Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 3 Set 3.3 Triangles Solutions

Question 1. Find the values of x and y using the information shown in the given figure. Find the measures of ∠ABD and ∠ACD.
ℹ️ चित्र व्याख्या (Diagram Explanation): चतुर्भुज ABCD दिखाया गया है, जिसमें त्रिभुज ABC और त्रिभुज BDC शामिल हैं। त्रिभुज ABC में, भुजा AB भुजा AC के बराबर है, और त्रिभुज BDC में, भुजा BD भुजा DC के बराबर है। कोण ACB का मान 50° और कोण DBC का मान 60° दिया गया है। कोण ABC को x और कोण DCB को y से दर्शाया गया है।
Answer:
Solution:
i. \( \angle ACB = 50° \) [Given]
In \( \triangle ABC \), seg \( AC \cong \text{seg } AB \) [Given]
\( \therefore \angle ABC = \angle ACB \) [Isosceles triangle theorem]
\( \therefore x = 50° \)
ii. \( \angle DBC = 60° \) [Given]
In \( \triangle BDC \), seg \( BD \cong \text{seg } DC \) [Given]
\( \therefore \angle DCB = \angle DBC \) [Isosceles triangle theorem]
\( \therefore y = 60° \)
iii. \( \angle ABD = \angle ABC + \angle DBC \) [Angle addition property]
\( = 50° + 60° \)
\( \therefore \angle ABD = 110° \)
iv. \( \angle ACD = \angle ACB + \angle DCB \) [Angle addition property]
\( = 50° + 60° \)
\( \therefore \angle ACD = 110° \)
\( \therefore x = 50° \), \( y = 60° \),
\( \angle ABD = 110° \), \( \angle ACD = 110° \)
In simple words: By using the properties of isosceles triangles, we found that x is 50° and y is 60°. Then, using the angle addition property, we calculated ∠ABD and ∠ACD to be 110° each.

🎯 Exam Tip: Remember that angles opposite to equal sides in an isosceles triangle are equal, and that angles can be added if they share a common arm and vertex.

Question 2. The length of hypotenuse of a right angled triangle is 15. Find the length of median on its hypotenuse.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दिखाया गया है। प्रश्न में इसके कर्ण पर खींची गई माध्यिका की लंबाई के बारे में बात की गई है।
Answer:
Solution:
Length of hypotenuse = 15 [Given]
Length of median on the hypotenuse = \( \frac{1}{2} \) x length of hypotenuse [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]
\( = \frac{1}{2} \times 15 = 7.5 \)
\( \therefore \) The length of the median on the hypotenuse is 7.5 units.
In simple words: The key property for a right-angled triangle is that the median to the hypotenuse is exactly half the length of the hypotenuse. Given the hypotenuse is 15, the median is 7.5.

🎯 Exam Tip: This is a fundamental theorem for right-angled triangles. State the theorem clearly when solving such problems to score full marks.

Question 3. In \( \triangle PQR \), \( \angle Q = 90° \), \( PQ = 12 \), \( QR = 5 \) and QS is a median. Find \( l(QS) \).
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज PQR दर्शाया गया है, जहाँ कोण Q 90° का है। भुजा PQ की लंबाई 12 इकाई और भुजा QR की लंबाई 5 इकाई है। QS, कर्ण PR पर खींची गई माध्यिका है।
Answer:
Solution:
i. \( PQ = 12 \), \( QR = 5 \) [Given]
In \( \triangle PQR \), \( \angle Q = 90° \) [Given]
\( \therefore PR^2 = QR^2 + PQ^2 \) [Pythagoras theorem]
\( = 25 + 144 \)
\( \therefore PR^2 = 169 \)
\( \therefore PR = 13 \) units [Taking square root of both sides]
ii. In right angled \( \triangle PQR \), seg QS is the median on hypotenuse PR.
\( \therefore QS = \frac{1}{2} PR \) [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]
\( = \frac{1}{2} \times 13 \)
\( \therefore l(QS) = 6.5 \) units
In simple words: First, we use the Pythagorean theorem to find the length of the hypotenuse PR. Then, we apply the property that the median to the hypotenuse of a right-angled triangle is half the length of the hypotenuse to find \( l(QS) \).

🎯 Exam Tip: Remember to clearly state the theorems used (Pythagoras theorem and median to hypotenuse theorem) to ensure your steps are justified and complete.

Question 4. In the given figure, point G is the point of concurrence of the medians of \( \triangle PQR \). If \( GT = 2.5 \), find the lengths of PG and PT.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज PQR दर्शाया गया है जिसमें माध्यिकाएँ खींची गई हैं जो बिंदु G पर प्रतिच्छेद करती हैं। माध्यिका PT को दर्शाया गया है, जिसमें बिंदु G, PT पर स्थित है।
Answer:
Solution:
i. In \( \triangle PQR \), G is the point of concurrence of the medians. [Given]
The centroid divides each median in the ratio 2: 1.
\( PG : GT = 2 : 1 \)
\( \implies \frac{PG}{GT} = \frac{2}{1} \)
\( \implies \frac{PG}{2.5} = \frac{2}{1} \)
\( \therefore PG = 2 \times 2.5 \)
\( \therefore PG = 5 \) units
ii. Now, \( PT = PG + GT \) [P – G – T]
\( = 5 + 2.5 \)
\( \therefore l(PG) = 5 \) units, \( l(PT) = 7.5 \) units
In simple words: The centroid (point G) divides the median in a 2:1 ratio. Using this, we found PG by doubling GT, and then PT by adding PG and GT.

🎯 Exam Tip: Clearly state the centroid property (2:1 ratio) and the collinearity property (P-G-T) in your solution steps.

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.3 Intext Questions And Activities

Question 1. Can the theorem of isosceles triangle be proved by doing a different construction? (Textbook pg. no.34)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समद्विबाहु त्रिभुज ABC दर्शाया गया है, जिसमें शीर्ष A से आधार BC पर एक लंब AD खींचा गया है, जो BC को D पर प्रतिच्छेद करता है।
Answer:
Solution:
Yes
Construction: Draw seg \( AD \perp \text{seg } BC \).
Proof:
In \( \triangle ABD \) and \( \triangle ACD \),
seg \( AB \cong \text{seg } AC \) [Given]
\( \angle ADB = \angle ADC \) [Each angle is of measure 90°]
seg \( AD \cong \text{seg } AD \) [Common side]
\( \therefore \triangle ABD \cong \triangle ACD \) [Hypotenuse side test]
\( \therefore \angle ABD = \angle ACD \) [c.a.c.t.]
\( \therefore \angle ABC \cong \angle ACB \) [B-D-C]
In simple words: Yes, the isosceles triangle theorem can be proven by constructing an altitude from the vertex angle to the base, which then divides the isosceles triangle into two congruent right-angled triangles using the Hypotenuse-Side (HS) test.

🎯 Exam Tip: When proving theorems, clearly mention the congruence test (e.g., HS, SAS, SSS, ASA) used for triangles.

Question 2. Can the theorem of isosceles triangle be proved without doing any construction?
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समद्विबाहु त्रिभुज ABC दर्शाया गया है, जिसमें भुजा AB भुजा AC के बराबर है।
Answer:
Solution:
Yes
Proof:
In \( \triangle ABC \) and \( \triangle ACB \),
seg \( AB \cong \text{seg } AC \) [Given]
\( \angle BAC = \angle CAB \) [Common angle]
seg \( AC \cong \text{seg } AB \) [Given]
\( \therefore \triangle ABC \cong \triangle ACB \) [SAS test]
\( \therefore \angle ABC \cong \angle ACB \) [c. a. c. t.]
In simple words: Yes, the isosceles triangle theorem can be proven without additional construction by considering the triangle itself and its reflection (or the triangle compared to itself) and applying the Side-Angle-Side (SAS) congruence test.

🎯 Exam Tip: This proof often involves comparing the triangle with itself by mapping A to A, B to C, and C to B, which highlights the symmetry.

Question 3. In the given figure, \( \triangle ABC \) is a right angled triangle, seg BD is the median on hypotenuse. Measure the lengths of the following segments.
(i) AD
(ii) DC
(iii) BD
From the measurements verify that \( BD = \frac{1}{2}AC \). (Textbook pg. no. 37)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दर्शाया गया है, जहाँ कोण B समकोण है। भुजा AC इस त्रिभुज का कर्ण है, और BD कर्ण AC पर खींची गई माध्यिका है।
Answer:
Solution:
AD = DC = BD= 1.9 cm
AC = AD + DC [A – D – C]
= 1.9 + 1.9
= 2 x 1.9 cm
\( \therefore AC = 2 \times BD \)
\( \therefore BD = \frac{1}{2} AC \)
In simple words: In a right-angled triangle, the median to the hypotenuse (BD) is half the length of the hypotenuse (AC), and it also divides the hypotenuse into two equal segments (AD = DC = BD). Here, if AD, DC, and BD are all 1.9 cm, then AC is 3.8 cm, verifying the property.

🎯 Exam Tip: This question tests the "median to the hypotenuse" theorem. Practicing with actual measurements helps solidify the understanding of this geometric property.