135" ar "2225

In a ABC,/_ABC=135o. Prove that AC^(2)=AB^(2)+BC^(2)+4ar(ABC)

If omega (ne 1) is a cube root of unity, then the value of tan[(omega^(2017) + omega^(2225)) pi - pi//3]

What value will come in place of question mark (?) in number-series given below- 144,73, 38.5, 22.25, ?, 12.5625

If the median of a △ABC intersect at G. show that ar (△AGC) = ar (△AGB) = ar (△BGC) = 1/3 ​ ar (△ABC)

Given that HCF (135,225)=45 find the LCM (135,225)

The value of sin135^@-cos135^@ will be

a+ar+ar^2+ . . . + oo=15 a^2+(ar)^2+(ar^2)^2+ . . . + oo=150 . Find ar^2+ar^4+ar^6+. . . oo

IN the figure, P is a point in the interior of a parallelogram ABCD. Show that (i) ar(APB) + ar(PCD) = 1/2 ar (ABCD) (ii) ar (APD) + ar(PBC) = ar(APB) + ar (PCD)

If the medians of a triangleABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)