ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
Solution: It can be observed that $\triangle \mathrm{ADX}$ and $\triangle \mathrm{ACX}$ lie on the same base $\mathrm{AX}$ and are between the same parallels $\mathrm{AB}$ and $\mathrm{DC}$. $\therefore$ Area $(\Delta \mathrm{ADX})=$ Area $(\triangle \mathrm{ACX}) \ldots(1)$ $\triangle A C Y$ and $\triangle A C X$ lie on the same base $A C$ and are between the same parallels $A C$ and $X Y$. $\therefore$ Area $(\triangle \mathrm{ACY})=$ Area $(\mathrm{ACX}) \ldots(2)$ From equations (1) and (2),...
Read More →Show that the diagonals of a square are equal and bisect each other at right angles.
Solution: Let $A B C D$ be a square. Let the diagonals $A C$ and $B D$ intersect each other at a point $O$. To prove that the diagonals of a square are equal and bisect each other at right angles, we have to prove $A C=B D, O A=O C, O B=O D$, and $\angle A O B=90^{\circ}$. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DCB}$, $A B=D C$ (Sides of a square are equal to each other) $\angle A B C=\angle D C B$ (All interior angles are of $90^{\circ}$ ) $\mathrm{BC}=\mathrm{CB}$ (Common side) $\t...
Read More →Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution: Let $A B C D$ be a quadrilateral, whose diagonals $A C$ and $B D$ bisect each other at right angle i.e., $O A=O C, O B=O D$, and $\angle A O B=\angle B O C=$ $\angle C O D=\angle A O D=90^{\circ} .$ To prove $A B C D$ a rhombus, we have to prove $A B C D$ is a parallelogram and all the sides of $A B C D$ are equal. In $\triangle \mathrm{AOD}$ and $\triangle C O D$, $O A=O C$ (Diagonals bisect each other) $\angle \mathrm{AOD}=\angle \mathrm{COD}$ (Given) OD = OD (Common) $\therefore \tr...
Read More →If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution: Let $A B C D$ be a parallelogram. To show that $A B C D$ is a rectangle, we have to prove that one of its interior angles is $90^{\circ}$. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DCB}$ $A B=D C$ (Opposite sides of a parallelogram are equal) $\mathrm{BC}=\mathrm{BC}$ (Common) $\mathrm{AC}=\mathrm{DB}$ (Given) $\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{DCB}$ (By SSS Congruence rule) $\Rightarrow \angle \mathrm{ABC}=\angle \mathrm{DCB}$ It is known that the sum ...
Read More →The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral.
Solution: Let the common ratio between the angles be $x$. Therefore, the angles will be $3 x, 5 x, 9 x$, and $13 x$ respectively. As the sum of all interior angles of a quadrilateral is $360^{\circ}$, $\therefore 3 x+5 x+9 x+13 x=360^{\circ}$ $30 x=360^{\circ}$ $x=12^{\circ}$ Hence, the angles are $3 x=3 \times 12=36^{\circ}$ $5 x=5 \times 12=60^{\circ}$ $9 x=9 \times 12=108^{\circ}$ $13 x=13 \times 12=156^{\circ}$...
Read More →Complete the hexagonal and star shaped rangolies (see the given figures) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
Solution: It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles in it. Area of $\Delta \mathrm{OAB}=\frac{\sqrt{3}}{4}(\text { side })^{2}=\frac{\sqrt{3}}{4}(5)^{2}$ $=\frac{\sqrt{3}}{4}(25)=\frac{25 \sqrt{3}}{4} \mathrm{~cm}^{2}$ Area of hexagonal-shaped rangoli $=6 \times \frac{25 \sqrt{3}}{4}=\frac{75 \sqrt{3}}{2} \mathrm{~cm}^{2}$ Area of equilateral triangle having its side as $1 \mathrm{~cm}=\frac{\sqrt{3}}{4}(1)^{2}=\frac{\sqrt{3}}{4} \mathrm{~cm}^{2}$ Number of equ...
Read More →In a triangle locate a point in its interior which is equidistant from all the sides of the triangle. ANSWER:
Solution: The point which is equidistant from all the sides of a triangle is called the incentre of the triangle. Incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle. Here, in ΔABC, we can find the incentre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of ΔABC....
Read More →ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
Solution: Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together. In ΔABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of ΔABC....
Read More →Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution: Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle. In $\triangle P N M$, $\angle \mathrm{N}=90^{\circ}$ $\angle \mathrm{P}+\angle \mathrm{N}+\angle \mathrm{M}=180^{\circ}$ (Angle sum property of a triangle) $\angle P+\angle M=90^{\circ}$ Clearly, $\angle M$ is an acute angle. $\therefore \angle \mathrm{M}<\angle \mathrm{N}$ $\Rightarrow$ PN < PM (Side opposite to the small...
Read More →In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
Solution: As PR $P Q$ $\therefore \angle \mathrm{PQR}\angle \mathrm{PRQ}$ (Angle opposite to larger side is larger) ... (1) $P S$ is the bisector of $\angle Q P R$. $\therefore \angle Q P S=\angle R P S \ldots(2)$ $\angle P S R$ is the exterior angle of $\triangle P Q S$. $\therefore \angle \mathrm{PSR}=\angle \mathrm{PQR}+\angle \mathrm{QPS} \ldots(3)$ $\angle P S Q$ is the exterior angle of $\triangle P R S$. $\therefore \angle \mathrm{PSQ}=\angle \mathrm{PRQ}+\angle \mathrm{RPS} \ldots(4)$ Ad...
Read More →AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
Solution: Let us join AC. $\ln \Delta \mathrm{ABC}$, $\mathrm{AB}\mathrm{BC}(\mathrm{AB}$ is the smallest side of quadrilateral $\mathrm{ABCD})$ $\therefore \angle 2\angle 1$ (Angle opposite to the smaller side is smaller) ... (1) In $\triangle \mathrm{ADC}$, AD < CD (CD is the largest side of quadrilateral ABCD) $\therefore \angle 4<\angle 3$ (Angle opposite to the smaller side is smaller)..(2) On adding equations (1) and (2), we obtain $\angle 2+\angle 4<\angle 1+\angle 3$ $\Rightarrow \angle ...
Read More →In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
Solution: In $\triangle \mathrm{AOB}$ $\angle B\angle A$ $\Rightarrow \mathrm{AO}\mathrm{BO}$ (Side opposite to smaller angle is smaller) ...(1) In $\triangle \mathrm{COD}$ $\angle C\angle D$ $\Rightarrow O D $ (Side opposite to smaller angle is smaller) ... (2) On adding equations (1) and (2), we obtain $\mathrm{AO}+\mathrm{OD}\mathrm{BO}+\mathrm{OC}$ $\mathrm{AD}<\mathrm{BC}$...
Read More →In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution: In the given figure, $\angle \mathrm{ABC}+\angle \mathrm{PBC}=180^{\circ}$ (Linear pair) $\Rightarrow \angle \mathrm{ABC}=180^{\circ}-\angle \mathrm{PBC} \ldots(1)$ Also, $\angle \mathrm{ACB}+\angle \mathrm{QCB}=180^{\circ}$ $\angle A C B=180^{\circ}-\angle Q C B \ldots(2)$ $\mathrm{As} \angle \mathrm{PBC}\angle \mathrm{QCB}$ $\Rightarrow 180^{\circ}-\angle \mathrm{PBC}180^{\circ}-\angle Q C B$ $\Rightarrow \angle A B C\angle A C B$ [From equations (1) and (2)] $\Rightarrow A CA B$ (Si...
Read More →Show that in a right angled triangle, the hypotenuse is the longest side.
Solution: Let us consider a right-angled triangle $\mathrm{ABC}$, right-angled at $\mathrm{B}$. In $\triangle \mathrm{ABC}$ $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$ (Angle sum property of a triangle) $\angle \mathrm{A}+90^{\circ}+\angle \mathrm{C}=180^{\circ}$ $\angle \mathrm{A}+\angle \mathrm{C}=90^{\circ}$ Hence, the other two angles have to be acute (i.e., less than $90^{\circ}$ ). $\therefore \angle B$ is the largest angle in $\triangle A B C$. $\Rightarrow \angle ...
Read More →ABC is an isosceles triangle with AB = AC. Drawn AP ⊥ BC to show that ∠B = ∠C.
Solution: In $\triangle \mathrm{APB}$ and $\triangle \mathrm{APC}$ $\angle \mathrm{APB}=\angle \mathrm{APC}\left(\right.$ Each $\left.90^{\circ}\right)$ $A B=A C($ Given $)$ $\mathrm{AP}=\mathrm{AP}($ Common $)$ $\therefore \triangle \mathrm{APB} \cong \triangle \mathrm{APC}(\cup$ sing RHS congruence rule $)$ $\Rightarrow \angle B=\angle C$ (By using CPCT)...
Read More →BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution: In $\triangle B E C$ and $\triangle C F B$, $\angle B E C=\angle C F B\left(\right.$ Each $\left.90^{\circ}\right)$ $\mathrm{BC}=\mathrm{CB}$ (Common) $\mathrm{BE}=\mathrm{CF}$ (Given) $\therefore \Delta \mathrm{BE} \mathrm{C} \cong \triangle \mathrm{CFB}(\mathrm{By} \mathrm{RHS}$ congruency $)$ $\Rightarrow \angle B C E=\angle C B F($ By CPCT $)$ $\therefore A B=A C$ (Sides opposite to equal angles of a triangle are equal) Hence, $\triangle \mathrm{ABC}$ is isosceles....
Read More →$A D$ is an altitude of an isosceles triangles $A B C$ in which $A B=A C$. Show that<br/><br/>(i) $A D$ bisects $B C$<br/><br/>(ii) $\mathrm{AD}$ bisects $\angle \mathrm{A}$.
Solution: (i) $\ln \triangle \mathrm{BAD}$ and $\triangle \mathrm{CAD}$, $\angle A D B=\angle A D C\left(\right.$ Each $90^{\circ}$ as $A D$ is an altitude $)$ $A B=A C$ (Given) $\mathrm{AD}=\mathrm{AD}$ (Common) $\therefore \triangle \mathrm{BAD} \cong \triangle \mathrm{CAD}$ (By RHS Congruence rule) $\Rightarrow \mathrm{BD}=\mathrm{CD}(\mathrm{By} \mathrm{CPCT})$ Hence, AD bisects BC. (ii) Also, by CPCT, $\angle \mathrm{BAD}=\angle \mathrm{CAD}$ Hence, AD bisects $\angle \mathrm{A}$....
Read More →Show that the angles of an equilateral triangle are 60º each.
Solution: Let us consider that $\mathrm{ABC}$ is an equilateral triangle. Therefore, $A B=B C=A C$ $A B=A C$ $\Rightarrow \angle C=\angle B$ (Angles opposite to equal sides of a triangle are equal) Also, $\mathrm{AC}=\mathrm{BC}$ $\Rightarrow \angle B=\angle A$ (Angles opposite to equal sides of a triangle are equal) Therefore, we obtain $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}$ In $\triangle \mathrm{ABC}$ $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$ $\Righta...
Read More →ABC is a right angled triangle in which ∠A = 90º and AB = AC. Find ∠B and ∠C.
Solution: It is given that $\mathrm{AB}=\mathrm{AC}$ $\Rightarrow \angle C=\angle B$ (Angles opposite to equal sides are also equal) $\ln \Delta \mathrm{ABC}$ $\angle A+\angle B+\angle C=180^{\circ}$ (Angle sum property of a triangle) $\Rightarrow 90^{\circ}+\angle B+\angle C=180^{\circ}$ $\Rightarrow 90^{\circ}+\angle B+\angle B=180^{\circ}$ $\Rightarrow 2 \angle B=90^{\circ}$ $\Rightarrow \angle B=45^{\circ}$ $\therefore \angle B=\angle C=45^{\circ}$...
Read More →ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.
Solution: In $\triangle A B C$ $\mathrm{AB}=\mathrm{AC}$ (Given) $\Rightarrow \angle A C B=\angle A B C$ (Angles opposite to equal sides of a triangle are also equal) In $\triangle \mathrm{ACD}$, $\mathrm{AC}=\mathrm{AD}$ $\Rightarrow \angle A D C=\angle A C D$ (Angles opposite to equal sides of a triangle are also equal) In $\triangle B C D$, $\angle A B C+\angle B C D+\angle A D C=180^{\circ}$ (Angle sum property of a triangle) $\Rightarrow \angle \mathrm{ACB}+\angle \mathrm{ACB}+\angle \mathr...
Read More →ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.
Solution: Let us join AD. In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACD}$ $\mathrm{AB}=\mathrm{AC}$ (Given) BD = CD (Given) $A D=A D$ (Common side) $\therefore \triangle \mathrm{ABD} \cong \triangle \mathrm{ACD}$ (By SSS congruence rule) $\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{ACD}(\mathrm{By} \mathrm{CPCT})$...
Read More →ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that<br/><br/>(i) $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACF}$<br/><br/>(ii) $A B=A C$, i.e., $A B C$ is an isosceles triangle.
Solution: (i) In $\triangle \mathrm{ABE}$ and $\triangle \mathrm{ACF}$, $\angle \mathrm{AEB}=\angle \mathrm{AFC}\left(\right.$ Each $\left.90^{\circ}\right)$ $\angle \mathrm{A}=\angle \mathrm{A}$ (Common angle) BE $=$ CF (Given) $\therefore \triangle \mathrm{ABE} \cong \triangle \mathrm{ACF}($ By $\mathrm{AAS}$ congruence rule $)$ (ii) It has already been proved that $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACF}$ $\Rightarrow \mathrm{AB}=\mathrm{AC}(\mathrm{By} \mathrm{CPCT})$...
Read More →ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.
Solution: In $\triangle \mathrm{AEB}$ and $\triangle \mathrm{AFC}$, $\angle A E B$ and $\angle A F C\left(\right.$ Each $\left.90^{\circ}\right)$ $\angle \mathrm{A}=\angle \mathrm{A}$ (Common angle) $\mathrm{AB}=\mathrm{AC}$ (Given) \ $\therefore \triangle \mathrm{AEB} \cong \triangle \mathrm{AFC}($ By $\mathrm{AAS}$ congruence rule $)$ $\Rightarrow \mathrm{BE}=\mathrm{CF}(\mathrm{By} \mathrm{CPCT})$...
Read More →In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC.
Solution: In $\triangle A D C$ and $\triangle A D B$, $A D=A D$ (Common) $\angle A D C=\angle A D B\left(\right.$ Each $\left.90^{\circ}\right)$ $C D=B D(A D$ is the perpendicular bisector of $B C)$ $\therefore \triangle \mathrm{ADC} \cong \triangle \mathrm{ADB}($ By $S A S$ congruence rule $)$ $\therefore \mathrm{AB}=\mathrm{AC}(\mathrm{By} \mathrm{CPCT})$ Therefore, $A B C$ is an isosceles triangle in which $A B=A C$....
Read More →In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:<br/><br/>(i) $\mathrm{OB}=\mathrm{OC}$<br/><br/>(ii) $\mathrm{AO}$ bisects $\angle \mathrm{A}$
Solution: (i) It is given that in triangle $A B C, A B=A C$ $\Rightarrow \angle A C B=\angle A B C$ (Angles opposite to equal sides of a triangle are equal) $\Rightarrow \frac{1}{2} \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{ABC}$ $\Rightarrow \angle \mathrm{OCB}=\angle \mathrm{OBC}$ $\Rightarrow O B=O C$ (Sides opposite to equal angles of a triangle are also equal) (ii) In $\triangle \mathrm{OAB}$ and $\triangle \mathrm{OAC}$, $\mathrm{AO}=\mathrm{AO}$ (Common) $\mathrm{AB}=\mathrm{AC}$ (Gi...
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