Students and Teachers Forum

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Given, Area of △ABC = 75 cm2  CB = 20 cm AB = ?  Now, Area of △ABC = ( 1 / 2 ) x CB x AB x sinB or, 75 = ( 1 / 2 ) x 20 x AB x sin30o or, 150/20 = AB x ( 1 / 2 ) (∴ sin30o = 1/2) ∴ AB = 15 .....

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Here, ∡PSQ = ∡PRQ = 35o ( angles on the same arc PQ) ​​​​​​​∡PQR = 90o ( inscribed angle made by lines extending from diameter ) In △PRQ x + y + 90o =180o ( sum of interior angles of a triangle .....

In the given circle, Triangle COB is an isosceles triangle because OC = OB ( radius of the circle ) ∡CAB = ∡CDB = 65o ( angles of an isosceles triangle ) ∡CAB + ∡CDB + ∡COB = 180o ( sum of interior angles .....

Here,  Exterior angle AOB + Interior angle AOB = 360o (complete angle) ∴  Exterior angle ∡AOB = 360o - 140o = 220o Now, ∡ACB = ( 1 / 2 ) of Exterior angle ∡AOB ( circumference angle is half of the centre .....

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Here,  PA = PB ( tangents to the circle originating from the same point are equal ) ∴△PAB is an isosceles triangle so, ∡PAB = ∡PBA = y In △PAB, ∡PAB + ∡PBA + ∡APB = 180o ( sum of interior .....

Given, DC = 8 cm DE = 8 cm  ∡CDE = 30o ∴ Area of triangle DCE = ( 1 / 2 ) x DC x DE x sin(∡CDE)                                       .....

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From the figure, OB = OC ( radius of a circle) ∴ OBC is an isosceles triangle i.e ∡OBC =  ∡OCB In ▲OBC, x + 64o + 64o  = 180o (sum of interior angles of a triangle) or, x = 180o - .....

The above figure consists of two shapes: A triangle at left and a rectangle at right. First for the triangle: Base ( b ) = 11 - 9 = 2 cm. Height ( h ) = 6 cm. Area of the triangle = 1/2 x b x h               .....

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Height of pyramid ( hp ) = 15 cm. Height of cuboid ( hc ) = 6 cm Length of cuboid ( lc ) = breadth of cuboid ( bc ) = 12 cm  Volume of upper pyramid ( Vp ) = ( 1 / 3 ) x length x breadth       .....

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