- Basic Proportionality Theorem Or Thales Theorem
- (Prove) If a Line is Drawn Parallel to One Side of a Triangle to Intersect the Other Two Sides in Distinct Points, the Other Two Sides Are Divided in the Same Ratio.
- Converse of Basic Proportionality Theorem Or Thales Theorem - (Motivate) If a Line Divides Two Sides of a Triangle in the Same Ratio, the Line is Parallel to the Third Side
In the given figure, ∠PQR = ∠PST = 90° ,PQ = 5cm and PS = 2cm.
(i) Prove that ΔPQR ~ ΔPST.
(ii) Find Area of ΔPQR : Area of quadrilateral SRQT.
ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ ∥ BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP : AB.
The following figure shows a triangle PQR in which XY is parallel to QR. If PX : XQ = 1 : 3 and QR = 9 cm. find the length of XY.
Further, if the area of Δ PXY =x cm2 ; find, in terms of x the area of:
(i) triangle PQR (ii) trapezium XQRY
In the adjoining figure, ABC is a right angled triangle with ∠BAC = 90°.
1) Prove ΔADB ~ ΔCDA.
2) If BD = 18 cm CD = 8 cm Find AD.
3) Find the ratio of the area of ΔADB is to an area of ΔCDA.
In the given triangle P, Q and R are the mid points of sides AB, BC and AC respectively. Prove that triangle PQR is similar to triangle ABC.
In the given figure ΔABC and ΔAMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
1) Prove ΔABC ~ ΔAMP
2) Find AB and BC.
In ΔPQR, MN is parallel to QR and `(PM)/(MQ) = 2/3`
1) Find `(MN)/(QR)`
2) Prove that ΔOMN and ΔORQ are similar.
3) Find, Area of ΔOMN : Area of ΔORQ
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- English 1 (English Language) 2006 to 2019
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