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Q.1. In the given figure, \Delta ABC is right angled at B in which BC = 15 cm and CA = 17 cm. Find the area of acute angled \Delta DBC , it being given that AD \parallel  BC 321

Answer:

AB= \sqrt{(17^2- 15^2 )}=8\ cm (since \Delta ABC is a right angled triangle)

Area of \Delta DBC =  \frac{1}{2} \times 15\times 8=60 \ cm^2

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Q.2. In the adjoining figure, area

(\parallel gm ABCD) = 48 cm^2 \ and\  FC \parallel AB   \ Find: \ i) \ Area \ of \parallel gm ABEF \ ii)\  Area \ of \Delta EAB 322

Answer:

Area of \parallel gm ABEF= Area \ of \parallel  gm \ ABCD= 48 cm^2 

This is because \parallel gm ABCD \ and\  \parallel  gm \ ABEF are between the same parallels and have the same base.

Also, since \Delta EAB \ and \ the \parallel gm \ ABCD are on the same base and between the same parallels, so the

Area of \Delta EAB= \frac{1}{2} \times  Area \ of \parallel  gm \ ABCD=24 \ cm^2 

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Q.3. In the adjoining figure, area \Delta ABC = 27 cm^2 \ and\  EF \parallel BC. \ Find : \ i) Area \ of \parallel gm \ ABCF \ ii) \ Area \ of rect. \ BCDE. 323

Answer:

Also, since \Delta ABC \ and \ the \parallel gm \ ABCF are on the same base and between the same parallels, so the

Area of \Delta ABC=  1/2\times  \ Area \ of \parallel  gm \ ABCF

\\Rightarrow  Area \ of \parallel  gm \ ABCF=2\times ( Area \ of \Delta ABC)=54cm^2

Area of \parallel gm \ BCDE= Area \ of \parallel  gm ABCF= 54 cm^2 

This is because \parallel gm \ ABCF \ and\ \parallel  gm \ BCDE are between the same parallels and have the same base.

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Q.4. In trapezium ABCD , it is being given that AB \parallel DC and diagonals AC \ and\  BD intersect at O. Prove that: \ i) \ Area (\Delta DAB) =Area (\Delta CAB) \ ii) \ Area (\Delta AOD) =Area (\Delta BOC) 324

Answer:

Since \Delta EAB \ and\ \Delta CAB are on the same base and between the same parallels, the area of the two \Delta ‘s will be equal.

\Rightarrow  \ Area (\Delta DAB) =Area (\Delta CAB)

Also Since \Delta ACD \ and\ \Delta CBD are on the same base and between the same parallels,the area of the \ two \Delta ‘s will be equal.

m9x\Rightarrow    \ Area (\Delta ACD)=Area (\Delta CBD)

Now, subtract the area of \Delta ODC from both sides we get

\Rightarrow  \ Area (\Delta AOD) =Area (\Delta BOC)

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Q.5. In the adjoining figure, ABCD is a parallelogram, P is a point on DC \ and \ Q \ is \ a \ point \ on \ BC . Prove that : i) \Delta APB \ and\ \Delta AQD are equal in area ii) Area (\Delta AQD) = Area (\Delta ADP) + Area (\Delta BCP) 325

Answer:

Since \Delta APB \ and \ the \parallel gm ABCD are on the same base and between the same parallels, so the

Area of \Delta APB=  \frac{1}{2} \times  \ Area \ of\ \parallel  gm \ ABCD

Similarly, \Delta AQD \ and \ the \ \parallel  gm \ ABCD are on the same base and between the same parallels, so the

Area of \Delta AQD= \frac{1}{2} \times  \ Area \ of \parallel  gm \ ABCD

\Rightarrow   \Delta APB \ and\ \Delta AQD \ are \ equal \ in \ area

Area (\Delta ADP)+ Area (\Delta BCP)= \frac{1}{2} \times DP\times (height)+\frac{1}{2} \times PC\times (height)

\Rightarrow  Area (\Delta ADP)+ Area (\Delta BCP)= \frac{1}{2} \times (DP+PC)\times (height)

= \frac{1}{2} \times (DC)\times (height)

= \frac{1}{2} \times  \ Area \ of\ \parallel  gm \ ABCD = Area \ of \Delta AQD

Hence Proved.

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Q.6. In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that Area (\Delta ABP) = Area (quad. ABCD) .326

Answer:

Area (quad. ABCD)= Area (\Delta ABC)+ Area (\Delta ACD)

Area (\Delta ACD)= \frac{1}{2} \times  \ Area \ of \parallel  gm \ ADPC

Similary, Area (\Delta PCD)= \frac{1}{2} \times  \ Area \ of \parallel  gm \ ADPC

\Rightarrow   Area (\Delta ACD)= Area (\Delta PCD)

Hence Area (quad. ABCD)= Area (\Delta ABC)+ Area (\Delta PCD)

But Area (\Delta PCD)= Area (\Delta PCA)

Hence Proved that Area (\Delta ABP) = Area (quad. ABCD)

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Q.7. PQRS is any quadrilateral. Line segments passing through the vertices are drawn parallel to the diagonals of this quadrilateral so as to obtain a parallelogram ABCD as shown in the adjoining figure. Prove that: Area (quad. PQRS) = \frac{1}{2}  \times  \ Area (\parallel gm ABCD) 327

Answer:

Area (quad. PQRS)=Area (\Delta PQR)+ Area (\Delta PRS)

= \frac{1}{2} \times  \ Area \ of \parallel  gm \ PRAB+ \frac{1}{2} \times  \ Area \ of \parallel  gm \ PRCD

Since \parallel  gm \ ABCD= \parallel  gm \ ABRP+ \parallel  gm \ PRCD

=\frac{1}{2}  \times  \ Area (\parallel gm \ ABCD)

Hence Proven.

m8x     m10x

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