The document RD Sharma Solutions -Ex-15.3 (Part - 1), Areas Of Parallelograms And Triangles, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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

**Solution:**

Given:

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

In Î”BCD we have

CD^{2} = BD^{2}+BC^{2}

â‡’ 17^{2} = BD^{2}+8^{2}

â‡’ BD^{2} = 289âˆ’64

â‡’ BD = 15

In Î”ABD we have

AB^{2}+AD^{2} = BD^{2}

â‡’ 15^{2} = AB^{2}+9^{2}

â‡’ AB^{2} = 225âˆ’81 = 144

â‡’ AB = 12

ar(quadABCD) = ar(Î”ABD)+ar(Î”BCD)

ar(quadABCD) = (12Ã—9)+(8Ã—17) = 54+68 = 122cm^{2}

ar(quadABCD) = (12Ã—9)+(8Ã—15) = 54+60 = 114cm^{2}

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

**Solution:**

From the figure,

T and U are mid points of PS and QR respectively

âˆ´TU||PQ

â‡’ TO||PQ

Thus , in Î”PQS , T is the mid point of PS and TO||PQ

âˆ´TO = PQ = 4cm

Also, TS = PS = 4cm

âˆ´ar(Î”OTS) = (TOÃ—TS) = (4Ã—4)cm^{2 }= 8cm^{2}

*Q3. Compute the area of trapezium PQRS in figure*

**Solution:**

We have,

ar(trap.PQRS) = ar(rect.PSRT)+ar(Î”QRT)

â‡’ ar(trap.PQRS) = PTÃ—RT+(QTÃ—RT)

= = 8Ã—RT+(8Ã—RT) = 12Ã—RT

In Î”QRT , we have

QR^{2} = QT^{2}+RT^{2}

â‡’ RT^{2} = QR^{2}âˆ’QT^{2}

â‡’ RT^{2} = 17^{2}âˆ’8^{2} = 225

â‡’ RT = 15

Hence , area of trapezium = 12Ã—15cm^{2} = 180cm^{2}

*Q4. In figure, âˆ AOB = 90 ^{âˆ˜} , AC = BC , OA = 12 cm and OC = 6.5 cm .Find the area of Î”AOB.*

**Solution:**

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

âˆ´CA = CB = OC

â‡’ CA = CB = 6.5cm

â‡’ AB = 13cm

In right angled triangle OAB , we have

AB^{2} = OB^{2}+OA^{2}

â‡’ 13^{2} = OB^{2}+12^{2}

â‡’ OB^{2} = 13^{2}âˆ’12^{2} = 169â€“144 = 25

â‡’ OB = 5

âˆ´ar(Î”AOB) = (12Ã—5) = 30cm^{2}

*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 âŠ¥ DC, BM âŠ¥ DC then ,

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

In Î” ADL , we have

AD^{2} = AL^{2}+DL^{2}

â‡’ 25 = 16+DL^{2}

â‡’ DL = 3cm

Similarly, MC =

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

ar(trap.ABCD) = (AB+CD)Ã—AL = (7+13)Ã—4cm^{2} = 40cm^{2}

*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

âˆ´OD^{2 }= OE^{2}+DE^{2}

â‡’ DE =

âˆ´ Area of rectangle OCDE = OEÃ—DE = = 40cm^{2}

*Q 7. In figure, ABCD is a trapezium in which AB || DC. Prove that ar(Î”AOD) = ar(Î”BOC)*

**Solution:**

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

To prove: ar(Î”AOD) = ar(Î”BOC)

Proof:- Since , Î”ADC and Î”BDC are on the same base DC and between same parallels AB and DC

Then, ar(Î”ADC) = ar(Î”BDC)

â‡’ ar(Î”AOD)+ar(Î”DOC) = ar(Î”BOC)+ar(Î”DOC)

â‡’ ar(Î”AOD) = ar(Î”BOC)

*Q 8. In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(Î”ADE) = ar(Î”BCF).*

**Solution:**

Given that

ABCD is parallelogram â‡’ AD = BC

CDEF is parallelogram â‡’ DE = CF

ABFE is parallelogram â‡’ AE = BF

Thus, in Î”s ADF and BCF , we have

AD = BC, DE = CF and AE = BF

So, by SSS criterion of congruence, we have

Î”ADEâ‰…Î”BCF

ar(Î”ADE) = ar(Î”BCF)

*Q 9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that : ar(Î”APB)Ã—ar(Î”CPD) = ar(Î”APD)Ã—ar(Î”BPC).*

**Solution:**

Construction: â€“ Draw BQ âŠ¥AC and DR âŠ¥ AC

Proof:-

L.H.S

= ar(Î”APB)Ã—ar(Î”CDP)

= [(APÃ—BQ)]Ã—(Ã—PCÃ—DR)

= (Ã—PCÃ—BQ)Ã—(Ã—APÃ—DR)

= ar(Î”APD)Ã—ar(Î”BPC).

= 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(Î”ABC) = ar(Î”ABD).*

**Solution:**

Given that CD is bisected by AB at O

To prove: ar(Î”ABC) = ar(Î”ABD).

Construction: Draw CP âŠ¥ AB and DQ âŠ¥ AB .

Proof:

ar(Î”ABC) = Ã—ABÃ—CPâ‹…â‹…â‹…â‹…â‹…â‹…â‹…(1)

ar(Î”ABD) = Ã—ABÃ—DQâ‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

In Î”CPO and Î”DQO

âˆ CPO = âˆ DQO [each90^{âˆ˜}]

Given that, CO = OD

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

Then , Î”CP0â‰…Î”DQO [By AAS condition]

âˆ´CP = DQ (3) [c.p.c.t]

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

âˆ´ ar(Î”ABC) = ar(Î”ABD).

*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 âŠ¥ AB and PM âŠ¥ AB

Now ,

ar(||^{gm}ABCD) = ABÃ—DN,ar(Î”APB) = (ABÃ—PM)

Now , PM < DN

â‡’ ABÃ—PM<ABÃ—DN

â‡’ (ABÃ—PM)<(ABÃ—DN)

â‡’ ar(Î”APB)<ar(||^{gm}ABCD)

*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(Î”BGC) = 2ar(Î”AGC).*

**Solution:**

Draw AM âŠ¥ BC

Since, AD is the median of Î”ABC

âˆ´ BD = DC

â‡’ BD = AM = DCÃ—AM

â‡’ (BDÃ—AM) = (DCÃ—AM)

â‡’ ar(Î”ABD) = ar(Î”ACD)â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(1)

In Î”BGC , GD is the median

âˆ´ ar(Î”BGD) = ar(Î”CGD)â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

In Î”ACD , CG is the median

âˆ´ ar(Î”AGC) = ar(Î”CGD)â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(3)

From (2) and (3) we have,

ar(Î”BGD) = ar(Î”AGC)

But, ar(Î”BGC) = 2ar(Î”BGD)

âˆ´ ar(Î”BGC) = 2ar(Î”AGC)

*Q 13. A point D is taken on the side BC of a Î”ABC , such that BD = 2DC . Prove that ar(Î”ABD) = 2ar(Î”ADC).*

**Solution:**

Given that,

In Î”ABC, BD = 2DC

To prove: ar(Î”ABD) = 2ar(Î”ADC).

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.

âˆ´ In Î”ABD , AE is the median .

Then , ar(Î”ABD) = 2ar(Î”AED) â‹…â‹…â‹…â‹…â‹…â‹…(1)

In Î”AEC , AD is the median .

Then, ar(Î”AED) = 2ar(Î”ADC) â‹…â‹…â‹…â‹…â‹…â‹…(2)

Compare equation 1 and 2

ar(Î”ABD) = 2ar(Î”ADC).

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

*(i) . ar(Î”ADO) = ar(Î”CDO).*

*(ii) . ar(Î”ABP) = 2ar(Î”CBP).*

**Solution:**

Given that ABCD is the parallelogram

To Prove: (i) ar(Î”ADO) = ar(Î”CDO).

(ii) ar(Î”ABP) = 2ar(Î”CBP).

Proof: we know that diagonals of parallelogram bisect each other

âˆ´ AO = OC and BO = OD

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

Then ar(Î”ADO) = ar(Î”CDO).

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

Then ar(Î”BAO) = ar(Î”BCO)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(1)

In Î” PAC , since PO is a median .

Then ar(Î”PAO) = ar(Î”PCO)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

Subtract equation 2 from 1.

â‡’ ar(Î”BAO)âˆ’ar(Î”PAO) = ar(Î”BCO)âˆ’ar(Î”PCO)

â‡’ ar(Î”ABP) = 2ar(Î”CBP).

*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(Î”ADF) = ar(Î”ECF).*

*(ii) . If the area of Î”DFB = 3cm ^{2 }, find the area of ||^{gm} ABCD .*

**Solution:**

In triangles ADF and ECF, we have

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

AD = EC [sinceAD = BC = CE]

And âˆ DFA = âˆ CFA [Vertically opposite angles]

So, by AAS congruence criterion, we have

Î”ADFâ‰…Î”ECF

â‡’ ar(Î”ADF) = ar(Î”ECF) and DF = CF .

Now, DF = CF

â‡’ BF is a median in Î” BCD.

â‡’ ar(Î”BCD) = 2ar(Î”BDF)

â‡’ ar(Î”BCD) = 2Ã—3cm^{2} = 6cm^{2}

Hence, area of a parallelogram = 2ar(Î”BCD) = 2Ã—6cm^{2} = 12cm^{2}

*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(Î”POA) = ar(Î”QOC).*

**Solution:**

In triangles POA and QOC, we have

âˆ AOP = âˆ COQ

AO = OC

âˆ PAC = âˆ QCA

So, by ASA congruence criterion , we have

Î”POAâ‰…Î”QOC

â‡’ ar(Î”POA) = ar(Î”QOC).

*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 âŠ¥ AB

We have,

BE = 2 EA and DF = 2FC

â‡’ AB â€“ AE = 2 AE and DC â€“ FC = 2 FC

â‡’ AB = 3 AE and DC = 3 FC

â‡’ AE = AB and FC = DCâ‹…â‹…â‹…â‹…â‹…â‹…â‹…â‹…(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 = AEÃ—FG

â‡’ ar(||^{gm}AECF) = 13ABÃ—FG from(1)

â‡’ 3ar(||^{gm}AECF) = ABÃ—FG â‹…â‹…â‹…â‹…â‹…(2)

Andar(||^{gm}ABCD) = ABÃ—FG â‹…â‹…â‹…â‹…â‹…(3)

Compare equation 2 and 3

â‡’ 3ar(||^{gm}AECF) = ar(||^{gm}ABCD)

â‡’ ar(||^{gm}AECF) = ar(||^{gm}ABCD)

*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(Î”PBQ) = ar(Î”ARC).*

* (ii) . ar(Î”PRQ) = **ar(Î”ARC).*

* (iii) . ar(Î”RQC) = **ar(Î”ABC).*

**Solution:**

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

(i) . Since CR is the median of Î” CAP

âˆ´ar(Î”CRA) = ar(Î”CAP)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(1)

Also , CP is the median of a Î” CAB

âˆ´ar(Î”CAP) = ar(Î”CPB)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

From 1 and 2 , we get

âˆ´ar(Î”ARC) = ar(Î”CPB)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(3)

PQ is the median of a Î” PBC

âˆ´ar(Î”CPB) = 2ar(Î”PBQ)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(4)

From 3 and 4, we get

âˆ´ar(Î”ARC) = ar(Î”PBQ)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(5)

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

âˆ´ar(Î”QAP) = ar(Î”QBP)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(6)

And ar(Î”QAP) = 2ar(Î”QRP)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(7)

From 6 and 7, we get

ar(Î”PRQ) = ar(Î”PBQ)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(8)

From 5 and 8, we get

ar(Î”PRQ) = ar(Î”ARC)

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

âˆ´ar(Î”ARC) = ar(Î”CAD)

= Ã—ar(Î”ABC)

= ar(Î”ABC)

Since RQ is the median of Î” RBC.

âˆ´ar(Î”RQC) = ar(Î”RBC)

= {ar(Î”ABC)âˆ’ar(Î”ARC)}

= {ar(Î”ABC)âˆ’ar(Î”ABC)}

= ar(Î”ABC)

*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(Î”EGB) = **ar(ABCD).*

* (iii) . ar(Î”EFC) = **ar(Î”EBF).*

* (iv) . ar(Î”EGB) = **Ã—ar(Î”EFC)*

* (v) . Find what portion of the area of parallelogram is the area of Î” 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(Î”EGB) = ar(ABCD).

(iii) . ar(Î”EFC) = ar(Î”EBF).

(iv) . ar(Î”EGB) = Ã—ar(Î”EFC)

(v) . Find what portion of the area of parallelogram is the area of Î” 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

â‡’ ED = CD = AB and EC = CD = 23AB

â‡’ 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(Î”EGB) = Ã—GBÃ—Height

ar(Î”EGB) = Ã—Ã—ABÃ—Height

ar(Î”EGB) = Ã—ABÃ—Height

ar(Î”EGB) = ar(ABCD).

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

ar(Î”EFC) = Ã—FCÃ—Height

ar(Î”EFC) = Ã—Ã—FBÃ—Height

ar(Î”EFC) = ar(EBF)

Hence , ar(Î”EFC) = ar(Î”EBF).

(iv) . Consider the trapezium in which

ar(EGBC) = ar(Î”EGB)+ar(Î”EBF)+ar(Î”EFC)

â‡’ ar(ABCD) = ar(ABCD)+2ar(Î”EFC)+ar(Î”EFC)

â‡’ ar(ABCD) = 3ar(Î”EFC)

â‡’ ar(Î”EFC) = ar(ABCD)

Now from (ii)part we have

ar(Î”EGB) = ar(Î”EFC)

ar(Î”EGB) = ar(ABCD)

ar(Î”EGB) = ar(Î”EFC)

âˆ´ar(Î”EGB) = ar(Î”EFC)

(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(Î”EFG) = ar(Î”ESF)+ar(Î”SGF)

ar(Î”EFG) = SFÃ—k+SFÃ—2k

ar(Î”EFG) = SFÃ—kâ‹…â‹…â‹…â‹…â‹…(i)

Now ,

ar(Î”EGBC) = ar(SGBF)+ar(ESFC)

ar(Î”EGBC) = (SF+GB)Ã—2k+(SF+EC)Ã—k

ar(Î”EGBC) = kÃ—SF+(GB+EC)Ã—k

ar(Î”EGBC) = kÃ—SF+(AB+Ã—AB)Ã—k

ar(Î”ABCD) = kÃ—SF+ABÃ—k

â‡’ ar(Î”ABCD) = 3kÃ—SF+ABÃ—k [Multiply both sides by 2]

â‡’ ar(Î”ABCD) = 3kÃ—SF+ar(ABCD)

â‡’ kÃ—SF = ar(ABCD)â‹…â‹…â‹…â‹…â‹…â‹…â‹…(2)

From 1 and 2 we have ,

ar(Î”EFG) =ar(ABCD)

ar(Î”EFG) = ar(ABCD)