Advertisements
Advertisements
प्रश्न
In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.
Find :
- `(AY)/(YC)`
- `(YC)/(AC)`
- XY
उत्तर
i. Given that XY || BC
So, ΔAXY ∼ ΔABC
`=> (AY)/(YC) = 9/4.5`
`=> (AY)/(YC) = 2/1`
`=> (AY)/(YC) = 2`
ii. Given that XY || BC
So, ΔAXY ∼ ΔABC
`=> (AX)/(AB) = (AY)/(AC)`
`=> (AY)/(AC) = 1/(2 + 1)`
`=> (YC)/(AC) = 1/3`
iii. In ΔAXY and ΔABC,
∠XAY = ∠BAC ...(Common angle)
∠AXY = ∠ABC ...(Corresponding angles for parallel lines, XY || BC)
∠AYX = ∠ACB ...(Corresponding angles for parallel lines, XY || BC)
Thus, ΔAXY ∼ ΔABC
Hence, `(AX)/(AB) = (XY)/(BC)` ...(Using similar triangle property)
`(AX)/(AX + XB) = (XY)/18`
`9/(9 + 4.5) = (XY)/18`
`XY = (18 xx 9)/(13.5)`
XY = 12 cm
APPEARS IN
संबंधित प्रश्न
State, true or false:
Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.
State, true or false:
The diagonals of a trapezium divide each other into proportional segments.
In quadrilateral ABCD, diagonals AC and BD intersect at point E such that
AE : EC = BE : ED. Show that: ABCD is a trapezium.
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E. Prove that: EL = 2BL.
In the following figure, DE || AC and DC || AP. Prove that : `(BE)/(EC) = (BC)/(CP)`.
In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate : AC
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
ΔADE ~ ΔACB.
Triangles ABC and DEF are similar.
If area (ΔABC) = 36 cm2, area (ΔDEF) = 64 cm2 and DE = 6.2 cm, find AB.
In figure ABC and DBC are two triangles on the same base BC. Prove that
`"Area (ΔABC)"/"Area (ΔDBC)" = "AO"/"DO"`.
In the adjoining figure, ΔACB ∼ ∆APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm find the area (∆ACB) : area (∆APQ).