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1.

In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that ``

Answer» Given, ABC and ABD are two triangle on the same base AB.
The line segment CD is bisected by AB at O. `( .: OC = OD)`
In `DeltaACD`, we have
`OC= OD` (given)
`:. AO` is the median.
Since, the median divides a triangle in two triangles of equal areas.
`:. ar(DeltaAOC) = ar(DeltaAOD)`...(1)
Similarly, in `DeltaBCD, " " ar(DeltaBOC) = ar(DeltaBOD)`...(2)
On adding (1) and (2), we get
`ar(DeltaAOC) + ar(DeltaBOC) = ar(DeltaAOD) + ar(DeltaBOD)`
`rArr ar(DeltaABC) = ar(ABD)`
Discussion
2.

If each diagonal of a quadrilateral separates itinto two triangles of equal area then show that the quadrilateral is aparallelogram.GIVEN : A quadrilateral `A B C D`such that its diagonals `A C`and `B D`are such that `a r( A B D)=a r( C D B`) and `a r( A B C)=a r( A C D)dot`TO PROVE: Quadrilateral `A B C D`is a parallelogram.

Answer» Since AC is the diagonal
`ar(/_ABC)=ar(/_ACD)`
`ar(/_ABC)+ar(/_ACD)=ar(ABCD)`
`ar(/_ABC)=1/2ar(ABCD)-(1)`
Since,BD is the diagonal
`ar(/_ABD)=ar(/_CBD)`
`ar(/_ABD)+ar(/_CBD)=ar(ABCD)`
`ar(/_ABD)=1/2ar(ABCD)-(2)`
AB||CD
AD||BC
It is a parallelogram.
Discussion
3.

In the figure, ABCD is a quadrilateral. A line DP drawn parallel to diagonal AC from point D, meet BC produced at P. Prove that: are of `Delta ABP = " area of " square ABCD`

Answer» Here `AC| |DP` (given)
`DeltaACP` and `DeltaACD` are on same base AC and between the same parallel lines AC and DP.
`therefore` area of `DeltaACP=` area of `squareABCD`
Adding area of `Delta ABC` on both sides.
area of `DeltaACP+` area of `DeltaABC=` area of `DeltaACD+` area of `DeltaABC`
`rArr` area of `DeltaABP=` area of `square ABCD`.
Discussion
4.

In the adjoining figure D, E and F are the mid-points of the sides BC, CA and AB respectively of `Delta ABC`. Prove that: (i) `square BDEF` is a parallelogram (ii) area of `Delta DEF = (1)/(4) xx " area of " Delta ABC` (iii) area of `square BDEF = (1)/(2) xx " area of " Delta ABC`

Answer» (i) In `Delta ABC`
Since, F is the mid-point of AB and E is the mid-point of AC
`{:( :." "FE ||BC),(rArr" "FE ||BD),("Similarly "ED||FB):}` (mid - point theorem)
`:. Square BDEF` is a parallelogram (`:.` pairs of opposite sides are parallel)
(ii) Since, `square BDEF` is a parallelogram and FD is its diagonal
`:.` area of `Delta FBD = " area of " Delta DEF`..(1)
Similarly, we can prove that
area of `Delta CDE = " area of " Delta DEF`...(2)
and area of `Delta AFE = " area of " Delta DEF`...(3)
`:. 4 xx " area of " Delta DEF = " area of "Delta DEF + " area of " Delta DEF + " area of "Delta DEF + " area of "Delta DEF`
`= " area of " Delta DEF + " area of "Delta AEF + " area of "Delta FBD + " area of "Delta CDE`
= area of `Delta ABC`
`rArr " area of " Delta DEF = (1)/(4) xx " area of " Delta ABC`
(iii) Now, area of `square BDEF = 2 xx " area of " Delta ABC`
(`:.` diagonal divides the parallelogram into two equal areas)
`= 2 xx (1)/(4) xx " area of " Delta ABC`
`= (1)/(2) xx " area of "Delta ABC`
Discussion
5.

`A B C D`is a parallelogram `X`and `Y`are the mid-points of `B C`and `C D`respectively. Prove that `a r( A X Y)=3/8a r(^(gm)A B C D)`GIVEN : A parallelogram `A B C D`in which `X`and `Y`are the mid-points of `B C`and `C D`respectively.TO PROVE : `a r( A X Y)=3/8a r(^(gm)a b c d)`CONSTRUCTION : Join `B Ddot`

Answer» `In /_BCD`
X and Y are midpoint of sides BC and CD
XY||BD and XY=1/2BD
`ar(/_CXY)=1/4*ar(/_ABC)`
`ar(/_CYX)=1/8*ar(paral l elogram ABCD)-(1)`
ParallelogramABCD and `/_ABX`
AD||BX
`BX=1/2BC`
`/_ABX=1/4(||gram ABCD)-(2)`
`ar(/_AYD)=1/4ar(||gram ABCD)-(3)`
`ar(||gram ABCD)=ar(/_ABX)+ar(/_AYB)+ar(CYX)+ar(/_AXY)`
`ar(AXY)=ar(||gram ABCD)-[ar(/_ABX)+ar(/_ATD)+ar(/_CYX)]`
`ar(/_AXY)=(1-5/8) ar(||gram ABCD)`
`ar(/_AXY)=3/8 ar(||gram ABCD)`.
Discussion
6.

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

Answer» Given : `CD ||AE and CY ||BA`
To prove: `ar (Delta CBX) = ar (Delta AXY0`
Proof: Since, triangle on the same base AB and between the same parallels AB and YC are equal in area, so we have
`ar(Delta ABC) = ar(Delta ABY)`
`rArr ar(Delta CBX) + ar(DeltaABX) = ar (Delta ABX) + ar (Delta AXY)`
Hence, `ar(Delta CBX) = ar(Delta AXY)` [cancelling `ar(Delta ABX)` from both sides]
Discussion
7.

Parallelogram `A B C D`and rectangle `A B E F`have the same base `A B`and also have equal areas. Show that the perimeter ofthe parallelogram is greater than that of the rectangle.GIVE : `A ^(gm)A B C D`and a rectangle `A B E F`with the same base `A B`and equal areas.TO PROVE : Perimeter of `^(gm)A B C D > P e r i m e t e rofr e c t a nge l`ABEFi.e. `A B+B C+C D+A D > A B+B E+E F+A F`

Answer» AB=DC-(1)
AB=EF-(2)
DC=EF
adding equation 1 and 2
AB+DC=AB+EF-(3)
`BEltBC`and`AFltAD`
`BC>BE` and`ADltAF`
BC+AD>BE+AF-(4)
adding equation 3 and 4
AB+BC+DC+AD>BE+AF+AB+EF
Perimeter of parallelogram > perimeter of rectangle.
Discussion
8.

In a parallelogram `A B C D ,E ,F`are any two points on the sides `A B`and `B C`respectively. Show that `a r( A D F)=a r( D C E)dot`

Answer» `In/_DCE` and parallelogram ABCD
have same base between same
`ar(/_DCE)=1/2ar(ABCD)-(1)`
`In /_AOF` and parallelogram ABCD
have same base AD and between same lines
`ar(/_ADF)=1/2ar(ABCD)-(2)`
`ar(/_DCE)=ar(/_ADF)`.
Discussion
9.

If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram. GIVEN : `A ABC and ||^(gm)BCDE` on the same base BC and between the same parallels `BC and AD`. TO PROVE : `ar(ABC=1/2 ar(||^(gm)BCDE)` CONSTRUCTION : Draw `AL _|_ BC and DM _|_ bc`, metting BC produced in M.

Answer» `ar(/_ABC)=1/2*BC*AC`
`=1/2*BC*DM`
`ar(/_ABC)=1/2ar (BCDE)`.
Discussion
10.

The area of a parallelogram is `150cm^(2)`. If the ratio of its base and corresponding altitude is `3 : 2`, find the length of base and altitude

Answer» Let base of the parallelogram = 3x
and corresponding altitude = 2x
`:.` area of parallelogram `= (3x) (2x)`
`rArr 6x^(2) = 150`
`rArr x^(2) = 25`
`rArr x = 5`
`:. 3x = 3 xx 5 = 15`
and `2x = 2 xx = 10`
`:.` Base of parallelogram `= 15 cm`
and corresponding altitude = 10 cm
Discussion
11.

A circular grassy plot of land, 42 m in diameter has a path of 3.5m wide running round it on the outside. Find the cost of gravelling the path at Rs.4 per square metre.

Answer» Area of path= area of ring`=piR^2-pir^2`
`=pi(R-r)(R+r)`
`=22/7(24.5-21.0)(24.5+21)`
`=500.5m^2`
Cost of 1`m^2`=Rs4
cost of 500.5`m^2`=Rs 2002.
Discussion