Maharashtra Board Class 10 Maths Chapter 2 Pythagoras Theorem Set 2 Solutions

Question 1. Some questions and their alternative answers are given. Select the correct alternative.

i. Out of the following which is the Pythagorean triplet?
(A) (1,5,10)
(B) (3,4,5)
(C) (2,2,2)
(D) (5,5,2)
Answer: (B) (3,4,5)
Hint: Refer Practice set 2.1 Q.1 (i)
In simple words: A Pythagorean triplet consists of three positive integers \(a, b, c\) such that \(a^2 + b^2 = c^2\). For option (B), \(3^2 + 4^2 = 9 + 16 = 25\), and \(5^2 = 25\), so (3,4,5) is a triplet.

🎯 Exam Tip: To identify a Pythagorean triplet, always check if the sum of the squares of the two smaller numbers equals the square of the largest number.

ii. In a right angled triangle, if sum of the squares of the sides making right angle is 169, then what is the length of the hypotenuse?
(A) 15
(B) 13
(C) 5
(D) 12
Answer: (B) 13
Hint:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज PQR है जिसमें कोण Q समकोण (\(90^\circ\)) है। भुजाएं PQ और QR समकोण बनाने वाली भुजाएं हैं, और PR कर्ण है।
ii. In ∆PQR, \(\angle Q = 90^\circ\)
\(\therefore\) \(PR^2 = PQ^2 + QR^2\) ...[Pythagoras theorem]
\(\therefore\) \(PR^2 = 169\)
\(\therefore\) \(PR = \sqrt{169} = 13\)
In simple words: According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Since the sum of the squares of the sides making the right angle is given as 169, the square of the hypotenuse is 169, making the hypotenuse 13.

🎯 Exam Tip: Understand the Pythagorean theorem thoroughly, as it's fundamental for solving problems involving right-angled triangles. Remember that the hypotenuse is the longest side.

iii. Out of the dates given below which date constitutes a Pythagorean triplet?
(A) 15/08/17
(B) 16/08/16
(C) 3/5/17
(D) 4/9/15
Answer: (A) 15/08/17
Hint:
Consider Option A.
\(\therefore\) Here, \(15^2 + 8^2 = 225 + 64 = 289\), and \(17^2 = 289\)
\(15^2 + 8^2 = 17^2\)
In simple words: A date can form a Pythagorean triplet if the day, month, and year (or parts of it) satisfy the Pythagorean theorem. For 15/08/17, we check if \(15^2 + 8^2 = 17^2\), which holds true (\(225 + 64 = 289\)), confirming it as a triplet.

🎯 Exam Tip: When given numbers in a set, always square the two smaller numbers and add them, then compare this sum to the square of the largest number to verify if it's a Pythagorean triplet.

iv. If a, b, c are sides of a triangle and \(a^2 + b^2 = c^2\), name the type of the triangle.
(A) Obtuse angled triangle
(B) Acute angled triangle
(C) Right angled triangle
(D) Equilateral triangle
Answer: (C) Right angled triangle
In simple words: The condition \(a^2 + b^2 = c^2\), where \(c\) is the longest side, is the definition of the Pythagorean theorem. This theorem applies exclusively to right-angled triangles.

🎯 Exam Tip: This is a direct application of the converse of the Pythagorean theorem. Knowing this fundamental relationship is crucial for identifying triangle types based on side lengths.

v. Find perimeter of a square if its diagonal is \(10\sqrt{2}\) cm.
(A) 10 cm
(B) \(40\sqrt{2}\) cm
(C) 20 cm
(D) 40 cm
Answer: (D) 40 cm
Hint:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वर्ग ABCD है जिसके विकर्ण की लंबाई दी गई है। एक वर्ग में, विकर्ण भुजा की लंबाई का \(\sqrt{2}\) गुना होता है, और एक समकोण त्रिभुज (जैसे \(\triangle\)ABC) में, विकर्ण कर्ण का काम करता है।
In \(\triangle\)ABC, \(\angle B = 90^\circ\), and \(\angle BAC = \angle BCA = 45^\circ\)
\(\therefore\) \(AB = \frac{1}{\sqrt{2}} AC\) ...[Theorem of \(45^\circ\) - \(45^\circ\) - \(90^\circ\) triangle]
\(\therefore\) \(=\frac{1}{\sqrt{2}} \times 10\sqrt{2}\)
\(\therefore\) \(AB = 10\) cm
\(\therefore\) Perimeter of square = \(4 (AB) = 4 \times 10 = 40\) cm
In simple words: In a square, the diagonal is \(\sqrt{2}\) times its side length. Given the diagonal is \(10\sqrt{2}\) cm, the side length is 10 cm. The perimeter of a square is 4 times its side, so \(4 \times 10 = 40\) cm.

🎯 Exam Tip: Remember the relationship between the side and diagonal of a square (diagonal = side \(\times \sqrt{2}\)). This property simplifies calculations for perimeter and area when only the diagonal is known.

vi. Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.
(A) 9 cm
(B) 4 cm
(C) 6 cm
(D) \(2\sqrt{6}\)
Answer: (C) 6 cm
Hint:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज ABC है, जहाँ B पर समकोण है। BD एक ऊंचाई है जिसे कर्ण AC पर डाला गया है, जो कर्ण को AD (4 cm) और DC (9 cm) दो भागों में विभाजित करता है।
In \(\triangle\)ABC,
\(BD^2 = AD \times DC\) ...[Theorem of geometric mean]
\(\therefore\) \(BD^2 = 4 \times 9\)
\(\therefore\) \(BD = \sqrt{36} = 6\) cm
In simple words: The geometric mean theorem states that the altitude drawn to the hypotenuse of a right triangle creates two segments, and the altitude's length squared is equal to the product of those segments. Here, \(BD^2 = 4 \times 9 = 36\), so the altitude \(BD\) is 6 cm.

🎯 Exam Tip: The geometric mean theorem (or altitude theorem) is crucial for finding the altitude when the segments of the hypotenuse are given. Memorize the formula: altitude\(^2\) = segment1 \(\times\) segment2.

vii. Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse.
(A) 24 cm
(B) 30 cm
(C) 15 cm
(D) 18 cm
Answer: (B) 30 cm
Hint:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज PQR है जहाँ Q पर समकोण है। PQ त्रिभुज की ऊँचाई (24 cm) और QR त्रिभुज का आधार (18 cm) है, तथा PR कर्ण है।
\(\therefore\) In \(\triangle\)PQR, \(\angle Q = 90^\circ\)
\(\therefore\) \(PR^2 = PQ^2 + QR^2\) ...[Pythagoras theorem]
\(\therefore\) \(PR^2 = 24^2 + 18^2\)
\(\therefore\) \(PR^2 = 576 + 324\)
\(\therefore\) \(PR^2 = 900\)
\(\therefore\) \(PR = \sqrt{900} = 30\) cm
In simple words: Using the Pythagorean theorem (\(a^2 + b^2 = c^2\)), where the height and base are the two shorter sides, we square them (\(24^2 = 576\), \(18^2 = 324\)) and add them (\(576 + 324 = 900\)). The square root of this sum gives the hypotenuse, which is 30 cm.

🎯 Exam Tip: For right-angled triangles, the Pythagorean theorem is your go-to formula. Practice recognizing common Pythagorean triplets (like 3-4-5, 5-12-13, 8-15-17, 7-24-25) and their multiples to quickly solve such problems.

viii. In \(\triangle\)ABC, \(AB = 6\sqrt{3}\) cm, \(AC = 12\) cm, \(BC = 6\) cm. Find measure of \(\angle A\).
(A) \(30^\circ\)
(B) \(60^\circ\)
(C) \(90^\circ\)
(D) \(45^\circ\)
Answer: (A) \(30^\circ\)
Hint:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC है जिसमें भुजाओं की लंबाई दी गई है: \(AB = 6\sqrt{3}\), \(AC = 12\), \(BC = 6\)। इस त्रिभुज में कोण A ज्ञात करना है।
We know that, \(6 = \frac{1}{2} (12)\) and \(6\sqrt{3} = \frac{\sqrt{3}}{2} (12)\)
\(\therefore\) \(BC = \frac{1}{2} AC\) and \(AB = \frac{\sqrt{3}}{2} AC\)
\(\therefore\) \(\angle A = 30^\circ\) ...[Converse of \(30^\circ\)-\(60^\circ\)-\(90^\circ\) theorem]
In simple words: By comparing the given side lengths with the properties of a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle, we find that side BC is half of AC (hypotenuse) and side AB is \(\frac{\sqrt{3}}{2}\) times AC. This means that BC is opposite to the \(30^\circ\) angle, so \(\angle A\) must be \(30^\circ\).

🎯 Exam Tip: Memorize the side ratios for \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangles: side opposite \(30^\circ\) is \(x\), opposite \(60^\circ\) is \(x\sqrt{3}\), and opposite \(90^\circ\) is \(2x\). The converse helps identify angles from known side ratios.

Question 2. Solve the following examples.

i. Find the height of an equilateral triangle having side \(2a\).
Answer: Solution:
Let \(\triangle\)ABC be the given equilateral triangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समबाहु त्रिभुज ABC है जिसकी प्रत्येक भुजा \(2a\) है। शीर्ष A से आधार BC पर एक लंब AD खींचा गया है, जो BC को D पर प्रतिच्छेद करता है। यह लंब त्रिभुज को दो समकोण त्रिभुजों में विभाजित करता है।
\(\therefore\) \(\angle B = 60^\circ\) [Angle of an equilateral triangle]
Let \(AD \perp BC\), \(B - D - C\).
In \(\triangle\)ABD, \(\angle B = 60^\circ\), \(\angle ADB = 90^\circ\)
\(\therefore\) \(\angle BAD = 30^\circ\) [Remaining angle of a triangle]
\(\therefore\) \(\triangle\)ABD is a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle.
\(\therefore\) \(AD = \frac{\sqrt{3}}{2} AB\) [Side opposite to \(60^\circ\)]
\(\therefore\) \(=\frac{\sqrt{3}}{2} \times 2a\)
\(\therefore\) \(= a\sqrt{3}\) units
The height of the equilateral triangle is \(a\sqrt{3}\) units.
In simple words: For an equilateral triangle with side \(2a\), drawing an altitude creates a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle. The altitude is the side opposite the \(60^\circ\) angle, which is \(\frac{\sqrt{3}}{2}\) times the hypotenuse (\(2a\)), resulting in a height of \(a\sqrt{3}\).

🎯 Exam Tip: The height of an equilateral triangle with side 's' can be calculated using the formula \(h = \frac{\sqrt{3}}{2}s\). Remember that the altitude in an equilateral triangle also acts as a median and angle bisector.

ii. Do sides 7 cm, 24 cm, 25 cm form a right angled triangle? Give reason.
Answer: The sides of the triangle are 7 cm, 24 cm and 25 cm.
The longest side of the triangle is 25 cm.
\(\therefore\) \((25)^2 = 625\)
Now, sum of the squares of the remaining sides is,
\((7)^2 + (24)^2 = 49 + 576\)
\(= 625\)
\(\therefore\) \((25)^2 = (7)^2 + (24)^2\)
\(\therefore\) Square of the longest side is equal to the sum of the squares of the remaining two sides.
\(\therefore\) The given sides will form a right angled triangle. [Converse of Pythagoras theorem]
In simple words: To check if sides form a right-angled triangle, we use the converse of the Pythagorean theorem: if the square of the longest side equals the sum of the squares of the other two sides, it's a right triangle. Here, \(25^2 = 625\) and \(7^2 + 24^2 = 49 + 576 = 625\), so it is a right-angled triangle.

🎯 Exam Tip: When proving if a triangle is right-angled, always clearly state the squares of all three sides and show the equality, citing the Converse of Pythagoras Theorem.

iii. Find the length of a diagonal of a rectangle having sides 11 cm and 60 cm.
Answer: Let JABCD be the given rectangle.
\(AB = 11\) cm, \(BC = 60\) cm
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आयत ABCD है जिसकी भुजाएँ AB (11 cm) और BC (60 cm) हैं। आयत का विकर्ण AC है, जिसे ज्ञात करना है। आयत के कोने पर समकोण होता है, जिससे \(\triangle\)ABC एक समकोण त्रिभुज बनता है।
In \(\triangle\)ABC, \(\angle B = 90^\circ\) [Angle of a rectangle]
\(\therefore\) \(AC^2 = AB^2 + BC^2\) [Pythagoras theorem]
\(\therefore\) \(= 11^2 + 60^2\)
\(\therefore\) \(= 121 + 3600\)
\(\therefore\) \(= 3721\)
\(\therefore\) \(AC = \sqrt{3721}\) [Taking square root of both sides]
\(\therefore\) \(= 61\) cm
The length of the diagonal of the rectangle is 61 cm.
In simple words: In a rectangle, the diagonal forms the hypotenuse of a right-angled triangle with the sides as legs. Using the Pythagorean theorem, \(diagonal^2 = side1^2 + side2^2\), which gives \(11^2 + 60^2 = 121 + 3600 = 3721\). The diagonal is \(\sqrt{3721} = 61\) cm.

🎯 Exam Tip: A rectangle's diagonal always creates two right-angled triangles. Remember that the Pythagorean theorem is directly applicable here, making the diagonal the hypotenuse.

iv. Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
Answer: Let \(\triangle\)PQR be the given right angled triangle.
In \(\triangle\)PQR, \(\angle Q = 90^\circ\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज PQR है जहाँ Q पर समकोण है। भुजाएँ PQ (9 cm) और QR (12 cm) समकोण बनाने वाली भुजाएँ हैं, और PR कर्ण है जिसे ज्ञात करना है।
\(\therefore\) \(PR^2 = PQ^2 + QR^2\) [Pythagoras theorem]
\(\therefore\) \(= 9^2 + 12^2\)
\(\therefore\) \(= 81 + 144\)
\(\therefore\) \(= 225\)
\(\therefore\) \(PR = \sqrt{225}\) [Taking square root of both sides]
\(\therefore\) \(= 15\) cm
\(\therefore\) The length of the hypotenuse of the right angled triangle is 15 cm.
In simple words: For a right-angled triangle, the hypotenuse is found using the Pythagorean theorem. With sides 9 cm and 12 cm, \(9^2 + 12^2 = 81 + 144 = 225\). The square root of 225 is 15, so the hypotenuse is 15 cm.

🎯 Exam Tip: Recognize common Pythagorean triplets like 9-12-15 (which is a multiple of 3-4-5) to quickly solve problems involving right-angled triangles. Showing the steps for the theorem is still important for full marks.

v. A side of an isosceles right angled triangle is x. Find its hypotenuse.
Answer: Let \(\triangle\)PQR be the given right angled isosceles triangle.
\(PQ = QR = x\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समद्विबाहु समकोण त्रिभुज PQR है जहाँ Q पर समकोण है। इसकी दो समान भुजाएँ PQ और QR हैं, दोनों की लंबाई x है, और PR कर्ण है जिसे ज्ञात करना है।
In \(\triangle\)PQR, \(\angle Q = 90^\circ\) [Pythagoras theorem]
\(\therefore\) \(PR^2 = PQ^2 + QR^2\)
\(\therefore\) \(PR^2 = x^2 + x^2\)
\(\therefore\) \(PR^2 = 2x^2\)
\(\therefore\) \(PR = \sqrt{2x^2}\) [Taking square root of both sides]
\(\therefore\) \(PR = x\sqrt{2}\) units
\(\therefore\) The hypotenuse of the right angled isosceles triangle is \(x\sqrt{2}\) units.
In simple words: For an isosceles right-angled triangle with equal sides \(x\), the Pythagorean theorem states \(hypotenuse^2 = x^2 + x^2 = 2x^2\). Taking the square root, the hypotenuse is \(x\sqrt{2}\).

🎯 Exam Tip: An isosceles right-angled triangle is also a \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle. Its sides are in the ratio \(1:1:\sqrt{2}\). This is a useful shortcut to remember.

vi. In \(\triangle\)PQR, \(PQ = \sqrt{8}\), \(QR = \sqrt{5}\), \(PR = \sqrt{3}\). Is \(\triangle\)PQR a right angled triangle? If yes, which angle is of \(90^\circ\)?
Answer: Longest side of \(\triangle\)PQR = \(PQ = \sqrt{8}\)
\(\therefore\) \(PQ^2 = (\sqrt{8})^2 = 8\)
Now, sum of the squares of the remaining sides is,
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR है जिसमें भुजाओं की लंबाई दी गई है: \(PQ = \sqrt{8}\), \(QR = \sqrt{5}\), \(PR = \sqrt{3}\)। हमें यह निर्धारित करना है कि क्या यह एक समकोण त्रिभुज है और यदि हाँ, तो \(90^\circ\) का कोण कौन सा है।
\(QR^2 + PR^2 = (\sqrt{5})^2 + (\sqrt{3})^2\)
\(= 5 + 3\)
\(= 8\)
\(\therefore\) \(PQ^2 = QR^2 + PR^2\)
\(\therefore\) Square of the longest side is equal to the sum of the squares of the remaining two sides.
\(\therefore\) \(\triangle\)PQR is a right angled triangle. [Converse of Pythagoras theorem]
Now, PQ is the hypotenuse.
\(\therefore\) \(\angle PRQ = 90^\circ\) [Angle opposite to hypotenuse]
\(\therefore\) \(\triangle\)PQR is a right angled triangle in which \(\angle PRQ\) is of \(90^\circ\).
In simple words: We square each side: \(PQ^2 = 8\), \(QR^2 = 5\), \(PR^2 = 3\). Since \(QR^2 + PR^2 = 5 + 3 = 8\), which equals \(PQ^2\), by the converse of the Pythagorean theorem, \(\triangle\)PQR is a right-angled triangle. The angle opposite the longest side (hypotenuse PQ) is \(\angle PRQ\), so \(\angle PRQ = 90^\circ\).

🎯 Exam Tip: To identify the right angle, always remember it is the angle opposite the hypotenuse (the longest side). Make sure to square all sides correctly when applying the converse of the Pythagorean theorem.

Question 3. In \(\triangle\)RST, \(\angle S = 90^\circ\), \(\angle T = 30^\circ\), \(RT = 12\) cm, then find RS and ST.
Answer: Solution:
in \(\triangle\)RST, \(\angle S = 90^\circ\), \(\angle T = 30^\circ\) [Given]
\(\therefore\) \(\angle R = 60^\circ\) [Remaining angle of a triangle]
\(\therefore\) \(\triangle\)RST is a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज RST है, जिसमें कोण S समकोण (\(90^\circ\)) है और कोण T \(30^\circ\) है। भुजा RT (जो कर्ण है) की लंबाई 12 cm है। भुजाएँ RS और ST ज्ञात करनी हैं।
\(\therefore\) \(RS = \frac{1}{2} RT\) [Side opposite to \(30^\circ\)]
\(\therefore\) \(=\frac{1}{2} \times 12 = 6\)cm
Also, \(ST = \frac{\sqrt{3}}{2} RT\) [Side opposite to \(60^\circ\)]
\(\therefore\) \(=\frac{\sqrt{3}}{2} \times 12 = 6\sqrt{3}\) cm
\(\therefore\) \(RS = 6\) cm and \(ST = 6\sqrt{3}\) cm
In simple words: Given a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle with hypotenuse 12 cm, the side opposite the \(30^\circ\) angle (RS) is half the hypotenuse, so 6 cm. The side opposite the \(60^\circ\) angle (ST) is \(\frac{\sqrt{3}}{2}\) times the hypotenuse, so \(6\sqrt{3}\) cm.

🎯 Exam Tip: Always identify the angles and corresponding sides (opposite \(30^\circ\), opposite \(60^\circ\), hypotenuse) correctly in a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle to apply the side ratios accurately.

Question 4. Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq. cm.
Answer: Solution:
Let JABCD be the given rectangle.
\(BC = 16\)cm
Area of rectangle = length \(\times\) breadth
Area of JABCD = \(BC \times AB\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आयत ABCD है। इसकी लंबाई BC 16 cm है। आयत का क्षेत्रफल 192 वर्ग cm है। हमें इस आयत का विकर्ण ज्ञात करना है।
\(\therefore\) \(192 = 16 \times AB\)
\(\therefore\) \(AB = \frac{192}{16}\)
\(\therefore\) \(= 12\)cm
Now, in \(\triangle\)ABC, \(\angle B = 90^\circ\) [Angle of a rectangle]
\(\therefore\) \(AC^2 = AB^2 + BC^2\) [Pythagoras theorem]
\(\therefore\) \(= 12^2 + 16^2\)
\(\therefore\) \(= 144 + 256\)
\(\therefore\) \(= 400\)
\(\therefore\) \(AC = \sqrt{400}\) [Taking square root of both sides]
\(\therefore\) \(= 20\)cm
\(\therefore\) The diagonal of the rectangle is 20 cm.
In simple words: First, use the area and given length to find the breadth: \(192 = 16 \times breadth \implies breadth = 12\) cm. Then, use the Pythagorean theorem for the right triangle formed by the length, breadth, and diagonal: \(diagonal^2 = 16^2 + 12^2 = 256 + 144 = 400\). So, the diagonal is \(\sqrt{400} = 20\) cm.

🎯 Exam Tip: Remember that the area of a rectangle is length \(\times\) breadth. You can use this to find an unknown side if the area is given, then apply Pythagoras theorem to find the diagonal.

Question 5. Find the length of the side and perimeter of an equilateral triangle whose height is \(\sqrt{3}\) cm.
Answer: Solution:
Let \(\triangle\)ABC be the given equilateral triangle.
\(\therefore\) \(\angle B = 60^\circ\) [Angle of an equilateral triangle]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समबाहु त्रिभुज ABC है। इसकी ऊंचाई AD \(\sqrt{3}\) cm है। आधार BC पर D बिंदु है। त्रिभुज की भुजा और परिधि ज्ञात करनी है। \(\triangle\)ADB एक \(30^\circ\)-\(60^\circ\)-\(90^\circ\) त्रिभुज है।
\(AD \perp BC\), \(B - D - C\).
In \(\triangle\)ABD, \(\angle B = 60^\circ\), \(\angle ADB = 90^\circ\)
\(\therefore\) \(\angle BAD = 30^\circ\) [Remaining angle of a triangle]
\(\therefore\) \(\triangle\)ABD is a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle.
\(\therefore\) \(AD = \frac{\sqrt{3}}{2} AB\) [Side opposite to \(60^\circ\)]
\(\therefore\) \(\sqrt{3} = \frac{\sqrt{3}}{2} AB\)
\(\therefore\) \(AB = \frac{2\sqrt{3}}{\sqrt{3}}\)
\(\therefore\) \(AB = 2\)cm
\(\therefore\) Side of equilateral triangle = 2cm
Perimeter of \(\triangle\)ABC = \(3 \times\) side
\(\therefore\) \(= 3 \times AB\)
\(\therefore\) \(= 3 \times 2\)
\(\therefore\) \(= 6\)cm
\(\therefore\) The length of the side and perimeter of the equilateral triangle are 2 cm and 6 cm respectively.
In simple words: For an equilateral triangle, its height is given by \(h = \frac{\sqrt{3}}{2} \times side\). Given \(h = \sqrt{3}\) cm, we can find the side: \(\sqrt{3} = \frac{\sqrt{3}}{2} \times side \implies side = 2\) cm. The perimeter is \(3 \times side = 3 \times 2 = 6\) cm.

🎯 Exam Tip: Know the formula for the height of an equilateral triangle \( (h = \frac{\sqrt{3}}{2}s)\) and its perimeter \((P = 3s)\). This allows for direct calculation if height or side is given.

Question 6. In \(\triangle\)ABC, seg AP is a median. If \(BC = 18\), \(AB^2 + AC^2 = 260\), find AP.
Answer: Solution:
\(PC = \frac{1}{2} BC\) [P is the midpoint of side BC]
\(\therefore\) \(=\frac{1}{2} \times 18 = 9\)cm
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC है। AP इस त्रिभुज की एक माध्यिका (median) है, जिसका अर्थ है कि P, BC का मध्यबिंदु है। BC की लंबाई 18 है, और \(AB^2 + AC^2 = 260\) दिया गया है। हमें AP की लंबाई ज्ञात करनी है।
in \(\triangle\)ABC, seg AP is the median,
Now, \(AB^2 + AC^2 = 2 AP^2 + 2 PC^2\) [Apollonius theorem]
\(\therefore\) \(260 = 2 AP^2 + 2 (9)^2\)
\(\therefore\) \(260 = 2 AP^2 + 2 (81)\)
\(\therefore\) \(260 = 2 AP^2 + 162\)
\(\therefore\) \(2 AP^2 = 260 - 162\)
\(\therefore\) \(2 AP^2 = 98\)
\(\therefore\) \(130 = AP^2 + 81\) [Dividing both sides by 2]
\(\therefore\) \(AP^2 = 130 - 81\)
\(\therefore\) \(AP^2 = 49\)
\(\therefore\) \(AP = \sqrt{49}\) [Taking square root of both sides]
\(\therefore\) \(AP = 7\) units
In simple words: Using Apollonius theorem, \(AB^2 + AC^2 = 2AP^2 + 2PC^2\). Given \(BC = 18\), so \(PC = 9\). Substitute the given values: \(260 = 2AP^2 + 2(9^2)\). Solve for \(AP^2\): \(260 = 2AP^2 + 162 \implies 98 = 2AP^2 \implies AP^2 = 49\). Thus, \(AP = 7\) units.

🎯 Exam Tip: Apollonius theorem is critical for median-related problems. Ensure you correctly identify the sides and the median in the formula. \(AB^2 + AC^2 = 2(median)^2 + 2(half\ of\ base)^2\).

Question 7. \(\triangle\)ABC is an equilateral triangle. Point P is on base BC such that \(PC = \frac{1}{3} BC\), if \(AB = 6\) cm find AP.
Answer: Given: \(\triangle\)ABC is an equilateral triangle.
\(PC = \frac{1}{3} BC\), \(AB = 6\)cm.
To find: AP
Construction: Draw seg \(AD \perp BC\), \(B - D - C\).
Solution:
\(\triangle\)ABC is an equilateral triangle.
\(\therefore\) \(AB = BC = AC = 6\)cm [Sides of an equilateral triangle]
\(PC = \frac{1}{3} BC\) [Given]
\(\therefore\) \(PC = \frac{1}{3} (6)\)
\(\therefore\) \(PC = 2\)cm
In \(\triangle\)ADC,
\(\angle D = 90^\circ\) [Construction]
\(\angle C = 60^\circ\) [Angle of an equilateral triangle]
\(\angle DAC = 30^\circ\) [Remaining angle of a triangle]
\(\therefore\) \(\triangle\)ADC is a \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समबाहु त्रिभुज ABC है जिसकी प्रत्येक भुजा 6 cm है। बिंदु P आधार BC पर इस प्रकार स्थित है कि \(PC = \frac{1}{3} BC\)। शीर्ष A से आधार BC पर लंब AD खींचा गया है, जो D पर मिलता है। हमें AP की लंबाई ज्ञात करनी है।
\(\therefore\) \(AD = \frac{\sqrt{3}}{2} AC\) [Side opposite to \(60^\circ\)]
\(\therefore\) \(AD = \frac{\sqrt{3}}{2} (6)\)
\(\therefore\) \(AD = 3\sqrt{3}\)cm
\(CD = \frac{1}{2} AC\) [Side opposite to \(30^\circ\)]
\(\therefore\) \(CD = \frac{1}{2} (6)\)
\(\therefore\) \(CD = 3\)cm
Now \(DP + PC = CD\) [D - P - C]
\(\therefore\) \(DP + 2 = 3\)
\(\therefore\) \(DP = 1\)cm
In \(\triangle\)ADP,
\(\angle ADP = 90^\circ\)
\(AP^2 = AD^2 + DP^2\) [Pythagoras theorem]
\(\therefore\) \(AP^2 = (3\sqrt{3})^2 + (1)^2\)
\(\therefore\) \(AP^2 = 9 \times 3 + 1 = 27 + 1\)
\(\therefore\) \(AP^2 = 28\)
\(\therefore\) \(AP = \sqrt{28}\)
\(\therefore\) \(AP = \sqrt{4 \times 7}\)
\(\therefore\) \(AP = 2\sqrt{7}\)cm
In simple words: Since \(\triangle\)ABC is equilateral with side 6 cm, altitude AD gives \(AD = 3\sqrt{3}\) cm and \(CD = 3\) cm. Given \(PC = \frac{1}{3} BC = \frac{1}{3}(6) = 2\) cm. From \(CD = DP + PC\), we get \(3 = DP + 2 \implies DP = 1\) cm. In right \(\triangle\)ADP, by Pythagoras theorem, \(AP^2 = AD^2 + DP^2 = (3\sqrt{3})^2 + 1^2 = 27 + 1 = 28\). So \(AP = \sqrt{28} = 2\sqrt{7}\) cm.

🎯 Exam Tip: For equilateral triangles, drawing an altitude is often the first step. This creates \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangles, allowing you to use their side ratios and eventually the Pythagorean theorem.

Question 8. From the information given in the adjoining figure, prove that \(PM = PN = \sqrt{3} \times a\)
Answer: Solution:
Proof:
In \(\triangle\)PMR,
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आकृति है जिसमें एक त्रिभुज PQR है और एक बिंदु S QR पर है। M, Q, S, R, N एक रेखा पर हैं। PS और PR माध्यिकाएं हैं या उनकी स्थिति दर्शाई गई है। QM = QR = a और अन्य खंडों की लंबाई भी a में दी गई है।
\(QM = QR = a\) [Given]
\(\therefore\) Q is the midpoint of side MR.
\(\therefore\) seg PQ is the median.
\(\therefore\) \(PM^2 + PR^2 = 2PQ^2 + 2QM^2\) [Apollonius theorem]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR है जिसमें Q, MR का मध्यबिंदु है (चूंकि QM = QR = a)। PQ माध्यिका है। यह आकृति Apollonius प्रमेय का उपयोग करके PM की लंबाई ज्ञात करने के लिए उपयोग की जाती है।
\(\therefore\) \(PM^2 + a^2 = 2a^2 + 2a^2\)
\(\therefore\) \(PM^2 + a^2 = 4a^2\)
\(\therefore\) \(PM^2 = 3a^2\)
\(\therefore\) \(PM = \sqrt{3}a\) (i) [Taking square root of both sides]
Similarly, in \(\triangle\)PNQ,
R is the midpoint of side QN.
\(\therefore\) seg PR is the median.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक अन्य त्रिभुज PNQ है जिसमें R, QN का मध्यबिंदु है। PR माध्यिका है। यह आकृति Apollonius प्रमेय का उपयोग करके PN की लंबाई ज्ञात करने के लिए उपयोग की जाती है।
\(\therefore\) \(PN^2 + PQ^2 = 2 PR^2 + 2 RN^2\) [Apollonius theorem]
\(\therefore\) \(PN^2 + a^2 = 2a^2 + 2a^2\)
\(\therefore\) \(PN^2 + a^2 = 4a^2\)
\(\therefore\) \(PN^2 = 3a^2\)
\(\therefore\) \(PN = \sqrt{3}a\) (ii) [Taking square root of both sides]
\(\therefore\) \(PM = PN = \sqrt{3}a\) [From (i) and (ii)]
In simple words: By applying Apollonius theorem to \(\triangle\)PMR with median PQ, and using the given lengths (QM=QR=a, PQ=a, PR=a), we find \(PM^2 = 3a^2\), so \(PM = \sqrt{3}a\). Similarly, applying Apollonius theorem to \(\triangle\)PNQ with median PR, we find \(PN^2 = 3a^2\), so \(PN = \sqrt{3}a\). Thus, \(PM = PN = \sqrt{3}a\).

🎯 Exam Tip: For problems involving medians and side lengths in triangles, Apollonius theorem is often the key. Clearly label triangles and medians, then apply the theorem carefully for each relevant triangle.

Question 9. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
Answer: Given: JABCD is a parallelogram, diagonals AC and BD intersect at point M.
To prove: \(AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + AD^2\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD है जिसके विकर्ण AC और BD बिंदु M पर प्रतिच्छेद करते हैं। इस आकृति का उपयोग यह सिद्ध करने के लिए किया गया है कि इसके विकर्णों के वर्गों का योग इसकी भुजाओं के वर्गों के योग के बराबर होता है।
Solution:
Proof:
JABCD is a parallelogram.
\(\therefore\) \(AB = CD\) and \(BC = AD\) (i) [Opposite sides of a parallelogram]
\(AM = \frac{1}{2} AC\) and \(BM = \frac{1}{2} BD\) (ii) [Diagonals of a parallelogram bisect each other]
\(\therefore\) M is the midpoint of diagonals AC and BD. (iii)
In \(\triangle\)ABC.
seg BM is the median. [From (iii)]
\(AB^2 + BC^2 = 2AM^2 + 2BM^2\) (iv) [Apollonius theorem]
\(\therefore\) \(AB^2 + BC^2 = 2(\frac{1}{2} AC)^2 + 2(\frac{1}{2} BD)^2\) [From (ii) and (iv)]
\(\therefore\) \(AB^2 + BC^2 = 2 \times \frac{BD^2}{4} + 2 \times \frac{AC^2}{4}\)
\(\therefore\) \(AB^2 + BC^2 = \frac{BD^2}{2} + \frac{AC^2}{2}\)
\(\therefore\) \(2AB^2 + 2BC^2 = BD^2 + AC^2\) [Multiplying both sides by 2]
\(\therefore\) \(AB^2 + AB^2 + BC^2 + BC^2 = BD^2 + AC^2\)
\(\therefore\) \(AB^2 + CD^2 + BC^2 + AD^2 = BD^2 + AC^2\) [From(i)]
i.e. \(AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + AD^2\)
In simple words: In a parallelogram, diagonals bisect each other. Apply Apollonius theorem to \(\triangle\)ABC with median BM, getting \(AB^2 + BC^2 = 2AM^2 + 2BM^2\). Substitute \(AM = \frac{1}{2}AC\) and \(BM = \frac{1}{2}BD\), simplify to get \(2AB^2 + 2BC^2 = AC^2 + BD^2\). Since \(AB=CD\) and \(BC=AD\), we can substitute and show that \(AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + AD^2\).

🎯 Exam Tip: This theorem (Parallelogram law) is a standard proof using Apollonius theorem. Break it down into applying Apollonius to one of the triangles formed by a diagonal and then substituting the relationships between the sides and halves of the diagonals.

Question 10. Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was \(15\sqrt{2}\) km. Find their speed per hour.
Answer: Solution:
Suppose Pranali and Prasad started walking from point A, and reached points B and C respectively after 2 hours.
Distance between them = \(BC = 15\sqrt{2}\) km
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज ABC को दर्शाता है। बिंदु A प्रारंभिक बिंदु है। प्रणाली पूर्व की ओर बिंदु B तक जाती है, और प्रसाद उत्तर की ओर बिंदु C तक जाता है। उनके बीच की दूरी BC \(15\sqrt{2}\) km है। AB और AC तय की गई दूरी का प्रतिनिधित्व करते हैं।
Since, their speed is same, both travel the same distance in the given time.
\(\therefore\) \(AB = AC\)
Let \(AB = AC = x\) km (i)
Now, in \(\triangle\)ABC, \(\angle A = 90^\circ\)
\(\therefore\) \(BC^2 = AB^2 + AC^2\) [Pythagoras theorem]
\(\therefore\) \((15\sqrt{2})^2 = x^2 + x^2\) [From (i)]
\(\therefore\) \(225 \times 2 = 2x^2\)
\(\therefore\) \(450 = 2x^2\)
\(\therefore\) \(x^2 = 225\)
\(\therefore\) \(x = \sqrt{225}\) [Taking square root of both sides]
\(\therefore\) \(x = 15\) km
\(\therefore\) \(AB = AC = 15\)km
Now, speed = \(\frac{\text{distance}}{\text{time}}\)
\(\therefore\) \(=\frac{15}{2}\)
\(\therefore\) \(= 7.5\) km/hr
\(\therefore\) The speed of Pranali and Prasad is 7.5 km/hour.
In simple words: Since they walk at the same speed, the distances covered (AB and AC) are equal, let's call it \(x\). Walking East and North forms a right angle at their starting point (A). The distance between them (\(15\sqrt{2}\) km) is the hypotenuse. Using Pythagoras theorem, \(x^2 + x^2 = (15\sqrt{2})^2 \implies 2x^2 = 225 \times 2 \implies x^2 = 225 \implies x = 15\) km. Speed = distance/time = \(15 \text{ km} / 2 \text{ hours} = 7.5\) km/hr.

🎯 Exam Tip: Visualizing the problem as a right-angled triangle is crucial. Remember that directions like East-North are perpendicular. Clearly define variables and apply the Pythagorean theorem, followed by the speed formula.

Question 11. In \(\triangle\)ABC, \(\angle BAC = 90^\circ\), seg BL and seg CM are medians of \(\triangle\)ABC. Then prove that \(4 (BL^2 + CM^2) = 5 BC^2\).
Answer: Given: \(\angle BAC = 90^\circ\)
seg BL and seg CM are the medians.
To prove: \(4(BL^2 + CM^2) = 5BC^2\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज ABC है, जहाँ कोण A समकोण है। BL और CM त्रिभुज की माध्यिकाएँ हैं। L, AC का मध्यबिंदु है, और M, AB का मध्यबिंदु है। हमें सिद्ध करना है कि \(4(BL^2 + CM^2) = 5BC^2\)।
Solution:
Proof:
In \(\triangle\)BAL, \(\angle BAL = 90^\circ\) [Given]
\(\therefore\) \(BL^2 = AB^2 + AL^2\) (i) [Pythagoras theorem]
In \(\triangle\)CAM, \(\angle CAM = 90^\circ\) [Given]
\(\therefore\) \(CM^2 = AC^2 + AM^2\) (ii) [Pythagoras theorem]
\(\therefore\) \(BL^2 + CM^2 = AB^2 + AC^2 + AL^2 + AM^2\) (iii) [Adding (i) and (ii)]
Now, \(AL = \frac{1}{2} AC\) and \(AM = \frac{1}{2} AB\) (iv) [seg BL and seg CM are the medians]
\(\therefore\) \(BL^2 + CM^2\)
\(\therefore\) \(= AB^2 + AC^2 + (\frac{1}{2} AC)^2 + (\frac{1}{2} AB)^2\) [From (iii) and (iv)]
\(\therefore\) \(= AB^2 + AC^2 + \frac{AC^2}{4} + \frac{AB^2}{4}\)
\(\therefore\) \(= AB^2 + \frac{AB^2}{4} + AC^2 + \frac{AC^2}{4}\)
\(\therefore\) \(= \frac{5AB^2}{4} + \frac{5AC^2}{4}\)
\(\therefore\) \(BL^2 + CM^2 = \frac{5}{4} (AB^2 + AC^2)\)
\(\therefore\) \(4(BL^2 + CM^2) = 5(AB^2 + AC^2)\) (v)
In \(\triangle\)BAC, \(\angle BAC = 90^\circ\) [Given]
\(\therefore\) \(BC^2 = AB^2 + AC^2\) (vi) [Pythagoras theorem]
\(\therefore\) \(4(BL^2 + CM^2) = 5BC^2\) [From (v) and (vi)]
In simple words: Apply Pythagoras theorem to \(\triangle\)BAL to get \(BL^2 = AB^2 + AL^2\) and to \(\triangle\)CAM to get \(CM^2 = AC^2 + AM^2\). Since L and M are midpoints, \(AL = \frac{1}{2}AC\) and \(AM = \frac{1}{2}AB\). Summing these equations and substituting for AL and AM yields \(BL^2 + CM^2 = \frac{5}{4}(AB^2 + AC^2)\). Since \(\triangle\)ABC is right-angled at A, \(BC^2 = AB^2 + AC^2\) by Pythagoras. Substituting this, we prove \(4(BL^2 + CM^2) = 5BC^2\).

🎯 Exam Tip: This is a common proof combining Pythagoras theorem with the properties of medians in a right-angled triangle. Break down the problem into smaller right triangles and express their hypotenuses (medians) in terms of other sides.

Question 12. Sum of the squares of adjacent sides of a parallelogram is 130 cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.
Answer: Solution:
Let JABCD be the given
parallelogram and its diagonals AC and BD intersect at point M.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD को दर्शाता है जिसके विकर्ण AC और BD बिंदु M पर प्रतिच्छेद करते हैं। बिंदु A, B, C, D चतुर्भुज के शीर्ष हैं और M विकर्णों का प्रतिच्छेदन बिंदु है।
.: AB2 + AD2 = 130cm, BD = 14cm
MD = \(\frac{1}{2}\) BD (i) [Diagonals of a parallelogram bisect each other]
= \(\frac{1}{2}\) x 14 = 7 cm
In ∆ABD, seg AM is the median. [From (i)]
.: AB2 + = 2AM² + 2MD² [Apollonius theorem]
.: 130 = 2 AM² + 2(7)2
.: 65 = AM² +49 [Dividing both sides by 2]
.: AM² = 65 - 49
.: AM² = 16 [Taking square root of both sides]
.: AM = \(\sqrt{16}\)
= 4cm
Now, AC =2 AM [Diagonals of a parallelogram bisect each other]
2 × 4 = 8 cm
.: The length of the other diagonal of the parallelogram is 8 cm.
In simple words: We used Apollonius theorem, which relates the sides of a triangle to the length of a median, to find the length of the diagonal. We first found half the length of one diagonal, then applied the theorem to determine the other half.

🎯 Exam Tip: Remember Apollonius theorem for problems involving medians and side lengths in triangles, especially within parallelograms.

Question 13. In ∆ABC, seg AD \(\perp\) seg BC and DB = 3 CD. Prove that: 2 AB2 = 2 AC2 + BC2.
Answer: Given: seg AD \(\perp\) seg BC
DB = 3CD
To prove: 2AB2 = 2AC2 + BC2

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसमें शीर्ष A से आधार BC पर एक लंब AD खींचा गया है। बिंदु D, BC पर स्थित है, जिससे दो समकोण त्रिभुज ADB और ADC बनते हैं।
Solution:
DB = 3CD (i) [Given]
In ∆ADB, \(\angle\)ADB = 90° [Given]
.: AB2 = AD2 + DB2 [Pythagoras theorem]
.: AB2 = AD² + (3CD)2 [From (i)]
.: AB2 = AD2 + 9CD² (ii)
In ∆ADC, \(\angle\)ADC = 90° [Given]
.: AC2 = AD2 + CD2 [Pythagoras theorem]
.: AD2 = AC2 - CD² (iii)
AB2 = AC2 - CD² + 9CD² [From (ii) and(iii)]
.: AB2 = AC2 + 8CD² (iv)
CD + DB = BC [C - D - B]
.: CD + 3CD = BC [From (i)]
.: 4CD = BC
.: CD = \(\frac{BC}{4}\) (v)
AB2 = AC2 + 8(\(\frac{BC}{4}\))2 [From (iv) and (v)]
.: AB2 = AC2 + 8 x \(\frac{BC^2}{16}\)
.: AB2 = AC2 + \(\frac{BC^2}{2}\)
.: 2AB2 = 2AC2 + BC2 [Multiplying both sides by 2]
In simple words: We used the Pythagorean theorem in both right-angled triangles formed by the altitude, then substituted the given relationship between DB and CD to derive the required equation relating the sides of the triangle.

🎯 Exam Tip: For proofs, clearly state the given, to prove, and construction steps. Use the Pythagorean theorem as a primary tool when dealing with right-angled triangles.

Question 14. In an isosceles triangle, length of the congruent sides is 13 em and its.base is 10 cm. Find the distance between the vertex opposite to the base and the centroid.
Answer: Given: ∆ABC is an isosceles triangle.
G is the centroid.
AB = AC = 13 cm, BC = 10 cm.
To find: AG
Construction: Extend AG to intersect side BC at D, B - D - C.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समद्विबाहु त्रिभुज ABC को दर्शाता है जहाँ AB और AC बराबर भुजाएँ हैं। शीर्ष A से आधार BC पर एक रेखा AD खींची गई है जो BC को D पर प्रतिच्छेद करती है। बिंदु G त्रिभुज का केंद्रक है जो AD पर स्थित है।
Solution:
Centroid G of ∆ABC lies on AD
.: seg AD is the median. (i)
.: D is the midpoint of side BC.
.: DC = \(\frac{1}{2}\) BC
= \(\frac{1}{2}\) x 10 = 5
In ∆ABC, seg AD is the median. [From (i)]
.: AB2 + AC2 = 2 AD² + 2 DC2 [Apollonius theorem]
.: 132 + 132 = 2 AD² + 2 (5)2
.: 2 x 132 = 2 AD² + 2 × 25
.: 169 = AD² + 25 [Dividing both sides by 2]
.: AD² = 169 - 25
.: AD² = 144
.: AD = \(\sqrt{144}\) [Taking square root of both sides]
= 12 cm
We know that, the centroid divides the median in the ratio 2: 1.
.: \(\frac{AG}{GD}\) = \(\frac{2}{1}\)
.: \(\frac{AG}{GD}\) = \(\frac{2}{1}\) [By invertendo]
.: \(\frac{AG}{GD+AG}\) = \(\frac{2}{1+2}\) [By componendo]
.: \(\frac{AG}{AD}\) = \(\frac{2}{3}\) [A - G - D]
.: \(\frac{AG}{12}\) = \(\frac{2}{3}\)
.: AG = \(\frac{12 \times 2}{3}\)
= 8cm
.: The distance between the vertex opposite to the base and the centroid is 8 cm.
In simple words: First, we used Apollonius theorem to find the length of the median AD. Then, knowing that the centroid divides the median in a 2:1 ratio, we calculated the distance AG.

🎯 Exam Tip: Remember the property that a centroid divides the median in a 2:1 ratio. This is crucial for solving problems involving centroids and medians.

Question 15. In a trapezium ABCD, seg AB || seg DC, seg BD \(\perp\) seg AD, seg AC \(\perp\) seg BC. If AD = 15, BC = 15 and AB = 25, find A (JABCD).
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समलंब चतुर्भुज ABCD को दर्शाता है, जिसमें AB और DC समांतर भुजाएँ हैं। AD और BC असमांतर भुजाएँ हैं, जिनकी लंबाई 15 इकाई है। आधार AB की लंबाई 25 इकाई है।
Construction: Draw seg DE \(\perp\) seg AB, A - E-B
and seg CF \(\perp\) seg AB, A - F- B.
Solution:
In ∆ACB, \(\angle\)ACB = 90° [Given]
.: AB2 = AC2 + BC2 [Pythagoras theorem]
.: 252 = AC2 + 152
.: AC2 = 625 - 225
= 400

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समलंब चतुर्भुज ABCD को दर्शाता है जिसमें AB और DC समांतर हैं। शीर्ष D से AB पर एक लंब DE और शीर्ष C से AB पर एक लंब CF खींचा गया है, जिससे दो समकोण त्रिभुज ADE और BCF बनते हैं और बीच में एक आयत ECDF बनता है।
.: AC = \(\sqrt{400}\) [Taking square root of both sides]
= 20 units
Now, A(∆ABC) = \(\frac{1}{2}\) x BC x AC
Also, A(∆ABC) = \(\frac{1}{2}\) x AB x CF
.: \(\frac{1}{2}\) x BC x AC = \(\frac{1}{2}\) x AB x CF
.: BC x AC = AB x CF
.: 15 x 20 = 25 × CF
.: CF = \(\frac{15 \times 20}{25}\) = 12 units
In ∆CFB, \(\angle\)CFB 90° [Construction]
.: BC2 = CF2 + FB2 [Pythagoras theorem]
.: 152 = 122 + FB2
.: FB2 = 225 - 144
.: FB2 = 81
.: FB = \(\sqrt{81}\) [Taking square root of both sides]
= 9 units
Similarly, we can show that, AE = 9 units
Now, AB = AE + EF + FB [A - E - F, E - F - B]
.: 25 = 9 + EF + 9
.: EF = 25 - 18 = 7 units
In JCDEF,
seg EF || seg DC [Given, A - E - F, E - F - B]
seg ED || seg FC [Perpendiculars to same line are parallel]
.: JCDEF is a parallelogram.
.: DC = EF 7 units [Opposite sides of a parallelogram]
A(JABCD) = \(\frac{1}{2}\) x CF x (AB + CD)
= \(\frac{1}{2}\) x 12 x (25 + 7)
= \(\frac{1}{2}\) x 12 x 32
.: A(JABCD) = 192 sq. units
In simple words: We used the given right angles and side lengths to find the diagonal AC using Pythagoras theorem. Then, by equating two expressions for the area of triangle ABC, we found the height CF. Finally, we found the lengths of segments AE, FB, and EF to calculate the length of DC, which allowed us to calculate the area of the trapezium.

🎯 Exam Tip: When dealing with areas of complex polygons like trapeziums, break them down into simpler shapes (triangles and rectangles). Use the area formula for triangles and Pythagoras theorem to find unknown lengths.

Question 16. In the adjoining figure, APQR is an equilateral triangle. Point S is on seg QR such that QS = \(\frac{1}{3}\) QR. Prove that: 9 PS2 = 7 PQ².
Answer: Given: APQR is an equilateral triangle.
QS = \(\frac{1}{3}\) QR
To prove: 9PS2 = 7PQ2

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समबाहु त्रिभुज PQR को दर्शाता है। भुजा QR पर एक बिंदु S इस प्रकार स्थित है कि QS, QR का एक-तिहाई है। बिंदु P से QR पर एक लंब PT खींचा गया है, जिससे दो समकोण त्रिभुज PTQ और PTS बनते हैं।
Solution:
Proof:
APQR is an equilateral triangle [Given]
.: \(\angle\)P = \(\angle\)Q = \(\angle\)R = 60° (i) [Angles of an equilateral triangle]
PQ = QR = PR (ii) [Sides of an equilateral triangle]
In APTS, \(\angle\)PTS = 90° [Given]
PS2 = PT2 + ST2 (iii) [Pythagoras theorem]
In APTQ,
\(\angle\)PTQ = 90° [Given]
\(\angle\)PQT = 60° [From (i)]
.: \(\angle\)QPT = 30° [Remaining angle of a triangle]
.: APTQ is a 30° - 60° - 90° triangle
.: PT = \(\frac{\sqrt{3}}{2}\) PQ (iv) [Side opposite to 60°]
QT = \(\frac{1}{2}\) PQ (v) [Side opposite to 30°]
QS + ST = QT [Q - S - T]
.: \(\frac{1}{3}\) QR + ST = \(\frac{1}{2}\) PQ [Given and from (v)]
.: \(\frac{1}{3}\) PQ + ST = \(\frac{1}{2}\) PQ [From (ii)]
.: ST = \(\frac{PQ}{2}\) - \(\frac{PQ}{3}\)
.: ST = \(\frac{3PQ-2PQ}{6}\)
.: ST = \(\frac{PQ}{6}\) (vi)
PS2 = \((\frac{\sqrt{3}}{2} PQ)^2\) + \((\frac{PQ}{6})^2\) [From (iii), (iv) and (vi)]
.: PS2 = \(\frac{3PQ^2}{4}\) + \(\frac{PQ^2}{36}\)
.: PS2 = \(\frac{27PQ^2}{36}\) + \(\frac{PQ^2}{36}\)
.: PS2 = \(\frac{28PQ^2}{36}\)
.: PS2 = \(\frac{7}{9}\) PQ2
.: 9PS2 = 7 PQ2
In simple words: By constructing an altitude from P to QR, we created a 30-60-90 triangle. We expressed PT and QT in terms of PQ. Using the given QS relationship, we found ST in terms of PQ. Finally, we applied the Pythagorean theorem to triangle PTS and substituted the expressions to prove the required relation.

🎯 Exam Tip: For geometry proofs, drawing auxiliary lines (like altitudes) can create special triangles (e.g., 30-60-90) that simplify calculations and help relate different side lengths.

Question 17. Seg PM is a median of APQR. If PQ = 40, PR = 42 and PM = 29, find QR.
Answer: Solution:
In APQR, seg PM is the median. [Given]
.: M is the midpoint of side QR.
.: PQ2 + PR2 = 2 PM² + 2 MR2 [Apollonius theorem]

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR को दर्शाता है, जिसमें शीर्ष P से भुजा QR के मध्यबिंदु M तक एक माध्यिका PM खींची गई है। यह माध्यिका त्रिभुज को दो छोटे त्रिभुजों में विभाजित करती है।
.: 402 + 422 = 2 (29)2 + 2 MR2
.: 1600 + 1764 = 2 (841) + 2 MR2
.: 3364 = 2 (841) + 2 MR2
.: 1682 = 841 +MR2 [Dividing both sides by 2]
.: MR2 = 1682 - 841
.: MR2 = 841
.: MR = \(\sqrt{841}\) [Taking square root of both sides]
= 29 units
Now, QR = 2 MR [M is the midpoint of QR]
= 2 × 29
.: QR = 58 units
In simple words: We applied Apollonius theorem, which relates the lengths of the sides of a triangle to the length of its median. By substituting the given values for PQ, PR, and PM, we calculated the length of MR, and then doubled it to find QR.

🎯 Exam Tip: Apollonius theorem is a direct way to solve problems involving medians and side lengths. Ensure you correctly identify the median and the sides it relates to.

Question 18. Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM.
Answer: Solution:
In ∆ABC, seg AM is the median. [Given]
.: M is the midpoint of side BC.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसमें शीर्ष A से भुजा BC के मध्यबिंदु M तक एक माध्यिका AM खींची गई है। यह माध्यिका त्रिभुज को दो छोटे त्रिभुजों में विभाजित करती है।
.: MC = \(\frac{1}{2}\) BC
= \(\frac{1}{2}\) x 24 = 12 units
Now, AB2 + AC2 = 2 AM² + 2 MC² [Apollonius theorem]
.: 222 + 342 = 2 AM² + 2 (12)2
.: 484 + 1156 = 2 AM² + 2 (144)
.: 1640 = 2 AM² + 2 (144)
.: 820 = AM² + 144 [Dividing both sides by 2]
.: AM² = 820 - 144
.: AM² = 676
.: AM = \(\sqrt{676}\) [Taking square root of both sides]
.: AM = 26 units
In simple words: We used Apollonius theorem, which connects the lengths of the sides of a triangle to its median. By plugging in the given values for sides AB, AC, and half of BC (MC), we directly calculated the length of the median AM.

🎯 Exam Tip: Clearly write down the Apollonius theorem and substitute the given values carefully to avoid calculation errors. Knowing the perfect squares helps in finding square roots quickly.