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Quadrilaterals · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 11 questions.

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Question 11 Mark
If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a _________. (rectangle, parallelogram, rhombus).
Answer
If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a parallelogram.
Explanation:

ABCD is a quadrilateral in which AB = CD and BC = DA. We need to show that ABCD is a parallelogram. In $\triangle\text{ACB}$ and $\triangle\text{CAD},$ we have AC = CA (Common) CB = AD (Given) AB = CD (Given) So, by SSS criterion of congruence, we have $\triangle\text{ACB}\cong\triangle\text{CAD}$ By corresponding parts of congruent triangles property. $\angle\text{CAB}=\angle\text{ACD}\ ....(\text{i})$ And $\angle\text{ACB}=\angle\text{CAD}$ Now lines AC intersects AB and DC at A and C,such that $\angle\text{CAB}=\angle\text{ACD}$ [From (i)] That is, alternate interior angles are equal. Therefore, AB || DC. Similarly, AD || BC. Therefore, ABCD is a parallelogram.
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Question 21 Mark
If one pair of opposite sides are equal and parallel, then the figure is _____________. (parallelogram, rectangle, trapezium).
Answer
If one pair of opposite sides are equal and parallel, then the figure is parallelogram.
Explanation:

In $\triangle\text{ABC}$ and $\triangle\text{CDA},$ AB = DC (Given) AC = AC (Common) $\angle\text{BAC}=\angle\text{DCA}$ (Because AB || CD, Alternate interior angle are equal) So, by SAS Congruence rule, we have  $\triangle\text{ABC}\cong\triangle\text{CDA}$ Also, $\angle\text{BCA}=\angle\text{DAC}$ (Corresponding parts of congruent triangles are equal) But, these are alternate interior angles, which are equal. AD || BC Thus, AB || CD and AD || BC. Hence, quadrilateral ABCD is parallelogram.
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Question 31 Mark
The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ____________.
Answer
The figure formed by joining the mid-points of consecutive sides of a quadrilateral is Parallelogram.
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Question 41 Mark
If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is _____________. (square, rectangle, trapezium).
Answer
If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is trapezium.
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Question 51 Mark
The triangle formed by joining the mid-points of the sides of a right triangle is ____________.
Answer
The triangle formed by joining the mid-points of the sides of a right triangle is Right triangle.
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Question 71 Mark
A line drawn from the mid-point of one side of a triangle _____________ another side intersects the third side at its mid-point. (perpendicular to parallel to, to meet).
Answer
A line drawn from the mid-point of one side of a triangle parallel to another side intersects the third side at its mid-point.
Explanation:
This is a theorem.
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Question 81 Mark
If one angle of a parallelogram is a right angle, then it is necessarily a _____________. (rectangle, square, rhombus).
Answer
If one angle of a parallelogram is a right angle, then it is necessarily a rectangle.
Explanation:

We have, $\angle\text{A}=90^\circ$ In a parallelogram, opposite angles are equal. Therefore, $\angle\text{C}=90^\circ$ Similarly, $\angle\text{A}+\angle\text{D}=180^\circ$ $90^\circ+\angle\text{D}=180^\circ$ $\angle\text{D}=90^\circ$ Also, $\angle\text{B}=90^\circ$ Thus, a parallelogram with all the angles being right angle and opposite sides being equal is a rectangle.
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Question 91 Mark
The triangle formed by joining the mid-points of the sides of an isosceles triangle is ____________.
Answer
The triangle formed by joining the mid-points of the sides of an isosceles triangle is Isosceles.
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Question 111 Mark
If opposite angles of a quadrilateral are equal, then it is necessarily a _________. (parallelogram, rhombus, rectangle).
Answer
If opposite angles of a quadrilateral are equal, then it is necessarily a parallelogram.
Explanation:

ABCD is a quadrilateral in which $\angle\text{A}=\angle\text{C}$ and $\angle\text{B}=\angle\text{D}.$ We need to show that ABCD is a parallelogram. In quadrilateral ABCD, we have $\angle\text{A}=\angle\text{C}$ $\angle\text{B}=\angle\text{D}$ Therefore, $\angle\text{A}+\angle\text{B}=\angle\text{C}+\angle\text{D}\ ....(\text{i})$ Since sum of angles of a quadrilateral is 360º $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$ From equation (i), we get: $(\angle\text{A}+\angle\text{B})+(\angle\text{A}+\angle\text{B})=360^\circ$ $2(\angle\text{A}+\angle\text{B})=360^\circ$ $\angle\text{A}+\angle\text{B}=180^\circ$ Similarly, $\angle\text{C}+\angle\text{D}=180^\circ$ Now, line AB intersects AD and BC at A and B respectively Such that $\angle\text{A}+\angle\text{B}=180^\circ$ That is, sum of consecutive interior angles is supplementary. Therefore, AD || BC. Similarly, we get AB || DC. Therefore, ABCD is a parallelogram.
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