Recall that two circles are congruent if they have the same radii.
[question] Question. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres [solution] Solution: A circle is a collection of points which are equidistant from a fixed point. This fixed point is called as the centre of the circle and this equal distance is called as radius of the circle. And thus, the shape of a circle depends on its radius. Therefore, it can be observed that if we try to superimpose tw...
Read More →Write True or False: Give reasons for your answers.
[question] Question. Write True or False: Give reasons for your answers. (i) Line segment joining the centre to any point on the circle is a radius of the circle. (ii) A circle has only finite number of equal chords. (iii) If a circle is divided into three equal arcs, each is a major arc. (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle. (v) Sector is the region between the chord and its corresponding arc. (vi) A circle is a plane figure. [/question] [s...
Read More →Fill in the blanks
[question] Question. Fill in the blanks (i) The centre of a circle lies in __________ of the circle. (exterior/ interior) (ii) A point, whose distance from the centre of a circle is greater than its radius lies in __________ of the circle. (exterior/ interior) (iii) The longest chord of a circle is a __________ of the circle. (iv) An arc is a __________ when its ends are the ends of a diameter. (v) Segment of a circle is the region between an arc and __________ of the circle. (vi) A circle divid...
Read More →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...
Read More →P and Q are respectively the mid-points of sides AB and BC of a triangle
[question] Question. P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that (i) $\operatorname{ar}($ PRQ $)=\frac{1}{2} \operatorname{ar}($ ARC $)$ (ii) $\operatorname{ar}(\mathrm{RQC})=\frac{3}{8} \operatorname{ar}(\mathrm{ABC})$ (iii) $\operatorname{ar}(\mathrm{PBQ})=\operatorname{ar}(\mathrm{ARC})$ [/question] [solution] Solution: Take a point $S$ on $A C$ such that $S$ is the mid-point of $A C$. Extend $P Q$ to $T$ such that $P Q=...
Read More →Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
[question] Question. Diagonals $A C$ and $B D$ of a quadrilateral $A B C D$ intersect each other at $P .$ Show that $\operatorname{ar}(A P B) \times \operatorname{ar}(C P D)=\operatorname{ar}(A P D) \times \operatorname{ar}(B P C)$ [/question] [solution] Solution: Let us draw $A M \perp B D$ and $C N \perp B D$ Area of a triangle $=\frac{1}{2} \times$ Base $\times$ Altitude $\operatorname{ar}(\mathrm{APB}) \times \operatorname{ar}(\mathrm{CPD})=\left[\frac{1}{2} \times \mathrm{BP} \times \mathrm...
Read More →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...
Read More →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...
Read More →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...
Read More →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...
Read More →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...
Read More →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...
Read More →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...
Read More →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 $\triangle \mathrm{ACX}$ ...
Read More →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...
Read More →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...
Read More →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 ...
Read More →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...
Read More →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]...
Read More →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...
Read More →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...
Read More →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...
Read More →Show that the diagonals of a parallelogram divide it into four triangles of equal area.
[question] Question. Show that the diagonals of a parallelogram divide it into four triangles of equal area. [/question] [solution] Solution: We know that diagonals of parallelogram bisect each other. Therefore, O is the mid-point of AC and BD. BO is the median in ΔABC. Therefore, it will divide it into two triangles of equal areas. $\therefore$ Area $(\Delta \mathrm{AOB})=$ Area $(\Delta \mathrm{BOC}) \ldots(1)$ In $\triangle B C D, C O$ is the median. $\therefore$ Area $(\Delta \mathrm{BOC})=$...
Read More →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...
Read More →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...
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