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Sia ? 5 years, 1 month ago
Given : ABCD is a parallelogram.
BC=CQ.
To prove :
ar(ΔBCP)=ar(ΔDPQ)
Construction :
Join AC
Proof :
ar(ΔBPC) = ar (ΔAPC) [Triangles on the same base and between same parallels ]
Similarly ,
ar(ΔADC) = ar(ΔADQ) [Triangles on the same base and between same parallels ]
ar(ΔADC) = ar(ΔADP) +ar(ΔAPC)
ar(ΔADQ) = ar(ΔADP) + ar(ΔDPQ)
ar(ΔADC) = ar(ΔADQ) and ,
ar(ΔADP) is common.
{tex}\therefore{/tex} ar(ΔAPC) = ar(ΔDPQ)
But ,
ar(ΔAPC) = ar(ΔBPC)
∴ ar(ΔBPC) = ar(ΔDPQ)
Hence ,proved.
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Yogita Ingle 5 years, 1 month ago
Base: ?
Area: 12 cm sq
side 1: 5 cm
The formula of area is :
1/2 × base × height
let the base be 2b and height be h.
b × h= 12
h= 12/b
use Pythagoras theorem:
b²+h²= 5²
so b²+ 12²/b²= 25
so next your equation will be: b( power 4) - 25b²+144=0
So next (b + 4)(b – 4)(b + 3)(b – 3) = 0
Neglecting the negative values, b could be 3 giving h = 4
or b could be 4 giving h = 3
This means the whole base could be 6 and the height is 4
OR the whole base could be 8 and the height is 3
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Gaurav Seth 5 years, 1 month ago
As ABCD is a square, so
AB = BC = CD = DA ...(1)
Also
AK = BL = CM = DN .....(2)
Subtracting (2) from (1)
We get
AB- AK = BC - BL = CD - CM = DA - DN
BK = CL = DM = AN ...(3)
So now we have
AK = BL = CM = DN
AN = BK= CL = DM
Squaring and adding
AK² + AN² = BL² + BK²= CM² + CL² = DN² + DM² ...(4)
But <A = <B = <C = <D = 90°
By Pythagorean theorem (4) Becomes
KN² = KL² = LM² = NM²
So
KN = KL = LM = NM
So KLMN is a rhombus
But <1 = <3 as triangles are congruent
And < 1 + <2 = 90
So <2 + <3 = 90
Hence <KNM = 90°
Therefore KLMN is a square.
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