Q) Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles. The height of Vijay’s house if 20m when Vijay’s house casts a shadow 10m long on the ground. At the same time, the tower casts a shadow 50m long on the ground […]
triangles
Q) In the given figure, ∠ CEF = ∠ CFE. F is the midpoint of DC. Prove that . Ans: Step 1: From given conditions: ∵ ∠ CEF = ∠ CFE ∴ CE = CF (sides of opp. Angles) 2. ∵ F is midpoint of CD ∴ FD = CF ∴ CE = CF =
Q) PA and PB are tangents drawn to a circle of centre O from an external point P. Chord AB makes and angle of 30 with the radius at the point of contact. If length of the chord of 6 cm, find the length of the tangent PA and the length of the radius OA. Ans:
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: Given that ABCD is a parallelogram. Therefore, AB ǁ CD and BC ǁ AD Since, Point P divides AB in the ratio 2:3 Therefore, if
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: Let’s join Point A, B,
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