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M.C.Q

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10 questions · timed · auto-graded

Question 11 Mark
Write the correct answer of the following:
In which of the following figures you find two polygons on the same base and between the same parallels?
Answer
Solution:
we find two polygons (parallelograms) on the same base and between the same parallels.
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MCQ 21 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:
  • $A$ rectangle of area $24\ cm^2.$
  • B
    $A$ square of area $25\ cm^2.$
  • C
    $A$ trapezium of area $24\ cm^2.$
  • D
    $A$ rhombus of area $24\ cm^2.$
Answer
Correct option: A.
$A$ rectangle of area $24\ cm^2.$
$\text{ABCD}$ is a rectangle and $\text{E, F, G}$ and $H$ are the mid $-$ points of the sides $\text{AB, BC, CD}$ 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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Question 31 Mark
Write the correct answer of the following:
In Fig. if parallelogram ABCD and rectangle ABEF are of equal area, then:
  1. Perimeter of ABCD = Perimeter of ABEM.
  2. Perimeter of ABCD < Perimeter of ABEM.
  3. Perimeter of ABCD > Perimeter of ABEM.
  4. Perimeter of ABCD $=\frac{1}{2}$ (perimeter of ABEM).
Answer
  1. Perimeter of ABCD > Perimeter of ABEM.
Solution:
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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Question 41 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:
  1. 1 : 2
  2. 1 : 1
  3. 2 : 1
  4. 3 : 1
Answer
  1. 1 : 1
Solution:
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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Question 51 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:
  1. a : b
  2. (3a + b) : (a + 3b)
  3. (a + 3b) : (3a + b)
  4. (2a + b) : (3a + b)
Answer
  1. (3a + b) : (a + 3b)
Solution:
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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Question 61 Mark
Write the correct answer of the following:
The median of a triangle divides it into two:
  1. Triangles of equal area.
  2. Congruent triangles.
  3. Right triangles.
  4. Isosceles triangles.
Answer
  1. Triangles of equal area.
Solution:
The median of a triangle divides it into two triangles of equal area.
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Question 71 Mark
Write the correct answer of the following:
In Fig. the area of parallelogram ABCD is:
  1. AB × BM
  2. BC × BN
  3. DC × DL
  4. AD × DL
Answer
  1. DC × DL
Solution:
Area of a parallelogram = Base × Corresponding altitude
= AB × DL = DC × DL
[ $\because$ AB = DC (opposite side of a parallelogram]
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Question 81 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:
  1. $\frac{1}{2}\ \text{ar}\ (\text{ABC})$
  2. $\frac{1}{3}\ \text{ar}\ (\text{ABC})$
  3. $\frac{1}{4}\ \text{ar}$
  4. $\text{ar}\ (\text{ABC})$
Answer
  1. $\frac{1}{2}\ \text{ar}\ (\text{ABC})$
Solution:
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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Question 91 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:
  1. 1 : 3
  2. 1 : 2
  3. 3 : 1
  4. 1 : 4
Answer
  1. 1 : 2
Solution:
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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Question 101 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:
  1. Is a rectangle.
  2. Is always a rhombus.
  3. Is a parallelogram.
  4. Need not be any of (A), (B) or (C).
Answer
  1. Need not be any of (A), (B) or (C).
Solution:
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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M.C.Q - MATHS STD 9 Questions - Vidyadip