RD Sharma Solutions Class 9 Areas Of Parallelograms And Triangles Exercise 15.3

RD Sharma Class 9 Solutions Chapter 15 Ex 15.3 Free Download

RD Sharma Solutions Class 9 Chapter 15 Ex 15.3

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

3

Solution:

Given:

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

In \(\Delta BCD\) we have

\(CD^{2}=BD^{2}+BC^{2}\)

\(\Rightarrow 17^{2}=BD^{2}+8^{2}\)

\(\Rightarrow BD^{2}=289-64\)

\(\Rightarrow BD=15\)

In \(\Delta ABD\) we have

\(AB^{2}+AD^{2}=BD^{2}\)

\(\Rightarrow 15^{2}=AB^{2}+9^{2}\)

\(\Rightarrow AB^{2}=225-81=144\)

\(\Rightarrow AB=12\)

\(ar\left ( quad \;ABCD \right )=ar\left ( \Delta ABD \right )+ar\left ( \Delta BCD \right )\)

\(ar\left ( quad \;ABCD \right )=\frac{1}{2}\left (12\times 9\right )+\frac{1}{2}\left ( 8\times 17 \right )=54+68=122\;cm^{2}\)

\(ar\left ( quad \;ABCD \right )=\frac{1}{2}\left (12\times 9\right )+\frac{1}{2}\left ( 8\times 15 \right )=54+60=114\;cm^{2}\)

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

4

Solution:

From the figure,

T and U are mid points of PS and QR respectively

\(∴ TU||PQ\)

\(\Rightarrow TO||PQ\)

Thus , in \(\Delta PQS\) , T   is the mid point of PS and TO||PQ

\(∴ TO=\frac{1}{2}PQ=4 \;cm\)

Also, TS = \(\frac{1}{2}PS=4 \;cm\)

\(∴ ar\left ( \Delta OTS \right )=\frac{1}{2}\left ( TO\times TS \right )=\frac{1}{2}\left ( 4\times 4 \right )\;cm^{2}=8\;cm^{2}\)

Q3. Compute the area of trapezium PQRS in figure

5

Solution:

We have,

\(ar\left ( trap.\;PQRS \right )=ar\left ( rect.\;PSRT \right )+ar\left ( \Delta QRT \right )\)

\(\Rightarrow ar\left ( trap.\;PQRS \right )=PT\times RT+\frac{1}{2}\left ( QT\times RT \right )\)

= \(=8\times RT+\frac{1}{2}\left ( 8\times RT \right )=12\times RT\)

In \(\Delta QRT\) , we have

\(QR^{2}=QT^{2}+RT^{2}\)

\(\Rightarrow RT^{2}=QR^{2}-QT^{2}\)

\(\Rightarrow RT^{2}=17^{2}-8^{2}=225\)

\(\Rightarrow RT=15\)

Hence , area of trapezium = \(12\times 15\;cm^{2}\) = \(180\;cm^{2}\)

Q4. In figure, \(\angle AOB=90^{\circ}\) , AC = BC , OA = 12 cm and OC = 6.5 cm .Find the area of \(\Delta AOB\).

6

Solution:

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

\(∴ CA = CB = OC\)

\(\Rightarrow CA = CB = 6.5\;cm\)

\(\Rightarrow AB = 13\;cm\)

In right angled triangle OAB , we have

\(AB^{2}=OB^{2}+OA^{2}\)

\(\Rightarrow 13^{2}=OB^{2}+12^{2}\)

\(\Rightarrow OB^{2}=13^{2}-12^{2} = 169 – 144 = 25\)

\(\Rightarrow OB=5\)

\(∴ ar\left ( \Delta AOB \right ) = \frac{1}{2}\left ( 12\times 5 \right ) = 30 \;cm^{2}\)

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.

7

Solution:

Draw AL \(\perp\) DC, BM \(\perp\) DC then ,

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

In \(\Delta\) ADL , we have

\(AD^{2} = AL^{2} + DL^{2}\)

\(\Rightarrow 25 = 16 + DL^{2}\)

\(\Rightarrow DL = 3\;cm\)

Similarly, MC = \(\sqrt{BC^{2}-BM^{2}}\) = \(\sqrt{25-16} = 3\;cm\)

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

\(ar\left ( trap . ABCD \right ) = \frac{1}{2}\left ( AB+CD \right )\times AL=\frac{1}{2}\left ( 7+13 \right )\times 4\;cm^{2} = 40\;cm^{2}\)

Q 6. In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = \(2\sqrt{5}\) cm , find the area of the rectangle .

8

Solution:

Given OD = 10 cm and OE = \(2\sqrt{5}\) cm

By using Pythagoras theorem

\(∴ OD^{2}=OE^{2}+DE^{2}\)

\(\Rightarrow DE = \sqrt{OD^{2}-OE^{2}}=\sqrt{10^{2}-\left ( 2\sqrt{5} \right )^{2}}=4\sqrt{5}\;cm\)

\(∴\) Area of rectangle OCDE = \(OE\times DE\)= \(2\sqrt{5}\times 4\sqrt{5}\;cm^{2}=40\;cm^{2}\)

Q 7. In figure, ABCD is a trapezium in which AB || DC. Prove that \(ar\left ( \Delta AOD \right ) = ar\left ( \Delta BOC \right )\)

9

Solution:

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

To prove: \(ar\left ( \Delta AOD \right ) = ar\left ( \Delta BOC \right )\)

Proof:- Since , \(\Delta ADC\) and  \(\Delta BDC\)  are on the same base DC and between same parallels AB and DC

Then, ar(\(\Delta ADC\)) = ar(\(\Delta BDC\))

\(\Rightarrow ar\left (\Delta AOD \right )+ar\left ( \Delta DOC \right )=ar\left ( \Delta BOC \right )+ar\left ( \Delta DOC \right )\)

\(\Rightarrow ar\left (\Delta AOD \right )=ar\left ( \Delta BOC \right )\)

Q 8. In figure, ABCD, ABFE and CDEF are parallelograms. Prove that \(ar\left ( \Delta ADE \right )=ar\left ( \Delta BCF \right )\).

10

Solution:

Given that

ABCD is parallelogram \(\Rightarrow AD=BC\)

CDEF is parallelogram \(\Rightarrow DE=CF\)

ABFE is parallelogram \(\Rightarrow AE=BF\)

Thus, in \(\Delta s\) ADF and BCF , we have

AD = BC, DE = CF and AE = BF

So, by SSS criterion of congruence, we have

\(\Delta ADE\cong \Delta BCF\)

\(ar\left ( \Delta ADE \right )=ar\left ( \Delta BCF \right )\)

Q 9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that : \(ar\left ( \Delta APB \right )\times ar\left ( \Delta CPD \right ) = ar\left ( \Delta APD \right )\times ar\left ( \Delta BPC \right )\).

Solution:

Construction: – Draw BQ \(\perp\)AC and DR \(\perp\) AC

Proof:-

L.H.S

= \(ar\left ( \Delta APB \right )\times ar\left ( \Delta CDP \right )\)

= \(\frac{1}{2}\left [ \left ( AP\times BQ \right ) \right ]\times \left ( \frac{1}{2}\times PC\times DR \right )\)

= \(\left ( \frac{1}{2}\times PC\times BQ \right )\times \left ( \frac{1}{2}\times AP\times DR \right )\)

= \(ar\left (\Delta APD \right )\times ar\left ( \Delta BPC \right )\).

= 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 \(ar\left ( \Delta ABC \right ) = ar\left ( \Delta ABD \right )\).

11

Solution:

Given that CD is bisected by AB at O

To prove: \(ar\left ( \Delta ABC \right ) = ar\left ( \Delta ABD \right )\).

Construction: Draw CP \(\perp\) AB and DQ  \(\perp\) AB .

Proof:

\(ar\left ( \Delta ABC \right )=\frac{1}{2}\times AB\times CP\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

\(ar\left ( \Delta ABD \right )=\frac{1}{2}\times AB\times DQ\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

In \(\Delta CPO\; and\;\Delta DQO\)

\(\angle CPO=\angle DQO\;\;\;\;\;\;\;\;\;\left [ each\;90^{\circ} \right ]\)

Given that, CO = OD

\(\angle COP=\angle DOQ\;\;\;\;\;\;\;\;\;\left [ Vertically\;opposite \;angles\; are\; equal \right ]\)

Then , \(\Delta CP0\cong \Delta DQO\;\;\;\;\;\;\;\;\;\left [By\;AAS \;condition \right ]\)

\(∴ CP=DQ\;\;\;\;\;\;\;\;\;\left ( 3 \right )\left [c.p.c.t \right ]\)

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

\(∴\) \(ar\left ( \Delta ABC \right ) = ar\left ( \Delta ABD \right )\).

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 \(\perp\) AB and PM \(\perp\) AB

Now ,

\(ar\left ( ||^{gm} ABCD\right )=AB\times DN , ar\left ( \Delta APB \right )=\frac{1}{2}\left ( AB\times PM \right )\)

Now , PM  < DN

\(\Rightarrow AB\times PM< AB\times DN\)

\(\Rightarrow \frac{1}{2}\left (AB\times PM \right )< \frac{1}{2}\left (AB\times DN \right )\)

\(\Rightarrow ar\left ( \Delta APB \right )< \frac{1}{2}ar\left ( ||^{gm}ABCD \right )\)

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 \(ar\left ( \Delta BGC \right ) = 2 ar\left ( \Delta AGC \right )\).

Solution:

Draw AM \(\perp\) BC

Since, AD is the median of \(\Delta ABC\)

\(∴\) BD = DC

\(\Rightarrow BD = AM = DC\times AM\)

\(\Rightarrow \frac{1}{2}\left ( BD \times AM \right )= \frac{1}{2}\left (DC\times AM \right )\)

\(\Rightarrow ar\left ( \Delta ABD \right ) = ar\left ( \Delta ACD \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

In \(\Delta BGC\) , GD is the median

\(∴\) \(ar\left ( \Delta BGD \right ) = ar\left ( \Delta CGD \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

In \(\Delta ACD\) , CG is the median

\(∴\) \(ar\left ( \Delta AGC \right ) = ar\left ( \Delta CGD \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 3 \right )\)

From (2) and (3) we have,

\(ar\left ( \Delta BGD \right ) = ar\left ( \Delta AGC \right )\)

But, \(ar\left ( \Delta BGC \right ) = 2 ar\left ( \Delta BGD \right )\)

\(∴\) \(ar\left ( \Delta BGC \right ) = 2 ar\left ( \Delta AGC \right )\)

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

Solution:

Given that,

In \(\Delta ABC\), BD = 2DC

To prove: \(ar\left ( \Delta ABD \right ) = 2 ar\left ( \Delta ADC \right )\).

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 \(\Delta ABD\) , AE is the median .

Then , \(ar\left ( \Delta ABD \right ) = 2 ar\left ( \Delta AED \right )\) \(\cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

In \(\Delta AEC\) , AD is the median .

Then, \(ar\left ( \Delta AED \right ) = 2 ar\left ( \Delta ADC \right )\) \(\cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

Compare equation 1 and 2

\(ar\left ( \Delta ABD \right ) = 2 ar\left ( \Delta ADC \right )\).

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

(i) . \(ar\left ( \Delta ADO \right ) = ar\left ( \Delta CDO \right )\).

(ii) . \(ar\left ( \Delta ABP \right ) = 2 ar\left ( \Delta CBP \right )\).

Solution:

Given that ABCD is the parallelogram

To Prove:    (i) \(ar\left ( \Delta ADO \right ) = ar\left ( \Delta CDO \right )\).

(ii) \(ar\left ( \Delta ABP \right ) = 2 ar\left ( \Delta CBP \right )\).

Proof:  we know that diagonals of parallelogram bisect each other

\(∴\) AO = OC and BO = OD

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

Then \(ar\left ( \Delta ADO \right ) = ar\left ( \Delta CDO \right )\).

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

Then \(ar\left ( \Delta BAO \right )=ar\left ( \Delta BCO \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

In \(\Delta\) PAC , since PO is a median .

Then \(ar\left ( \Delta PAO \right )=ar\left ( \Delta PCO \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

Subtract equation 2 from 1.

\(\Rightarrow ar\left ( \Delta BAO \right )-ar\left ( \Delta PAO \right )=ar\left ( \Delta BCO \right )-ar\left ( \Delta PCO \right )\)

\(\Rightarrow\) \(ar\left ( \Delta ABP \right ) = 2 ar\left ( \Delta CBP \right )\).

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 \(ar\left ( \Delta ADF \right ) = ar\left ( \Delta ECF \right )\).

                (ii)  . If the area of  \(\Delta DFB\)\(3 \;cm^{2}\) , find the area of \(||^{gm}\) ABCD .

Solution:

In triangles ADF and ECF, we have

\(\angle ADF=\angle ECF\;\;\;\;\;\;\;\;\;\left [ Alternate\;interior\;angles,Since\;AD||BE \right ]\)

AD = EC                                                \(\left [ since\;AD = BC = CE \right ]\)

And \(\angle DFA = \angle CFA \;\;\;\;\;\;\;\;\;\left [ Vertically\;opposite\;angles \right ]\)

So, by AAS congruence criterion, we have

\(\Delta ADF\cong \Delta ECF\)

\(\Rightarrow\) \(ar\left ( \Delta ADF \right ) = ar\left ( \Delta ECF \right )\) and DF = CF .

Now, DF = CF

\(\Rightarrow\) BF is a median in \(\Delta\) BCD.

\(\Rightarrow\) \(ar\left ( \Delta BCD \right )=2ar\left ( \Delta BDF \right )\)

\(\Rightarrow ar\left (\Delta BCD \right )=2\times 3\;cm^{2} = 6\;cm^{2}\)

Hence, area of a parallelogram = \(2 ar\left ( \Delta BCD \right )=2\times 6\;cm^{2} = 12\;cm^{2}\)

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 \(ar\left ( \Delta POA \right ) = ar\left ( \Delta QOC \right )\).

Solution:

In triangles POA and QOC, we have

\(\angle AOP = \angle COQ\)

AO = OC

\(\angle PAC = \angle QCA\)

So, by ASA congruence criterion  , we have

\(\Delta POA\cong \Delta QOC\)

\(\Rightarrow\)  \(ar\left ( \Delta POA \right ) = ar\left ( \Delta QOC \right )\).

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 \(\perp\) AB

We have,

BE = 2 EA and DF = 2FC

\(\Rightarrow\) AB – AE = 2 AE and DC – FC = 2 FC

\(\Rightarrow\) AB = 3 AE and DC = 3 FC

\(\Rightarrow AE = \frac{1}{3 }AB\;and\;FC= \frac{1}{3}DC\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

But AB = DC

Then, AE = FC                                    \(\left [opposite \;sides\; of\; ||^{gm} \right ]\)

Thus, AE = FC and AE || FC

Then, AECF is a parallelogram

Now, area of parallelogram AECF = \(AE\times FG\)

\(\Rightarrow ar\left ( ||^{gm} \;AECF\right )=\frac{1}{3}AB\times FG\)          from(1)

\(\Rightarrow 3\;ar\left ( ||^{gm} \;AECF\right )=AB\times FG\cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

\(And\;ar\left ( ||^{gm} \;ABCD\right )=AB\times FG\cdot \cdot \cdot \cdot \cdot \left ( 3 \right )\)

Compare equation 2 and 3

\(\Rightarrow 3ar\left ( ||^{gm} \;AECF\right )=ar\left ( ||^{gm} \;ABCD\right )\)

\(\Rightarrow ar\left ( ||^{gm} \;AECF\right )=\frac{1}{3}ar\left ( ||^{gm} \;ABCD\right )\)

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)  . \(ar\left ( \Delta PBQ \right ) = ar\left ( \Delta ARC \right )\).

                (ii)  . \(ar\left ( \Delta PRQ \right )\)= \(\frac{1}{2}ar\left ( \Delta ARC \right )\).

                (iii) . \(ar\left ( \Delta RQC \right )\)= \(\frac{3}{8}ar\left ( \Delta ABC \right )\).

Solution:

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

(i)  .  Since CR is the median of \(\Delta\) CAP

\(∴ ar\left ( \Delta CRA\right ) = \frac{1}{2}ar\left ( \Delta CAP \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

Also , CP is the median of a \(\Delta\) CAB

\(∴ar\left ( \Delta CAP\right ) = ar\left ( \Delta CPB \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

From 1 and 2 , we get

\(∴ ar\left ( \Delta ARC\right ) = \frac{1}{2}ar\left ( \Delta CPB \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 3 \right )\)

PQ is the median of a \(\Delta\) PBC

\(∴ ar\left ( \Delta CPB\right ) =2  ar\left ( \Delta PBQ \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 4 \right )\)

From 3 and 4, we get

\(∴ ar\left ( \Delta ARC\right ) =  ar\left ( \Delta PBQ \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 5 \right )\)

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

\(∴ ar \left ( \Delta QAP \right ) =  ar \left ( \Delta QBP \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 6 \right )\)

And \(ar\left ( \Delta QAP \right ) = 2 ar\left ( \Delta QRP \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 7 \right )\)

From 6 and 7, we get

\(ar\left ( \Delta PRQ \right ) = \frac{1}{2}ar\left ( \Delta PBQ \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 8 \right )\)

From 5 and 8, we get

\(ar\left ( \Delta PRQ \right ) = \frac{1}{2}ar\left ( \Delta ARC \right )\)

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

\(∴ ar\left ( \Delta ARC \right ) = \frac{1}{2}ar\left ( \Delta CAD \right )\)

\(= \frac{1}{2}\times \frac{1}{2}ar\left ( \Delta ABC \right )\)

\(= \frac{1}{4}ar\left ( \Delta ABC \right )\)

Since RQ is  the median of  \(\Delta\) RBC.

\(∴ ar\left ( \Delta RQC \right ) = \frac{1}{2}ar\left ( \Delta RBC \right )\)

\(= \frac{1}{2}\left \{ ar\left ( \Delta ABC \right )- ar\left ( \Delta ARC \right ) \right \}\)

\(= \frac{1}{2}\left \{ ar\left ( \Delta ABC \right )-\frac{1}{4} ar\left ( \Delta ABC \right ) \right \}\)

\(= \frac{3}{8} ar\left ( \Delta ABC \right )\)

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)  . \(ar\left ( ADEG \right ) = ar\left (GBCE\right )\).

                (ii)  . \(ar\left ( \Delta EGB \right )\)= \(\frac{1}{6}ar\left ( ABCD \right )\).

                (iii) . \(ar\left ( \Delta EFC \right )\)= \(\frac{1}{2}ar\left ( \Delta EBF \right )\).

                (iv)  . \(ar\left ( \Delta EGB \right )=\frac{3}{2}\times ar\left ( \Delta EFC \right )\)

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

Solution:

Given: ABCD is a parallelogram

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

To prove:

(i)  . \(ar\left ( ADEG \right ) = ar\left (GBCE\right )\).

(ii)  . \(ar\left ( \Delta EGB \right )\)= \(\frac{1}{6}ar\left ( ABCD \right )\).

(iii) . \(ar\left ( \Delta EFC \right )\)= \(\frac{1}{2}ar\left ( \Delta EBF \right )\).

(iv)  . \(ar\left ( \Delta EGB \right )=\frac{3}{2}\times ar\left ( \Delta EFC \right )\)

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

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

Proof:

12

(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

\(\Rightarrow ED =\frac{1}{3}CD =\frac{1}{3}AB\;and\;EC = \frac{2}{3}CD=\frac{2}{3}AB\)

\(\Rightarrow AG =\frac{2}{3}AB \;and\;BG = \frac{1}{3}AB\)

So , \(DE+ AG =\frac{1}{3}AB +\frac{2}{3}AB=AB\;and\;EC +BG = \frac{2}{3}AB+\frac{1}{3}AB =AB\)

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

Since Area of trapezium = \(\frac{sum\;of\;parallel\;sides}{2}\times height\)

So, \(ar\left ( ADEG \right ) = ar\left (GBCE\right )\).

(ii)  . Since we know from above that

\(BG = \frac{1}{2}AB\). So

\(ar\left (\Delta EGB \right ) = \frac{1}{2}\times GB\times Height\)

\(ar\left (\Delta EGB \right ) =\frac{1}{2}\times \frac{1}{3}\times AB\times Height\)

\(ar\left (\Delta EGB \right ) = \frac{1}{6}\times AB\times Height\)

\(ar\left ( \Delta EGB \right )\)= \(\frac{1}{6}ar\left ( ABCD \right )\).

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

\(ar\left (\Delta EFC \right ) = \frac{1}{2}\times FC\times Height \)

\(ar\left ( \Delta EFC \right )=\frac{1}{2}\times \frac{1}{2}\times FB\times Height\)

\(ar\left ( \Delta EFC \right )=\frac{1}{2}ar\left ( EBF \right )\)

Hence , \(ar\left ( \Delta EFC \right )\)= \(\frac{1}{2}ar\left ( \Delta EBF \right )\).

(iv) . Consider the trapezium in which

\(ar\left ( EGBC \right )=ar\left (\Delta EGB \right )+ar\left (\Delta EBF \right )+ar\left (\Delta EFC \right )\)

\(\Rightarrow \frac{1}{2}ar\left ( ABCD \right )=\frac{1}{6}ar\left (ABCD \right )+2ar\left (\Delta EFC \right )+ar\left (\Delta EFC \right )\)

\(\Rightarrow \frac{1}{3}ar\left ( ABCD \right )=3ar\left (\Delta EFC \right )\)

\(\Rightarrow ar\left (\Delta EFC \right )=\frac{1}{9}ar\left ( ABCD \right )\)

Now from (ii)part we have

\(ar\left (\Delta EGB \right )=\frac{1}{6}ar\left ( \Delta EFC\right )\)

\(ar\left (\Delta EGB \right )=\frac{3}{2}\times \frac{1}{9}ar\left ( ABCD\right )\)

\(ar\left (\Delta EGB \right )=\frac{3}{2}ar\left ( \Delta EFC\right )\)

\(∴ ar\left (\Delta EGB \right )=\frac{3}{2}ar\left ( \Delta EFC\right )\)

(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

\(ar\left (\Delta EFG \right )=ar\left ( \Delta ESF\right )+ar\left ( \Delta SGF \right )\)

\(ar\left (\Delta EFG \right )=\frac{1}{2}SF \times k+\frac{1}{2}SF\times 2k\)

\(ar\left (\Delta EFG \right )=\frac{3}{2}SF \times k\cdot \cdot \cdot \cdot \cdot \left ( i \right )\)

Now ,

\(ar\left (\Delta EGBC \right )=ar\left ( SGBF \right )+ar\left ( ESFC \right )\)

\(ar\left (\Delta EGBC \right )=\frac{1}{2}\left ( SF+GB \right )\times 2k+\frac{1}{2}\left ( SF+EC \right )\times k\)

\(ar\left (\Delta EGBC \right )=\frac{3}{2}k\times SF+\left ( GB+\frac{1}{2}EC \right )\times k\)

\(ar\left (\Delta EGBC \right )=\frac{3}{2}k\times SF+\left ( \frac{1}{3}AB+\frac{1}{2}\times \frac{2}{3}AB \right )\times k\)

\(\frac{1}{2} ar\left (\Delta ABCD \right )=\frac{3}{2}k\times SF+\frac{2}{3}AB \times k\)

\(\Rightarrow ar\left (\Delta ABCD \right )=3k\times SF+\frac{4}{3}AB \times k\)                         [Multiply both sides by 2]

\(\Rightarrow ar\left (\Delta ABCD \right )=3k\times SF+\frac{4}{9}ar\left ( ABCD \right )\)

\(\Rightarrow k\times SF=\frac{5}{27}ar\left ( ABCD \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

From 1 and 2 we have ,

\(ar\left ( \Delta EFG \right )=\frac{3}{2}\times \frac{5}{27}ar\left ( ABCD \right )\)

\(ar\left ( \Delta EFG \right )=\frac{5}{18}ar\left ( ABCD \right )\)

Q 20. In figure, CD || AE and CY || BA.

                (i)  . Name a triangle equal in area of \(\Delta CBX\)

                (ii) . Prove that \(ar\left ( \Delta ZDE \right ) = ar\left ( \Delta CZA \right )\).

                (iii) . Prove that \(ar\left (BCZY\right ) = ar\left ( \Delta EDZ \right )\).  

13

Solution:

Since, triangle BCA and triangle BYA are on the same base BA and between same parallel s BA and CY.

Then \(ar\left ( \Delta BCA \right ) = ar\left ( \Delta BYA \right )\)

\(\Rightarrow ar\left ( \Delta CBX \right )+ ar\left ( \Delta BXA \right )= ar\left ( \Delta BXA \right )+ar\left ( \Delta AXY \right )\)

\(\Rightarrow ar\left ( \Delta CBX \right )=ar\left ( \Delta AXY \right )\cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

Since, triangles  ACE and ADE are on the same base AE and between same parallels CD and AE

Then, \(ar\left ( \Delta ACE \right ) = ar\left ( \Delta ADE \right )\)

\(ar\left ( \Delta CZA \right )+ ar\left ( \Delta AZE \right ) = ar\left ( \Delta AZE \right )+ ar\left ( \Delta DZE \right )\)

\(ar\left ( \Delta CZA \right )=ar\left ( \Delta DZE \right )\cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

Adding \(ar\left ( \Delta CYG \right )\) on both sides , we get

\(\Rightarrow ar\left ( \Delta CBX \right )+ar\left ( \Delta CYZ \right )= ar\left ( \Delta CAY \right )+ar\left ( \Delta CYZ \right )\)

\(\Rightarrow ar\left ( BCZY \right )=ar\left ( \Delta CZA \right )\cdot \cdot \cdot \cdot \cdot \left ( 3 \right )\)

Compare equation 2 and 3

\(\Rightarrow ar\left ( BCZY \right )=ar\left ( \Delta DZE \right )\)

Q 21. In figure, PSDA is a parallelogram in which PQ = QR =RS and AP || BQ ||CR. Prove that \(ar\left ( \Delta PQE \right ) = ar\left ( \Delta CFD \right )\).

14

Solution:

Given that PSDA is a parallelogram

Since, AP || BQ || CR ||DS and AD ||PS

Therefore, PQ = CD                                         (equ.  1)

In triangle BED, C is the midpoint of BD and CF || BE

Therefore, F is the midpoint of ED

\(\Rightarrow\) EF = PE

Smiliarly,

EF = PE

Therefore, PE = FD                                          (equ .  2)

In triangles PQE and CFD, we have

PE = FD

Therefore, \(\angle\) EPQ = \(\angle\)FDC                                                [Alternate angles]

So, by SAS criterion , we have

\(\Delta PQE\cong \Delta DCF\)

\(\Rightarrow ar\left ( \Delta PQE \right )=ar\left ( \Delta DCF \right )\)

Q 22.  In figure, ABCD  is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm .If X and Y are , respectively , the mid points of AD and BC , prove that :

                (i) . XY = 50 cm

                (ii) . DCYX is a trapezium

                (iii) . \(ar\left (trap. DCYX\right )\)= \(\frac{9}{11}ar\left ( XYBA \right )\).

15

Solution:

(i) Join DY and produce it to meet AB produced at P.

In triangles BYP and CYD we have,

\(\angle\)BYP = \(\angle\) CYD                                       [Vertically opposite angles]

\(\angle\)DCY = \(\angle\) PBY                                       [Since , DC ||AP]

And BY = CY

So, by ASA congruence criterion, we have

\(\left ( \Delta BYP \right )\cong \left ( \Delta CYD \right )\)

\(\Rightarrow\) DY = Yp and DC = BP

\(\Rightarrow\) Y is the midpoint of DP

Also , x is the mid point of AD

Therefore, XY ||AP and XY ||\(\frac{1}{2}AP\)

\(\Rightarrow XY=\frac{1}{2}\left ( AB +BP \right )\)

\(\Rightarrow XY=\frac{1}{2}\left ( AB +DC \right )\)

\(\Rightarrow XY=\frac{1}{2}\left ( 60 + 40 \right )\)

= 50 cm

(ii) We have, XY ||AP

\(\Rightarrow\) XY || AB and AB ||DC

\(\Rightarrow\) XY || DC

\(\Rightarrow\) DCYX is a trapezium

(iii) Since x and y are the mid points of Ad and BC respectively.

Therefore, trapezium DCYX and ABYX are of the same height say h cm

Now,

\(ar\left ( trap . DCXY \right )=\frac{1}{2}\left ( DC + XY \right )\times h\)

\(\Rightarrow ar\left ( trap . DCXY \right )=\frac{1}{2}\left ( 50 +40 \right )\times h\;cm^{2} = 45h\;cm^{2}\)

\(\Rightarrow ar\left ( trap . ABYX \right )=\frac{1}{2}\left ( AB+XY \right )\times h\)

\(\Rightarrow ar\left ( trap . ABYX \right )=\frac{1}{2}\left ( 60 +50 \right )\times h\;cm^{2}=55h\;cm^{2}\)

\(\frac{ar\left ( trap. DCYX \right )}{ar\left ( trap. ABYX \right )}=\frac{45h}{55h}=\frac{9}{11}\)

\(\Rightarrow ar\left ( trap. DCYX \right )=\frac{9}{11}ar\left ( trap. ABYX \right )\)

Q 23. In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that:

                (i) . \(ar\left ( \Delta BDE \right )\)= \(\frac{1}{4}ar\left ( \Delta ABC \right )\).

                (ii) . \(ar\left ( \Delta BDE\right )\)= \(\frac{1}{2}ar\left ( \Delta BAE \right )\).

                (iii) . \(ar\left ( \Delta BFE \right )\)= \( ar\left ( \Delta AFD \right )\).

                (iv) . \(ar\left ( \Delta ABC \right )\)= 2 \( ar\left ( \Delta BEC \right )\).

                (v) . \(ar\left ( \Delta FED \right )\)= \(\frac{1}{8}ar\left ( \Delta AFC \right )\).

                (vi) . \(ar\left ( \Delta BFE \right )\)= 2 \( ar\left ( \Delta EFD \right )\).

16

Solution:

Given that ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x. Then, BD = \(\frac{x}{2}\) = DE = BE

(i) We have,

\(ar\left ( \Delta ABC \right )=\frac{\sqrt{3}}{4}x^{2}\;and\;ar\left ( \Delta BDE \right )=\frac{\sqrt{3}}{4}\left ( \frac{x}{2} \right )^{2}=\frac{1}{4}\times \frac{\sqrt{3}}{4}x^{2}\)

Therefore , \(ar\left ( \Delta BDE \right )\)= \(\frac{1}{4}ar\left ( \Delta ABC \right )\).

(ii) . It is given that triangles, ABC and BED are equilateral triangles

\(\angle\) ACB = \(\angle\)DBE = \(60^{\circ}\)

\(\Rightarrow\) BE || AC         (Since , alternative angles are equal)

Trinagles BAF and BEC are on the same base BE and between same parallels BF and AC .

Therefore, \(ar\left ( \Delta BAE \right )=ar\left ( \Delta BEC \right )\)

\(\Rightarrow ar\left ( \Delta BAE \right )=2 ar\left ( \Delta BDE \right )\)                         [Since , ED is a median of triangle EBC ; \(ar\left ( \Delta BEC \right )=2 ar\left ( \Delta BDE \right )\) ]

\(∴ ar\left ( \Delta BDE \right )=\frac{1}{2} ar\left ( \Delta BAE \right )\)

(iii) Since, triangles ABC and BDE are equilateral triangles

\(∴ \angle ABC = 60^{\circ} \;and \;\angle BDE = 60^{\circ}\)

\(\angle ABC =\angle BDE\)

\(\Rightarrow\) AB ||DE                                         (since, alternate angles are equal)

Triangles BED and AED are on the same base ED and between same parallels AB and DE.

Therefore, \(ar\left ( \Delta BED \right ) = ar\left ( \Delta AED \right )\)

\(\Rightarrow ar\left (\Delta BED \right)-ar\left ( \Delta EFD \right ) = ar\left ( \Delta AED \right ) – ar\left ( \Delta EFD \right )\)

\(\Rightarrow ar\left (\Delta BEF \right )= ar\left ( \Delta AFD \right )\)

(iv) Since ED is the median of triangle BEC

Therefore, \(ar\left ( \Delta BEC \right )= 2ar\left ( \Delta BDE \right )\)

\(\Rightarrow ar\left ( \Delta BEC \right )= 2\times \frac{1}{4}ar\left ( \Delta ABC \right )\)                      [From 1 , \(ar\left ( \Delta BDE \right )= \frac{1}{4}ar\left ( \Delta ABC \right )\)]

\(\Rightarrow ar\left ( \Delta BEC \right )= \frac{1}{2}ar\left ( \Delta ABC \right )\)

\(\Rightarrow ar\left ( \Delta ABC \right ) = 2ar\left ( \Delta BEC \right )\)

(v) \(ar\left ( \Delta AFC \right )=ar\left ( \Delta AFD \right )+ar\left ( \Delta ADC \right )\)

\(\Rightarrow ar\left ( \Delta BFE \right )+\frac{1}{2}ar\left ( \Delta ABC \right )\)                        [using part (iii) , and AD is the median of triangle ABC ]

\(= ar\left ( \Delta BFE \right )+\frac{1}{2}\times 4ar\left ( \Delta BDE \right )\)                                             (using part (i))

\(= ar\left ( \Delta BFE \right )= 2ar\left ( \Delta FED \right )\cdot \cdot \cdot \cdot \left ( 3 \right )\)

\(ar\left ( \Delta BDE \right )= ar\left ( \Delta BFE \right )+ar\left ( \Delta FED \right )\)

\(\Rightarrow 2ar\left (\Delta FED \right )+ar\left ( \Delta FED \right )\)

\(\Rightarrow 3ar\left (\Delta FED \right )\cdot \cdot \cdot \cdot \cdot \cdot \left ( 4 \right )\)

From 2, 3 and 4 we get ,

\(ar\left ( \Delta AFC \right ) = 2ar\left ( \Delta FED \right )+2\times 3ar\left ( \Delta FED \right )=8ar\left ( \Delta FED \right )\)

\(ar\left ( \Delta FED\right ) = \frac{1}{8}ar\left (\Delta AFC \right )\)

(vi) Let h be the height of vertex E, corresponding to the side BD in triangle BDE.

Let H be the height of vertex A, corresponding to the side BC in triangle ABC

From part (i)

\(ar\left ( \Delta BDE \right ) = \frac{1}{4}ar\left ( \Delta ABC \right )\)

\(\Rightarrow \frac{1}{2}\times BD\times h=\frac{1}{4}\left ( \frac{1}{2}\times BC\times h \right )\)

\(\Rightarrow BD\times h=\frac{1}{4}\left ( 2BD\times H \right )\)

\(\Rightarrow h=\frac{1}{2}H\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

From part (iii)

\(ar\left ( \Delta BFE \right )=ar\left ( \Delta AFD \right )\)

\(ar\left ( \Delta BFE \right )=\frac{1}{2}\times FD\times H\)

\(ar\left ( \Delta BFE \right )=\frac{1}{2}\times FD\times 2h\)

\(ar\left ( \Delta BFE \right )=2\left (\frac{1}{2}\times FD\times h \right )\)

\(ar\left ( \Delta BFE \right )=2ar\left ( \Delta EFD \right )\)

Q 24. D is the midpoint of side BC of \(\Delta ABC\) and E is the midpoint of BD. If O is the midpoint of AE, Prove that \(ar\left ( \Delta BOE \right )\)= \(\frac{1}{8}ar\left ( \Delta ABC \right )\).

Solution:

Given that

D is the midpoint of sides BC of triangle ABC

E is the midpoint of BD and O is the midpoint of AE

Since AD and AE are the medians of triangles, ABC and ABD respectively

\(∴ ar\left (\Delta ABD\right) = \frac{1}{2}ar\left (\Delta ABC \right)\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

\(∴ ar\left (\Delta ABE\right) = \frac{1}{2}ar\left (\Delta ABD \right)\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

OB is the median of triangle ABE

Therefore, \(∴ ar\left ( \Delta BOE\right ) = \frac{1}{2}ar\left (\Delta ABE \right )\)

From 1, 2 and 3 , we have

\(∴ ar\left ( \Delta BOE\right ) = \frac{1}{8}ar\left (\Delta ABC \right )\)

Q 25. In figure, X and Y are the mid points of AC and AB respectively, QP ||BC and CYQ and BXP are straight lines . Prove that \(ar\left ( \Delta ABP \right )\)= \( ar\left ( \Delta ACQ \right )\).

17

Solution:

Since X and Y are the mid points of AC and AB respectively.

Therefore, XY || BC

Clearly, triangles BYC and BXC are on the same base BC and between the same parallels Xy and BC

\(∴ ar\left ( \Delta BYC\right ) = ar\left (\Delta BXC \right )\)

\(\Rightarrow ar\left ( \Delta BYC\right ) -ar\left ( \Delta BOC\right )= ar\left (\Delta BXC \right )-ar\left ( \Delta BOC\right )\)

\(\Rightarrow ar\left ( \Delta BOY\right ) = ar\left ( \Delta COX\right )\)

\(\Rightarrow ar\left ( \Delta BOY\right )+ar\left ( \Delta XOY\right ) = ar\left ( \Delta COX\right )+ar\left ( \Delta XOY\right )\)

\(\Rightarrow ar\left ( \Delta BXY\right )=ar\left ( \Delta CXY\right )\)                               (2)

We observed that the quadrilaterals XYAP and XYAQ are on the same base XY and between same parallels XY and PQ.

\(∴ ar\left (quad . XYAP\right )=ar\left ( quad XYQA\right )\cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)

Adding 1 and 2 , we get

\(∴ ar\left ( \Delta BXY \right ) +ar\left (quad . XYAP\right )=ar\left ( \Delta CXY \right )+ar\left ( quad XYQA\right )\)

\(\Rightarrow ar\left ( \Delta ABP \right ) = ar\left ( \Delta ACQ \right )\)

Q 26. In figure, ABCD and AEFD are two parallelograms. Prove that

                (i) . PE = FQ

                (ii) . \(ar\left ( \Delta APE \right ):ar\left ( \Delta PFA \right )=ar\left ( \Delta QFD \right ):ar\left ( \Delta PFD \right )\)

                (iii) . \(ar\left ( \Delta PEA \right )\)\( ar\left ( \Delta QFD \right )\).

18

Solution:

Given that, ABCD and AEFD are two parallelograms

(i)  . In triangles, EPA and FQD

\(\angle PEA=\angle QFD\)                    [corresponding angles]

\(\angle EPA=\angle FQD\)                    [corresponding angles]

PA = QD                                                                               [opposite sides of parallelogram]

Then, \(\Delta EPA\cong \Delta FQD\)                              [By AAS condition]

Therefore, EP = FQ                                                          [C.P.C.T]

(ii) . Since triangles, PEA and QFD stand on equal bases PE and FQ lies between the same parallels EQ and AD

Therefore, \(ar\left ( \Delta PEA \right )=ar\left ( \Delta QFD \right )\)                              (1)

Since, triangles PEA and PFD stand on the same base PF and between same parallels PF and AD

Therefore, \(ar\left ( \Delta PFA \right )=ar\left ( \Delta PFD \right )\)                               (2)

Divide the equation 1 by equation 2

\(\frac{ar\left ( \Delta PEA \right )}{ar\left ( \Delta PFA \right )}=\frac{ar\left ( \Delta QFD \right )}{ar\left ( \Delta PFD \right )}\)

(iii) . From part (i), \(\Delta EPA\cong \Delta FQD\)

Then, \(ar\left ( \Delta PEA \right )\)\( ar\left ( \Delta QFD \right )\).

Q 27. In figure, ABCD  is a parallelogram . O is any point on AC. PQ ||AB and LM || AD. Prove that : \(ar\left ( ||^{gm}DLOP \right ) = ar\left ( ||^{gm}BMOQ \right )\).

19

Solution:

Since a diagonal of a parallelogram divides it into two triangles of equal area

Therefore, \(ar\left ( \Delta ADC \right )=ar\left ( \Delta ABC \right )\)

\(\Rightarrow ar\left ( \Delta APO \right )+ar\left ( ||^{gm}DLOP \right )+ar\left ( \Delta OLC \right )\)

\(\Rightarrow ar\left ( \Delta AOM \right )+ar\left ( ||^{gm}BMOQ \right )+ar\left ( \Delta OQC \right )\)                        (1)

Since AO and Oc are diagonals of parallelograms AMOP and OQCL respectively.

\(∴ ar\left ( \Delta APO \right ) = ar\left ( \Delta AMO \right )\)                         (2)

And \(ar\left ( \Delta OLC \right ) = ar\left ( \Delta OQC \right )\)                                         (3)

Subtracting 2 and 3 from 1, we get

\(ar\left ( ||^{gm}DLOP \right ) = ar\left ( ||^{gm}BMOQ \right )\).

Q 28. In a triangle ABC, if L and M are points on AB and AC respectively such that LM || BC. Prove that:

                (i)  . \(ar\left ( \Delta LCM \right ) = ar\left ( \Delta LBM \right )\).

                (ii) . \(ar\left ( \Delta LBC \right ) = ar\left ( \Delta MBC \right )\).

                (iii)  . \(ar\left ( \Delta ABM \right ) = ar\left ( \Delta ACL \right )\).

                (iv)  . \(ar\left ( \Delta LOB \right ) = ar\left ( \Delta MOC \right )\).

Solution:

(i)  . Clearly triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.

\(∴ ar\left ( \Delta LMB \right ) = ar\left ( \Delta LMC \right )\)                           (1)

(ii) . We observe that triangles LBC and MBC are on the same base BC and between same parallels LM and BC.

\(∴ ar\left ( \Delta LBC \right ) = ar\left ( \Delta MBC \right )\)                            (2)

(iii) . We have,

\(ar\left ( \Delta LMB \right ) = ar\left ( \Delta LMC \right )\)                  [From 1]

\(\Rightarrow ar\left ( \Delta ALM \right ) +ar\left ( \Delta LMB \right )=ar\left ( \Delta ALM \right )+ ar\left ( \Delta LMC \right )\)

\(\Rightarrow ar\left ( \Delta ABM \right )= ar\left ( \Delta ACL \right )\)

(iv) . We have,

\(ar\left ( \Delta LBC \right )= ar\left ( \Delta MBC \right )\)                    [From 1]

\(\Rightarrow ar\left ( \Delta LBC \right )-ar\left ( \Delta BOC \right )= ar\left ( \Delta MBC \right )-ar\left ( \Delta BOC \right )\)

\(\Rightarrow ar\left ( \Delta LOB \right )= ar\left ( \Delta MOC \right )\).

Q 29. In figure, D and E are two points on BC such that BD = DE = EC. Show that \(ar\left ( \Delta ABD \right ) = ar\left ( \Delta ADE \right ) = ar\left ( \Delta AEC \right )\).

20

Solution:

Draw a line l through A parallel to BC.

Given that, BD = DE = EC

We observed that the triangles ABD and AEC are on the equal bases and between the same parallels l and BC. Therefore, their areas are equal.

Hence,  \(ar\left ( \Delta ABD \right ) = ar\left ( \Delta ADE \right ) = ar\left ( \Delta AEC \right )\).

Q 30. In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX \(\perp\) DE meets BC at Y. Show that

                (i) . \(\Delta MBC\cong \Delta ABD\)

                (ii) . \(ar\left ( BYXD \right ) = 2 ar \left ( \Delta MBC \right )\)

                (iii) . \(ar\left ( BYXD \right ) = ar \left ( ABMN \right )\)

                (iv) . \(\Delta FCB\cong \Delta ACE\)

                (v) . \(ar\left ( CYXE \right ) = 2 ar \left ( \Delta FCB \right )\)

                (vi) . \(ar\left ( CYXE \right ) = ar \left ( ACFG \right )\)

                (vii) . \(ar\left ( BCED \right ) = ar\left ( ABMN \right )+ar\left ( ACFG \right )\)

21

Solution:

(i) . In \(\Delta\)MBC  and  \(\Delta\) ABD , we have

MB = AB

BC = BD

And \(\angle\) MBC = \(\angle\) ABD                          [since ,\(\angle\)MBC and \(\angle\)ABC are obtained by adding \(\angle\)ABC to a right angle.]

So, by SAS congruence criterion, we have

\(\Delta MBC\cong \Delta ABD\)

\(\Rightarrow ar\left ( \Delta MBC \right )= ar\left (\Delta ABD \right )\cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )\)

(ii) . Clearly, triangle ABC and rectangle BYXD are on the same base BD and between the same parallels AX and BD .

\(∴ ar\left ( \Delta ABD \right )= \frac{1}{2}ar\left (rect \;BYXD \right )\)

\(\Rightarrow ar\left (rect \;BYXD \right )=2ar\left ( \Delta ABD \right )\)

\(\Rightarrow ar\left (rect \;BYXD \right )=2ar\left ( \Delta MBC \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2 \right )\)                                       [From  equ .1]

(iii) . Since triangles MBC and square MBAN are on the same base Mb and between the same parallels MB and NC.

\(∴2ar\left ( \Delta MBC \right )=ar\left ( MBAN \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 3 \right )\)

From equ. 2 and 3, we have

\(ar\left ( sq . MBAN \right )=ar\left ( rect BYXD \right )\)

(iv) . In triangles FCB and ACE, we have

FC = AC

CB = CE

And ,  \(\angle\)FCB = \(\angle\) ACE                           [since ,\(\angle\)FCB and \(\angle\)ACE  are obtained by adding \(\angle\)ACB to a right angle.]

So, by SAS congruence criterion, we have

\(\Delta FCB\cong \Delta ACE\)

(v) . We have,

\(\Delta FCB\cong \Delta ACE\)

\(\Rightarrow ar\left ( \Delta FCB \right )=ar \left (\Delta ACE \right )\)

Clearly, triangle ACE and rectangle CYXE are on the same base CE and between same parallels CE and AX.

\(∴ 2ar\left ( \Delta ACE \right )=ar\left ( CYXE \right )\)

\(\Rightarrow 2ar\left ( \Delta FCB \right )=ar \left (\Delta CYXE \right )\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 4 \right )\)

(vi) . Clearly , triangle FCb and rectangle FCAG are on the same base FC and between the same parallels FC and BG.

\(∴ 2ar\left ( \Delta FCB \right )=ar\left ( FCAG \right )\cdot \cdot \cdot \cdot \cdot \cdot \left ( 5 \right )\)

From 4 and 5, we get

\(ar\left ( CYXE \right ) = ar\left ( ACFG \right )\)

(vii) . Applying Pythagoras theorem in triangle ACB, we have

\(BC^{2} = AB^{2}+AC^{2}\)

\(\Rightarrow BC\times BD= AB\times MB+AC\times FC\)

\(\Rightarrow ar\left ( BCED \right ) = ar\left ( ABMN \right )+ar\left ( ACFG \right )\)

Leave a Comment

Your email address will not be published. Required fields are marked *