In the following figure, ABC is a right triangle right angled at A.
[question] Question. In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that: (i) $\triangle \mathrm{MBC} \cong \triangle \mathrm{ABD}$ (ii) $\operatorname{ar}(\mathrm{BYXD})=2 \operatorname{ar}(\mathrm{MBC})$ (iii) $\operatorname{ar}(\mathrm{BYXD})=2 \operatorname{ar}(\mathrm{ABMN})$ (iv) $\Delta \mathrm{FCB} \cong \triangle \mathrm{ACE}$ (v) $\operatorname{ar}(\mat...
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC.
[question] Question. In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that (i) $\operatorname{ar}(\mathrm{BDE})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC})$ (ii) $\operatorname{ar}(\mathrm{BDE})=\frac{1}{2} \operatorname{ar}(\mathrm{BAE})$ (iii) $\operatorname{ar}(\mathrm{ABC})=2 \operatorname{ar}(\mathrm{BEC})$ (iv) $\operatorname{ar}(\mathrm{BFE})=\operatorname{ar}(\mathrm{AFD})$ (v) $\operatorname{ar}(\mathrm...
In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ.
[question] Question. In the following figure, ABCD is parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [/question] [solution] Solution: It is given that ABCD is a parallelogram. AD || BC and AB || DC(Opposite sides of a parallelogram are parallel to each other) Join point A to point C. Consider $\triangle A P C$ and $\triangle B P C$ $\triangle A P C$ and $\triangle B P C$ are lying on the same base $P C$ and between the sam...
In the following figure, ABCD, DCFE and ABFE are parallelograms.
[question] Question. In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF). [/question] [solution] Solution: It is given that ABCD is a parallelogram. We know that opposite sides of a parallelogram are equal. $\therefore A D=B C \ldots(1)$ Similarly, for parallelograms DCEF and ABFE, it can be proved that $D E=C F \ldots(2)$ And, $E A=F B \ldots(3)$ In $\triangle \mathrm{ADE}$ and $\triangle B C F$, $A D=B C$ [Using equation (1)] $D E=C F[U \sin g$ equati...
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas.
[question] Question. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle. [/question] [solution] Solution: As the parallelogram and the rectangle have the same base and equal area, therefore, these will also lie between the same parallels. Consider the parallelogram ABCD and rectangle ABEF as follows. Here, it can be observed that parallelogram ABCD and rectangle ABEF are between th...
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC).
[question] Question. In the given figure, $\operatorname{ar}(\mathrm{DRC})=\operatorname{ar}(\mathrm{DPC})$ and ar $(\mathrm{BDP})=\operatorname{ar}(\mathrm{ARC}) .$ Show that both the quadrilaterals $A B C D$ and $D C P R$ are trapeziums. [/question] [solution] Solution: It is given that Area $(\Delta \mathrm{DRC})=$ Area $(\Delta \mathrm{DPC})$ As $\triangle \mathrm{DRC}$ and $\triangle \mathrm{DPC}$ lie on the same base $D C$ and have equal areas, therefore, they must lie between the same par...
Diagonals AC and BD of a quadrilateral ABCD intersect at
[question] Question. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium. [/question] [solution] Solution: It is given that Area $(\triangle A O D)=$ Area $(\triangle B O C)$ Area $(\triangle \mathrm{AOD})+$ Area $(\triangle \mathrm{AOB})=$ Area $(\triangle \mathrm{BOC})+$ Area $(\triangle \mathrm{AOB})$ Area $(\triangle \mathrm{ADB})=$ Area $(\triangle \mathrm{ACB})$ We know that triangles on the same base having area...
In the given figure, AP || BQ || CR.
[question] Question. In the given figure, $A P\|B Q\| C R .$ Prove that $\operatorname{ar}(A Q C)=\operatorname{ar}(P B R) .$ [/question] [solution] Solution: Since $\triangle A B Q$ and $\triangle P B Q$ lie on the same base $B Q$ and are between the same parallels $A P$ and $B Q$, $\therefore$ Area $(\triangle A B Q)=$ Area $(\triangle P B Q) \ldots(1)$ Again, $\triangle B C Q$ and $\triangle B R Q$ lie on the same base $B Q$ and are between the same parallels $B Q$ and $C R$. $\therefore$ Are...
ABCD is a trapezium with AB || DC.
[question] Question. $A B C D$ is a trapezium with $A B \| D C$. A line parallel to $A C$ intersects $A B$ at $X$ and $B C$ at $Y$. Prove that ar $(A D X)=\operatorname{ar}(A C Y)$. [/question] [solution] Solution: It can be observed that $\triangle A D X$ and $\triangle A C X$ lie on the same base $A X$ and are between the same parallels $A B$ and $D C$. $\therefore$ Area $(\triangle \mathrm{ADX})=$ Area $(\triangle \mathrm{ACX}) \ldots(1)$ $\triangle \mathrm{ACY}$ and $\triangle \mathrm{ACX}$ ...
In the given figure, ABCDE is a pentagon.
[question] Question. In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) $\operatorname{ar}(\mathrm{ACB})=\operatorname{ar}(\mathrm{ACF})$ (ii) $\operatorname{ar}(\mathrm{AEDF})=\operatorname{ar}(\mathrm{ABCDE})$ [/question] [solution] Solution: (i) $\triangle \mathrm{ACB}$ and $\triangle \mathrm{ACF}$ lie on the same base $\mathrm{AC}$ and are between The same parallels AC and BF. $\therefore$ Area $(\triangle \mathrm{ACB})=$ Area $(\t...
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.
[question] Question. Diagonals $A C$ and $B D$ of a trapezium $A B C D$ with $A B \| D C$ intersect each other at $O$. Prove that $\operatorname{ar}(A O D)=\operatorname{ar}(B O C)$. [/question] [solution] Solution: It can be observed that ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD. $\therefore$ Area $(\Delta \mathrm{DAC})=$ Area $(\Delta \mathrm{DBC})$ $\Rightarrow$ Area $(\Delta D A C)-$ Area $(\Delta D O C)=$ Area $(\Delta D B C)-$ Area $(\Delta D O C)$ $\R...
The side AB of a parallelogram ABCD is produced to any point P.
[question] Question. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that ar (ABCD) = ar (PBQR). [/question] [solution] Solution: Let us join AC and PQ. $\triangle A C Q$ and $\triangle A Q P$ are on the same base $A Q$ and between the same parallels $A Q$ and $C P$. $\therefore$ Area $(\triangle A C Q)=$ Area $(\triangle A P Q)$ $\Rightarrow$ Area ...
XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and E respectively
[question] Question. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and E respectively, show that ar (ABE) = ar (ACF) [/question] [solution] Solution: It is given that $X Y\|B C \Rightarrow E Y\| B C$ $B E\|A C \Rightarrow B E\| C Y$ Therefore, $\mathrm{EBCY}$ is a parallelogram. It is given that $X Y\|B C \Rightarrow X F\| B C$ $\mathrm{FC}\|\mathrm{AB} \Rightarrow \mathrm{FC}\| \mathrm{XB}$ Therefore, BCFX is a parallelogram. Parallelograms EBCY and B...
D and E are points on sides AB and AC respectively of ΔABC such that
[question] Question. D and E are points on sides AB and AC respectively of ΔABC such that ar (DBC) = ar (EBC). Prove that DE || BC. [/question] [solution] Solution: Since ΔBCE and ΔBCD are lying on a common base BC and also have equal areas, ΔBCE and ΔBCD will lie between the same parallel lines. $\therefore \mathrm{DE} \| \mathrm{BC}$ [/solution]...
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.
[question] Question. In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that: (i) $\operatorname{ar}(D O C)=\operatorname{ar}(A O B)$ (ii) $\operatorname{ar}(\mathrm{DCB})=\operatorname{ar}(\mathrm{ACB})$ (iii) DA II CB or ABCD is a parallelogram. [/question] [solution] Solution: Let us draw $D N \perp A C$ and $B M \perp A C$. (i) In $\triangle \mathrm{DON}$ and $\triangle \mathrm{BOM}$, $\angle D N O=\angle B M O$ (By construc...
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.
[question] Question. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that (i) BDEF is a parallelogram. (ii) $\operatorname{ar}(D E F)=\frac{1}{4} \operatorname{ar}(A B C)$ (iii) $\operatorname{ar}(\mathrm{BDEF})=\frac{1}{2} \operatorname{ar}(\mathrm{ABC})$ [/question] [solution] Solution: (i) $\ln \triangle A B C$, $E$ and $F$ are the mid-points of side $A C$ and $A B$ respectively. Therefore, $E F \| B C$ and $E F=\frac{1}{2} B C$ (Mid-point theorem) Howeve...
In the given figure, ABC and ABD are two triangles on the same base AB.
[question] Question. In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD). [/question] [solution] Solution: Consider ΔACD. Line-segment CD is bisected by AB at O. Therefore, AO is the median of $\triangle \mathrm{ACD}$ $\therefore$ Area $(\Delta \mathrm{ACO})=$ Area $(\triangle \mathrm{ADO}) \ldots$(1) Considering $\triangle B C D, B O$ is the median. $\therefore$ Area $(\Delta \mathrm{BCO})=$ Area $(\Del...
In a triangle ABC, E is the mid-point of median AD.
[question] Question. In a triangle $A B C, E$ is the mid-point of median $A D .$ Show that $\operatorname{ar}(B E D)=\frac{1}{4} \operatorname{ar}(A B C)$ [/question] [solution] Solution: $\mathrm{AD}$ is the median of $\triangle \mathrm{ABC}$. Therefore, it will divide $\triangle \mathrm{ABC}$ into two triangles of equal areas. $\therefore$ Area $(\triangle \mathrm{ABD})=$ Area $(\triangle \mathrm{ACD})$ $\Rightarrow$ Area $(\Delta A B D)=\frac{1}{2}$ Area $(\triangle A B C) \ldots$(1) In $\tri...
In the given figure, E is any point on median AD of a ΔABC
[question] Question. In the given figure, $E$ is any point on median $A D$ of a $\triangle A B C$. Show that $\operatorname{ar}(A B E)=\operatorname{ar}(A C E)$ [/question] [solution] Solution: AD is the median of ΔABC. Therefore, it will divide ΔABC into two triangles of equal areas. $\therefore$ Area $(\triangle \mathrm{ABD})=$ Area $(\triangle \mathrm{ACD}) \ldots(1)$ ED is the median of ΔEBC. $\therefore$ Area $(\Delta \mathrm{EBD})=$ Area $(\Delta \mathrm{ECD}) \ldots(2)$ On subtracting equ...
In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR.
[question] Question. In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that (i) $\operatorname{ar}(\mathrm{PQRS})=\operatorname{ar}(\mathrm{ABRS})$ (ii) $\operatorname{ar}(A X S)=\frac{1}{2} \operatorname{ar}(P Q R S)$ [/question] [solution] Solution: (i) It can be observed that parallelogram PQRS and ABRS lie on the same base SR and also, these lie in between the same parallel lines SR and PB. $\therefore$ Area $(P Q R S)=$ Area (ABRS) ... (1) (ii) Consid...
In the given figure, P is a point in the interior of a parallelogram ABCD.
[question] Question. In the given figure, P is a point in the interior of a parallelogram ABCD. Show that (i) $\operatorname{ar}(A P B)+\operatorname{ar}(P C D)=\frac{1}{2} \operatorname{ar}(A B C D)$ (ii) $\operatorname{ar}(A P D)+\operatorname{ar}(P B C)=\operatorname{ar}(A P B)+\operatorname{ar}(P C D)$ [/question] [solution] Solution: (i) Let us draw a line segment EF, passing through point $P$ and parallel to line segment $A B$. In parallelogram $A B C D$, $A B \| E F$ (By construction) ......
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
[question] Question. P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC). [/question] [solution] Solution: It can be observed that ΔBQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC. $\therefore$ Area $(\triangle B Q C)=\frac{1}{2}$ Area $(A B C D) \ldots$(1) Similarly, ΔAPB and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC. $...
If E, F, G and H are respectively the mid-points of the sides of a parallelogram
[question] Question. If $\mathrm{E}, \mathrm{F}, \mathrm{G}$ and $\mathrm{H}$ are respectively the mid-points of the sides of a parallelogram $\mathrm{ABCD}$ show that $\operatorname{ar}(\mathrm{EFGH})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$ [/question] [solution] Solution: Let us join HF. In parallelogram ABCD, AD = BC and AD || BC (Opposite sides of a parallelogram are equal and parallel) AB = CD (Opposite sides of a parallelogram are equal) $\Rightarrow \frac{1}{2} \mathrm{AD}=\frac{1}{...
In the given figure, ABCD is parallelogram
[question] Question. In the given figure, $A B C D$ is parallelogram, $A E \perp D C$ and $C F \perp A D$. If $A B=16 \mathrm{~cm}, A E=8 \mathrm{~cm}$ and $C F=10 \mathrm{~cm}$, find $A D$. [/question] [solution] Solution: In parallelogram $A B C D, C D=A B=16 \mathrm{~cm}$ [Opposite sides of a parallelogram are equal] We know that Area of a parallelogram $=$ Base $\times$ Corresponding altitude Area of parallelogram $A B C D=C D \times A E=A D \times C F$ $16 \mathrm{~cm} \times 8 \mathrm{~cm}...