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B = D (opposite angles of a parallelogram)
But B + D = 180^o (opposite angles of a cyclic quad. are supplementary)
B = D = 90^o
We know that, a parallelogram with one angle as 90^o is a rectangle.
So, ABCD is a rectangle.
AC = BD (diagonals of a rectangle are equal)

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Since 5,6 are the only numbers ina die greater than 4.

HenceP(>4)=possible outcomes/total numberof outcomes

=2/6=1/3which is the probability

To write it as a linear equation in two variables, 

0.y + 2.x =5 

Solution :

You can take any values of y but you will always get x= 5/2  

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cone cylinder hemisphere sphere or frustum

The central angle in a semicirle is 180^o since it is a straight line. Inscribed angle is half of the central angle so 1/2 of 180^o is 90^o. 

Hence angle in a semicircle is 90^o.

Ans. When y Varies directly as x, the eqaution will be in form of 

y = kx, where k is constant of proportionality.

Putting y = 12 and x = 4, we get 

=> 12 = 4k 

=> k = 3 

So, y = 3x 

Linear Equation : y - 3x = 0

When x = 5 

y - 3*5 = 0

=> y = 15 

Ans. Degree Measure Theorem states that angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the circle.

Given : An arc PQ of a circle C with center O and radius r with a point R in arc \(\stackrel\frown{QP}\) other than P or Q.To Prove : \(\angle POQ = 2\angle PRQ\)
Construction : Join RO and draw the ray ROM.

Proof : There will be three cases as

(i) \(\stackrel\frown{PQ}\)is a minor arc

(ii)\(\stackrel\frown{PQ}\) is a semi-circle

(iii)\(\stackrel\frown{PQ}\) is a major arc

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In each of these three cases, the exterior angle of a triangle is equal to the sum of two interior opposite angles.

Therefore, \(\angle POM = \angle PRO + \angle RPO \)  ............ (1) 

\(\angle MOQ = \angle ORQ + \angle RQO \).......... (2)  

\(In\space \triangle OPR \space and \space \triangle OQR \)

Now, OP = OR and OR = OQ  (radii of the same circle)

\(\angle PRO = \angle RPO\) and \(\angle ORQ = \angle RQO \)  (angles opposite to the equal sides are equal)

Hence, \(\angle POM = 2\angle PRO\)      .......... (3)

And \(\angle MOQ = 2 \angle ORQ\)     ............ (4)

Case (1) : adding equations (3) and (4) we get

\(\angle POM + \angle MOQ = 2\angle PRO + 2\angle ORQ \)

\(\angle POQ = 2(\angle PRO + \angle ORQ) = 2 \angle PRQ \)

\(\angle POQ = 2\angle PRQ\)

Hence Proved.

Similarly, you can proceed for case (ii) and case (iii)

CSA of hollow cylinder ÷ 2πh(R + r)

TSA of hollow cylinder = 2π(R + r)(h + R - r)

The circle circumscribing the rectangle will have diameter along the diagonal of the rectangle and centre as the midpoint of the diagonal. Since the angle subtended by diameter at any point above or below it is 90 degree and here it is the interior angle of the rectange . THerefore radius = 10/2 = 5cm

Area = \(\pi 5^2 = 25\pi cm^2\)

Ans.

Given : ABCD is a trapezium with AB || DC, DC = 30cm , AB = 50cm

To Prove:  \(ar(DCYX)={7\over 9} ar(XYBA)\)
Construction: Join DY and produce it to meet AB produced at P.

Proof : \(In\space \triangle BYP \space and\space \triangle CYD \)

\(\angle BYP = \angle CYD\)      (vertically opposite angles)

\(\angle DCY = \angle PBY\)      (Alternate opposite angles as DC || AP and BC is the transversal)

BY = CY (Y is the mid point of BC)

Thus \(\triangle BYP \cong \triangle CYD\) (by ASA criterion)

=>  DY = YP and DC = BP    [by CPCT]  ......... (2)

=> Y is the mid point of AD

XY || AP and XY = \({1\over 2}\)AP   [Mid-point theorem]     ........ (2)

=> AP = AB + BP 

=> AP = AB + DC   [From (1)]

=> AP = 50 + 30 = 80 

\(=> XY = {1\over 2} \times 80 = 40 cm\)

Since X and Y are the mid points of AD and BC respectively.
Trapezium DCYX and ABYX are of same height, say h cm
Now, 

\(=> {ar(DCYX) \over ar(XYBA)} = {{1\over 2} (DC+XY)\times h \over {1\over 2} (AB+XY)\times h} = {30+40\over 50+40} = {70\over 90} = {7\over 9}\)

\(=> {ar(DCYX) ={7\over 9} ar(XYBA)} \)

Hence Proved

Ans.

Given: ABCD is a parallelogram in which points P and Q trisects the BC.

\(=> BP = PQ = QC  = {1\over 3}BC\)

To prove: \(ar (\triangle APQ) = ar (\triangle DPQ) = {1\over 6} ar(\|gm \space ABCD)\)

Construction : Through P and Q, draw PR and QS parallel to AB. Now PQRS is a parallelogram and its base \( PQ = {1\over 3} BC\)

Proof : 

Now, \(\triangle APD \space and \space \triangle AQD\) lie on the same base AD and between the same parallel AD and BC.
\(ar (\triangle APD) \space = ar \space( \triangle AQD)\)        .......... (1)

Subtracting \(ar (\triangle AOD)\) from both sides, we get

\(=> ar (\triangle APD) - ar (\triangle AOD) \space = ar \space( \triangle AQD) -ar (\triangle AOD) \)

\(=> ar (\triangle APO) = ar (\triangle OQD) \)      ............... (2)

Adding \(ar (\triangle OPQ)\) on both sides in (2), we get

\(=> ar (\triangle APO) + ar (\triangle OPQ) = ar (\triangle OQD) + ar (\triangle OPQ)\)

\(=> ar (\triangle APQ) = ar (\triangle DPQ)\)    ....... (3)

Again, triangle APQ and parallelogram PQSR are on the same base PQ and between same parallels PQ and AD

\(ar (\triangle APQ) ={1\over2} ar (\|gm \space PQRS)\)   ........ (4)

Now, 

\({ar(\| gm \space ABCD )\over ar(\| gm \space PQRS )} = {BC \times height\over PQ\times height } = {3PQ\over PQ} = 3\)

\(=> ar(\| gm \space PQRS ) = {1\over 3} ar(\| gm \space ABCD ) \)     ....... (5)

\(=> ar (\triangle APQ) ={1\over2} \times {1\over 3}ar (\|gm \space ABCD)\)

\(=> ar (\triangle APQ) ={1\over6}ar (\|gm \space ABCD)\)   ....... (6)

From (3) and (6)

\(=> ar (\triangle APQ) = ar (\triangle DPQ)={1\over6}ar (\|gm \space ABCD)\)

Hence Proved

Ans. \(Area \space of \space Rhombus \space = {1\over 2} (d_1\times d_2)\)

Where d₁ and d₂ are length of two diagonals.

Ans. Depth/Height of Cylinderical Vessel = 10m

Total Cost =Rs. 2200

Rate of Painting =Rs 20/m²

i) Inner Curved Surface Area Area= 2200/20 = 110m²

ii) As we know

Curved surface area = \(2\pi rh\)

\(=> 2\times {22\over 7}\times r\times 10 = 110\)

\(=> r = {110 \times 7 \over 2\times 22\times 10} = {7\over 4}\)

=> r = 1.75m

Ans. Radius of Cone = 4x

Slant Height of Cone = 7x 

Curved Surface Area of Cone = 792 cm²

We Know

\(CSA \space of \space Cone = \pi r l \)

\(=> 792 = {22\over 7 } \times 4x \times 7x \)

\(=> 4x^2 = 36 \)

\(=> x^2 = 9\)

\(=> x = 3 \)

i) Radius of Cone = \(4\times 3 = 12 cm\)

Slant Height of Cone = \(7\times 3 = 21 cm\)

ii) Enviornment Care, Creativity

r : l = 4 : 7 

Let r = 4x

    l = 7x 

Curved Surface Area = 792

                        π r l = 792 

        \(22/7\) × 4x × 7x = 792

                       x² = \(792/22 × 4\)

                             x = 3

                      radius = 4 × 3 = 12 cm

        slant height = 7 × 3 = 21 cm