2 Mark Question

Question 12 Marks

In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O.
Prove that $\text{ar}(\triangle\text{AOD})=\text{ar}\triangle\text{BOC}.$

Answer

$\triangle\text{CDA}$ and $\triangle\text{CBD}$ lies on the same base and between the same parallel lines.So,
$\text{ar}(\triangle\text{CDA})=\text{ar}(\triangle\text{CDB})\dots(1)$
Subtracting $\text{ar}(\triangle\text{OCD})$ from both sides of equation (1), we get:
$\text{ar}(\triangle\text{CDA})-\text{ar}(\triangle\text{OCD})=\text{ar}(\triangle\text{CDB})-\text{ar}(\triangle\text{OCD})$
$\Rightarrow\ \text{ar}(\triangle\text{AOD})=\text{ar}\triangle\text{BOC}.$

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Question 22 Marks

Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12cm and 16cm.

Answer

Area of rhombus $=\frac{1}{2}\times\text{Product of diagonal}$$=\frac{1}{2}\times12\times16$
$=96\text{cm}^2$
Area of figure formed by joining the mid-points of rhombus $=\frac{1}{2}\times\text{Area of rhombus}$$=\frac{1}{2}\times96$
$=48\text{cm}^2$

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Question 32 Marks

In the adjoining figure, ABC and ABD are two triangles on the same base AB. If line segment CD is
bisected by AB at O, show that $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$

Answer

We know that median of a triangle divides it into two triangles of equal area. Now, AO is the median of $\triangle\text{ACD}.$$\Rightarrow\ \text{A}(\triangle\text{COA})=\text{A}(\triangle\text{DOA})\dots(1)$
And, Bo is the median of $\triangle\text{BCD}.$$\Rightarrow\ \text{A}(\triangle\text{COB})=\text{A}(\triangle\text{DOB})\dots(2)$
Adding (1) and (2), we get:$\text{A}(\triangle\text{COA})+\text{A}(\triangle\text{COB})=\text{A}(\triangle\text{DOA})+\text{A}(\triangle\text{DOB})$
$\Rightarrow\ \text{A}(\triangle\text{ABC})=\text{A}(\triangle\text{ABD})$

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Question 42 Marks

Prove that a median divides a triangle into two triangles of equal area.

Answer

Given: ABC is a triangle in which AD is the median. To prove:$\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{ACD})$
Construction:
Draw $\text{AE}\perp\text{BC}$

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Question 52 Marks

Find the area of a trapezium whose parallel sides are 9cm and 6cm respectively and the distance between these sides is 8cm.

Answer

ABCD is a trapezium in which, AB || CD. AB = 9cm and CD = 6cm. CE is a perpendicular drawn to AB through C and CE = 8cm.

Area of trapezium $=\frac{1}{2}(\text{Sum of parallel sides})\times\text{Distance between them}$
$=\Big[\frac{1}{2}(9+6)\times8\Big]\text{cm}^2$
$=\Big(\frac{1}{2}\times15\times8\Big)\text{cm}^2$
$=60\text{cm}^2$
$\therefore$ Area of trapezium = $60cm^2.$

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Question 62 Marks

In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that $\text{ar}(\triangle\text{RSC})=\text{ar}(\triangle\text{PQB}).$

Answer

In $\triangle\text{RSC}$ and $\triangle\text{PQB},$$\angle\text{CRS}=\angle\text{BPQ}$ [RC || PB, corresponding angles]
$\angle\text{RSC}=\angle\text{PQB}$ [RC || PB, corresponding angles]
$\text{SC}=\text{QB}$ [Opposite sides of a parallelogram BQSC]
$\therefore\ \triangle\text{RSC}\cong\triangle\text{PQB}$ [by AAS congruence criterion]
$\Rightarrow\ \text{ar}(\triangle\text{RSC})=\text{ar}(\triangle\text{PQB})$

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Question 72 Marks

In a trapezium ABCD, AB || DC and M is the midpoint of BC. Through M, a line PQ || AD has been drawn which meets AB in P and DC produced in Q, as shown in the adjoining figure.
Prove that $\text{A}\text{(ABCD)}=\text{A}(\text{APQD}).$

Answer

In $\triangle\text{MCQ}$ and $\triangle\text{MPB},$$\angle\text{QCM}=\angle\text{PBM}$ [alternate angles]
CM = BM [M is the mid-point of BC]$\angle\text{CMQ}=\angle\text{PMB}$ [vertically opposite angles]
$\therefore\ \triangle\text{MCQ}\cong\triangle\text{MPB}$
$\Rightarrow\ \text{A}(\triangle\text{MCQ})=\text{A}(\triangle\text{MPB})$
Now,$\text{A}(\text{ABCD})=\text{A}(\text{APQD})+\text{A}(\text{DMPB})-\text{A}(\triangle\text{MCQ})$
$\Rightarrow\ \text{A}\text{(ABCD)}=\text{A}(\text{APQD})$

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Question 82 Marks

D and E are points on sides AB and AC respectively of $\triangle\text{ABC}$ such that$\text{ar}(\triangle\text{BCD})=\text{ar}(\triangle\text{BCE}).$ Prove that DE || BC.

Answer

Since $\triangle\text{BCD}$ and $\triangle\text{BCE}$ are equal in area and have a same base BC. Therefore, Altitude from D of $\triangle\text{BCD}$ = Altitude from E of $\triangle\text{BCE}.$$\Rightarrow\ \triangle\text{BCD}$ and $\triangle\text{BCE}.$ are between the same parallel lines.
⇒ DE || BC.

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Question 92 Marks

In the adjoining figure, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ are on the same base BC with A and D on opposite sides of BC such that $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{DBC}).$
Show that BC bisects AD.

Answer

Given:
Two triangle, $\text{i.e.}\triangle\text{ABC}$ and $\triangle\text{DBC}$ which have same base BC and point A
and D lie on opposite sides of BC and $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{DBC}).$

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