M.C.Q

MCQ 11 Mark

Write the correct answer of the following: The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $8\ cm$ and $6\ cm$, is:

A
A rectangle of area $24 \mathrm{~cm}^2$.
B
A square of area $25 \mathrm{~cm}^2$.
C
A trapezium of area $24 \mathrm{~cm}^2$.
✓
A rhombus of area $24 \mathrm{~cm}^2$.

Answer

Correct option: D.

A rhombus of area $24 \mathrm{~cm}^2$.

$ABCD$ is a rectangle and $E, F, G$ and $H$ are the mid-points of the sides $AB, BC, CD$ and $DA$ respectively.
The figure obtained is rhombus whose area
$=\frac{1}{2}\times\text{EG}\times\text{FH}=\frac{1}{2}\times6\text{cm}\times8\text{cm}=24\text{cm}^2$

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MCQ 21 Mark

Write the correct answer of the following: In Fig. if parallelogram $ABCD$ and rectangle $ABEF$ are of equal area, then:

A
Perimeter of $ABCD$ = Perimeter of $ABEM$.
B
Perimeter of $ABCD$ < Perimeter of $ABEM$.
✓
Perimeter of $ABCD$ > Perimeter of $ABEM$.
D
Perimeter of $ABCD$ $=\frac{1}{2}$ (perimeter of $ABEM$).

Answer

Correct option: C.

Perimeter of $ABCD$ > Perimeter of $ABEM$.

If parallelogram $ABCD$ and rectangle $ABEM$ are of equal area, then perimeter of $ABCD$ > Perimeter of $ABEM$ because of all the line segments to a given line from a point outside it, the perpendicular is the least.

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MCQ 31 Mark

Write the correct answer of the following: Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is:

A
$1 : 2$
✓
$1 : 1$
C
$2 : 1$
D
$3 : 1$

Answer

Correct option: B.

$1 : 1$

We know that parallelogram on the same or equal bases and between the same parallel are equal in area.
So, the ratio of their area is $1 : 1$.

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MCQ 41 Mark

Write the correct answer of the following: The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to:

✓
$\frac{1}{2}\ \text{ar}\ (\text{ABC})$
B
$\frac{1}{3}\ \text{ar}\ (\text{ABC})$
C
$\frac{1}{4}\ \text{ar}$
D
$\text{ar}\ (\text{ABC})$

Answer

Correct option: A.

$\frac{1}{2}\ \text{ar}\ (\text{ABC})$

We know that, if $D, E$ and $F$ are respectively the mid - point of the sides $BC, CA$ and $AB$ of a $\triangle\text{ABC}$, then all four triangles has equal area i.e.,
$\text{ar}(\triangle\text{AFE})=\text{ar}(\triangle\text{BFD})=\text{ar}(\triangle\text{DEF})\ ...(\text{i})$
$\therefore\text{Area of}\ \triangle\text{DEF}=\frac{1}{4}\text{Area of}\ \triangle\text{ABC}\ ...(\text{ii})$
if we take $D$ as the fourth vertex, then area of the parallelogram $AFDE$
$=\text{Area of}\triangle\text{AFE}+\text{Area of}\triangle\text{DEF}$
$=\text{Area of}\ \triangle\text{DEF}+\text{Area of}\ \triangle\ \text{DEF}=2\text{Area of}\ \triangle\ \text{DEF}$ [using eq. $(i)$]
$=2\times\frac{1}{4}\ \text{Area of}\ \triangle\text{ABC}$ [using eq. $(ii)$]
$=\frac{1}{2}\ \text{Area of}\ \triangle\text{ABC}$

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MCQ 51 Mark

Write the correct answer of the following: The median of a triangle divides it into two:

✓
Triangles of equal area.
B
Congruent triangles.
C
Right triangles.
D
Isosceles triangles.

Answer

Correct option: A.

Triangles of equal area.

The median of a triangle divides it into two triangles of equal area.

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MCQ 61 Mark

Write the correct answer of the following: If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is:

A
$1 : 3$
✓
$1 : 2$
C
$3 : 1$
D
$1 : 4$

Answer

Correct option: B.

$1 : 2$

We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
i.e., $\text{Area of triangle}=\frac{1}{2}\ \text{Area of parallelogram}$
$\Rightarrow\frac{\text{Area of triangle }}{\text{Area of parallelogram}}=\frac{1}{2}$
$\therefore$ Area of triangle : Area of parallelogram = $1 : 2$

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MCQ 71 Mark

Write the correct answer of the following: $ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts, equal in area, then $ABCD$:

A
Is a rectangle.
B
Is always a rhombus.
C
Is a parallelogram.
✓
Need not be any of $(A), (B)$ or $(C)$.

Answer

Correct option: D.

Need not be any of $(A), (B)$ or $(C)$.

Since diagonal of a parallelogram divides it into two triangles of equal area and rectangle and a rhombus are also parallelograms. It may be a $KITE$ as diagonal of a kite divides it into two triangles of equal areas. Then $ABCD$ need not be any of $(a), (b)$ or $(c)$.

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MCQ 81 Mark

Write the correct answer of the following: $ABCD$ is a trapezium with parallel sides $AB = a\ cm$ and $DC = b\ cm$ (Fig). $E$ and $F$ are the mid-points of the non-parallel sides. The ratio of ar $(ABFE)$ and ar $(EFCD)$ is:

A
$a : b$
✓
$(3a + b) : (a + 3b)$
C
$(a + 3b) : (3a + b)$
D
$(2a + b) : (3a + b)$

Answer

Correct option: B.

$(3a + b) : (a + 3b)$

$ABCD$ is a trapizium in which $AB || DC$. $E$ and $F$ are the mid - point of $AD$ and $BC$, so
$\text{EF}=\frac{1}{2}(\text{a}+\text{b})$
$ABEF$ and $EFCD$ are also trapeziums.
$\text{ar}(\text{ABEF})=\frac{1}{2}\Big[\frac{1}{2}(\text{a}+\text{b})+\text{a}\Big]\times\text{h}=\frac{\text{h}}{4}(3\text{a}+\text{b})$
$\text{ar}(\text{EFCD})=\frac{1}{2}\Big[\text{b}+\frac{1}{2}(\text{a}+\text{b})\Big]\times\text{h}=\frac{\text{h}}{4}(\text{a}+3\text{b})$
$\therefore\frac{\text{ar}(\text{ABEF})}{\text{ar}(\text{EFCD})}=\frac{\frac{\text{h}}{4}(3\text{a}+\text{b})}{\frac{\text{h}}{4}(\text{a}+3\text{b})}=\frac{(3\text{a}+\text{b})}{(\text{a}+3\text{b})}$
So, the required ratio is $(3a + b) : (a + 3b)$.

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