In Fig. 9.15, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
In Fig. 9.17, PQRS, and ABRS are parallelograms and X is any point on side BR. Show that
ar (PQRS) = ar (ABRS)
ar (AXS) =1/2 ar (PQRS)
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
D, E, and F are respectively the mid-points of the sides BC, CA, and AB of a ΔABC.
(i) BDEF is a parallelogram.
(ii) ar(DEF) = 1/4 ar(ABC)
(iii) ar (BDEF) =1/2 ar(ABC)
D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC).
Prove that DE || BC.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that
ar (AOD) = ar (BOC).
In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
(i) ar(△ACB) = ar(△ACF)
(ii) ar(AEDF) = ar(ABCDE)
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given an equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (△ADX) = ar (△ACY).
[Hint: Join CX.]
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(△AOD) = ar(△BOC). Prove that ABCD is a trapezium.
In Fig. 9.32, ABCD is a parallelogram, and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).
[Hint: Join AC.]
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint: From A and C, draw perpendiculars to BD.]
In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA, and AB respectively. Line segment AX ^ DE meets BC at Y.
(i) ΔMBC ≅ ΔABD
(ii) ar(BYXD) = 2ar(MBC)
(iii) ar(BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2ar(FCB)
(vi) ar(CYXE) = ar(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof
of this theorem in Class X.
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