In the given figure, the area of $\triangle ABC$ is $64 cm^2$. State giving reasons :
(i) ar ( $\| gm ABCD$ )
(ii) ar (rect. ABEF).
View full solution →(i) ar ( $\| gm ABCD$ )
(ii) ar (rect. ABEF).
Question types
Areas of Parallelograms and Triangles question types
103 questions across 5 question groups — pick any mix to generate a MATHEMATICS paper with step-by-step answer keys.
103
Questions
5
Question groups
5
Question types
01
[2 Mark Question Answer]
15 Q→02[3 marks sum]
19 Q→03[4 marks sum]
22 Q→04[5 marks sum]
12 Q→05MCQ
35 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.
In the adjoining figure, ABCD is a parallelogram. P is a point on DC such that ar $(\triangle APD )=25 cm^2$ and ar $(\triangle BPC )$ $=15 cm^2$. Calculate :(i) ar ( $\| gm ABCD$ )
(ii) $DP : PC$.
View full solution →(ii) $DP : PC$.
In the adjoining figure, ABCD is a parallelogram and O is any point on its diagonal AC.
Show that : $\operatorname{ar}(\triangle AOB )=\operatorname{ar}(\triangle AOD )$.
View full solution →Show that : $\operatorname{ar}(\triangle AOB )=\operatorname{ar}(\triangle AOD )$.
In a $\| gm ABCD$, it is given that $AB =16 cm$ and the altitudes corresponding to the sides AB and AD are 6 cm and 8 cm respectively.
Find the length of AD .
View full solution →Find the length of AD .
ABCD is a quadrilateral. If $AL \perp BD$ and $CM \perp BD$, prove that : $\operatorname{ar}($ quad. ABCD $)=\frac{1}{2} \times BD \times( AL + CM )$.
View full solution →In the given figure, the area of $\| gm ABCD$ is $90 cm^2$. State giving reasons :
(i) ar ( $\| gm$ ABEF)
(ii) ar ( $\triangle ABD$ )
(iii) ar ( $\triangle BEF$ ).
View full solution →(i) ar ( $\| gm$ ABEF)
(ii) ar ( $\triangle ABD$ )
(iii) ar ( $\triangle BEF$ ).
In the given figure, squares ABDE and AFGC are drawn on the side AB and hypotenuse AC of right triangle ABC and $BH \perp FG$. Prove that :
(i) $\triangle EAC \cong \triangle BAF$.
(ii) ar (sq. ABDE) = ar (rect. ARHF).
View full solution →(i) $\triangle EAC \cong \triangle BAF$.
(ii) ar (sq. ABDE) = ar (rect. ARHF).
Find the area of a rhombus, the lengths of whose diagonals are 18 cm and 24 cm respectively.
In the adjoining figure, CE is drawn parallel to DB to meet AB produced at E .
Prove that : ar(quad. ABCD $)=\operatorname{ar}(\triangle DAE )$.
View full solution →Prove that : ar(quad. ABCD $)=\operatorname{ar}(\triangle DAE )$.
In the given figure, ABCD is a quadrilateral. A line through D , parallel to AC , meets BC produced in P .
Prove that : ar $(\triangle ABP )=$ ar (quad. ABCD $)$.
View full solution →Prove that : ar $(\triangle ABP )=$ ar (quad. ABCD $)$.
Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram, divides it into two equal parallelograms.
Find the area of a trapezium whose parallel sides measure 10 cm and 8 cm respectively and the distance between these sides is 6 cm.
Construct a quadrilateral ABCD in which AB = 3.2 cm, BC = 2.8 cm CD = 4 cm, DA = 4.5 cm and BD = 5.3 cm. Also construct a triangle equal in area to this quadrilateral.
In the given figure, $AB \| DC \| EF , AD \| BE$ and $DE \| AF$.
Prove that : $\operatorname{ar}(\| gm$ DEFH $)=\operatorname{ar}(\| gm ABCD )$.
View full solution →Prove that : $\operatorname{ar}(\| gm$ DEFH $)=\operatorname{ar}(\| gm ABCD )$.
In the adjoining figure, ABCD is a parallelogram. Any line through A cuts DC at a point P and BC produced at Q. Prove that :ar $(\triangle BPC )=\operatorname{ar}(\triangle DPQ )$.
View full solution →In the given figure, the side AB of $\| gm ABCD$ is produced to a point P. A line through A drawn parallel to CP meets CB produced in Q and the parallelogram PBQR is completed.
Prove that : $\operatorname{ar}(\| gm ABCD )=\operatorname{ar}(\| gm$ BPRQ $)$.
View full solution →Prove that : $\operatorname{ar}(\| gm ABCD )=\operatorname{ar}(\| gm$ BPRQ $)$.
A farmer was having a field in the form of a parallelogram ABCD. He divided the field into several parts by taking a point X on the side CD and joining it to vertices A and B. The farmer sowed wheat and pulses in equal portions of the field separately.
Based on the above information, answer the following questions:
Q.1. By joining XA and XB, the field has been divided into how many parts?
(a) 2 (b) 3 (c) 4 (d) 5
Q.2. The shapes of the parts obtained above are:
(a) triangles
(b) rectangles
(c) one triangle two squares
(d) none of these
Q.3. Area of $\triangle XAB$ is equal to:
(a) area of parallelogram ABCD
(b) $\frac{1}{2}$ area of parallelogram ABCD
(c) area of $\triangle ADX +$ area of $\triangle BCX$
(d) both (b) and (c)
Q.4. $\triangle ABX$ and parallelogram ABCD are:
(a) On the same base DC
(b) On the same base AB and between the same parallels BC and AD .
(c) On the same base AB and between the same parallels AB and CD .
(d) On the same base CD and between the same parallels AB and CD .
Q.5. If instead of taking point $X$ on side $C D$, the farmer takes a point $Y$ on side $B C$ and joins YA and YD, then:
(a) area of $\triangle ADY =$ area of $\triangle ABY +$ area of $\triangle DCY$
(b) area of $\triangle ADY =\frac{1}{3}$ area of parallelogram ABCD
(c) area of $\triangle ADY =$ area of $\triangle ABY$ (d) area of $\triangle ADY =$ area of $\triangle DCY$
View full solution →Based on the above information, answer the following questions:
Q.1. By joining XA and XB, the field has been divided into how many parts?
(a) 2 (b) 3 (c) 4 (d) 5
Q.2. The shapes of the parts obtained above are:
(a) triangles
(b) rectangles
(c) one triangle two squares
(d) none of these
Q.3. Area of $\triangle XAB$ is equal to:
(a) area of parallelogram ABCD
(b) $\frac{1}{2}$ area of parallelogram ABCD
(c) area of $\triangle ADX +$ area of $\triangle BCX$
(d) both (b) and (c)
Q.4. $\triangle ABX$ and parallelogram ABCD are:
(a) On the same base DC
(b) On the same base AB and between the same parallels BC and AD .
(c) On the same base AB and between the same parallels AB and CD .
(d) On the same base CD and between the same parallels AB and CD .
Q.5. If instead of taking point $X$ on side $C D$, the farmer takes a point $Y$ on side $B C$ and joins YA and YD, then:
(a) area of $\triangle ADY =$ area of $\triangle ABY +$ area of $\triangle DCY$
(b) area of $\triangle ADY =\frac{1}{3}$ area of parallelogram ABCD
(c) area of $\triangle ADY =$ area of $\triangle ABY$ (d) area of $\triangle ADY =$ area of $\triangle DCY$
Construct a quadrilateral ABCD in which $AB =3.2 cm, BC =2.8 cm, CD =4 cm$, $DA =4.5 cm$ and $BD =5.3 cm$. Also construct a triangle equal in area to this quadrilateral.
View full solution →In the figure, ABCD is a parallelogram in which BC is produced to E such that $CE = BC$. AE intersects $C D$ at $F$. If area of $\triangle D F B$ is $3 cm^2$, then area of the parallelogram ABCD is :
- A$9 cm^2$
- B$10 cm^2$
- ✓$12 cm^2$
- D$15 cm^2$
Answer: C.
View full solution →In the figure, ABCD is a trapezium with parallel sides $AB =x$ and $CD =y . E$ and F are mid-points of the non-parallel sides AD and BC respectively. The ratio of $\operatorname{ar}( ABFE )$ and $\operatorname{ar}( EFCD )$ is :
- A$x: y$
- ✓$(3 x+y):(x+3 y)$
- C$(x+3 y):(3 x+y)$
- D$(2 x+y): 3 x+y)$
Answer: B.
View full solution →In which of the following, you find two polygons on the same base and between the same parallels?
Assertion (A) : In $\triangle A B C$, if $D$ is the mid-point of side AB , then, area of $\triangle BCD =$ area of $\triangle ACD$.
Reason (R) : A triangle and a parallelogram on the same base and between the same parallels are equal in area.
Reason (R) : A triangle and a parallelogram on the same base and between the same parallels are equal in area.
- ✓A is true, R is false
- BA is false, R is true
- CBoth A and R are true
- DBoth A and R are false.
Answer: A.
View full solution →
- AA is true, R is false
- BA is false, R is true
- ✓Both A and R are true
- DBoth A and R are false.
Answer: C.
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