CBSE Class 9 Mathematics Plane Geometry I MCQs Set 21

Answer: (c) 5.4 cm

Explanation:
1. For similar triangles, the ratio of corresponding sides = ratio of perimeters = 25 : 15 = 5 : 3.
2. Required side = 9 × 3/5 = 5.4 cm.

Teacher's Note:
1. Perimeters are in the same ratio as corresponding sides.
2. Set up the proportion carefully: first triangle ↔ 25.

Answer: (b) \( 2\sqrt{5} \) cm

Explanation:
1. Triangle ADE ~ triangle ABC.
2. (DE/BC)² = area ratio = 1/5, so DE = 10/√5.
3. DE = 2√5 cm.

Teacher's Note:
1. Take the square root of the area ratio to get the ratio of sides.
2. Rationalise 10/√5 = 2√5.

Answer: (c) 100 cm²

Explanation:
1. Ratio of areas = (EF/BC)² = (5/4)² = 25/16.
2. Area of DEF = 64 × 25/16 = 100 cm².

Teacher's Note:
1. Square the ratio of corresponding sides to get the ratio of areas.
2. Be careful to put the new triangle on top.

Answer: (b) a + b

Explanation:
1. Draw CE ∥ DA with E on AB. Then AECD is a parallelogram: AE = CD = b and CE = AD = a.
2. ∠CEB = ∠DAB = 180° − ∠ADC = 180° − 2∠B, so in triangle CEB, ∠ECB = 180° − ∠B − ∠CEB = ∠B.
3. So EB = EC = a, and AB = AE + EB = a + b.

Teacher's Note:
1. Drawing a line parallel to a leg turns a trapezium into a parallelogram plus a triangle.
2. The isosceles triangle CEB gives EB = a.

Answer: (c) 1 : 5

Explanation:
1. Let the area of triangle ABD = S.
2. Area of △DEF = (DE/DB) × (DF/DA) × S = ½ × ⅓ × S = S/6.
3. Area of ABEF = S − S/6 = 5S/6.
4. Ratio = 1 : 5.

Teacher's Note:
1. Two triangles sharing an angle have areas in the ratio of the products of the adjacent sides.
2. Quadrilateral ABEF = triangle ABD minus triangle DEF.

Answer: (b) ⅛ cm²

Explanation:
1. The side of the square is 2 cm, so DC = 2 and E is 2 cm above DC.
2. F is the mid-point of DE, so its height above DC is 1. G is the mid-point of CF: height ½. H is the mid-point of DG: height ¼. K is the mid-point of CH: height ⅛.
3. Area of △KDC = ½ × 2 × ⅛ = ⅛ cm².

Teacher's Note:
1. Triangles on the same base DC have areas proportional to heights.
2. Each mid-point halves the height above DC.

Answer: (a) 1/2

Explanation:
1. Let the rectangle have sides a and b. Each corner of the rectangle gets a right-angled triangle with legs ka and (1 + k)b, or kb and (1 + k)a.
2. Total extra area = 2 × ½ × ka(1 + k)b + 2 × ½ × kb(1 + k)a = 2k(1 + k)ab.
3. So ab[1 + 2k(1 + k)] = 2.5 ab, giving 2k² + 2k − 1.5 = 0, that is 4k² + 4k − 3 = 0.
4. (2k − 1)(2k + 3) = 0, and k = ½ (k is positive).

Teacher's Note:
1. Sum the four corner triangles added to the rectangle.
2. Solve the quadratic and reject the negative root.

Answer: (c) 91

Explanation:
1. The segment joining the mid-points of the diagonals of a trapezium equals half the difference of the parallel sides.
2. (97 − s)/2 = 3, so s = 91.

Teacher's Note:
1. Memorise: length = ½ (longer base − shorter base).
2. The segment lies along the mid-line of the trapezium.

Answer: (b) only II

Explanation:
1. In a parallelogram, a² + b² = 2(p² + q²) = 2(9 + 16) = 50, so II is determined.
2. The individual values of a and b depend on the angle between the sides, so a + b and a³ + b³ change as the angle changes.

Teacher's Note:
1. The parallelogram law fixes only the sum of the squares of the diagonals.
2. Check by taking a rectangle (a = b = 5) and a thinner parallelogram.

Answer: (c) 50

Explanation:
1. EF = ½ (DC − AB), since DC is the longer base.
2. 5 = ½ (60 − AB), so AB = 50.

Teacher's Note:
1. The segment joining the mid-points of the diagonals is parallel to the bases.
2. The result uses the difference of the bases.

Answer: (c) \( \sqrt{5} + 1 \)

Explanation:
1. Let AQ = x, so AB = x + 2, and let AD = BC = h. Area of QBC = ½ × 2 × h = h.
2. Area of PCD = ½ × AB × DP = h gives DP = 2h/(x + 2); so AP = h − 2h/(x + 2) = hx/(x + 2).
3. Area of PAQ = ½ × x × AP = x²h/(2(x + 2)) = h, so x² = 2x + 4.
4. x² − 2x − 4 = 0 gives x = 1 + √5 (positive root).

Teacher's Note:
1. Write each area in terms of x and h; h cancels.
2. Choose the positive root of the quadratic.

Answer: (c) 12 units

Explanation:
1. The three mid-point joins divide the triangle into four congruent triangles.
2. Area of LMN = 48 ÷ 4 = 12 square units.

Teacher's Note:
1. The medial triangle has exactly one-fourth the area of the original.
2. All four small triangles are congruent.

Answer: (b) \( 5\sqrt{10} \) cm

Explanation:
1. Parallelogram law: d₁² + d₂² = 2(15² + 10²) = 650.
2. d₂² = 650 − 400 = 250, so d₂ = 5√10 cm.

Teacher's Note:
1. The sum of squares of the diagonals equals twice the sum of squares of the sides.
2. Simplify 250 = 25 × 10.

Answer: (b) 4

Explanation:
1. The ratio of corresponding sides is 2 : 1 (same as the ratio of perimeters).
2. The ratio of areas = 2² : 1² = 4 : 1.

Teacher's Note:
1. Areas scale as the square of the linear ratio.
2. Perimeter is a length, so it scales like a side.

Answer: (b) any rectangle

Explanation:
1. Diagonals that bisect each other make it a parallelogram.
2. A parallelogram with equal diagonals is a rectangle.

Teacher's Note:
1. Bisecting diagonals ⇒ parallelogram; equal diagonals ⇒ right angles at the corners.
2. A rhombus has perpendicular, not equal, diagonals.

Answer: (a) parallelogram

Explanation:
1. AP ∥ CQ since AD ∥ BC.
2. AP = ⅓ AD = ⅓ BC = CQ.
3. One pair of opposite sides is equal and parallel, so AQCP is a parallelogram.
4. It need not be a rhombus or rectangle because the angles and side lengths are not specified.

Teacher's Note:
1. One pair of opposite sides equal and parallel is enough for a parallelogram.
2. Do not assume extra properties that are not given.

Answer: (d) 6 : 7 : 8

Explanation:
1. The angles add to 180°, so the sum of the ratio terms must divide 180 to give integer angles.
2. 2 + 3 + 4 = 9 and 180/9 = 20; 3 + 4 + 5 = 12 and 180/12 = 15; 5 + 6 + 7 = 18 and 180/18 = 10 all work.
3. 6 + 7 + 8 = 21 and 180/21 is not an integer, so the angles cannot all be integers.

Teacher's Note:
1. Divide 180° by the sum of the ratio terms.
2. If the quotient is an integer, every angle is an integer.

Answer: (d) x² + y² = (5/4)AC²

Explanation:
1. E is the mid-point of BC and D is the mid-point of AB.
2. x² = AE² = AB² + (BC/2)² and y² = CD² = BC² + (AB/2)².
3. x² + y² = (5/4)(AB² + BC²) = (5/4)AC².

Teacher's Note:
1. Use Pythagoras in the right triangles ABE and CBD.
2. AB² + BC² = AC².

Answer: (d) 31 cm

Explanation:
1. Let the legs be p and q: p² + q² = 41² = 1681 and ½pq = 180, so pq = 360.
2. (p − q)² = p² + q² − 2pq = 1681 − 720 = 961.
3. p − q = 31 cm (the legs are 40 − 9 = 31).

Teacher's Note:
1. Use the identity (p − q)² = (p² + q²) − 2pq.
2. Check with the triple 9-40-41.

Answer: (b) it is an acute-angled triangle

Explanation:
1. Take sides 6k, 8k, 9k. The longest side is 9k.
2. Compare squares: 36 + 64 = 100 > 81.
3. The square of the longest side is less than the sum of the squares of the others, so the largest angle is acute and the triangle is acute-angled.

Teacher's Note:
1. Compare c² with a² + b² using the longest side as c.
2. Angles are not proportional to sides.

Answer: (c) greater than (3/2)BC

Explanation:
1. Let G be the centroid; BG = ⅔ BE and CG = ⅔ CF.
2. In triangle BGC, BG + CG > BC, so ⅔(BE + CF) > BC.
3. Therefore BE + CF > (3/2)BC.

Teacher's Note:
1. The centroid divides each median in the ratio 2 : 1.
2. Apply the triangle inequality in triangle BGC.

Answer: (c) 2003

Explanation:
1. Let AC = x, an integer. The triangle inequality requires 2001 − 1002 < x < 2001 + 1002, that is 999 < x < 3003.
2. x can take the values 1000, 1001, …, 3002.
3. Number of values = 3002 − 1000 + 1 = 2003.

Teacher's Note:
1. The third side lies strictly between the difference and the sum of the other two.
2. Count the integers carefully, including both ends.

Answer: (b) 6 : 4 : 3

Explanation:
1. Side × altitude = 2 × area is constant, so the sides are inversely proportional to the altitudes.
2. Sides ∝ 1/2 : 1/3 : 1/4 = 6 : 4 : 3.

Teacher's Note:
1. Clear fractions by multiplying by 12.
2. The shortest altitude belongs to the longest side.

Answer: (a) only 1

Explanation:
1. By the basic proportionality theorem, AD/BD = AE/CE: (3x − 2)(5x − 3) = (5x − 4)(7x − 5).
2. 15x² − 19x + 6 = 35x² − 53x + 20, so 10x² − 17x + 7 = 0, i.e. (10x − 7)(x − 1) = 0.
3. x = 7/10 gives AE = 5x − 4 = −0.5, which is not a length, so reject it.
4. x = 1 gives AD = 1, AE = 1, BD = 2, CE = 2, which is valid. So x = 1 only.

Teacher's Note:
1. Always check that the solutions make every length positive.
2. The proportionality theorem is applied as AD : DB = AE : EC.