Home
Class 10
MATHS
In the adjoing figure, squareABCD is a s...

In the adjoing figure, `squareABCD` is a square. `DeltaBCE`on side BC and `DeltaACF` on the diagonal AC are similar to each other. Then, show that `A(DeltaBCE)=(1)/(2)A(DeltaACF)`

Text Solution

Verified by Experts

The correct Answer is:
`thereforeA(DeltaBCE)=(1)/(2)A(DeltaACF)`
Promotional Banner

Topper's Solved these Questions

  • CHALLENGING QUESTIONS

    CHETAN PUBLICATION|Exercise Theorem of Pythagoras|7 Videos

  • CHALLENGING QUESTIONS

    CHETAN PUBLICATION|Exercise Circle|9 Videos

  • CHALLENGING QUESTIONS

    CHETAN PUBLICATION|Exercise Statistics|4 Videos

  • ARITHMETIC PROGRESSION

    CHETAN PUBLICATION|Exercise ASSIGENMENT -3|10 Videos

  • CIRCLE

    CHETAN PUBLICATION|Exercise Assignment - 3 (Solve any two of the following questions):|3 Videos

Similar Questions

Explore conceptually related problems

In squareABCD, "seg " AD||"seg " BC. Diagonal AC and diagonal BC intersect each other in point P. Then show that (AP)/(PD)=(PC)/(BP)

In a quadrilateral ABCD, it is given that AB |\|CD and the diagonals AC and BD are perpendicular to each other. Show that AD.BC >= AB. CD .

squareABCD is a trapezium in which AB|| DC and its diagonals intersect each other at points O . Show that AO : BO = CO : DO.

In the figure, ABCD is a quadrilateral. AC is the diagonal and DE || AC and also DE meets BC produced at E. Show that ar(ABCD) = ar (DeltaABE).

In squareABCD," side "BC|| side AD. Diagonal AC and diagonal BD intersects in point Q. If AQ=(1)/(3)AC, then show that DQ=(1)/(2)BQ.

Show that the diagonals of a square are equal and right bisectors of each other.

Attempt any Two of the following: In squareABCD ,seg AB ||seg CD . Diagonal AC and BD intersect each other at point P . Prove : (A(DeltaABP))/(A(DeltaCPD))=(AB^(2))/(CD^(2))

D is a point on side BC of DeltaABC such that , angleADC=angleBAC . Show that AC^(2)= BCxxDC .

In the adjacent figure, we have BX = 1/2 AB, BY = 1/2 BC and AB=BC. Show that BX = BY.

In the adjoining figure, seg BD bot "side" AC, C-D-A. "Prove that" : AB^(2) = BC^(2) + AC^(2) - BC.AC