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Areas of Parallelograms and Triangles MATHS - Areas of Parallelograms and Triangles

Rajasthan Board (RBSE) - STD 9 - MATHS (Rajasthan - English Medium). 29 questions in this chapter.

Question types

Areas of Parallelograms and Triangles question types

29 questions across 4 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.

29
Questions
4
Question groups
5
Question types
Sample Questions

Areas of Parallelograms and Triangles questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

Q 1M.C.Q1 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?
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Q 2M.C.Q1 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: A.

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Q 3M.C.Q1 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).
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Q 4M.C.Q1 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
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Q 5M.C.Q1 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)
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Q 6True False[1 Marks ]1 Mark
Write True or False and justify your answer:
In the figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then, $\text{ar}(\triangle\text{DPC})=\frac{1}{2}\text{ar}(\text{EFGD}).$
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Q 7True False[1 Marks ]1 Mark
Write True or False and justify your answer:
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then $\text{ar}(\triangle\text{BDE})=\frac{1}{4}\text{ar}(\triangle\text{ABC}).$
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Q 83 Marks Question3 Marks
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See Fig.) Prove that ar (LZY) = ar (MZYX)
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Q 93 Marks Question3 Marks
ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF, prove th at $\text{ar}(\triangle\text{AER})= \text{ar}(\triangle\text{AFR}).$
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Q 124 Marks Questions4 Marks
In $\triangle\text{ABC},$ D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (shown in figure), then prove that $\text{ar}(\triangle\text{BPQ})=\frac{1}{2}\text{ar}(\triangle\text{ABC})$
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Q 144 Marks Questions4 Marks
ABCD is a trapezium in which AB || DC, DC = 30cm and AB = 50cm. If X and Y are, respectively the mid-points of AD and BC, prove that $\text{ar}(\text{DCYX})=\frac{7}{9}\ \text{ar}(\text{XYBA})$
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