Q) In the given figure, ABCD is a parallelogram. AE divides the line segment BD in the ratio 1:2. If BE = 1.5 cm, then find the length of BC. Ans: Since AD ǁ BC, and EA cuts these lines, ∠DAE = ∠AEB or (∠OEB) Similarly, Line DB cuts these parallel lines, ∠ADB = ∠DBC […]

# August 2023

Q) If tan θ = , then show that = Ans: Given that, tan θ = cot θ = √7 Let’s start from numerator of LHS: cosec2 θ – sec2 θ = (1 + cot2 θ) – (1 + tan2 θ) = cot2 θ – tan2 θ = (√7)2 – = 7 – = …………………

Q) If sin θ + sin2 θ = 1, then prove that cos2 θ + cos4 θ = 1. Ans: Given that sin θ + sin2 θ = 1 sin θ = 1 – sin2 θ LHS: cos2 θ + cos4 θ = (1- sin2 θ) + (1- sin2 θ)2 = sin θ + (sin

If sin θ + sin^2 θ = 1, then prove that cos^2 θ + cos^4 θ = 1. Read More »

Q) India meteorological department observes seasonal and annual rainfall every year in different sub-divisions of our country. It helps them to compare and analyse the results. The table given below shows sub-division wise seasonal (monsoon) rainfall (mm) in 2018: Based on the above information, answer the following questions: (I) Write the modal class. (II) Find

Q) In the given figure, CD and RS are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR then prove that: (i) Δ ADC ~ Δ PSR (ii) AD x PR = AC x PS Ans: (i) Its given that: CD is median of ΔABC RS is median of ΔPQR ΔABC∼ΔPQR AB =

Q) PA, QB and RC are each perpendicular to AC. If AP = x, QB = z, RC = Y, AB = a and BC = b, then prove that + = Ans: Let’s look at Δ CQB & Δ CPA, By AA similarity theorem, ∠PAC = ∠QBC (perpendicular to AC) ∠PCA = ∠QCB

Q) A ladder set against a wall at an angle 45° to the ground. If the foot of the ladder is pulled away from the wall through a distance of 4 m, its top slides a distance of 3 m down the wall making an angle 30° with the ground. Find the final height of

Q)Solve the pair of equations x = 5 and y = 7 graphically. Ans: Plot both the equations on the graph. Increase till intersection point (5,7)

Solve the pair of equations x = 5 and y = 7 graphically. Read More »

Q) Prove that (cosec A – sin A) (sec A- cos A) = Ans: Let’s start with LHS: (cosec A – sin A) (sec A- cos A) = ( – A) ( – A) = () () = () () = cos A . sin A Now, Let’s solve RHS: = = =

Prove that (cosecA – sinA) (secA-cosA) = 1/(cotA+tanA) Read More »

Q) The centre of a circle is (2a, a-7). Find the values of ‘a’ if the circle passes through the point (11,-9). Radius of the circle is 5√2cm. Ans: Let’s find out radius of the circle by co-ordinate method. Therefore, (2a – 11)2 + (a -7 + 9)2 = (5√2 )2 or (4a2 + 121