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1.

In a parallelogram ABCD, it is being given that AB = 10 cm and the altitudes corresponding to the sides AB and AD are DL = 6 cm and BM = 8 cm, respectively. Find AD.

Answer» Correct Answer - 7.5 cm
`ar("||gm ABCD")=ABxxDL=ADxxBM`.
Discussion
2.

In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that `a r (A D E) = a r(B C F)dot`

Answer» In parallelogram, `ABCD`, `AD = BC`
In parallelogram, `DCFE`, `DE = CF`
In parallelogram, `ABFE`, `AE = AF`
Now, in `Delta ADE` and `Delta BCF`
`AD = BC`
`DE = CF`
`AE = AF`
So, from SSS criterion, `Delta ADE ~= Delta BCF`
As these two triangles are congruent, their areas will be equal.
`ar(ADE) = ar(BCF)`
Discussion
3.

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that`a r (A P B) xx a r (C P D) = a r (A P D) xx a r (B P C)dot`

Answer» Let us draw a point `X` and `N` on `BD` such that
`AN_|_BD` and `CX_|_BD`.
Then,
`L.H.S. = ar(APB)xxar(CPD)`
`1/2(AN)(PB)xx1/2(CX)(PD)`
`1/2(AN)(PD)xx1/2(CX)(PB)`
`ar(APD)xxar(BPC) = R.H.S.`
Discussion
4.

In a triangle ABC, E is the mid-point of median AD. Show that `a r (B E D)=1/4a r (A B C)```

Answer» We know, the median of a triangle, divides the triangle into two triangles of equal areas.
So, in `Delta ABC`
`ar(ACD)=ar(ABD) = 1/2ar(ABC)->(1)`
Also, we are given,`E` is the mid-point of `AD` that means `BE` is the median of `Delta ABD`
So,
`ar(BED)=ar(AEB) = 1/2ar(ABD)`
From (1),
`ar(BED) = 1/2(1/2ar(ABC))`
`ar(BED) = 1/4ar(ABC)`
Discussion
5.

XY is a line parallel to side BC of a triangle ABC. If `B E || A C`and `C F || A B`meet XY at E and F respectively, show that `a r (A B E) = a r (A C F)`

Answer» From diagram we can see that
`ar(EBCY)=ar(BXFC)`
`ar(/_AEB)=1/2ar(EBLY)`
`arBXFC=2ar(/_AFC)`
from both the equations
ar(/_AEB)=ar(/_AFC)
Discussion
6.

In the adjoining figure, ABCD is a parallelogram and P is any points on BC. Prove that `ar(triangleABP)+ar(triangleDPC)=ar(trianglePDA)`.

Answer» Draw `ALbotBC and PMbotAD`.
Since BC||AD, so distance between them remain the same.
`therefore AL=PM`.
`=(1)/(2)xxBPxxAL+(1)/(2)xxPCxxAL=(1)/(2)xxALxx(BP+PC)`
`=(1)/(2)xxALxxBC=(1)/(2)xxPMxxAD" "[therefore AL=PM and BC = AD]`
`=ar(triangle PDA)`.
Discussion
7.

In figure, ABCD and AEFD are two parallelograms. Prove that `ar (DeltaPEA) = ar (DeltaQFD)`.

Answer» `triangleCPD` and ||gm ABCD are on the same base DC and between the same prallels CD and AB.
`therefore ar(triangleCPD)=(1)/(2)ar("||gm ABCD").`
Similarly, `ar(triangleAQD)=(1)/(2)ar("||gm ABCD").`
` therefore ar(triangleCPD)=ar(triangleAQD).`
Discussion
8.

In figure, ABCD and AEFD are two parallelograms. Prove that `ar (DeltaPEA) = ar (DeltaQFD)`.

Answer» Given, ABCD and AEFD are two parallelograms.
To prove `" " ar(DeltaPEA) = ar (DeltaQFD)`
Proof In quadrilateral PQDA,
`AP|| DQ" "` [since, in parallelogram ABCD, `AB ||CD`]
and `" " PQ || AD" "` [since, in parallelogram AEFD, `FE || AD`]
Then, quadrilateral PQDA is a parallelogram.
Also, parallelogram PQDA and AEFD are on the same base AD and between the same parallels AD and EQ.
`therefore" "` ar (parallelogram PQDA) = ar (parallelogram AEFD)
On subtracting ar (quadrilateral APFD) from both sides, we get
ar (parallelogram PQDA)- ar (quadrilateral APFD)
= ar (parallelogram AEFD) - ar (quadrilateral APFD)
`rArr" "` `ar (DeltaQFD) = ar (DeltaPEA)" "` Hence proved.
Discussion
9.

In an equilateral `Delta ABC, D ` is the midpoint of AB and E is the midpoint of AC. Then, `ar (Delta ABC) : ar (Delta ADE)=?` A. `1 : 2`B. `1 : 4`C. `sqrt3 : 2`D. `3 : 4`

Answer» Correct Answer - B
Let BC = a Then, `BD = (a)/(2)`.
`therefore (ar(triangleBDE))/(ar(triangleABC))=((sqrt3)/(4)*((a)/(2))^(2))/((sqrt3)/(4)*a^(2))=(1)/(4)`.
So, the required ratio is `1 : 4`.
Discussion
10.

A farmer was having a field in the form of a parallelogram PQRS. She took any point Aon RS and joined it to points P and Q. In how many parts the fields is divided? Whatare the shapes of these parts? The farmer wants to sow wheat and pulses in equalportions of the field separately. How should she do it?

Answer» `/_APQ`
PQRS is a parallelogram
PQ||SR
`area(/_APQ)=1/2area(PQRS)`
`area(APQ)+area(PAS)+area(AQR)=area(PQRS)`
`area(PAS)+area(AQR)=1/2(PQR)`
Discussion
11.

In figure, ABCD and AEFD are two parallelograms. Prove that `ar (DeltaPEA) = ar (DeltaQFD)`.

Answer» We have:
AC = BD (opp. Sides of ||gm ABDC)
CE = DF (opp. Sides of ||gm CDFE)
AE = BF (opp. Sides of ||gm ABFE)
`therefore triangleACE cong triangleBDF` [SSS-criterion].
And so, `ar(triangleACE)=ar(triangleBDF)`
" " [`therefore` congruent figures have equal areas].
Discussion
12.

In a ||gm ABCD, if P and Q are midpoints of AB and CD respectively and ar(||gm ABCD) `= 16 cm^(2)` then ar(||gm APQD) = ? A. `8 cm^(2)`B. `12 cm^(2)`C. `6 cm^(2)`D. `9 cm^(2)`

Answer» Correct Answer - A
Let the distance between AB and CD be h cm. Then, ar(||gm APQD)`=AP xx h = (1)/(2)xx AB xx h =(1)/(2)` ar(||gm ABCD)
`=((1)/(2)xx16)cm^(2)= 8cm^(2)`.
Discussion
13.

Calculate the area of quad. ABCD, given in til(ii) Calculate the area of trap. PQRS,

Answer» Correct Answer - `(i) 114 cm^(2)" " (ii) 180 cm^(2)`
Discussion
14.

Let P, Q, R, S be respectively the midpoints of the sides AB, BC, CD and DA of quad. ABCD Show that PQRS is a parallelogram such that `ar("||gm PQRS")=(1)/(2)ar("quad. ABCD")`.

Answer» Join AC and AR.
In `triangleABC`, P and Q are midpoints of AB and BC respectively.
`therefore` PQ||AC and PQ `=(1)/(2) AC`.
In `triangleDAC`, S and R are midpoints of AD and DC respectively.
`therefore` SR||AC SR`=(1)/(2) AC`.
Thus, PQ||SR and PQ = SR.
`therefore` PQRS is a ||gm.
Now, median AR divides `triangle`ACD into two `triangle` of equal area.
`therefore ar(triangleARD)=(1)/(2)ar(triangleACD)." "...(i)`
Median RS divides `triangle`ARD into two `triangle` of equal area.
`therefore ar(triangleDSR)=(1)/(2)ar(triangleARD)." "...(ii)`
From (i) and (ii), we get `ar(triangleDSR)=(1)/(4)ar(triangleACD)`.
Similarly, `ar(triangleBQP)=(1)/(4)ar(triangleABC)`.
`therefore ar(triangleDSR)+ar(triangleBQP)=(1)/(4)[ar(triangleACD)+ar(triangleABC)]`
`rArrar(triangle DSR)+ar(triangleBQP)=(1)/(4)ar["quad.ABCD"]." "...(iii)`
Similarly, `ar(triangleCRQ)+ar(triangleASP)=(1)/(4)ar("quad. ABCD")." "...(iv)`
Adding (iii) and (iv), we get
`ar(triangleDSR)+ar(triangleBQP)+ar(triangleCRQ)+ar(triangleASP)`
`=(1)/(2)ar("quad. ABCD")" "...(v)`
But, `ar(triangleDSR)+ar(triangleBQP)+ar(triangleCRQ)+ar(triangleASP)`
`+ar("||gm PQRS)=ar("quad. ABCD")." "...(vi)`
Substracting (v) from (vi), we get
`ar("||gm PQRS") =(1)/(2)ar("quad. ABCD")`.
Discussion
15.

In figure, `CD || AE` and `CY || BA`. Prove that `ar (DeltaCBX) = ar (DeltaAXY)`.

Answer» Given In figure, `" " CD || AE`
and `" " CY || BA`
To prove `" " ar (DeltaCBX) = ar (DeltaAXY)`
Proof We know that, triangles on the same base and between the same parallels are equal in areas.
Here, `DeltaABY` and `DeltaABC` both lie on the same base AB and between the same parallels CY and BA.
`therefore" "` `ar (DeltaABY) = ar (DeltaABC)`
`rArr" "` `ar (ABX) + ar (AXY) = ar (ABX) + ar (CBX)`
`rArr" "` `ar (AXY) = ar (CBX)" "` [eliminating ar (ABX) from both sides]
`" "` Hence proved.
Discussion
16.

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If `ar (DeltaDFB) = 3 cm^(2)`, then find the area of the parallelogram ABCD.

Answer» Given, ABCD is a parallelogram and CE = BC i.e., C is the mid-point of BE.
Also, `" " ar (DeltaDFB) = 3 cm^(2)`
Now, `DeltaADF` and `DeltaDFB` are on the same base DF and between parallels CD and AB. Then,
`ar(DeltaADF) = ar (DeltaDFB) = 3 cm^(2)" "` ...(i)
In `DeltaABE`, by the converse of mid-point theorem,
`EF = AF" "` [since, C is mid-point of BE] ...(ii)
In `DeltaADF` and `DeltaECF`,
`angleAFD = angleCFE" "` [Vertically opposite angles]
`AF = EF" "` [from Eq. (ii)]
and `" " angleDAF = angleCEF`
[since, `BE || AD` and AE is transversal, then alternate interior angles are equal]
`therefore" "` `DeltaADF ~=DeltaECF" "` [by ASA congruence rule]
Then, `" " ar (DeltaADF) = ar (DeltaCFE)" "` [since, congruent figures have equal area]
`therefore" "` `ar (DeltaCFE) = ar (DeltaADF) = 3 cm^(2)" "` [from Eq. (i)] ...(iii)
Now, in `DeltaBFE`, C is the mid-point of BE then CF is median of
`DeltaBFE`,
`therefore" "` `ar (DeltaCEF) = ar (DeltaBFC)" "` [since, median of a triangle divides it into two triangles of equal area]
`rArr" "` `ar (DeltaBFC) = 3 cm^(2)" "` ...(iv)
Now, `" " ar (DeltaBDC) = ar (DeltaDFB) + ar (DeltaBFC)" "` [from Eqs. (i) and (iv)]
We know that, diagonal of a parallelogram divides it into two congruent triangles of equal areas.
`therefore" "` Area of parallelogram `ABCD = 2 xx "Area of " DeltaBDC`
`= 2 xx 6 = 12 cm^(2)`
Hence, the area of parallelogram ABCD is `12 cm^(2)` .
Discussion
17.

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If `ar (DeltaDFB) = 3 cm^(2)`, then find the area of the parallelogram ABCD.

Answer» Correct Answer - `28 cm^(2)`
`triangleMDA congMCP" " [therefore angleDMA = angleCMP, angleMDA=angleMCP, AD = CP " since " AD = BC and CP = BC]`
`therefore DM = MC` (c.p.c.t) and so BM is a median of `triangleBDC`.
Thus, `ar(DMB)=(1)/(2)ar(BDC)`.
But, `ar(BDC)=(1)/(2)ar(ABCD)` [`therefore` BD is a diagonal of ||gm ABCD].
`therefore ar(DMB)=(1)/(4)ar(ABCD)` and so `ar(ABCD)= 28 cm^(2).
Discussion
18.

In `Delta` ABC, D is the mid-point of AB and P is any point on BC. If `CQ || PD` meets AB and Q (shown in figure), then prove that `ar (DeltaBPQ) = (1)/(2) ar (DeltaABC)`.

Answer» GIVEN D is the midpoint of side AB of `triangle` ABC and P is any point on BC. CQ||PD meets AB in Q.
TO PROVE `ar(triangleBPQ)=(1)/(2)ar(triangleABC)`.
CONSTRUCTION Join CD and PQ.
PROOF We know that a median of a triangle divides it into two triangles of equal area.
And, in `triangle`ABC, CD is a median.
`therefore ar(triangleBCD)=(1)/(2)ar(triangleABC)`
`rArr ar(triangleBPD)+ar(triangleDPC)=(1)/(2)ar(triangleABC)" "...(i)`
But, `triangle`DPC and DPQ being on the same base DP and between the same parallels DP and CQ, we have
`ar(triangleDPC)=ar(triangleDPQ)" "...(ii)`
Using (ii) in(i), we get
`ar(triangleBPQ)+ar(triangleDPQ)=(1)/(2)ar(triangleABC)`
`therefore ar(triangleBPQ)=(1)/(2)ar(triangleABC).`
Discussion
19.

In Figure, `A B C D`is aquadrilateral and `B E|A C`and also `B E`meets `D C`produced at`Edot`Show thatarea of ` A D E`is equal tothe area of the quadrilateral `A B C D`

Answer» We have
`ar(triangleABC)=ar(triangleAEC)`
`[therefore triangle " on the same base and between the same parallels are equal in area"].`
And so, `ar(triangleABC)+ar(triangleADC)=ar(triangleAEC)+ar(triangleADC)`
[adding same areas on both sides]
`rArr ar ("quad.ABCD")=ar(triangleADE)`.
Discussion
20.

In Fig. 9.22, ABCD is a quadrilateral and `B E || A C`and also BE meets DC produced at E. Show that area of `DeltaA D E`is equal to the area of the quadrilateral ABCD.

Answer» Lets proof BE||AC and take other things given.
area(ABCD)=area(ACD)+area(ABC)
area(ADE)=area(ACD)+area(AEC)
So, area(ABC)=area(AEC)
According theorem 9.2
BE||AC.
Discussion
21.

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (see figure). Prove that `ar (DeltaLZY) = ar (MZYX)`.

Answer» Given X and Y are points on the side LN such as that
LX = XY = YN and `XZ || LM`
To prove `" " ar (DeltaLZY) = ar (MZYX)`
Proof Since, `DeltaXMZ` and `DeltaXLZ` are on the same base XZ and between the same parallel lines LM and XZ.
Then, `" " ar (DeltaXMZ) = ar (DeltaXLZ)" "` ...(i)
On adding `ar (DeltaXYZ)` both sides of Eq. (i), we get
`ar (DeltaXMZ) + ar (DeltaXYZ) = ar (DeltaXLZ) + ar (DeltaXYZ)`
`rArr` `" " ar (MZYX) = ar (DeltaLZY)" "` Hence proved.
Discussion
22.

In Figure, `A B C D E `is a pentagon. A line through `B`parallel to`A C`meets `D C`produced at`Fdot`Show that:`a r ( A C B)=a r ( A C F)``a r (A E D F)=a r (A B C D E)`

Answer» Join AF.
`triangleACB and triangleACF` have the same base AC and lie between the same parallels AC and BF.
`therefore ar(triangleACB)=ar(triangleACF)`.
And so, `ar(triangleACB)+ar(triangleACDF)=ar(triangleACF)+ar(triangleACDE)`
`rArr ar(ABCDE)=ar(AEDF)`.
Discussion
23.

ABC is a triangle in which D is the midpoint of BC and E is the midpoint of AD. Prove that `ar(triangleBED)=(1)/(4)ar(triangleABC)`.

Answer» GIVEN A `triangle`ABC in which D is the midpoint of BC and E is the important of AD.
To PROVE `ar(triangleBED)=(1)/(4)ar(triangleABC)`.
PROOF D is the midpoint of BC `rArr` AD is a median of `triangle` ABC
`rArr ar(triangleABD=ar(triangleACD)`
`[therefore " a median divides a " triangle " into two "triangle " of equal area"]`.
`rArr ar(triangleABD)=(1)/(2)ar(triangleABC)." "...(i)`
E is the midpoint of AD `rArr` BE is a median of `triangle` ABD
`rArr ar(triangleBED)=ar(triangleBEA)`
`[therefore " a median divides a " triangle " into two "triangle " of equal area"]`.
`rArr ar(triangleBED)=(1)/(2)ar(triangleABD)=(1)/(2){(1)/(2)ar(triangleABC)}" " ["using (i)"]`
`rArr ar(triangleBED)=(1)/(4)ar(triangleABC)`.
Discussion
24.

Diagonals AC and BD of `square ABCD` intersect each other at point P. Show that `ar(triangle APB)xxar(triangle CPD)=ar(triangle APD)xxar(triangle BPC)`

Answer» `AM _|_ BD & CN _|_ BD `
`L.H.S.= area(/_APB)xxarea(/_CPD)`
`=1/2xx(BPxxAM)xx1/2(PDxxCN)`
`=1/2(BPxxCN)xx1/2(PDxxAM)`
`=area(/_BPC)xxarea(/_APD)`
`L.H.S=R.H.S`
Discussion
25.

In `triangleABC`, if D is the midpoint of BC and E is the midpoint of AD then `ar(triangleBED)` = ? A. `(1)/(2)ar(triangleABC)`B. `(1)/(3)ar(triangleABC)`C. `(1)/(4)ar(triangleABC)`D. `(2)/(3)ar(triangleABC)`

Answer» Correct Answer - C
Median AD divides `triangleABC` into two triangles of equal area.
`thereforear(triangleABD)=ar(triangleADC)=(1)/(2)ar(triangleABC).`
In `triangleABD`, median BE divides it into two `triangle` of equal area.
`therefore ar(triangleBED)=ar(triangleABE)=(1)/(2)ar(triangleABD)=(1)/(4)ar(triangleABC)`.
Discussion
26.

In `triangleABC`, it is given that D is the midpoint of BC, E is the midpoint of BD and O is the midpoint of AE. Then, `ar(triangleBOE)` = ? A. `(1)/(3)ar(triangleABC)`B. `(1)/(4)ar(triangleABC)`C. `(1)/(6)ar(triangleABC)`D. `(1)/(8)ar(triangleABC)`

Answer» Correct Answer - D
Median AD divides `triangleABC` into two `triangle` of equal area.
`therefore ar(triangleABD)=(1)/(2)ar(triangleABC)`.
Median AE divides `triangleABD` into two `triangle` of equal area.
`therefore ar(triangleABE)=(1)/(2)ar(triangleABD)=(1)/(4)ar(triangleABC)`.
Median OB divides `triangleABE` into two `triangle` of equal area.
`therefore ar(triangleBOE)=(1)/(2)ar(triangleABE)=(1)/(8)ar(triangleABC)`.
Discussion
27.

In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that `ar (ABCDE) = ar (DeltaAPQ)`.

Answer» Given ABCDE is a pentagon.
`BP || AC` and `EQ ||AD`.
To prove `" " ar(ABCDE) = ar (APQ)`
Proof We know that, triangles on the same base and between the same parallels are equal in area.
Here, `DeltaADQ` and `DeltaADE` lie on the same base AD and between the same parallels AD and EQ.
So, `" " ar (DeltaADQ) = ar (DeltaADE)" "` ...(i)
Similarly, `DeltaACP` and `DeltaACB` lie on the same base AC and between the same parallels AC and BP.
So, `" " ar (DeltaACP) = ar (DeltaACB)" "` ...(ii)
On adding Eqs. (i) and (ii), we get
`ar (DeltaADQ) + ar (DeltaACP) = ar (DeltaADE) + ar (DeltaACB)`
On adding `ar (DeltaACD)` both sides, we get
`ar (DeltaADQ) + ar (DeltaACP) + ar (DeltaACD) = ar (DeltaADE) + ar (DeltaACB) + ar (DeltaACD)`
`rArr" "` `ar (DeltaAPQ) = ar (ABCDE)" "` Hence proved.
Discussion
28.

If the medians of a `triangle`ABC intersect at G, show that `ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)` `=(1)/(3)ar(triangleABC)`.

Answer» GIVEN A `triangle`ABC. Its medians AD, BE and CF intersect at G.
TO PROVE `ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)=(1)/(3)ar(triangleABC)`
PROOF We know that a median of a triangle divides it into two triangle of equal area.
In `triangle`ABC,AD is the median.
`therefore ar(triangleABD)=ar(triangleACD)." "...(i)`
In `triangle`GBC, GD is the median.
`therefore ar(triangleGBD)=ar(triangleGCD)." "...(ii)`
From (i) and (ii), we get
`ar(triangleABD)-ar(triangleGBD)=ar(triangleACD)-ar(triangleGCD)`
`rArr ar(triangleAGB)=ar(triangleAGC)`.
Similarly, `ar(triangleAGB)=ar(triangleBGC)`.
`rArr ar(triangleAGB)=ar(triangleAGC)`.
But, `ar(triangleABC)=ar(triangleAGB)+ar(triangleAGC)+ar(triangleBGC)`
`=3ar(triangleAGB)" "["using (iii)"]`.
`therefore ar(triangleAGB)=(1)/(3)ar(triangleABC)`.
Hence, `ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)=(1)/(3)ar(triangleABC).`
Discussion
29.

ABCD is a parallelogram and O is a point in its interior. Prove that (i) `ar(triangleAOB)+ar(triangleCOD)` `=(1)/(2)ar("||gm ABCD").` (ii) `ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC)`.

Answer» GIVEN A||gm ABCD and O is a point in its interior.
TO PROVE (i) `ar(triangleAOB)+ar(triangleCOD)=(1)/(2)ar("||gm ABCD"),`
(ii) `ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC)`.
CONSTRUCTION Draw EOF||AB||DC and GOH||AD||BC.
PROOF `triangle`AOB and ||gm EABF being on the same base AB and between the same parallels AB and EF, we have
`ar(triangleAOB)=(1)/(2)ar("||gm EABF")." "...(i)`
Similarly, `ar(triangle COD)=(1)/(2)ar("||gm EFCD")," "...(ii)`
`ar(triangleAOD)=(1)/(2)ar("||gm AHGD")" "...(iii)`
and `ar(triangleBOC)=(1)/(2)ar("||gm BCGH")" "...(iv)`
Adding (i) and (ii), we get
`ar(triangleAOB)+ar(triangleCOD)=(1)/(2)ar("||gm EABF")+(1)/(2)ar("||gm EFCD")`
`=(1)/(2)ar("||gm ABCD")." "...(v)`
Adding (iii) and (iv), we get
`aar(triangleAOD)+ar(triangleBOC)=(1)/(2)ar("||gm AHGD")+(1)/(2)ar("||gm BCGH")`
`=(1)/(2)ar("||gm ABCD")." "...(vi)`
`therefore ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC)`
[from (v) and (vi)].
Discussion
30.

In the adjoining figure, ABCD is a quadrilateral in which diag. BD = 14 cm. If `ALbotBD and CMbotBD` such that AL = 8 cm and CM = 6 cm, find the area of quad. ABCD.

Answer» Correct Answer - `98 cm^(2)`
Discussion
31.

In the figure, the area of parallelogram ABCD is A. `AB xx BM`B. `BC xx BN`C. `DC xx DL`D. `AD xx DL`

Answer» Correct Answer - C
We know that, area of parallelogram is the product of its any side and the corresponding altitude ( or height).
Here, when AB is base, then height is DL.
`therefore` Area of parallelogram = `AB xx DL`
and when AD is base, then height is BM.
`therefore` Area of parallelogram = `AD xx BM`
when DC is base, then height is DL.
`therefore` Area of parallelogram `= DC xx DL`
and when BC is base, then height is not given.
Hence, option (c) is correct.
Discussion
32.

In which of the following figures, you find two polygons on the same base and between the same parallels?A. B. C. D.

Answer» Correct Answer - D
In figures (a), (b) and (c) there are two polygons on the same base but theyare not between the same parallels.
In figure (d), there are two polygons (PQRA and BQRS) on the same base and between the same parallels.
Discussion
33.

Thediagonals of a parallelogram `A B C D`intersectat `Odot`A linethrough `O`meets `A B `in `x a n d C D`in `Ydot`Show that`a r (A X Y X)=1/2(a r|""|^(gm) A B C D)`

Answer» Diagonal AC of ||gm ABCD divides it into two triangles of equal area.
`therefore ar(squareACD)=(1)/(2)ar("||"gm ABCD)" "...(i)`
In `triangle` OAP and OCQ, we have:
OA = OC (diagonals of a ||gm bisect each other)
`angleAOP=angleCOQ " "("vert. opp." angle)`
`angle PAO=angleQCO " " ("alt. interior " angle)`
`therefore triangle OAPcongtriangleOCQ.`
`therefore ar(triangleOAP)=ar(triangleOCQ)`
`rArr ar(triangleOAP)+ar("quad. AOQD")`
`=ar(triangleOCQ)+ar("quad. AOQD")`
`rArr ar("quad. APQD")=ar(triangleACD)`
`=(1)/(2)ar("||gm ABCD") " " ["using (i)"]`.
`therefore ar(squareAPQD)=(1)/(2)ar("||gm ABCD")`.
Discussion
34.

The median of a triangle divides it into twoA. triangles of equal areaB. congruent trianglesC. right angled trianglesD. isosceles triangles

Answer» Correct Answer - A
We know that , a median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Thus, a median of triangle divides it into two triangles of equal area.
Discussion
35.

The area of ||gm ABCD is A. `ABxxBM`B. `BCxxBN`C. `DCxxDL`D. `ADxxDL`

Answer» Correct Answer - C
Discussion
36.

The median of a triangle divides it into twoA. triangles of equal areaB. congruent trianglesC. isosceles trianglesD. right triangles

Answer» Correct Answer - A
Discussion
37.

In the given figure, ABCD is a ||gm in which `AB = CD = 5 cm and BD bot DC` such that BD = 6.8 cm. Then, the area of ABCD = ? A. `17 cm^(2)`B. `25 cm^(2)`C. `34 cm^(2)`D. `68 cm ^(2)`

Answer» Correct Answer - C
Discussion
38.

Assertion (A) : In the given figlure, ABCD is a ||gm in which `DEbotAD and BF bot AD`. If AB = 16 cm, DE = 8 cm and BF = 10 cm then what is the length of AD?.A. Both Assertion (A) and Reason (R ) are true and Reason (R ) is a correct explansion of Assertion (A).B. Both Assertion (A) and Reason (R ) are true but Reason (R ) is not a correct explansion of Assertion (A).C. Assertion (A) is true and Reason (R ) is false.D. Assertion (A) is false and Reason (R ) is true.

Answer» Correct Answer - D
Reason (R ) is clearly true.
Area of ||gm `=Abxx DE =AD xx BF`.
Let AD = x cm. Then,
`16xx8=xx x xx10rArrx=(16xx8)/(10)=12.8 cm.`
But, AD is 12 cm.
`therefore` Assertion (A) is false.
Thus, Assertion (A) is false and Reason (R ) is true.
So, the correct answer is (d).
Discussion
39.

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas isA. `1 : 2`B. `1 : 1`C. `2 : 1`D. `3 : 1`

Answer» Correct Answer - B
We know that, parallelogram on the equal bases and between the same parallels are equal in area. So, ratio of their areas is `1 : 1`.
Discussion
40.

ABCD is a parallelogram and O is the point of intersection of its diagonals. If `ar(triangle AOD)= 4cm^2` find `ar(triangle AOB)`

Answer» ABCD is a parallelogram
Diagonal AC and BD bisects each other.
Mid point of AC=Midpoint of BD
O is bisect BD in two part.
AO is themedian of`/_ABD`
Median divides the area of triangle `/_ABD` in two equals parts.
area of`/_AOD= area of`/_AOB``
A=area of`/_AOB`
`area(/_ADB)=4cm^2`.
Discussion
41.

Which of the following is a false statement? A. A median of a triangle divides it into two triangle of equal area.B. The diagonals of a ||gm divide it into four triangle of equal area.C. In a `triangleABC`, if E is the important of mdian AD then `ar(triangleBED)=(1)/(4)ar(triangleABC)`.D. In a trap. ABCD, it is given that AB||DC and the diagonals AC and BD intersect at O. Then, `ar(triangleAOB)= ar(triangleCOD)`.

Answer» Correct Answer - D
Discussion
42.

ABCD is a rhombus in which `angleC= 60^(@)`then AC:BD=. A. `sqrt3 : 1`B. `sqrt3 : sqrt2`C. `3 : 1`D. `3 : 2`

Answer» Correct Answer - A
ABCD is a rhombus. So its all sides are equal.
Now, `BC = DC rArr angleBDC = angleDBC =x^(@)` (say).
Also, `angleBCD = 60^(@)` (given).
`therefore x^(@)+x^(@)+60^(@)=180^(@)rArr2x=120rArrx = 60`.
`therefore angle BDC=angleDBC=angleBCD=60^(@).` So, `triangleBCD` is an equilateral triangle.
`therefore BD = BC =a`(say).
`AB^(2)=OA^(2)+OB^(2)rArrOA^(2)=AB^(2)-OB^(2)=a^(2)-((a)/(2))^(2)=a^(2)-(a^(2))/4=(3a^(2))/(4)`
`therefore AC : BD =sqrt3a : a=sqrt3 : 1`.
Discussion
43.

The lengths of the diagonals of a rhombus are 12 cm and 16 cm. Find the area of rhombusA. `192 cm^(2)`B. `96 cm^(2)`C. `64 cm^(2)`D. `80 cm^(2)`

Answer» Correct Answer - B
Area of the rhombus `=((1)/(2)xxd_(1)xxd_(2))=((1)/(2)xx12xx16)cm^(2)=96 cm^(2)`
Discussion
44.

Which of the following is a false statement?A. If the diagonals of a rhombus are 18 cm and 14 cm then its area is 126 `cm^(2)`B. Area of a ||gm `=(1)/(2)xx "base" xx "corresponding height"`.C. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.D. If the area of a ||gm with one side 24 cm and corresponding height h cm is `192 cm^(2)` then h =8 cm.

Answer» Correct Answer - B
Discussion
45.

The area of the rhombus is 360 sq cm. If one of its diagonals is 40 cm find the other diagonal

Answer» Area of rhombus=`1/2*d_1*d_2=360`
`=1/2*4*d_2=360`
`d_2=18cm`.
Discussion
46.

Assertion (A) : If ABCD is a rhombus whose one angle is `60^(@)` then the ratio of the lengths of its diagonals is `sqrt3 : 1` Reason (R ) : Median of a triangle divides it into two triangle of equal area.A. Both Assertion (A) and Reason (R ) are true and Reason (R ) is a correct explansion of Assertion (A).B. Both Assertion (A) and Reason (R ) are true but Reason (R ) is not a correct explansion of Assertion (A).C. Assertion (A) is true and Reason (R ) is false.D. Assertion (A) is false and Reason (R ) is true.

Answer» Correct Answer - B
Discussion
47.

Asserion (A) : The diagonals of a ||gm divide it into four triangle of equal area. Reason (R ) : A diagonal of a ||gm divides it into two triangle of equal area.A. Both Assertion (A) and Reason (R ) are true and Reason (R ) is a correct explansion of Assertion (A).B. Both Assertion (A) and Reason (R ) are true but Reason (R ) is not a correct explansion of Assertion (A).C. Assertion (A) is true and Reason (R ) is false.D. Assertion (A) is false and Reason (R ) is true.

Answer» Correct Answer - A
Discussion
48.

In the centre of a rectangular lawn of dimensioms `50 mxx40 m` a rectangular pond has to be constructed so that the area of the grass surroundng the pond would be `1184 m^2` Find the length and breadth of the pond

Answer» Area of lawn=50*40=2000`m^2`.
Area of pond=2000-1184=816`m^2.`
(50-2x)(40-2x)=816
`4x^2-180x+1184-0`
`x=(45pmsqrt(2025-1184))/2`
`x=(45pm29)/2=74/2,16/2`
`x=8m`
`l=50-2*8=50-16=34m`
`b=40-2*8=40-16=24m`.
Discussion
49.

In the given figure, ABCD is a trapezium such that `AL bot DC and BM bot DC`. If AB = 7 cm, BC = AD = 5 cm and AL = BM = 4 cm then ar(trap. ABCD) = ? A. `24 cm^(2)`B. `40 cm^(2)`C. `55 cm^(2)`D. `27.5 cm^(2)`

Answer» Correct Answer - B
`CM^(2)=BC^(2)-BM^(2)=(5)^(2)-(4)^(2)=9 rArrCM=3 cm.`
Similarly, DL = 3 cm.
`therefore CD = DL + LM + CM =(3+7+3)cm = 13 cm.`
ar(trap. ABCD) `=[(1)/(2)xx(7+13)xx4]cm^(2)=40 cm^(2)`.
Discussion
50.

In Fig. 9.13, ABCD is a parallelogram and EFCD is a rectangle.Also, `A L_|_D C`. Prove that(i) `a r (A B C D) = a r (E F C D)`(ii) `a r (A B C D) = D C xx A L`

Answer» As per the given figure, both `ABCD` and `EFCD` lies on the same parallel line `EB` and have common base `CD`.
We know, if two parallelograms lie on the same parallel line and have a common base, then they have equal area.
Thus, `ar(ABCD) = ar(EFCD)`
Also, we can say that ,`ar(ABCD) = ar(EFCD) = DCxxDE`
As, `AL` is a perpendicular to `CD` in rectangle `EFCD`,
`AL=DE`
So, `ar(ABCD) =DCxxAL`
Discussion