[5 marks sum]

Question 15 Marks

In the adjoining figure, ABCD is a parallelogram. Any line through A cuts DC at a point P and BC produced at Q. Prove that :ar $(\triangle BPC )=\operatorname{ar}(\triangle DPQ )$.

Answer

[Hint. Join AC.
$\begin{array}{l}
\operatorname{ar}(\triangle B P C)=\operatorname{ar}(\triangle A P C) \\
=\operatorname{ar}(\triangle A Q C)-\operatorname{ar}(\triangle P Q C) \\
=\operatorname{ar}(\triangle D Q C)-\operatorname{ar}(\triangle P Q C)=\operatorname{ar}(\triangle D P Q) .]\end{array}$

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Question 25 Marks

In the given figure, the side AB of $\| gm ABCD$ is produced to a point P. A line through A drawn parallel to CP meets CB produced in Q and the parallelogram PBQR is completed.
Prove that : $\operatorname{ar}(\| gm ABCD )=\operatorname{ar}(\| gm$ BPRQ $)$.

Answer

[Hint. Join AC and QP.
$\begin{array}{l}
\operatorname{ar}(\triangle A Q C)=\operatorname{ar}(\triangle A Q P) \\
\Rightarrow \operatorname{ar}(\triangle A Q C)-\operatorname{ar}(\triangle A Q B)=\operatorname{ar}(\triangle A Q P)
\operatorname{ar}(\triangle A Q B) \\
\Rightarrow \operatorname{ar}(\triangle A B C)=\operatorname{ar}(\triangle Q P B) \\
\Rightarrow 2 \operatorname{ar}(\triangle A B C)=2 \operatorname{ar}(\triangle Q P B)]\end{array}$

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Question 35 Marks

If the medians of a $\triangle A B C$ intersect at $G$, show that :
$\operatorname{ar}(\triangle AGB )=\operatorname{ar}(\triangle AGC )=\operatorname{ar}(\triangle BGC )=\frac{1}{3} \operatorname{ar}(\triangle ABC )$

Answer

[Hint. Median $A D$ divides $\triangle A B C$ into two $\triangle s$ of equal area.
\[\therefore \operatorname{ar}(\triangle A B D)=\operatorname{ar}(\triangle A C D)\]
Median GD divides $\triangle G B C$ into two $\triangle s$ of equal area.
\[\begin{array}{l}
\therefore \operatorname{ar}(\triangle G B D)=\operatorname{ar}(\triangle G C D) \\
\therefore \operatorname{ar}(\triangle A B D)-\operatorname{ar}(\triangle G B D)=\operatorname{ar}(\triangle A C D)
\operatorname{ar}(\triangle G C D) \\
\Rightarrow \operatorname{ar}(\triangle A G B)=\operatorname{ar}(\triangle A G C) .\end{array}\]
Similarly, ar $(\triangle A G C)=\operatorname{ar}(\triangle B G C)$.

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Question 45 Marks

A farmer was having a field in the form of a parallelogram ABCD. He divided the field into several parts by taking a point X on the side CD and joining it to vertices A and B. The farmer sowed wheat and pulses in equal portions of the field separately.
Based on the above information, answer the following questions:
Q.1. By joining XA and XB, the field has been divided into how many parts?
(a) 2 (b) 3 (c) 4 (d) 5
Q.2. The shapes of the parts obtained above are:
(a) triangles
(b) rectangles
(c) one triangle two squares
(d) none of these
Q.3. Area of $\triangle XAB$ is equal to:
(a) area of parallelogram ABCD
(b) $\frac{1}{2}$ area of parallelogram ABCD
(c) area of $\triangle ADX +$ area of $\triangle BCX$
(d) both (b) and (c)
Q.4. $\triangle ABX$ and parallelogram ABCD are:
(a) On the same base DC
(b) On the same base AB and between the same parallels BC and AD .
(c) On the same base AB and between the same parallels AB and CD .
(d) On the same base CD and between the same parallels AB and CD .
Q.5. If instead of taking point $X$ on side $C D$, the farmer takes a point $Y$ on side $B C$ and joins YA and YD, then:
(a) area of $\triangle ADY =$ area of $\triangle ABY +$ area of $\triangle DCY$
(b) area of $\triangle ADY =\frac{1}{3}$ area of parallelogram ABCD
(c) area of $\triangle ADY =$ area of $\triangle ABY$ (d) area of $\triangle ADY =$ area of $\triangle DCY$

Answer

1. (b) 2. (a) 3. (d) 4. (c) 5. (a)

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Question 55 Marks

Construct a quadrilateral ABCD in which $AB =3.2 cm, BC =2.8 cm, CD =4 cm$, $DA =4.5 cm$ and $BD =5.3 cm$. Also construct a triangle equal in area to this quadrilateral.

Answer

self

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Question 65 Marks

In the given figure, $A D$ is a median of $\triangle A B C$ and $P$ is a point on AC such that :
$\operatorname{ar}(\triangle ADP ): \operatorname{ar}(\triangle ABD )=2: 3$.
Find: (i) $AP : PC$ (ii) ar ( $\triangle PDC$ ) : ar ( $\triangle ABC$ ).

Answer

(i) $2: 1$
(ii) $1: 6$

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Question 75 Marks

$D$ is a point on base $B C$ of a $\triangle A B C$ such that $2 B D=D C$.
Prove that : ar $(\triangle ABD )=\frac{1}{3}$ ar $(\triangle ABC )$.

Answer

self

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Question 85 Marks

In the given figure, D is the mid-point of BC and E is the mid-point of AD .
Prove that: $\operatorname{ar}(\triangle ABE )=\frac{1}{4} \operatorname{ar}(\triangle ABC )$.

Answer

self

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Question 95 Marks

ABCD is a quadrilateral. If $AL \perp BD$ and $CM \perp BD$, prove that : $\operatorname{ar}($ quad. ABCD $)=\frac{1}{2} \times BD \times( AL + CM )$.

Answer

self

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Question 105 Marks

In a $\|$ gm $A B C D$, it is given that $A B=16 cm$ and the altitudes corresponding to the sides AB and AD are 6 cm and 8 cm respectively.
Find the length of AD .

Answer

12 cm

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Question 115 Marks

In the adjoining figure, BD is a diagonal of quad. $A B C D$. Show that $A B C D$ is a parallelogram and calculate the area of $\| gm ABCD$.

Answer

$48 cm^2$

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Question 125 Marks

Case Study : A farmer was having a field in the form of a parallelogram $A B C D$. He divided the field into several parts by taking a point $X$ on the side $C D$ and joining it to vertices $A$ and B. The farmer sowed wheat and pulses in equal portions of the field separately.
Based on the above information, answer the following questions :
1. By joining $X A$ and $X B$, the field has been divided into how many parts?
(a) 2
(b) 3
(c) 4
(d) 5
2. The shapes of the parts obtained above are:
(a) triangles
(b) rectangles
(c) one triangle two squares
(d) none of these
3. Area of $\triangle XAB$ is equal to:
(a) area of parallelogram ABCD
(b) $\frac{1}{2}$ area of parallelogram ABCD
(c) area of $\triangle ADX +$ area of $\triangle BCX$
(d) both (b) and (c)
4. $\triangle ABX$ and parallelogram ABCD are:
(a) On the same base DC
(b) On the same base $A B$ and between the same parallels $B C$ and $A D$.
(c) On the same base $A B$ and between the same parallels $A B$ and CD.
(d) On the same base CD and between the same parallels $A B$ and CD.
5. If instead of taking point $X$ on side $C D$, the farmer takes a point $Y$ on side $B C$ and joins YA and YD, then:
(a) area of $\triangle A D Y=$ area of $\triangle A B Y+$ area of $\triangle D C Y$
(b) area of $\triangle ADY =\frac{1}{3}$ area of parallelogram ABCD
(c) area of $\triangle A D Y=$ area of $\triangle A B Y$
(d) area of $\triangle ADY =$ area of $\triangle DCY$

Answer

(1. b), (2. a), (3. d) (4. c), (5. a)

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MATHEMATICS [5 marks sum]

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