Question 14 Marks
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Case study (4 Marks)
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Question 24 Marks
Read the following text carefully and answer the questions that follow:
In the middle of the city, there was a park $\text{ABCD}$ in the form of a parallelogram form so that $\ce{AB=CD, AB \| CD}$ and $\ce{AD = BC , AD \| BC}$.
Municipality converted this park into a rectangular form by adding land in the form of $\Delta APD$ and $\Delta BCQ$. Both the triangular shape of land were covered by planting flower plants.
$i.$ Show that $\Delta APD$ and $\Delta BQC$ are congruent.
$ii. PD$ is equal to which side?
$iii.$ Show that $\Delta ABC$ and $\Delta CDA$ are congruent.
OR
What is the value of $\angle m ?$
In the middle of the city, there was a park $\text{ABCD}$ in the form of a parallelogram form so that $\ce{AB=CD, AB \| CD}$ and $\ce{AD = BC , AD \| BC}$.
Municipality converted this park into a rectangular form by adding land in the form of $\Delta APD$ and $\Delta BCQ$. Both the triangular shape of land were covered by planting flower plants.
$i.$ Show that $\Delta APD$ and $\Delta BQC$ are congruent.
$ii. PD$ is equal to which side?
$iii.$ Show that $\Delta ABC$ and $\Delta CDA$ are congruent.
OR
What is the value of $\angle m ?$
Answer
View full question & answer→$i.$ In $\triangle A P D$ and $\triangle B Q C$
$AD = BC ($given$)$
$AP = CQ ($opposite sides of rectangle$)$
$\angle APD =\angle BQC =90^{\circ}$
By $\text{RHS}$ criteria $\triangle APD \cong \triangle CQB$
$ii.$ $\triangle APD \cong \triangle CQB$
Corresponding part of congruent triangle
side $PD =$ side $BQ$
$iii.$ In $\triangle ABC$ and $\triangle CDA$
$AB = CD ($given$)$
$BC = AD ($given$)$
$AC = AC ($common$)$
By $\text{SSS}$ criteria $\triangle ABC \cong \triangle CDA$
OR
In $\triangle APD$
$\angle APD+\angle PAD+\angle ADP=180^{\circ}$
$\Rightarrow 90^{\circ}+(180^{\circ}-110^{\circ})+\angle ADP =180^{\circ} ($ angle sum property of $\triangle)$
$\Rightarrow \angle ADP = m =180^{\circ}-90^{\circ}-70^{\circ}=20^{\circ}$
$\angle ADP = m =20^{\circ}$
$AD = BC ($given$)$
$AP = CQ ($opposite sides of rectangle$)$
$\angle APD =\angle BQC =90^{\circ}$
By $\text{RHS}$ criteria $\triangle APD \cong \triangle CQB$
$ii.$ $\triangle APD \cong \triangle CQB$
Corresponding part of congruent triangle
side $PD =$ side $BQ$
$iii.$ In $\triangle ABC$ and $\triangle CDA$
$AB = CD ($given$)$
$BC = AD ($given$)$
$AC = AC ($common$)$
By $\text{SSS}$ criteria $\triangle ABC \cong \triangle CDA$
OR
In $\triangle APD$
$\angle APD+\angle PAD+\angle ADP=180^{\circ}$
$\Rightarrow 90^{\circ}+(180^{\circ}-110^{\circ})+\angle ADP =180^{\circ} ($ angle sum property of $\triangle)$
$\Rightarrow \angle ADP = m =180^{\circ}-90^{\circ}-70^{\circ}=20^{\circ}$
$\angle ADP = m =20^{\circ}$
Question 34 Marks
Read the following text carefully and answer the questions that follow:
Ajay lives in Delhi, The city of Ajay's father in laws residence is at Jaipur is $600 \ km$ from Delhi.
Ajay used to travel this $600 \ km$ partly by train and partly by car.
He used to buy cheap items from Delhi and sale at Jaipur and also buying cheap items from Jaipur and sale at Delhi.
Once From Delhi to Jaipur in forward journey he covered $2x \ km$ by train and the rest $y \ km$ by taxi.
But, while returning he did not get a reservation from Jaipur in the train.
So first $2y \ km$ he had to travel by taxi and the rest $x \ km$ by Train.
From Delhi to Jaipur he took $8 hrs$ but in returning it took $10\ hrs.$
$i.$ Write the above information in terms of equation.
$ii.$ Find the value of $x$ and $y$?
$iii.$ Find the speed of Taxi?
OR
Find the speed of Train?
Ajay lives in Delhi, The city of Ajay's father in laws residence is at Jaipur is $600 \ km$ from Delhi.
Ajay used to travel this $600 \ km$ partly by train and partly by car.
He used to buy cheap items from Delhi and sale at Jaipur and also buying cheap items from Jaipur and sale at Delhi.
Once From Delhi to Jaipur in forward journey he covered $2x \ km$ by train and the rest $y \ km$ by taxi.
But, while returning he did not get a reservation from Jaipur in the train.
So first $2y \ km$ he had to travel by taxi and the rest $x \ km$ by Train.
From Delhi to Jaipur he took $8 hrs$ but in returning it took $10\ hrs.$
$i.$ Write the above information in terms of equation.
$ii.$ Find the value of $x$ and $y$?
$iii.$ Find the speed of Taxi?
OR
Find the speed of Train?
Answer
View full question & answer→$i.$ Delhi to Jaipur: $2 x+y=600$
Jaipur to Delhi: $2 y + x =600$
Let $S_1$ and $S_2$ be the speeds of Train and Taxi respectively, then
Dehli to Jaipur: $\frac{2 x}{S_1}+\frac{y}{S_2}=8\ldots (i)$
Jaipur to Delhi: $\frac{x}{S_1}+\frac{2 y}{S_2}=10\ldots (ii)$
$ii. 2 x+y=600 \ldots (1)$
$x+2 y=600 \ldots (2)$
Solving $(1)$ and $(2) \times 2$
$2 x+y-2 x-4 y=600-1200$
$\Rightarrow-3 y=-600$
$\Rightarrow y=200$
Put $y=200$ in $(1)$
$2 x+200=600$
$\Rightarrow x=\frac{400}{2}=200$
$iii.$ We know that speed $=\frac{\text { Distance }}{\text { Time }}$
$\Rightarrow$ Time $=\frac{\text { Distance }}{ \text { Speed }}$
Let $S_1$ and $S_2$ are speeds of train and taxi respectively.
Delhi to Jaipur: $\frac{2 x}{S_1}+\frac{y}{S_2}=8\ldots (i)$
Jaipur to Delhi: $\frac{x}{S_1}+\frac{2 y}{S_2}=10\ldots (ii)$
Solving $(i)$ and $(ii) \times 2$
$\Rightarrow \frac{2 x}{S_1}+\frac{y}{S_2}-\frac{2 x}{S_1}-\frac{4_y}{S_2}=8-20=-12$
$\Rightarrow \frac{-3 y}{S_2}=-12$
We know that $y = 200 \ km$
$\Rightarrow S_2=\frac{3 \times 200}{12}=50 \ km / hr$
Hence speed of Taxi $=50 \ km / hr$
OR
We know that $x =200 \ km$
Put $S_2=50 \ km / hr \ldots (i)$
$\frac{400}{S_1}+\frac{200}{50}=8$
$\Rightarrow \frac{400}{S_1}=8-4=4$
$\Rightarrow S_1=\frac{400}{4}=100 \ km / hr $
Hence speed of Train $=100 \ km / hr$
Jaipur to Delhi: $2 y + x =600$
Let $S_1$ and $S_2$ be the speeds of Train and Taxi respectively, then
Dehli to Jaipur: $\frac{2 x}{S_1}+\frac{y}{S_2}=8\ldots (i)$
Jaipur to Delhi: $\frac{x}{S_1}+\frac{2 y}{S_2}=10\ldots (ii)$
$ii. 2 x+y=600 \ldots (1)$
$x+2 y=600 \ldots (2)$
Solving $(1)$ and $(2) \times 2$
$2 x+y-2 x-4 y=600-1200$
$\Rightarrow-3 y=-600$
$\Rightarrow y=200$
Put $y=200$ in $(1)$
$2 x+200=600$
$\Rightarrow x=\frac{400}{2}=200$
$iii.$ We know that speed $=\frac{\text { Distance }}{\text { Time }}$
$\Rightarrow$ Time $=\frac{\text { Distance }}{ \text { Speed }}$
Let $S_1$ and $S_2$ are speeds of train and taxi respectively.
Delhi to Jaipur: $\frac{2 x}{S_1}+\frac{y}{S_2}=8\ldots (i)$
Jaipur to Delhi: $\frac{x}{S_1}+\frac{2 y}{S_2}=10\ldots (ii)$
Solving $(i)$ and $(ii) \times 2$
$\Rightarrow \frac{2 x}{S_1}+\frac{y}{S_2}-\frac{2 x}{S_1}-\frac{4_y}{S_2}=8-20=-12$
$\Rightarrow \frac{-3 y}{S_2}=-12$
We know that $y = 200 \ km$
$\Rightarrow S_2=\frac{3 \times 200}{12}=50 \ km / hr$
Hence speed of Taxi $=50 \ km / hr$
OR
We know that $x =200 \ km$
Put $S_2=50 \ km / hr \ldots (i)$
$\frac{400}{S_1}+\frac{200}{50}=8$
$\Rightarrow \frac{400}{S_1}=8-4=4$
$\Rightarrow S_1=\frac{400}{4}=100 \ km / hr $
Hence speed of Train $=100 \ km / hr$