Price/Cost of a pen = Rs X

Price/Cost of a copy = Rs y

Total cost = 3 X + Y = Rs 110

a pen costs Rs 20. => X = Rs 20

So Y = 110 - 3 * 20 = Rs 50.

cost of a copy = Rs 50.

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see diagram.

When we join the mid points of the sides of the triangle DEF, we have the triangle DEF divided into 4 equal triangles. Thus the triangles ABC, BAF, AEC and BCD are all identical congruent triangles. So area of the triangle BAF = 24 cm² = 1/4 th of area of triangle DEF.

Area of parallelogram AFBC = 2 * area of triangle BAF = 48 cm²

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see diagram.

AC and BD are chords bisecting each other at O.

Hence, AO = OC and BO = OD. Also, the angles AOB and COD are same (vertical angles at O). Hence, the two triangles AOB and COD are congruent.

Hence, AB = CD.

Similarly, in triangles AOD and BOC, the angles AOD = angle BOC (vertical angles). As, AO = OC and BO = OD, the two triangles are congruent.

Hence, AD = BC.

The two triangles ACB and ADB are congruent as all the three corresponding sides are equal to one another.

So angle D = angle B.

Similarly, angle C = angle A.

In a cyclic quadrilateral ABCD, the opposite angles are supplementary. Hence, angle B = angle D = 90°,

same way, angle C = angle A = 90°.

In a circle, if a chord AC (or BD) subtends an angle 90° at a point B (or A) on the circumference, then it is the diameter of the circle.

hence, AC and BD are diameters.

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Draw a line segment SR of length 10 cm horizontally using the ruler. Then using the compass, measure about 6 to 7 cm. With S as the center draw an arc above SR and an arc below SR. With R as the center and with the same radius, draw an arc above SR and an arc below SR , intersecting the earlier arcs drawn.

The points are intersection of arcs can be joined to get the perpendicular bisector of SR. Let it intersect SR at P. So P is the mid point of SR.

Follow the above procedure for the line segments SP and PR. Measure 3 to 4 cm with the compass. Draw arcs with centers S and P above and below SP. Join their points of intersection. Similarly, draw arcs with centers as P and R above and below PR. Join their intersection points.

Join/Draw the perpendicular bisectors for SP and PR. Now the line segment SR is divided into 4 equal parts.

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construct ABC:

draw a line segment horizontally BC of length 3.2 cm.

Now draw a line BD (long one) making the angle CBD = 45°.

We need to find the point A such that AC = AB + 2.1 cm.

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see diagram.

AB || CD. AB and CD are chords of the circle. Let R be the radius of the circle.

AB = 10 cm CD = 24 cm. Center of circle is O. Draw perpendicular bisectors for AB and CD. They will join at center O. Given, EF = EO + OF = 17 cm.

we have, AE = EB = 5 cm. CF = FD = 12 cm.

Using Pythagoras theorem in triangle AEO, EO = √(R² - 5²)

Using Pythagoras theorem in triangle BOF, OF = √(R² - 12²)

Hence, EO + OF = √(R² - 25) + √(R² - 144) = 17

√(R² - 144) = 17 - √(R² - 25)

R² - 144 = 17² + (R² - 25) - 2 * 17 * √(R² - 25)

- 408 = - 2 * 17 * √(R² - 25)

12 = √(R² - 25)

R² = 144+25 = 169

** R = 13 cm.** = radius of the circle.

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