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Areas Of Parallelograms And Triangles MATHS - Areas Of Parallelograms And Triangles

Rajasthan Board (RBSE) - STD 9 - MATHS (Rajasthan - English Medium). 68 questions in this chapter.

Question types

Areas Of Parallelograms And Triangles question types

68 questions across 5 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.

68
Questions
5
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
ABCD is a trapezium with parallel sides AB = a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is:
  1. a : b
  2. (a + 3b) : (3a + b)
  3. (3a + b) : (a + 3b)
  4. (2a + b) : (3a + b)
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Q 2M.C.Q1 Mark
If AD is median of $\triangle\text{ABC}$ and P is a point on AC such that $\text{ar}(\triangle\text{ADP}):\text{ar}(\triangle\text{ABD})=2:3,$ then $\text{ar}(\triangle\text{PDC}):\text{ar}(\triangle\text{ABC})$ is:
  1. 1 : 5
  2. 1 : 5
  3. 1 : 6
  4. 3 : 5
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Q 3M.C.Q1 Mark
The median of a triangle divides it into two:
  1. Congruent triangle.
  2. Isosceles triangles.
  3. Right triangles.
  4. Triangles of equal areas.
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Q 4M.C.Q1 Mark
Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of the sides of $\triangle\text{ABC}.$ Then the area of $\triangle\text{PQR}$ is:
  1. 12sq. units
  2. 6sq. units
  3. 4sq. units
  4. 3sq. units
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Q 5M.C.Q1 Mark
The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals $16\ cm$ and $12\ cm$ is:
  • A
    $28\ cm^2$
  • ✓
    $48\ cm^2$
  • C
    $96\ cm^2$
  • D
    $24\ cm^2$

Answer: B.

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Q 113 Marks Question3 Marks
Let $\text{ABCD}$ be a parallelogram of area $124\ cm^2.$ If $E$ and $F$ are the mid$-$points of sides $AB$ and $CD$ respectively, then find the area of parallelogram $\text{AEFD}.$
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Q 133 Marks Question3 Marks
In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $\text{AX}\perp\text{DE}$ meets BC at Y. Show that:

$\text{ar}(\text{BYXD})=\text{ar}(\text{ABMN})$
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Q 143 Marks Question3 Marks
In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment $\text{AX}\perp\text{DE}$ meets BC at Y. Show that:

$\text{ar}(\text{CYXE})=\text{ar}(\text{ACFG})$
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Q 153 Marks Question3 Marks
In figure, D and E are two points on BC such that BD = DE = EC. Show that $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{AEC}).$
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Q 165 Marks Questions5 Marks
In figure, ABCD is a trapezium in which AB || DC and DC = 40cm and AB = 60cm. If X and Y are, respectively, the mid-points of AD and BC, prove that:
  1. XY = 50cm
  2. DCYX is a trapezium
  3. $\text{ar}(\text{trap}.\ \text{DCYX})=\Big(\frac{9}{11}\Big)\text{ar}(\text{XYBA}).$
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Q 175 Marks Questions5 Marks
ABCD is a parallelogram whose diagonals intersect at O .If P is any point on BO, prove that:
  1. $\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO})$
  2. $\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP})$
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Q 185 Marks Questions5 Marks
If $\text{ABCD}$ is a parallelogram, then prove that
$\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD}) \ =\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar} (\|^{ gm} \text{ABCD})$
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Q 204 Marks Questions4 Marks
$\text{ABCD}$ is a parallelogram. $P$ is the mid$-$point of $\text{AB. BD}$ and $CP$ intersect at $Q$ such that $\text{CQ : QP} = 3 : 1.$ If $\text{ar}(\triangle\text{PBQ})=10\text{ cm}^2,$ find the area of parallelogram $\text{ABCD}.$
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Q 214 Marks Questions4 Marks
$\text{PQRS}$ is a rectangle inscribed in a quadrant of a circle of radius $13\ cm.$ A is any point on $PQ.$ If $PS = 5\ cm,$ then find $\text{ar}(\triangle\text{RAS}).$
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Q 224 Marks Questions4 Marks
In figure, ABCD and AEFD are two parallelograms. Prove that:
  1. $\text{PE}=\text{FQ}$
  2. $\text{ar}(\triangle\text{APE}) : \text{ar}(\triangle\text{PFA})\\=\text{ar}(\triangle\text{QFD}): \text{ar}(\triangle\text{PFD})$
  3. $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD})$
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Q 234 Marks Questions4 Marks
In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$
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