw BQ ⊥AC and DR ⊥ AC

Proof:-

L.H.S

=  ar(ΔAPB)×ar(ΔCDP)

=  [(AP×BQ)]×(×PC×DR)

=  (×PC×BQ)×(×AP×DR)

=  ar(ΔAPD)×ar(ΔBPC).

= R.H.S

Hence proved.

Q 10.  In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD).

Solution:

Given that CD is bisected by AB at O

To prove: ar(ΔABC) = ar(ΔABD).

Construction: Draw CP ⊥ AB and DQ  ⊥ AB .

Proof:

ar(ΔABC) = ×AB×CP⋅⋅⋅⋅⋅⋅⋅(1)

ar(ΔABD) = ×AB×DQ⋅⋅⋅⋅⋅⋅⋅(2)

In ΔCPO and ΔDQO

∠CPO = ∠DQO                     [each90^∘]

Given that, CO = OD

∠COP = ∠DOQ                          [Vertically opposite angles are equal]

Then , ΔCP0≅ΔDQO                  [By AAS condition]

∴CP = DQ                          (3)    [c.p.c.t]

Compare equation (1), (2) and (3)

∴ ar(ΔABC) = ar(ΔABD).

Q 11. If P is any point in the interior of a parallelogram ABCD , then prove that area of the triangle APB is less than half the area of parallelogram.

Solution:

Draw DN ⊥ AB and PM ⊥ AB

Now ,

ar(||^gmABCD) = AB×DN,ar(ΔAPB) = (AB×PM)

Now , PM  < DN

⇒ AB×PM<AB×DN

⇒ (AB×PM)<(AB×DN)

⇒ ar(ΔAPB)<ar(||^gmABCD)

Q 12. If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of the median AD, prove that ar(ΔBGC) = 2ar(ΔAGC).

Solution:

Draw AM ⊥ BC

Since, AD is the median of ΔABC

∴ BD = DC

⇒ BD = AM = DC×AM

⇒ (BD×AM) = (DC×AM)

⇒ ar(ΔABD) = ar(ΔACD)⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)

In ΔBGC , GD is the median

∴ ar(ΔBGD) = ar(ΔCGD)⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)

In ΔACD , CG is the median

∴ ar(ΔAGC) = ar(ΔCGD)⋅⋅⋅⋅⋅⋅⋅⋅⋅(3)

From (2) and (3) we have,

ar(ΔBGD) = ar(ΔAGC)

But, ar(ΔBGC) = 2ar(ΔBGD)

∴ ar(ΔBGC) = 2ar(ΔAGC)

Q 13. A point D is taken on the side BC of a ΔABC , such that BD = 2DC . Prove that   ar(ΔABD) = 2ar(ΔADC).

Solution:

Given that,

In ΔABC, BD = 2DC

To prove: ar(ΔABD) = 2ar(ΔADC).

Construction:

Take a point E on BD such that BE = ED

Proof: Since, BE = ED and BD = 2 DC

Then, BE = ED = DC

We know that median of triangle divides it into two equal triangles.

∴ In ΔABD , AE is the median .

Then , ar(ΔABD) = 2ar(ΔAED) ⋅⋅⋅⋅⋅⋅(1)

In ΔAEC , AD is the median .

Then, ar(ΔAED) = 2ar(ΔADC) ⋅⋅⋅⋅⋅⋅(2)

Compare equation 1 and 2

ar(ΔABD) = 2ar(ΔADC).

Q 14. ABCD is a parallelogram whose diagonals intersect at O .If P is any point on BO, prove that :

(i) . ar(ΔADO) = ar(ΔCDO).

(ii) . ar(ΔABP) = 2ar(ΔCBP).

Solution:

Given that ABCD is the parallelogram

To Prove:    (i) ar(ΔADO) = ar(ΔCDO).

(ii) ar(ΔABP) = 2ar(ΔCBP).

Proof:  we know that diagonals of parallelogram bisect each other

∴ AO = OC and BO = OD

(i)  . In Δ DAC , since DO is a median .

Then ar(ΔADO) = ar(ΔCDO).

(ii) . In Δ BAC , since BO is a median .

Then ar(ΔBAO) = ar(ΔBCO)⋅⋅⋅⋅⋅⋅⋅(1)

In Δ PAC , since PO is a median .

Then ar(ΔPAO) = ar(ΔPCO)⋅⋅⋅⋅⋅⋅⋅(2)

Subtract equation 2 from 1.

⇒ ar(ΔBAO)−ar(ΔPAO) = ar(ΔBCO)−ar(ΔPCO)

⇒  ar(ΔABP) = 2ar(ΔCBP).

Q 15. ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.

(i)  . Prove that ar(ΔADF) = ar(ΔECF).

(ii)  . If the area of  ΔDFB  =   3cm² , find the area of ||^gm ABCD .

Solution:

In triangles ADF and ECF, we have

∠ADF = ∠ECF                 [Alternate interior angles,SinceAD||BE]

AD = EC             [sinceAD = BC = CE]

And ∠DFA = ∠CFA         [Vertically opposite angles]

So, by AAS congruence criterion, we have

ΔADF≅ΔECF

⇒  ar(ΔADF) = ar(ΔECF) and DF = CF .

Now, DF = CF

⇒  BF is a median in Δ BCD.

⇒  ar(ΔBCD) = 2ar(ΔBDF)

⇒ ar(ΔBCD) = 2×3cm² = 6cm²

Hence, area of a parallelogram =  2ar(ΔBCD) = 2×6cm² = 12cm²

Q 16. ABCD is a parallelogram whose diagonals AC and BD intersect at O . A line through O intersects AB at P and DC at Q. Prove that ar(ΔPOA) = ar(ΔQOC).

Solution:

In triangles POA and QOC, we have

∠AOP = ∠COQ

AO = OC

∠PAC = ∠QCA

So, by ASA congruence criterion  , we have

ΔPOA≅ΔQOC

⇒   ar(ΔPOA) = ar(ΔQOC).

Q 17. ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.

Solution:

Draw FG ⊥ AB

We have,

BE = 2 EA and DF = 2FC

⇒  AB – AE = 2 AE and DC – FC = 2 FC

⇒  AB = 3 AE and DC = 3 FC

⇒ AE = AB and FC = DC⋅⋅⋅⋅⋅⋅⋅⋅(1)

But AB = DC

Then, AE = FC                                    [opposite sides of ||^gm]

Thus, AE = FC and AE || FC

Then, AECF is a parallelogram

Now, area of parallelogram AECF =  AE×FG

⇒ ar(||^gmAECF) = 13AB×FG          from(1)

⇒ 3ar(||^gmAECF) = AB×FG       ⋅⋅⋅⋅⋅(2)

Andar(||^gmABCD) = AB×FG        ⋅⋅⋅⋅⋅(3)

Compare equation 2 and 3

⇒ 3ar(||^gmAECF) = ar(||^gmABCD)

⇒ ar(||^gmAECF) = ar(||^gmABCD)

Q 18. In a triangle ABC, P and Q are respectively the mid points of AB and BC and R is the mid point of AP. Prove that :

                (i)  . ar(ΔPBQ) = ar(ΔARC).

                (ii)  . ar(ΔPRQ) =  ar(ΔARC).

                (iii) . ar(ΔRQC) =  ar(ΔABC).

Solution:

We know that each median of a triangle divides it into two triangles of equal area.

(i)  .  Since CR is the median of Δ CAP

∴ar(ΔCRA) = ar(ΔCAP)⋅⋅⋅⋅⋅⋅⋅(1)

Also , CP is the median of a Δ CAB

∴ar(ΔCAP) = ar(ΔCPB)⋅⋅⋅⋅⋅⋅⋅(2)

From 1 and 2 , we get

∴ar(ΔARC) = ar(ΔCPB)⋅⋅⋅⋅⋅⋅⋅(3)

PQ is the median of a Δ PBC

∴ar(ΔCPB) = 2ar(ΔPBQ)⋅⋅⋅⋅⋅⋅⋅(4)

From 3 and 4, we get

∴ar(ΔARC) = ar(ΔPBQ)⋅⋅⋅⋅⋅⋅⋅(5)

(ii) . Since QP and QR medians of triangles QAB and QAP respectively

∴ar(ΔQAP) = ar(ΔQBP)⋅⋅⋅⋅⋅⋅⋅(6)

And ar(ΔQAP) = 2ar(ΔQRP)⋅⋅⋅⋅⋅⋅⋅(7)

From 6 and 7, we get

ar(ΔPRQ) = ar(ΔPBQ)⋅⋅⋅⋅⋅⋅⋅(8)

From 5 and 8, we get

ar(ΔPRQ) = ar(ΔARC)

(iii) . Since, LR is a median of Δ CAP

∴ar(ΔARC) = ar(ΔCAD)

= ×ar(ΔABC)

= ar(ΔABC)

Since RQ is  the median of  Δ RBC.

∴ar(ΔRQC) = ar(ΔRBC)

= {ar(ΔABC)−ar(ΔARC)}

= {ar(ΔABC)−ar(ΔABC)}

= ar(ΔABC)

Q 19. ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

                (i)  . ar(ADEG) = ar(GBCE).

                (ii)  . ar(ΔEGB) =  ar(ABCD).

                (iii) . ar(ΔEFC) =  ar(ΔEBF).

                (iv)  . ar(ΔEGB) = ×ar(ΔEFC)

                (v) . Find what portion of the area of parallelogram is the area of   Δ EFG.

Solution:

Given: ABCD is a parallelogram

AG = 2 GB, CE = 2 DE and BF = 2 FC

To prove:

(i)  . ar(ADEG) = ar(GBCE).

(ii)  . ar(ΔEGB) =  ar(ABCD).

(iii) . ar(ΔEFC) =  ar(ΔEBF).

(iv)  . ar(ΔEGB) = ×ar(ΔEFC)

(v) . Find what portion of the area of parallelogram is the area of   Δ EFG.

Construction: Draw a parallel line to AB through point F and a perpendicular line to AB from C

Proof:

(i)  .Since ABCD is a parallelogram

So, AB = CD and AD = BC

Consider the two trapezium s ADEG and GBCE

Since AB = DC, EC = 2DE, AG = 2GB

⇒ ED = CD = AB and EC =  CD = 23AB

⇒ AG = AB and BG = AB

So , DE+AG = AB +AB = AB and EC+BG = AB +AB = AB

Since the two trapezium ADEG and GBCE have same height and their sum of two parallel sides are equal

Since Area of trapezium =  

So, ar(ADEG) = ar(GBCE).

(ii)  . Since we know from above that

BG = AB. So

ar(ΔEGB) = ×GB×Height

ar(ΔEGB) = ××AB×Height

ar(ΔEGB) = ×AB×Height

ar(ΔEGB) =  ar(ABCD).

(iii) . Since height if triangle EFC and EBF are equal.So

ar(ΔEFC) = ×FC×Height

ar(ΔEFC) = ××FB×Height

ar(ΔEFC) = ar(EBF)

Hence , ar(ΔEFC) =  ar(ΔEBF).

(iv) . Consider the trapezium in which

ar(EGBC) = ar(ΔEGB)+ar(ΔEBF)+ar(ΔEFC)

⇒ ar(ABCD) = ar(ABCD)+2ar(ΔEFC)+ar(ΔEFC)

⇒ ar(ABCD) = 3ar(ΔEFC)

⇒ ar(ΔEFC) = ar(ABCD)

Now from (ii)part we have

ar(ΔEGB) = ar(ΔEFC)

ar(ΔEGB) = ar(ABCD)

ar(ΔEGB) = ar(ΔEFC)

∴ar(ΔEGB) = ar(ΔEFC)

(v) . In the figure it is given that FB = 2CF .Let CF = x and FB = 2x.

Now consider the two triangles CFI and CBH which are similar triangle.

So by the property of similar triangle CI = k and IH = 2k

Now consider the triangle EGF in which

ar(ΔEFG) = ar(ΔESF)+ar(ΔSGF)

ar(ΔEFG) = SF×k+SF×2k

ar(ΔEFG) = SF×k⋅⋅⋅⋅⋅(i)

Now ,

ar(ΔEGBC) = ar(SGBF)+ar(ESFC)

ar(ΔEGBC) = (SF+GB)×2k+(SF+EC)×k

ar(ΔEGBC) = k×SF+(GB+EC)×k

ar(ΔEGBC) = k×SF+(AB+×AB)×k

ar(ΔABCD) = k×SF+AB×k

⇒ ar(ΔABCD) = 3k×SF+AB×k                         [Multiply both sides by 2]

⇒ ar(ΔABCD) = 3k×SF+ar(ABCD)

⇒ k×SF = ar(ABCD)⋅⋅⋅⋅⋅⋅⋅(2)

From 1 and 2 we have ,

ar(ΔEFG) =ar(ABCD)

ar(ΔEFG) = ar(ABCD)