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This periodic test covers three core chapters: Triangles (similarity, Basic Proportionality Theorem, criteria for similarity), Coordinate Geometry (distance formula, section formula, midpoint, collinearity), and Introduc…
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Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of a triangle PQR (see figure). Show that triangle ABC is similar to triangle PQR.
Use SSS similarity: since AB/PQ = BC/QR = AD/PM, and D, M are midpoints, we get AB/PQ = BC/QR = AC/PR, hence △ABC ~ △PQR.
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In the given figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
Apply Basic Proportionality Theorem in △BAE (DF∥AE gives BF/FE = BD/DA) and in △BAC (DE∥AC gives BE/EC = BD/DA); hence BF/FE = BE/EC.
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Prove the following trigonometric identity: cosθ/(1 + sinθ) + (1 + sinθ)/cosθ = 2secθ
Take LCM: [cos²θ + (1+sinθ)²] / [cosθ(1+sinθ)] = [cos²θ + 1 + 2sinθ + sin²θ] / [cosθ(1+sinθ)] = [2 + 2sinθ] / [cosθ(1+sinθ)] = 2/cosθ = 2secθ.
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State 'Basic Proportionality Theorem' and use it to prove the following: In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that AO/BO = CO/DO as shown in the given figure. Prove that ABCD is a trapezium.
BPT: In a triangle, a line parallel to one side divides the other two sides proportionally. Given AO/BO = CO/DO ⇒ AO/CO = BO/DO. Consider △AOB and △COD: ∠AOB = ∠COD (vertically opposite) and AO/CO = BO/DO, so △AOB ~ △COD (SAS similarity). Hence ∠OAB = ∠OCD, so AB ∥ CD (alternate angles equal). Therefore ABCD is a trape…
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Prove that: tan³θ / (1+tan²θ) + cot³θ / (1+cot²θ) = sec θ cosec θ - 2 sin θ cos θ.
Simplify LHS: tan³θ/sec²θ + cot³θ/cosec²θ = sin³θ/cosθ + cos³θ/sinθ = (sin⁴θ + cos⁴θ)/(sinθ cosθ) = [(sin²θ + cos²θ)² - 2sin²θcos²θ]/(sinθ cosθ) = (1 - 2sin²θcos²θ)/(sinθ cosθ) = secθcosecθ - 2sinθcosθ.
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Prove that (1 + cosecA)/cosecA = cos²A/(1 - sinA)
LHS = (1 + 1/sinA)/(1/sinA) = (sinA + 1)/1 = 1 + sinA; RHS = cos²A/(1 - sinA) = (1 - sin²A)/(1 - sinA) = 1 + sinA; hence proved.
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In the given figure, MNOP is a parallelogram and AB || MP. Prove that QC || PO.
Use BPT in triangles: In △ABN, MP∥AB gives ratios; in △ABC, use converse of BPT to show QC∥PO. (Full proof requires applying BPT in two triangles and equating ratios.)
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The centre of a circle is (2a, a - 7). Find the values of a, if the circle passes through the point (11, -9) and has diameter 10√2 units.
Radius = 5√2; distance from centre to (11,-9) = 5√2 gives (11-2a)² + (-9-a+7)² = 50 → solve to get a = 3 or a = 5.
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Find the ratio in which the segment joining the points (2, -5) and (5,3) is divided by x-axis. Also, find coordinates of the point on x-axis.
Ratio is 5:3 (using y-coordinate 0 in section formula); point is (35/8, 0).
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Using distance formula, prove that the points A(3,1), B(6,4) and C(8,6) are collinear.
AB = √18 = 3√2, BC = √8 = 2√2, AC = √50 = 5√2; since AB + BC = AC, points are collinear.
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If sin3 A = 1, then find the value of cos2 A - tan²45°.
sin3A = 1 ⇒ 3A = 90° ⇒ A = 30°; cos60° - tan²45° = 1/2 - 1 = -1/2.
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Point R(4, y) divides the line segment OP in a certain ratio, find the ratio. Hence, find the value of y.
Using section formula: 4 = (k·8)/(k+1) ⇒ k=1 (ratio 1:1); then y = (1·6 + 1·0)/2 = 3.
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Using distance formula, show that PQ/OS = 2/3.
PQ = √[(12-8)² + (2-6)²] = √32 = 4√2; OS = √[(-6)² + 6²] = √72 = 6√2; hence PQ/OS = 4√2/6√2 = 2/3.
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If △ PQR ~ △ ABC; PQ = 6 cm, AB = 8 cm and the perimeter of △ ABC is 36 m, then the perimeter of △ PQR is
27 cm (since ratio of perimeters = ratio of corresponding sides = 6/8, so perimeter of PQR = 36 × 6/8 = 27 cm)
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PQ is a diameter of a circle with centre O(2, -4). If the coordinates of the point P are (-4,5), then the coordinates of the point Q will be:
(8, -13) (option d) using midpoint formula: O is midpoint of PQ.
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The distance between the points (-6,9) and (2,7) is:
2√17 (option c) using distance formula: √[(2+6)² + (7-9)²] = √(64+4) = √68 = 2√17.
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Which of the following is a trigonometric identity?
sec²θ = 1 + tan²θ (option b) is the standard Pythagorean identity.
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(sinθ + cosθ)² + (sinθ - cosθ)² =
2 (option b) because expansion gives 2(sin²θ + cos²θ) = 2.
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Gurveer and Arushi built a robot that can paint a path as it moves on a graph paper. Some co-ordinate of points are marked on it. It starts from (0,0), moves to the points listed in order (in straight lines) and ends at (0, 0). Arushi entered the points P(8,6), Q(12,2) and S(-6,6) in order. The path drawn by robot is shown in the figure. Based on the above, answer the following questions: i. Determine the distance OP.
OP = √(8² + 6²) = √100 = 10 units.
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In △ DEF and △ PQR, it is given that ∠ D = ∠ Q and ∠ R = ∠ E, then which of the following is not true?
Option (c) DE/PQ = EF/RP is not true; correct correspondence gives DE/QR = EF/RP = DF/QP.
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What values of x and y will make △ ABC similar to △ QRP in the figures given below?
x = 6, y = 5 (option d) by SSS similarity: 2/4 = 2.5/y = 3/x → y=5, x=6.
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The mid-point of the line segment joining the points P(-4,5) and Q(4,6) lies on:
y-axis (option c) because midpoint is (0, 5.5).
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If tanθ = x/y, then cos θ is equal to
y / √(x²+y²) (option c) using right triangle: cosθ = adjacent/hypotenuse = y/√(x²+y²).
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Assertion (A): sec²θ - tan²θ = cosec²θ - cot²θ. Reason (R): As 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ.
Option (a) Both A and R are true and R is the correct explanation of A.
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QS is represented by equation 2x + 9y = 42. Find the co-ordinates of the point where it intersects y-axis.
On y-axis, x=0 ⇒ 9y = 42 ⇒ y = 14/3, so point is (0, 14/3).
The full exam-length answer is in the app.
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