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we know that within two parallel lines.. if we keep the base of triangle fixed. then the triangle formed between those two lines will always have same area... ( height will always be same.)

so ∆APB amd ∆ADB are between the same parallel lines.. and have same base AB SO AREA of ∆ABD = 10cm²

and area of reactangle ABCD = 2*10 = 20cm².

10cm2

any other answers

these are correct answers check it.

kk tnk uu

Since the lengths of tangents drawn from an exterior point to a circle are equal.∴ AP = AR , BP = BQ, CQ = CRNow, AB+CQ = AP+PB+CQ = AR+BQ+CQ (∵ AP=AR and PB=BQ) =(AR+CR)+BQ (∵CQ=CR) = AC+BQ∴ AB+ CQ = AC+BQ

Area (tri . ABC) = Area (tri .AOB) + Area (tri .BOC) + Area (tri . COA)

= 1/2 AB x r + 1/2 BC x r + 1/2 CA x r

= 1/2 r (AB + BC + CA)

-= 1/2 r ( perimeter of triangle ABC )

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2(x-2)=12x-4=12x=1+42x=5x=5/2

y+3=1y=1-3y=-2

I hope your doubt is solved!

धन्यवाद

3457+7654222 answer

Area= length× breath= a×bgiven, a^2+b^2=(34)^2a+b=92/2=46therefore (a+b)^2--2ab=(34)^22ab=(46)^2--(34)^2ab=480Area 480m^2

we knowsin (n ∙ 360° + θ) = sin θsin(765°)sin ( 360 × 2 + 45°)sin (45°)1/√2

यदि किसी शंकु की त्रिज्या तथा ऊंचाई दी गई हो तब उसका आयतन होगा1/3πr^2h

462+(765×588)-574622+449820-57=450225

value of sin 45

the value of sin 45 is 1/√2

4

188.374

Kg Gram 578 139 - 389 765----------- ----------- 188. 374

proceed as in section 1.4 for2.665

what is your age.

also, find the area of rectangle ABCD

in recatngle the side bisect each other soOC = OB In ∆ OCB , let OCB = x44° + x + x = 180°2x = 180-44 = 136x = 136/2 = 68°so OAD = 68°

Yes it's Parallel to it becaUse parallel lines never meet And here line NM is exactly ahead to line RQ so It's obvious that they would never meet.

Ans :- 72283782827×87 =6.289 × 10^12

Step1:- Draw a line segment AB of length 6 cm.Step2:- At B, draw BE perpendicular to AB.Step3:- With B as centre position and radius 5.5 cm draw an arc cutting BE at C.Step4:- With A as centre position and radius 5.5 cm, draw an arc.Step5:- With C as centre position ad radius 6 cm draws another arc cutting the previous arc at D.Step6:- Join AD and CD. Then, ABCD is the required rectangle.

please send the answer with pictures

x-13 = 87

x = 87-13

x = 74

87/√3 = 87√3 * √3/√3 = 29√3

87/root(3) = 29×3/root(3) = 29*root(3)*root(3)/root(3) = 29root(3)

-2.15926676539

87 + 8x = 98x = 9 - 878x = -78x = -78/8x = - 39/4

Yes it's Parallel to it because parallel lines never meet And here line NM is exactly ahead to line RQ.So It's obvious that they would never meet.

Using basic proportionality theorem