Q10 of 77 Page 15

In Fig. 15.82, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, Show that ar(Δ ABC) = ar(Δ ABD).

Given that,

CD bisected AB at O


To prove: Area () = Area ()


Construction: CP perpendicular to AB and DQ perpendicular to AB


Proof: Area ( = (AB * CP) (i)


Area ( = (AB * DQ) (ii)


In , we have


∠CPO = ∠DQO (Each 90o)


Given that,


CO = DO


∠COP = ∠DOQ (Vertically opposite angle)


Then, by AAS congruence rule



Therefore,


CP = DQ (By c.p.c.t)


Thus,


Area ( = Area ()


Hence, proved


More from this chapter

All 77 →
8

In Fig. 15.81, ABCD and CDEF are parallelograms. Prove that

ar(Δ ADE) = ar(Δ BCF).


 9

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:

ar(Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(Δ BPC).

 11

If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.

 12

If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar(Δ BGC) = 2ar(Δ AGC).