Question types

AreaAreas of Parallelograms and Triangles question types

76 questions across 6 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.

76
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6
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5
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Sample Questions

AreaAreas 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
Look at the statements given below:
$i.$ A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
$ii.$ In a $\|gm \  \text{ABCD,}$ it is given that $AB = 10\ cm.$ The altitudes $DE$ on $AB$ and $BF$ on $AD$ being $6\ cm$ and $8\ cm$ respectively, then $AD = 7.5\ cm.$
$iii.$ Area of a $\|gm =\frac{1}{2}\times\text{base}\times\text{altitude}.$
Which is true?
  • A
    $I$ only
  • B
    $II$ only
  • $I$ and $II$
  • D
    $II$ and $III$

Answer: C.

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Q 2M.C.Q1 Mark
In the given figure ABCD is a trapezium in which AB || DC such that AB = acm and DC = bcm. If E and F are the midpoints of AD and BC respectively. Then, ar(ABFE) : ar(EFCD) = ?
  1. a : b
  2. (a + 3b) : (3a + b)
  3. (3a + b) : (a + 3b)
  4. (2a + b) : (3a + b)
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Q 3M.C.Q1 Mark
If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is:
  1. 1 : 2
  2. 1 : 3
  3. 1 : 4
  4. 3 : 4
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Q 4M.C.Q1 Mark
In the given figure, $\text{ABCD}$ is a $\|gm$ in which diagonals $AC$ and $BD$ intersect at $O$. If $ar ( \|gm\ \text{ABCD)}$ is $52\ cm^2,$ then the ar $(\triangle\text{AOB})=?$
  • A
    $26\ cm^2$
  • B
    $18.5\ cm^2$
  • C
    $39\ cm^2$
  • $13\ cm^2$

Answer: D.

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Q 5M.C.Q1 Mark
Which of the following is a false statement?
  1. A median of a triangle divides it into two triangles of equal area.
  2. The diagonals of a ||gm divide it into four triangles of equal area.
  3.  In a $\triangle\text{ABC},$ if E is the midpoint of median AD, then $\text{ar}(\triangle\text{BED})=\frac{1}{4}\text{ar}(\triangle\text{ABC}).$
  1. In a trapezium ABCD, it is given that AB || DC and the diagonals AC and BD intersect at O. Then, $\text{ar}(\triangle\text{AOB})=\text{ar}(\triangle\text{ABC}).$
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In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O.
Prove that $\text{ar}(\triangle\text{AOD})=\text{ar}\triangle\text{BOC}.$
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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 $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$
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Q 163 Marks Question3 Marks
In the adjoining figure, ABCD is a quadrilateral. A line through D, parallel to AC, meets BC produced in P.
Prove that $\text{ar}(\triangle\text{ABP})=\text{ar}(\text{quadrilateral ABCD}).$
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Q 193 Marks Question3 Marks
In the adjoining figure, BD || CA, E is the midpoint of CA and $\text{BD}=\frac{1}{2}\text{CA}.$ Prove that $\text{ar}(\triangle\text{ABC})=2\text{ar}(\triangle\text{DBC}).$
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Q 203 Marks Question3 Marks
In the adjoining figure, D and E are respectively the midpoints of sides AB and AC of $\triangle\text{ABC}.$ If PQ || BC and CDP and BEQ are straight lines then prove that$\text{ar}(\triangle\text{ABQ})=\text{ar}(\triangle\text{ACP}).$
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P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of ||gm ABCD. Show that PQRS is a parallelogram and also show that
$\text{ar}(||\text{gm PQRS})=\frac{1}{2}\times\text{ar}(||\text{gm ABCD}).$
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In a trapezium ABCD, AB || DC, AB = acm, and DC = bcm. If M and N are the midpoints ofthe nonparallel sides, AD and BC respectively then find the ratio of
ar(DCNM) and ar(MNBA).
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ABCD is a trapezium in which AB || CD, AB = 16cm and DC = 24cm. If E and F are respectively the midpoints of AD and BC, prove that $\text{ar(ABFE)}=\frac{9}{11}\text{ar(EFCD)}.$
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The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F. Show that:
 $\text{ar}(\text{pentagon ABCDE})=\text{ar}(\triangle\text{DGF}).$
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In the adjoining figure, the point D divides the side BC of $\triangle\text{ABC}$ in the ratio m : n. Prove that $\text{ar}(\triangle\text{ABD}):\text{ar}(\triangle\text{ADC})=\text{m}:\text{n}$
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BD is one of the diagonals of a quadrilaterl ABCD. If $\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD},$show that
ar(quadrilaterl ABCD) $=\frac{1}{2}\times\text{BD}\times(\text{AL}+\text{CM}).$
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The vertex A of $\triangle\text{ABC}$ is joined to a point D on the side BC. The mid-point of AD is E.
Prove that $\text{ar}(\triangle\text{BEC})=\frac{1}{2}\text{ar}(\triangle\text{ABC}).$
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