Q) Find the area of the unshaded region shown in the given figure. Ans: Let’s redraw the diagram: As we can see in the diagram, in the center area: diameter of semicircle (2R) = side of the inside square (S) or S = 2R ………………. (i) Also we see that Side of larger square = Gap […]
Find the area of the unshaded region shown in the given figure. Read More »
Q) With vertices A, B and C of Δ ABC as centres, arcs are drawn with radii 14 cm and the three portions of the triangle so obtained are removed. Find the total area removed from the triangle. Ans: We know that the area made by an arc of θ angle is given by = r2
Q) ABCD is a parallelogram. Point P divides AB in the ratio 2:3 and point Q divides DC in the ratio 4:1. Prove that OC is half of OA. Ans: VIDEO SOLUTION STEP BY STEP SOLUTION Given that ABCD is a parallelogram. Therefore, AB ǁ CD and BC ǁ AD Step 1: Since, Point P
Q) A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that area ▲ ABC = 90 cm². Ans: Step 1: Let’s join
Q) In the given figure, ABCD is a parallelogram. BE bisects CD at M and intersects AC at L. Prove that EL = 2BL. Ans: VIDEO SOLUTION STEP BY STEP SOLUTION Given that: BE bisects CD at M, DM = MC Let’s look at Δ ALE and Δ CLB: ∠ ALE = ∠ CLB (vertically
Q) In the given figure, CD is perpendicular bisector of AB. EF is perpendicular to CD. AE intersects CD at G. Prove that CF/CD = FG/DG. Ans: Given that: CD is perpendicular bisector of AB, AD = BD, ∠ CDB = ∠ GDA = 900 EF is perpendicular bisector of CD, ∠ EFC = ∠
In the given figure, CD is perpendicular bisector of AB. EF is perpendicular to CD. Read More »
Q) In the given figure, Δ ABC and ΔDBC are on the same base BC. If AD intersects BC at O, prove that Ans: Let’s draw perpendicular from points A and D on line BC: In Δ AON and Δ DOM, ∠ AON = ∠ DOM (interior angles) ∠ ANO = ∠ DMO (given that AN and
Q) In a Δ PQR, N is a point on PR, such that QN is to PR. If PN x NR = QN2, prove that ∠ PQR = 900. Ans: Ans: Given that, PN x NR = QN2 Therefore, In Δ PNQ and Δ QNR, ∠ QNP = ∠ RNQ (given that QN is to
Q) In the given figure, ∠ ADC = ∠ BCA, prove that Δ ACB ~ Δ ADC. Hence find if AC = 8 cm and AD = 3 cm. Ans: Ans: In Δ ACB and Δ ADC, ∠ ADC = ∠ BCA (given) ∠ A = ∠ A (common angle) Δ ACB ~ Δ
Q) If AD and PM are medians of triangle ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that AB / PQ = AD / PM. Ans: Step 1: Given that, Δ ABC ~ Δ PQR, therefore ∠ B = ∠ Q …………. (i) and …………. (ii) Step 2: Since AD is median