The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $8cm$ and $6cm$ is:
- A$A rhombus of area 24cm^2$
- B$A rectangle of area 24cm^2$
- C$A square of area 26cm^2$
- D$A trapezium of area 14cm^2$
Question types
Areas Of Parallelograms And Triangles question types
73 questions across 5 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
73
Questions
5
Question groups
5
Question types
01
MCQ(1M)
12 Q→022 Mark Question
6 Q→033 Mark Question
23 Q→045 Mark Question
29 Q→05Activity Based Questions.
3 Q→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.
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:
- A$1 : 4$
- B$1 : 5$
- ✓$1 : 6$
- D$3 : 5$
Answer: C.
View full solution →The median of a triangle divides it into two:
- ACongruent triangle.
- BIsosceles triangles.
- CRight triangles.
- ✓Triangles of equal areas.
Answer: D.
View full solution →$A, B, C, D$ are mid $-$ points of sides of parallelogram $\text{PQRS}$. If $ar\text{(PQRS)} = 36\ cm^2$, then $ar\text{(ABCD)}=$
- A$24\ cm^2$
- ✓$18\ cm^2$
- C$30\ cm^2$
- D$36\ cm^2$
Answer: B.
View full solution →In figure, $\text{ABCD}$ and $\text{FECG}$ are parallelograms equal in area. If $\text{ar} (\triangle\text{AQE})=12\text{cm}^2,$ then $\text{ar}(||^{\text{gm}}\text{FGBQ}) =$
- A$12\ cm^2$
- B$20\ cm^2$
- ✓$24\ cm^2$
- D$36\ cm^2$
Answer: C.
View full solution →Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:
In figure, ABCD, ABFE and CDEF are parallelograms. Prove that $\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{BCF}).$
View full solution →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: 
$\text{ar}(\text{BYXD})=\text{ar}(\text{ABMN})$
View full solution →$\text{ar}(\text{BYXD})=\text{ar}(\text{ABMN})$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: 
$\text{ar}(\text{CYXE})=\text{ar}(\text{ACFG})$
View full solution →$\text{ar}(\text{CYXE})=\text{ar}(\text{ACFG})$In figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of $\triangle\text{OTS}$ if PQ = 8cm.
View full solution →In figure, ABCD is a trapezium in which AB || DC. Prove that $\text{ar}(\triangle\text{AOD}) =\text{ar}(\triangle\text{BOC}).$
View full solution →PQRS is a rectangle inscribed in a quadrant of a circle of radius 13cm. A is any point on PQ. If PS = 5cm, then
find $\text{ar}(\triangle\text{RAS}).$
View full solution →find $\text{ar}(\triangle\text{RAS}).$
In figure, ABCD and AEFD are two parallelograms. Prove that:
View full solution →- $\text{PE}=\text{FQ}$
- $\text{ar}(\triangle\text{APE}) : \text{ar}(\triangle\text{PFA})\\=\text{ar}(\triangle\text{QFD}): \text{ar}(\triangle\text{PFD})$
- $\text{ar}(\triangle\text{PEA})=\text{ar}(\triangle\text{QFD})$
In figure, find the area of $\triangle\text{GEF}.$
View full solution →In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABD}).$
View full solution →In figure, ABCD is a trapezium in which $AB = 7cm, AD = BC = 5cm, DC = x cm$, and distance between AB and DC is $4cm$. Find the value of x and area of trapezium ABCD.
View full solution →If ABCD is a parallelogram, then prove that$\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})\\ \ =\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}$ $(||^{gm} ABCD)$
View full solution →ABCD is a parallelogram whose diagonals intersect at O .If P is any point on BO, prove that:
View full solution →- $\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO})$
- $\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP})$
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:
View full solution →- XY = 50cm
- DCYX is a trapezium
- $\text{ar}(\text{trap}.\ \text{DCYX})=\Big(\frac{9}{11}\Big)\text{ar}(\text{XYBA}).$
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