eSaral - JEE | NEET | Class 8-10 Exam Prep App

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}...

Which of the following figures lie on the same base and between the same parallels.

[question] Question. Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels. [/question] [solution] Solution: Yes. It can be observed that trapezium ABCD and triangle PCD have a common base CD and these are lying between the same parallel lines AB and CD. No. It can be observed that parallelogram PQRS and trapezium MNRS have a common base RS. However, their vertices, (i.e., opposite to the common base) P, Q ...

ABC is a triangle right angled at C.

[question] Question. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) $\mathrm{MD} \perp \mathrm{AC}$ (iii) $\mathrm{CM}=\mathrm{MA}=\frac{1}{2} \mathrm{AB}$ [/question] [solution] Solution: (i) $\ln \triangle \mathrm{ABC}$, It is given that $M$ is the mid-point of $A B$ and $M D \| B C$. Therefore, $D$ is the mid-point of $A C$. (Converse of mid-point theorem) (ii) $A s D M \| ...

Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

[question] Question. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. [/question] [solution] Solution: Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD. In ΔABD, S and P are the mid-points of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that $S P \| B D$ and $S P=\frac{1}{2} B D \ldots$(1) Similarly in $\triangl...

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

[question] Question. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD. [/question] [solution] Solution: $A B C D$ is a parallelogram. $\therefore \mathrm{AB} \| \mathrm{CD}$ And hence, $A E \|$ FC Again, $A B=C D$ (Opposite sides of parallelogram $A B C D$ ) $\frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{CD}$ $\mathrm{AE}=\mathrm{FC}(\mathrm{E}$ and $\mathrm{F}$ are mid-points o...

ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid - point of AD.

[question] Question. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid - point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC. [question] [solution] Solution: Let EF intersect DB at G. By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side. In $\triangle \mathrm{ABD}$, $E F \| A B$ and $E$...

ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

[question] Question. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. [/question] [solution] Solution: Let us join $\mathrm{AC}$ and $\mathrm{BD}$. In $\triangle \mathrm{ABC}$, $P$ and $Q$ are the mid-points of $A B$ and $B C$ respectively. $\therefore P Q \| A C$ and $P Q=\frac{1}{2} A C$ (Mid-point theorem) $\ldots$(1) Similarly in $\triangle \mathrm{ADC}$, $S R \| A C$ and $S R=\frac{1}{2} A C$ (Mi...

ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

[question] Question. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. [/question] [solution] Solution: In ΔABC, P and Q are the mid-points of sides AB and BC respectively. $\therefore \mathrm{PQ} \| \mathrm{AC}$ and $\mathrm{PQ}=\frac{1}{2} \mathrm{AC}$ (Using mid-point theorem) $\ldots$(1) In $\triangle \mathrm{ADC}$ $R$ and $S$ are the mid-points of $C D$ and $A D$ respectively. $\therefore R S ...

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA

[question] Question. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that: (i) $S R \| A C$ and $S R=\frac{1}{2} A C$ (ii) $P Q=S R$ (iii) $\mathrm{PQRS}$ is a parallelogram. [/question] [solution] Solution: (i) In ΔADC, S and R are the mid-points of sides AD and CD respectively. In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is ha...

ABCD is a trapezium in which AB || CD and AD = BC

[question] Question. ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that (i) $\angle \mathrm{A}=\angle \mathrm{B}$ (ii) $\angle \mathrm{C}=\angle \mathrm{D}$ (iii) $\triangle \mathrm{ABC} \cong \triangle \mathrm{BAD}$ (iv) diagonal $A C=$ diagonal $B D$ [/question] [solution] Solution: Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram. (i) $A D=C E$ (Opposite sides of parall...

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF.

[question] Question. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the given figure). Show that (i) Quadrilateral ABED is a parallelogram (ii) Quadrilateral BEFC is a parallelogram (iii) $A D \| C F$ and $A D=C F$ (iv) Quadrilateral ACFD is a parallelogram (v) $\mathrm{AC}=\mathrm{DF}$ (vi) $\triangle \mathrm{ABC} \cong \triangle \mathrm{DEF}$ [/question] [solution] Solution: (i) It is given that $A B=D E$ and $...

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ

[question] [/question] Solution: (i) In $\triangle \mathrm{APB}$ and $\triangle \mathrm{CQD}$, $\angle \mathrm{APB}=\angle \mathrm{CQD}\left(\right.$ Each $\left.90^{\circ}\right)$ $A B=C D$ (Opposite sides of parallelogram $A B C D$ ) $\angle A B P=\angle C D Q$ (Alternate interior angles for $A B \| C D$ ) $\therefore \triangle \mathrm{APB} \cong \triangle \mathrm{CQD}$ (By AAS congruency) (ii) By using the above result $\triangle \mathrm{APB} \cong \triangle \mathrm{CQD}$, we obtain $\mathrm{...

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D.

Solution: (i) It is given that $\mathrm{ABCD}$ is a rectangle. $\therefore \angle \mathrm{A}=\angle \mathrm{C}$ $\Rightarrow \frac{1}{2} \angle \mathrm{A}=\frac{1}{2} \angle \mathrm{C}$ $\Rightarrow \angle \mathrm{DAC}=\angle \mathrm{DCA} \quad($ AC bisects $\angle \mathrm{A}$ and $\angle \mathrm{C})$ $C D=D A$ (Sides opposite to equal angles are also equal) However, $D A=B C$ and $A B=C D$ (Opposite sides of a rectangle are equal) $\therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ $A ...

ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Solution: Let us join AC. In $\triangle \mathrm{ABC}$, $\mathrm{BC}=\mathrm{AB}$ (Sides of a rhombus are equal to each other) $\therefore \angle 1=\angle 2$ (Angles opposite to equal sides of a triangle are equal) However, $\angle 1=\angle 3$ (Alternate interior angles for parallel lines $A B$ and $C D$ ) $\Rightarrow \angle 2=\angle 3$ Therefore, $A C$ bisects $\angle C$. Also, $\angle 2=\angle 4$ (Alternate interior angles for $\|$ lines $B C$ and $D A$ ) $\Rightarrow \angle 1=\angle 4$ Theref...

Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). Show that (i) It bisects $\angle \mathrm{C}$ also, (ii) $A B C D$ is a rhombus.

Solution: (i) $A B C D$ is a parallelogram. $\therefore \angle \mathrm{DAC}=\angle \mathrm{BCA}$ (Alternate interior angles) $\ldots$ (1) And, $\angle B A C=\angle D C A$ (Alternate interior angles) $\ldots$ (2) However, it is given that $A C$ bisects $\angle A$. $\therefore \angle \mathrm{DAC}=\angle \mathrm{BAC} \ldots(3)$ From equations $(1),(2)$, and $(3)$, we obtain $\angle \mathrm{DAC}=\angle \mathrm{BCA}=\angle \mathrm{BAC}=\angle \mathrm{DCA} \ldots(4)$ $\Rightarrow \angle \mathrm{DCA}=\...

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

[question] Question: Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. [/question] [solution] Solution: Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles. Therefore, AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. To prove ABCD is a square, we have to prove that ABCD is a parallelogr...

In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).

Solution: It is given that $A B C D$ is a parallelogram. We know that opposite sides of a parallelogram are equal. $\therefore \mathrm{AD}=\mathrm{BC} \ldots(1)$ Similarly, for parallelograms DCEF and ABFE, it can be proved that $\mathrm{DE}=\mathrm{CF} \ldots(2)$ And, $E A=F B \ldots(3)$ In $\triangle \mathrm{ADE}$ and $\triangle \mathrm{BCF}$, AD = BC [Using equation (1)] DE = CF [Using equation (2)] EA = FB [Using equation (3)] $\therefore \triangle \mathrm{ADE} \cong \mathrm{BCF}(\mathrm{SSS...

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.

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 $A B C D$ and rectangle $A B E F$ are between the same parallels $A B$ and $C F$. We know that opposite sides of a parallelogram or a rectangle are of equal lengths. Therefore, $\mathrm{AB}=\mathrm{EF}$ (For rectangle) $A B=C D$ (For parallelogram) $...

In the given figure, ar (DRC) $=\operatorname{ar}(\mathrm{DPC})$ and ar (BDP) $=\operatorname{ar}(\mathrm{ARC})$. Show that both the quadrilaterals $\mathrm{ABCD}$ and $\mathrm{DCPR}$ are trapeziums.

Solution: It is given that Area $(\Delta D R C)=$ Area $(\Delta D P C)$ As $\triangle \mathrm{DRC}$ and $\triangle \mathrm{DPC}$ lie on the same base $\mathrm{DC}$ and have equal areas, therefore, they must lie between the same parallel lines. $\therefore \mathrm{DC} \| \mathrm{RP}$ Therefore, DCPR is a trapezium. It is also given that Area $(\triangle \mathrm{BDP})=$ Area $(\triangle \mathrm{ARC})$ $\Rightarrow$ Area $(B D P)-$ Area $(\Delta D P C)=$ Area $(\Delta A R C)-$ Area $(\Delta D R C)$...

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.

Solution: It is given that Area $(\triangle \mathrm{AOD})=$ Area $(\Delta \mathrm{BOC})$ Area $(\triangle \mathrm{AOD})+$ Area $(\triangle \mathrm{AOB})=$ Area $(\triangle \mathrm{BOC})+$ Area $(\triangle \mathrm{AOB})$ Area $(\triangle A D B)=$ Area $(\triangle A C B)$ We know that triangles on the same base having areas equal to each other lie between the same parallels. Therefore, these triangles, $\triangle \mathrm{ADB}$ and $\triangle \mathrm{ACB}$, are lying between the same parallels. i.e...

In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

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 $(\Delta \mathrm{ABQ})=$ Area $(\triangle \mathrm{PBQ}) \ldots(1)$ 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 $C R$. $\therefore$ Area $(\Delta B C Q)=$ Area $(\Delta B R Q) \ldots(2)$ On adding equations (1) and (2), we obtain Area $(\Delta \mathrm{ABQ})+$ Area $(\Delta...

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