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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 $AD \parallel BC$

$AB= \sqrt{(17^2- 15^2 )}=8\ cm$ (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

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

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 $\Delta ABC = 27 cm^2 \ and\ EF \parallel BC. \ Find : \ i) Area \ of \parallel gm \ ABCF \ ii) \ Area \ of rect. \ BCDE.$

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: $\ i) \ Area (\Delta DAB) =Area (\Delta CAB) \ ii) \ Area (\Delta AOD) =Area (\Delta BOC)$

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.

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

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 $ii) Area (\Delta AQD) = Area (\Delta ADP) + Area (\Delta BCP)$

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 $Area (\Delta ABP) = Area (quad. ABCD)$.

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

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

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: $Area (quad. PQRS) = \frac{1}{2} \times \ Area (\parallel gm ABCD)$

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

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

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

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

Hence Proven.