Areas Class 9th Mathematics AP Board Solution

Class 9th Mathematics AP Board Solution
Exercise 11.1
  1. In ∆ABC, ∠ABC = 90°, AD = DC, AB = 12 cm and BC = 6.5 cm. Find the area of…
  2. Find the area of a quadrilateral PQRS in which ∠QPS = ∠SQR = 90°, PQ = 12 cm,…
  3. Find the area of trapezium ABCD as given in the figure in which ADCE is a…
  4. ABCD is a parallelogram. The diagonals AC and BD intersect each other at ‘O’.…
Exercise 11.2
  1. The area of parallelogram ABCD is 36 cm^2 . Calculate the height of…
  2. ABCD is a parallelogram. AE is perpendicular on DC and CF is perpendicular on…
  3. If E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD…
  4. What figure do you get, if you join ∆APM, ∆DPO, ∆OCN and ∆MNB in the example 3.…
  5. P and Q are any two points lying on the sides DC and AD respectively of a…
  6. P is a point in the interior of a parallelogram ABCD. Show that (i) ar(∆APB) +…
  7. Prove that the area of a trapezium is half the sum of the parallel sides…
  8. PQRS and ABRS are parallelograms and X is any point on the side BR. Show that…
  9. A farmer has a field in the form of a parallelogram PQRS as shown in the…
  10. Prove that the area of a rhombus is equal to half of the product of the…
Exercise 11.3
  1. In a triangle ABC (see figure), E is the midpoint of median AD, show that (i)…
  2. Show that the diagonals of a parallelogram divide it into four triangles of…
  3. In the figure, ∆ABC and ∆ABD are two triangles on the same base AB. If line…
  4. In the figure, ∆ABC, D, E, F are the midpoints of sides BC, CA and AB…
  5. In the figure D, E are points on the sides AB and AC respectively of ∆ABC such…
  6. In the figure, XY is a line parallel to BC is drawn through A. If BE || CA and…
  7. In the figure, diagonals AC and BD of a trapezium ABCD with AB || DC intersect…
  8. In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC…
  9. In the figure, if ar ∆RAS = ar ∆RBS and [ar (∆QRB) = ar(∆PAS) then show that…
  10. A villager Ramayya has a plot of land in the shape of a quadrilateral. The…

Exercise 11.1
Question 1.

In ∆ABC, ∠ABC = 90°, AD = DC, AB = 12 cm and BC = 6.5 cm. Find the area of ∆ADB.



Answer:

Given: ∠ABC = 90°


AD = DC


AB = 12 cm and BC = 6.5 cm


Area of ∆ABC = 1/2 × BC × AB


= 1/2 × 6.5 × 12 …(given)


= 6 × 6.5


= 39 sq. cm


Area of ∆ABC = 39 sq. cm …(i)


AD = DC which means BD is the median


Median divides area of triangle in two equal parts


Therefore area(∆ABD) = area(∆CDB) …(ii)


From figure area(∆ABC) = area(∆ABD) + area(∆CDB)


Using equation (i) and (ii) we can write


39 = area(∆ABD) + area(∆ABD)


39 = 2 area(∆ABD)


Therefore area(∆ABD) = 19.5 sq. cm



Question 2.

Find the area of a quadrilateral PQRS in which ∠QPS = ∠SQR = 90°, PQ = 12 cm, PS = 9 cm, QR = 8 cm and SR = 17 cm (Hint: PQRS has two parts)



Answer:

Area of quadrilateral PQRS = area(ΔSQR) + area(ΔPQS)…(i)


Let us find area(ΔPQS)


Base = PQ = 12 cm


Height = PS = 9 cm


area of triangle =  × base × height


⇒ area(ΔPQS) =  × PQ × PS


⇒ area(ΔPQS) =  × 12 × 9


⇒ area(ΔPQS) = 6 × 9


⇒ area(ΔPQS) = 54 cm2


Using pythagoras theorem


SQ = 


⇒ SQ = 


⇒ SQ = 


⇒ SQ = 


⇒ SQ = 15 …(ii)


Now let us find area(ΔSQR)


Base = QR = 8 cm


Height = SQ = 15 cm …from (ii)


area of triangle =  × base × height


⇒ area(ΔSQR) =  × QR × SQ


⇒ area(ΔSQR) =  × 8 × 15


⇒ area(ΔSQR) = 4 × 15


⇒ area(ΔSQR) = 60 cm2


Therefore from (i)


Area of quadrilateral PQRS = area(ΔSQR) + area(ΔPQS)


= 60 + 54


= 114 cm2


Hence area of quadrilateral PQRS = 114 cm2



Question 3.

Find the area of trapezium ABCD as given in the figure in which ADCE is a rectangle. (Hint: ABCD has two parts)



Answer:

area of trapezium ABCD = area of rectangle ADCE + area(ΔBEC)…(i)


let us find area of rectangle ADCE


length = AD = 8 cm


breadth = AE = 3 cm


area of rectangle = length × breadth


⇒ area of rectangle ADCE = length × breadth


= AD × AE


= 8 × 3


= 24 sq. cm


Therefore, area of rectangle ADCE = 24 sq. cm


From figure EC || AD


⇒ ∠BEC = ∠EAD = 90° …corresponding angles


⇒ ∠BEC = 90°


And since ADCE is a rectangle EC = AD


⇒ EC = 8 cm


Now let us find area(ΔBEC)


area of triangle =  × base × height


⇒ area(ΔBEC) =  × EC × BE


⇒ area(ΔBEC) =  × 8 × 3


⇒ area(ΔBEC) = 4 × 3


⇒ area(ΔBEC) = 12 cm2


From (i)


area of trapezium ABCD = area of rectangle ADCE + area(ΔBEC)…(i)


= 24 + 12


= 36 sq. cm


therefore, area of trapezium ABCD = 36 sq. cm



Question 4.

ABCD is a parallelogram. The diagonals AC and BD intersect each other at ‘O’. Prove that ar(∆AOD) = ar(∆BOC). (Hint: Congruent figures have equal area)



Answer:


Extend AB to F and draw perpendiculars from point D and point C on line AF as shown in the figure


As ABCD is a parallelogram DC || AF


The perpendicular distances between parallel lines i.e. DE and CG are equal DE = CG = h


Therefore, the perpendicular distance DE and CG are equal


Consider ΔABD


Base = AB


Height = DE = h


area of triangle =  × base × height


⇒ area(ΔABD) =  × AB × DE


⇒ area(ΔABD) = 


From figure


area(ΔAOD) = area(ΔABD) - area(ΔAOB)


⇒ area(ΔAOD) =  - area(ΔAOB) …(i)


Consider ΔABC


Base = AB


Height = CG


area of triangle =  × base × height


⇒ area(ΔABC) =  × AB × CG


⇒ area(ΔABC) = 


From figure


area(ΔBOC) = area(ΔABC) - area(ΔAOB)


⇒ area(ΔBOC) =  - area(ΔAOB) …(ii)


From (i) and (ii)


area(ΔAOD) = area(ΔBOC)




Exercise 11.2
Question 1.

The area of parallelogram ABCD is 36 cm2. Calculate the height of parallelogram ABEF if AB = 4.2 cm.



Answer:


Extend BA to H and drop a perpendicular from D on AH mark intersection point I as shown in the figure


DI is the height of parallelogram ABCD


Given base = AB = 4.2 cm


Area of parallelogram ABCD = 36 sq. cm


Area of parallelogram = base × height


⇒ Area of parallelogram ABCD = AB × DI


⇒ 36 = 4.2 × DI


⇒ DI =  =  =  = 


⇒ DI = 8.57 cm


Now as seen in the figure points E and F of the parallelogram ABEF lie on the same line as that of D and C


Therefore DE || AB


Perpendicular distance between parallel lines is constant


Therefore, for parallelogram ABEF the perpendicular distance between EF and AB will be DI i.e. 8.57 cm


Therefore, height of parallelogram ABEF is 8.57 cm



Question 2.

ABCD is a parallelogram. AE is perpendicular on DC and CF is perpendicular on AD.

If AB = 10 cm, AE = 8 cm and CF = 12 cm. Find AD.



Answer:

Let us start by finding the area of parallelogram ABCD


If we consider DC as the base of the parallelogram the height will be AE


Area of parallelogram = base × height


⇒ area of parallelogram ABCD = DC × AE


Given is AB = 10 cm


As it is a parallelogram opposite sides are equal i.e. DC = 10 cm


AE = 8 cm …(given)


Therefore, area of parallelogram ABCD = 10 × 8 = 80 cm2


As for the same shape area won’t change even if we find area by other terms


Now consider AD as the base of parallelogram ABCD then the height will be FC


Area of parallelogram = base × height


⇒ area of parallelogram ABCD = AD × CF


CF = 12 cm …(given)


⇒ 80 = AD × 12


⇒ AD =  = 


Therefore, AD = 6.67 cm



Question 3.

If E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a parallelogram ABCD, show that ar(EFGH)  ar(ABCD).



Answer:


Construct line HF as shown and construct perpendiculars EJ and GK on HF as shown


The line HF divides the parallelogram ABCD into two parallelograms ABFH and parallelogram HFCD


Consider parallelogram ABFH


EJ is the perpendicular distance between AB and HF therefore EJ is the height of parallelogram ABFH and also EJ is height of ΔEFH


Area of ΔEFH =  × HF × EJ


But area of parallelogram ABFH = HF × EJ


Therefore, area of ΔEFH =  × area of parallelogram ABFH …(i)


Consider parallelogram HFCD


GK is the perpendicular distance between DC and HF therefore GK is the height of parallelogram HFCD and also GK is height of ΔGFH


Area of ΔGFH =  × HF × GK


But area of parallelogram HFCD = HF × GK


Therefore, area of ΔGFH =  × area of parallelogram HFCD …(ii)


Add equation (i) and (ii)


⇒ area of ΔEFH + area of ΔGFH =  × area of parallelogram ABFH +  × area of parallelogram HFCD


⇒ area of ΔEFH + area of ΔGFH =  × (area of parallelogram ABFH + area of parallelogram HFCD) …(iii)


From figure area of ΔEFH + area of ΔGFH = area of parallelogram EFGH and


area of parallelogram ABFH + area of parallelogram HFCD = area of parallelogram ABCD


therefore equation (iii) becomes


area of parallelogram EFGH =  × area of parallelogram ABCD


hence proved



Question 4.

What figure do you get, if you join ∆APM, ∆DPO, ∆OCN and ∆MNB in the example 3.


Answer:

We get a figure like this




Question 5.

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD show that ar(∆APB) = ar ∆(BQC).



Answer:


Extend CB to G and drop perpendiculars from point P and Q on AB and BG respectively as shown


If we consider AB as the base of parallelogram ABCD then PF is the height and if we consider BC as the base of parallelogram ABCD then BG is the height


So we can write area of parallelogram ABCD in two ways


Area of parallelogram = base × height


Considering AB as base


⇒ Area of parallelogram ABCD = AB × PF …(i)


Considering BC as base


⇒ Area of parallelogram ABCD = BC × QH …(ii)


Now consider ΔABP


PF is the height


Base = AB


Area of triangle =  × base × height


⇒ Area of ΔABP =  × AB × PF


Using (i)


⇒ Area of ΔABP =  × area of parallelogram ABCD …(iii)


Now consider ΔCQB


QH is the height


Base = BC


Area of triangle =  × base × height


⇒ Area of ΔCQB =  × BC × QH


Using (ii)


⇒ Area of ΔCQB =  × area of parallelogram ABCD …(iv)


From (iii) and (iv)


Area of ΔABP = Area of ΔCQB



Question 6.

P is a point in the interior of a parallelogram ABCD. Show that

(i) ar(∆APB) + ar(∆PCD)  ar(ABCD)

(ii) ar(∆APD) + ar(∆PBC) = ar(∆APB) + ar(∆PCD)

(Hint : Through P, draw a line parallel to AB)



Answer:


i) Construct segment GH parallel to AB and CD passing through point P as shown


Also construct perpendiculars PI and PJ on segments CD and AB respectively


Consider parallelogram DCGH


Base = CD …from figure


Height = PI …from figure


Area of parallelogram = base × height


Area of parallelogram DCGH = CD × PI …(i)


Consider parallelogram ABGH


Base = AB …from figure


Height = PJ …from figure


Area of parallelogram = base × height


Area of parallelogram ABGH = AB × PJ …(ii)


Area of triangle =  × base × height


For ΔPCD


Base = CD


Height = PI


Area of ΔPCD =  × CD × PI


Using (i)


⇒ Area of ΔPCD =  × Area of parallelogram DCGH …(iii)


For ΔAPB


Base = AB


Height = PJ


Area of ΔAPB =  × AP × PJ


Using (ii)


⇒ Area of ΔAPB =  × Area of parallelogram ABGH …(iv)


Add (iii) and (iv)


⇒ Area of ΔPCD + Area of ΔAPB =  × Area of parallelogram DCGH +  × Area of parallelogram ABGH


⇒ Area of ΔPCD + Area of ΔAPB =  × (Area of parallelogram DCGH + Area of parallelogram ABGH)


From figure Area of parallelogram DCGH + Area of parallelogram ABGH = area of parallelogram ABCD


Therefore, Area of ΔPCD + Area of ΔAPB =  × area of parallelogram ABCD


⇒ area of parallelogram ABCD = 2 × Area of ΔPCD + 2 × Area of ΔAPB …(*)


ii) from figure


area(ΔDPC) = area(DCGH) – area(ΔDHP) – area(CPG) …(i)


area(ΔAPB) = area(ABGH) – area(ΔAPH) – area(BPG) …(ii)


add equation (i) and (ii)


⇒ area(ΔDPC) + area(ΔAPB) = [area(DCGH) + area(ABGH)] – [area(ΔDHP) + area(ΔAPH)] – [area(CPG) + area(BPG)]


⇒ area(ΔDPC) + area(ΔAPB) = area(ABCD) – area(APD) – area(BPC)


Using equation (*) from first part of question


⇒ area(ΔDPC) + area(ΔAPB) = 2 × Area of ΔPCD + 2 × Area of ΔAPB – area(APD) – area(BPC)


Rearranging the terms we get


area(APD) + area(BPC) = 2 × Area of ΔPCD + 2 × Area of ΔAPB - area(ΔDPC) - area(ΔAPB)


therefore, area(APD) + area(BPC) = area(ΔPCD) + area(ΔAPB)



Question 7.

Prove that the area of a trapezium is half the sum of the parallel sides multiplied by the distance between them.


Answer:


ABCD be trapezium with CD || AB


CF and DH are perpendiculars to segment AB from C and D respectively


From figure


Area of trapezium ABCD = area(ΔAFC) + area of rectangle CDFH + area(ΔBHD) …(i)


Consider rectangle CDHF


Length = FH


Breadth = CF


Area of rectangle = length × breadth


area of rectangle CDFH = FH × CF …(ii)


Consider ΔAFC


base = AF


height = CF


⇒ area(ΔAFC) =  × AF × CF …(iii)


Consider ΔDBH


base = BH


height = HD


⇒ area(ΔDBH) =  × BH × HD …(iv)


Substitute (ii), (iii) and (iv) in (i) we get


Area of trapezium ABCD = FH×CF + ×AF×CF + ×BH×HD


Since CDHF is rectangle


CF = HD = h


⇒ Area of trapezium ABCD = FH×h + ×AF×h + ×BH×h


⇒ Area of trapezium ABCD = h × (FH + ×AF + ×BH)


= h × [FH +  × (AF + BH)]


= h × [FH +  × (AB – FH)]


= h × (FH + ×AB - ×FH)


= h × (×FH + ×AB)


 × h × (FH + AB)


Since CDHF is rectangle


FH = CD


⇒ Area of trapezium ABCD =  × h × (CD + AB)


h is the distance between parallel sides AB and CD


Therefore, area of a trapezium is half the sum of the parallel sides multiplied by the distance between them



Question 8.

PQRS and ABRS are parallelograms and X is any point on the side BR. Show that

(i) ar(PQRS) = ar(ABRS)

(ii) ar(∆AXS) =  ar(PQRS)



Answer:


Constructions:


Extend the common base SR to C


Drop perpendicular from point B on the extended line mark intersection point as D


BD will be the height of both the parallelograms PQRS and ABRS with common base SR


Drop perpendicular on AS from point X thus XF will be the height for ΔAXS and also height for parallelogram ABRS if we consider AS as the base


i) consider parallelogram ABSR


base = SR


height = BD


area of parallelogram = base × height


area(ABSR) = SR × BD …(i)


consider parallelogram PQRS


base = SR


height = BD


area of parallelogram = base × height


area(PQRS) = SR × BD …(ii)


from (i) and (ii)


area(ABSR) = area(PQRS) …(*)


ii) Consider parallelogram ABRS


Let base = AS


Then Height = XF


Area of parallelogram = base × height


Area of parallelogram ABRS = AS × XF …(i)


For ΔAXS


Base = AS


Height = XF


Area of ΔAXS =  × AS × XF


Using (i)


⇒ Area of ΔAXS =  × Area of parallelogram ABRS


Using equation (*) from first part of question


Area of ΔAXS =  × area of parallelogram PQRS



Question 9.

A farmer has a field in the form of a parallelogram PQRS as shown in the figure. He took the mid- point A on RS and joined it to points P and Q. In how many parts of field is divided? What are the shapes of these parts?

The farmer wants to sow groundnuts which are equal to the sum of pulses and paddy. How should he sow? State reasons?



Answer:

It can be seen from the figure that the field is divided in three triangular parts ΔSPA, ΔAPQ and ΔARQ


Extend the segment PQ to B and drop a perpendicular from point R on the extended line


Thus the segment RC becomes the height of ΔAPQ and also the height of parallelogram PQRS



consider parallelogram PQRS


base = PQ


height = RC


area of parallelogram = base × height


area(PQRS) = PQ × RC …(i)


For ΔAPQ


Base = PQ


Height = RC ...(because even if we drop a perpendicular from point


A on base PQ it would be of the same length as RC


since SR||PB)


Area of ΔAPQ =  × AP × RC


Using (i)


⇒ Area of ΔAPQ =  × Area of parallelogram PQRS


⇒ 2 × area(ΔAPQ) = Area of parallelogram PQRS …(ii)


Since Area(PQRS) = area(ΔPSA) + area(ΔAPQ) + area(ΔAQR)


Using equation (ii) we get


2 × area(ΔAPQ) = area(ΔPSA) + area(ΔAPQ) + area(ΔAQR)


⇒ area(ΔAPQ) = area(ΔPSA) + area(ΔAQR) …(iii)


let the number of groundnuts be g, pulses be pu and paddy be pa


given g = pu + pa


compare this with equation (iii) we get


area(ΔAPQ) = g


area(ΔPSA) = pu


area(ΔAQR) = pa


therefore, the farmer must sow ground nuts in the region under the area(ΔAPQ), the pulses in the region under the area(ΔPSA) and the paddy in the region under the area(ΔAQR)



Question 10.

Prove that the area of a rhombus is equal to half of the product of the diagonals.


Answer:

Consider rhombus PQRS as shown with diagonals intersecting at point A


Property of rhombus diagonals intersect at 90°



From figure area(PQRS) = area(ΔPQS) + area(ΔRQS) …(i)


Consider ΔPQS


Base = SQ


Height = PA


area(ΔPQS) =  × SQ × PA …(ii)


Consider ΔSQR


Base = SQ


Height = RA


area(ΔSQR) =  × SQ × RA …(iii)


substitute (ii) and (iii) in (i)


⇒ area(PQRS) =  × SQ × PA +  × SQ × RA


 × SQ × (PA + RA)


From figure PA + RA = PR


Therefore, area(PQRS) =  × SQ × PR


Hence, the area of a rhombus is equal to half of the product of the diagonals




Exercise 11.3
Question 1.

In a triangle ABC (see figure), E is the midpoint of median AD, show that

(i) ar ∆ABE = ar∆ACE

(ii) ar ∆ABE =  ar(∆ABC)



Answer:

i) Consider ΔABC


AD is the median which will divide area(ΔABC) in two equal parts


⇒ area(ΔABD) = area(ΔADC) …(i)


Consider ΔEBC


ED is the median which will divide area(ΔEBC) in two equal parts


⇒ area(ΔEBD) = area(ΔEDC) …(ii)


Subtract equation (ii) from (i) i.e perform equation (i) – equation (ii)


⇒ area(ΔABD) - area(ΔEBD) = area(ΔADC) - area(ΔEDC)


⇒ area(ΔABE) = area(ΔACE) …(iii)


ii) consider ΔABD


BE is the median which will divide area(ΔABD) in two equal parts


⇒ area(ΔEBD) = area(ΔABE) …(iv)


Using equation (iv), (iii) and (ii) we can say that


area(ΔABE) = area(ΔEBD) = area(ΔEDC) = area(ΔACE) …(v)


from figure


⇒ area(ΔABC) = area(ΔABE) + area(ΔEBD) + area(ΔEDC) + area(ΔACE)


using (v)


⇒ area(ΔABC) = area(ΔABE) + area(ΔABE) + area(ΔABE) + area(ΔABE)


⇒ area(ΔABC) = 4 × area(ΔABE)


⇒ area(ΔABE) =  × area(ΔABC)



Question 2.

Show that the diagonals of a parallelogram divide it into four triangles of equal area.


Answer:


Consider parallelogram PQRS whose diagonals intersect at point A


Property of parallelogram is that its diagonal bisect each other


⇒ SA = AQ and PA = AR


Consider ΔPQS


PA is the median which divides the area(ΔPQS) into two equal parts


⇒ area(ΔPAS) = area(ΔPAQ) …(i)


Consider ΔRQS


RA is the median which divides the area(ΔRQS) into two equal parts


⇒ area(ΔRAS) = area(ΔRAQ) …(ii)


Consider ΔQPR


QA is the median which divides the area(ΔQPR) into two equal parts


⇒ area(ΔPAQ) = area(ΔRAQ) …(iii)


Using equations (i), (ii) and (iii)


area(ΔPAS) = area(ΔPAQ) = area(ΔRAQ) = area(ΔRAS)


hence, the diagonals of a parallelogram divide it into four triangles of equal area



Question 3.

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



Answer:

The Figure given in question does not match what the question says here is the correct figure according to the question



Consider ΔACD


AO is the median which divides the area(ΔACD) into two equal parts


⇒ area(ΔAOD) = area(ΔAOC) …(i)


Consider ΔBCD


BO is the median which divides the area(ΔBCD) into two equal parts


⇒ area(ΔBOD) = area(ΔBOC) …(ii)


Add equation (i) and (ii)


⇒ area(ΔAOD) + area(ΔBOD) = area(ΔAOC) + area(ΔBOC) …(iii)


From figure


area(ΔAOD) + area(ΔBOD) = area(ΔABD) and


area(ΔAOC) + area(ΔBOC) = area(ΔABC)


therefore equation (iii) becomes


area(ΔABD) = area(ΔABC)



Question 4.

In the figure, ∆ABC, D, E, F are the midpoints of sides BC, CA and AB respectively. Show that

(i) BDEF is a parallelogram

(ii) ar(∆DEF) = ar(∆ABC)

(iii) ar(BDEF) = ar(∆ABC)



Answer:

i) consider ΔABC


E and F are midpoints of the sides AB and AC


The line joining the midpoints of two sides of a triangle is parallel to the third side and half the third side


⇒ EF || BC


⇒ EF || BD …(i)


And EF =  × BC


But D is the midpoint of BC therefore  × BC = BD


⇒ EF = BD …(ii)


E and D are midpoints of the sides AC and BC


⇒ ED || AB


⇒ ED || FB …(iii)


And ED =  × AB


But F is the midpoint of AB therefore  × AB = FB


⇒ ED = FB …(iv)


Using (i), (ii), (iii) and (iv) we can say that BDEF is a parallelogram


Similarly we can prove that AFDE and FECD are also parallelograms


ii) as BDEF is parallelogram with FD as diagonal


the diagonal divides the area of parallelogram in two equal parts


⇒ area(ΔBFD) = area(ΔDEF) …(v)


as AFDE is parallelogram with FE as diagonal


the diagonal divides the area of parallelogram in two equal parts


⇒ area(ΔAFE) = area(ΔDEF) …(vi)


as CEFD is parallelogram with DE as diagonal


the diagonal divides the area of parallelogram in two equal parts


⇒ area(ΔEDC) = area(ΔDEF) …(vii)


From (v), (vi) and (vii)


area(ΔDEF) = area(ΔBFD) = area(ΔAFE) = area(ΔEDC) …(*)


from figure


⇒ area(ΔABC) = area(ΔDEF) + area(ΔBFD) + area(ΔAFE) + area(ΔEDC)


Using (*)


⇒ area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)


⇒ area(ΔABC) = 4 × area(ΔDEF)


⇒ area(ΔDEF) =  × area(ΔABC)


iii) from figure


⇒ area(ΔABC) = area(ΔDEF) + area(ΔBFD) + area(ΔAFE) + area(ΔEDC)


Using (*)


⇒ area(ΔABC) = area(ΔDEF) + area(ΔBFD) + area(ΔFED) + area(ΔEDC) …(viii)


From figure


area(ΔDEF) + area(ΔBFD) = area(BDEF) …(ix)


using (*)


area(ΔDEF) + area(ΔDEF) = area(BDEF)


area(ΔFED) + area(ΔEDC) = area(DCEF) …(x)


using (*)


area(ΔDEF) + area(ΔDEF) = area(DCEF)


therefore area(BDEF) = area(DCEF) …(xi)


substituting equation (ix), (x) and (xi) in equation (viii)


⇒ area(ΔABC) = area(BDEF) + area(BDEF)


⇒ area(ΔABC) = 2 × area(BDEF)


⇒ area(BDEF) =  × area(ΔABC)



Question 5.

In the figure D, E are points on the sides AB and AC respectively of ∆ABC such that ar(∆DBC) = ar(∆EBC). Prove that DE || BC.



Answer:


Consider h1 and h2 as heights of ∆DBC and ∆EBC from points D and E respectively


Given area(∆DBC) = area(∆EBC)


The base of both the triangles is common i.e. BC


Height of ∆DBC = h1


Height of ∆EBC = h2


⇒  × BC × h1 =  × BC × h2


⇒ h1 = h2


Which means points D and E are on the same height from segment BC which implies that line passing through both the points i.e. D and E is parallel to the BC


Therefore DE || BC



Question 6.

In the figure, XY is a line parallel to BC is drawn through A. If BE || CA and CF || BA are drawn to meet XY at E and F respectively. Show that ar(∆ABE) = ar (∆ACF).



Answer:

Given XY || BC and BE || CA and CF || BA


XY || BC implies EA || BC and AF || BC as points E, A and F lie on XY line


Consider quadrilateral ACBE


AC || EB and EA || BC opposite sides are parallel


Therefore, quadrilateral ACBE is a parallelogram with AB as the diagonal


the diagonal divides the area of parallelogram in two equal parts


⇒ area(ΔABE) = area(ΔABC) …(i)


Consider quadrilateral ABCF


AB || FC and AF || BC opposite sides are parallel


Therefore, quadrilateral ABCF is a parallelogram with AC as the diagonal


the diagonal divides the area of parallelogram in two equal parts


⇒ area(ΔACF) = area(ΔABC) …(ii)


From (i) and (ii)


area(∆ABE) = area(∆ACF)



Question 7.

In the figure, diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(∆AOD) = ar(∆BOC).



Answer:

Drop perpendiculars from points D and C on segment AB as shown



Given CD || AB


Therefore the perpendicular distance between the parallel lines I equal


⇒ DG = CH = h


Consider ΔABD


Base = AB


Height = GD = h


Area(ΔABD) =  × AB × h …(i)


Consider ΔABC


Base = AB


Height = CH = h


Area(ΔABC) =  × AB × h …(ii)


From (i) and (ii)


Area(ΔABD) = Area(ΔABC) …(*)


Consider ΔAOD


Area(ΔAOD) = area(ΔABD) - area(ΔABO) …(iii)


Consider ΔBOC


Area(ΔBOC) = area(ΔABC) - area(ΔABO)


But Area(ΔABD) = Area(ΔABC) from (*)


⇒ Area(ΔBOC) = area(ΔABD) - area(ΔABO) …(iv)


Using (iii) and (iv)


Area(ΔAOD) = Area(ΔBOC)



Question 8.

In the figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (∆ACB) = ar (∆ACF)

(ii) ar (AEDF) = ar (ABCDE)



Answer:

i) Given AC || BF


Distance between two parallel lines is constant therefore if we consider AC as common base of ΔABC and ΔFAC then perpendicular distance between lines AC and BF will be same i.e height of triangles ΔABC and ΔFAC will be same


As ΔABC and ΔFAC are triangles with same base and equal height


⇒ area(ΔABC) = area(ΔFAC)


ii) since area(ΔABC) = area(ΔFAC)


add area(ACDE) to both sides


⇒ area(ΔABC) + area(ACDE) = area(ΔFAC) + area(ACDE) …(i)


From figure


area(ΔABC) + area(ACDE) = area(ABCDE) …(ii)


area(ΔFAC) + area(ACDE) = area(AFDE) …(iii)


using (ii) and (iii) in (i)


⇒ area(ABCDE) = area(AFDE)



Question 9.

In the figure, if ar ∆RAS = ar ∆RBS and [ar (∆QRB) = ar(∆PAS) then show that both the quadrilaterals PQSR and RSBA are trapeziums.



Answer:

Extend lines R and S to points J and K as shown



Given that area(∆RAS) = area(∆RBS) …(i)


Common base is RS


Let height of ∆RAS be h1 and ∆RBS be h2 as shown


area(∆RAS) =  × RS × h1


area(∆RBS) =  × RS × h2


by given  × RS × h1 =  × RS × h2


⇒ h1 = h2


As the distance between two lines is constant everywhere then lines are parallel


⇒ RS || AB …(*)


Therefore, ABSR is a trapezium


Given area(∆QRB) = area(∆PAS) …(ii)


area(∆QRB) = area(∆RBS) + area(∆QRS) …(iii)


area(∆PAS) = area(∆RAS) + area(∆RPS) …(iv)


subtract (iii) from (iv)


area(∆QRB) - area(∆PAS) = area(∆RBS) + area(∆QRS) - area(∆RAS) - area(∆RPS)


using (i) and (ii)


⇒ 0 = area(∆QRS) - area(∆RPS)


⇒ area(∆QRS) = area(∆RPS)


Common base for ∆QRS and ∆RPS is RS


Let height of ∆RPS be h3 and ∆RQS be h4 as shown


area(∆RPS) =  × RS × h3


area(∆RQS) =  × RS × h4


by given  × RS × h3 =  × RS × h4


⇒ h3 = h4


As the distance between two lines is constant everywhere then lines are parallel


⇒ RS || PQ


⇒ PQSR is a trapezium



Question 10.

A villager Ramayya has a plot of land in the shape of a quadrilateral. The grampanchayat of the village decided to take over some portion of his plot from one of the corners to construct a school. Ramayya agrees to the above proposal with the condition that he should be given equal amount of land in exchange of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented. (Draw a rough sketch of plot).


Answer:

The shape of plot is quadrilateral but actual shape is not mentioned so we can take any quadrilateral


Here let us consider shape of plot to be square as shown



Consider O as midpoint of AB and join DO as shown



Thus AO = OB …(i)


Area(ΔAOD) is the area given by ramayya to construct school


Now extend DO and CB so that they meet at point R as shown



Area(ΔBOR) is given to Ramayya so that now his plot is ΔDRC


We have to prove that Area(ΔAOD) = Area(ΔBOR)


∠DAO = 90° and ∠OBR = 90° …(ABCD is a square)


∠DOA = ∠BOR …(opposite air of angles)


By AA criteria


ΔDOA ~ ΔROB …(ii)


Area(ΔDOA) = 1/2 × DA × OA …(iii)


Area(ΔROB) = 1/2 × BR × OB


But from (i) OA = OB


⇒ Area(ΔROB) = 1/2 × BR × OA …(iv)


Now looking at (iii) and (iv) if we prove DA = BR then it would imply Area(ΔAOD) = Area(ΔBOR)


Using (ii)


 = 


But from (i) OA = OB


⇒  = 1


⇒ DA = BR


⇒ Area(ΔAOD) = Area(ΔBOR)


PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

HINDI ENTIRE PAPER SOLUTION

MARATHI PAPER SOLUTION

SSC MATHS I PAPER SOLUTION

SSC MATHS II PAPER SOLUTION

SSC SCIENCE I PAPER SOLUTION

SSC SCIENCE II PAPER SOLUTION

SSC ENGLISH PAPER SOLUTION

SSC & HSC ENGLISH WRITING SKILL

HSC ACCOUNTS NOTES

HSC OCM NOTES

HSC ECONOMICS NOTES

HSC SECRETARIAL PRACTICE NOTES

2019 Board Paper Solution

HSC ENGLISH SET A 2019 21st February, 2019

HSC ENGLISH SET B 2019 21st February, 2019

HSC ENGLISH SET C 2019 21st February, 2019

HSC ENGLISH SET D 2019 21st February, 2019

SECRETARIAL PRACTICE (S.P) 2019 25th February, 2019

HSC XII PHYSICS 2019 25th February, 2019

CHEMISTRY XII HSC SOLUTION 27th, February, 2019

OCM PAPER SOLUTION 2019 27th, February, 2019

HSC MATHS PAPER SOLUTION COMMERCE, 2nd March, 2019

HSC MATHS PAPER SOLUTION SCIENCE 2nd, March, 2019

SSC ENGLISH STD 10 5TH MARCH, 2019.

HSC XII ACCOUNTS 2019 6th March, 2019

HSC XII BIOLOGY 2019 6TH March, 2019

HSC XII ECONOMICS 9Th March 2019

SSC Maths I March 2019 Solution 10th Standard11th, March, 2019

SSC MATHS II MARCH 2019 SOLUTION 10TH STD.13th March, 2019

SSC SCIENCE I MARCH 2019 SOLUTION 10TH STD. 15th March, 2019.

SSC SCIENCE II MARCH 2019 SOLUTION 10TH STD. 18th March, 2019.

SSC SOCIAL SCIENCE I MARCH 2019 SOLUTION20th March, 2019

SSC SOCIAL SCIENCE II MARCH 2019 SOLUTION, 22nd March, 2019

XII CBSE - BOARD - MARCH - 2019 ENGLISH - QP + SOLUTIONS, 2nd March, 2019

HSC Maharashtra Board Papers 2020

(Std 12th English Medium)

HSC ECONOMICS MARCH 2020

HSC OCM MARCH 2020

HSC ACCOUNTS MARCH 2020

HSC S.P. MARCH 2020

HSC ENGLISH MARCH 2020

HSC HINDI MARCH 2020

HSC MARATHI MARCH 2020

HSC MATHS MARCH 2020

SSC Maharashtra Board Papers 2020

(Std 10th English Medium)

English MARCH 2020

HindI MARCH 2020

Hindi (Composite) MARCH 2020

Marathi MARCH 2020

Mathematics (Paper 1) MARCH 2020

Mathematics (Paper 2) MARCH 2020

Sanskrit MARCH 2020

Sanskrit (Composite) MARCH 2020

Science (Paper 1) MARCH 2020

Science (Paper 2)

MUST REMEMBER THINGS on the day of Exam

Are you prepared? for English Grammar in Board Exam.

Paper Presentation In Board Exam

How to Score Good Marks in SSC Board Exams

Tips To Score More Than 90% Marks In 12th Board Exam

How to write English exams?

How to prepare for board exam when less time is left

How to memorise what you learn for board exam

No. 1 Simple Hack, you can try out, in preparing for Board Exam

How to Study for CBSE Class 10 Board Exams Subject Wise Tips?

JEE Main 2020 Registration Process – Exam Pattern & Important Dates

NEET UG 2020 Registration Process Exam Pattern & Important Dates

How can One Prepare for two Competitive Exams at the same time?

8 Proven Tips to Handle Anxiety before Exams!

BUY FROM PLAY STORE

DOWNLOAD OUR APP

HOW TO PURCHASE OUR NOTES?

S.P. Important Questions For Board Exam 2021

O.C.M. Important Questions for Board Exam. 2021

Economics Important Questions for Board Exam 2021

Chemistry Important Question Bank for board exam 2021

Physics – Section I- Important Question Bank for Maharashtra Board HSC Examination

Physics – Section II – Science- Important Question Bank for Maharashtra Board HSC 2021 Examination