Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 5 Set 5.4 Quadrilaterals Solutions

Question 1. In IJKL, side IJ || side KL, \( \angle I = 108^\circ \) and \( \angle K = 53^\circ \), then find the measures of \( \angle J \) and \( \angle L \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समलम्ब चतुर्भुज IJKL का चित्र है जिसमें भुजा IJ भुजा KL के समांतर है। कोण I को 108 डिग्री और कोण K को 53 डिग्री दर्शाया गया है। यह चित्र चतुर्भुज के अंदर कोणों और भुजाओं के संबंध को समझने में मदद करता है।
Answer: Solution:
i. \( \angle I = 108^\circ \) [Given]
side IJ || side KL and side IL is their transversal. [Given]

\( \implies \angle I + \angle L = 180^\circ \) [Interior angles]

\( \implies 108^\circ + \angle L = 180^\circ \)

\( \implies \angle L = 180^\circ - 108^\circ = 72^\circ \)
ii. \( \angle K = 53^\circ \) [Given]
side IJ || side KL and side JK is their transversal. [Given]

\( \implies \angle J + \angle K = 180^\circ \) [Interior angles]

\( \implies \angle J + 53^\circ = 180^\circ \)

\( \implies \angle J = 180^\circ - 53^\circ = 127^\circ \)

\( \implies \angle L = 72^\circ, \angle J = 127^\circ \)
In simple words: Given a trapezoid IJKL with IJ parallel to KL, we use the property that interior angles on the same side of a transversal add up to 180 degrees. This helps us find the measures of angles L and J by subtracting the given angles I and K from 180 degrees.

🎯 Exam Tip: Remember to correctly identify parallel sides and transversals in quadrilaterals to apply the properties of interior angles and alternate angles accurately for scoring full marks.

Question 2. In ABCD, side BC || side AD, side AB = side DC. If \( \angle A = 72^\circ \), then find the measures of \( \angle B \) and \( \angle D \).
Construction: Draw seg BP \( \perp \) side AD, A - P - D, seg CQ \( \perp \) side AD, A - Q - D.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समलम्ब चतुर्भुज ABCD को दर्शाता है जहाँ भुजा BC भुजा AD के समांतर है और भुजा AB भुजा DC के बराबर है, जो इसे एक समद्विबाहु समलम्ब चतुर्भुज बनाता है। कोण A को 72 डिग्री दिया गया है। बिंदुओं P और Q के साथ AD पर लंब BP और CQ खींचे गए हैं।
Answer: Solution:
i. \( \angle A = 72^\circ \) [Given]
In ABCD, side BC || side AD and side AB is their transversal. [Given]

\( \implies \angle A + \angle B = 180^\circ \) [Interior angles]

\( \implies 72^\circ + \angle B = 180^\circ \)

\( \implies \angle B = 180^\circ - 72^\circ = 108^\circ \)
ii. In \( \Delta BPA \) and \( \Delta CQD \),
\( \angle BPA = \angle CQD \) [Each angle is of measure \( 90^\circ \)]
Hypotenuse AB \( \cong \) Hypotenuse DC [Given]
seg BP = seg CQ [Perpendicular distance between two parallel lines]

\( \implies \Delta BPA \cong \Delta CQD \) [Hypotenuse-side test]

\( \implies \angle BAP \cong \angle CDQ \) [c. a. c. t.]

\( \implies \angle A = \angle D \)

\( \implies \angle D = 72^\circ \)

\( \implies \angle B = 108^\circ, \angle D = 72^\circ \)
In simple words: For an isosceles trapezoid ABCD with BC parallel to AD and AB = DC, angle A and angle B are consecutive interior angles, summing to 180 degrees. By drawing altitudes and proving triangle congruence, we establish that base angles A and D are equal.

🎯 Exam Tip: When dealing with isosceles trapezoids, remember that base angles are equal and consecutive interior angles between parallel sides sum to 180 degrees. Using auxiliary constructions like altitudes can help prove congruence and find unknown angles.

Question 3. In ABCD, side BC < side AD, side BC || side AD and if side BA \( \cong \) side CD, then prove that \( \angle ABC = \angle DCB \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समद्विबाहु समलम्ब चतुर्भुज ABCD को दर्शाता है जिसमें भुजा BC, भुजा AD से छोटी है, भुजा BC भुजा AD के समांतर है, और भुजा BA भुजा CD के सर्वांगसम है। यह आकृति समद्विबाहु समलम्ब चतुर्भुज की विशेषताओं को दर्शाती है जिसका उपयोग यह सिद्ध करने के लिए किया जाएगा कि कोण ABC कोण DCB के बराबर है।
Answer: Solution:
Proof:
Given: side BC < side AD, side BC || side AD, side BA \( \cong \) side CD
To prove: \( \angle ABC = \angle DCB \)
Construction: Draw seg BP \( \perp \) side AD, A - P - D
seg CQ \( \perp \) side AD, A - Q - D
In \( \Delta BPA \) and \( \Delta CQD \),
\( \angle BPA = \angle CQD \) [Each angle is of measure \( 90^\circ \)]
Hypotenuse BA \( \cong \) Hypotenuse CD [Given]
seg BP = seg CQ [Perpendicular distance between two parallel lines]

\( \implies \Delta BPA \cong \Delta CQD \) [Hypotenuse-side test]

\( \implies \angle BAP \cong \angle CDQ \) [c. a. c. t.]

\( \implies \angle A = \angle D \) ....(i)
Now, side BC || side AD and side AB is their transversal. [Given]

\( \implies \angle A + \angle B = 180^\circ \) ....(ii) [Interior angles]
Also, side BC || side AD and side CD is their transversal. [Given]

\( \implies \angle C + \angle D = 180^\circ \) ....(iii) [Interior angles]

\( \implies \angle A + \angle B = \angle C + \angle D \) [From (ii) and (iii)]

\( \implies \angle A + \angle B = \angle C + \angle A \) [From (i)]

\( \implies \angle B = \angle C \)

\( \implies \angle ABC = \angle DCB \)
In simple words: To prove that the base angles of an isosceles trapezoid are equal, we first use altitude constructions to show congruence between the triangles formed at the non-parallel sides, implying that the base angles at AD are equal. Then, by using the property of interior angles between parallel lines, we show that angles ABC and DCB are equal.

🎯 Exam Tip: For proving properties in trapezoids, especially isosceles ones, drawing perpendiculars (altitudes) from vertices to the opposite base is a common and effective strategy. Remember to cite congruence criteria and properties of parallel lines correctly.