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PROPOSITION XX. PROBLEM.

392. To construct a parallelogram equivalent to a given square, and having the sum of its base and altitude equal to a given line.

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Let R be the given square, and let the sum of the base and altitude of the required parallelogram be equal to the given line MN.

To construct a equivalent to R, and having the sum of its base and altitude equal to MN.

Construction. Upon MN as a diameter, describe a semicircle. At Merect a MP, equal to a side of the given square R. Draw PQ to MN, cutting the circumference at S.

Draw SCL to MN.

Any having CM for its altitude and CN for its base is equivalent to R.

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(a let fall from any point in the circumference to the diameter is a mean proportional between the segments of the diameter).

Then

SC2 MCX CN.

Q. E. F.

NOTE. This problem may be stated: To construct two straight lines the sum and product of which are known.

PROPOSITION XXI. PROBLEM.

393. To construct a parallelogram equivalent to a given square, and having the difference of its base and altitude equal to a given line.

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Let R be the given square, and let the difference of the base and altitude of the required parallelogram be equal to the given line MN.

To construct a☐ equivalent to R, with the difference of the base and altitude equal to MN.

Construction. Upon the given line MN as a diameter, describe a circle.

From M draw MS, tangent to the O, and equal to a side of the given square R.

Through the centre of the O draw SB intersecting the circumference at Cand B.

Then any □, as R', having SB for its base and SC for its altitude, is equivalent to R.

Proof.

SB: SM=SM: SC,

§ 348 (if from a point without a O a secant and a tangent are drawn, the tangent is a mean proportional between the whole secant and the part without the O).

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and the difference between SB and SC is the diameter of the

O, that is, MN.

Q. E. F.

NOTE. This problem may be stated: To construct two straight lines the difference and product of which are known.

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394. To construct a polygon similar to a given polygon P, and equivalent to a given polygon Q.

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Let P and Q be two polygons, and AB a side of P. To construct a polygon similar to P and equivalent to Q. Construction. Find squares equivalent to P and Q, $ 391 and let m and n respectively denote their sides.

Find A'B', a fourth proportional to m, n, and AB. § 351 Upon A'B', homologous to AB, construct P' similar to P. Then P' is the polygon required.

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(similar polygons are to each other as the squares of their homologous sides).

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.. P' is equivalent to Q, and is similar to P by construction.

Q. E. F.

PROBLEMS OF COMPUTATION.

Ex. 307. To find the area of an equilateral triangle in terms of its side.

Denote the side by a, the altitude by h, and the area by S.

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Ex. 308. To find the area of a triangle in terms of its sides.

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Ex. 309. To find the area of a triangle in terms of the radius of the circumscribing circle.

If R denote the radius of the circumscribing circle, and h the altitude of the triangle, we have, by Ex. 222,

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NOTE. The radius of the circumscribing circle is equal to

abc

4 S

THEOREMS.

310. In a right triangle the product of the legs is equal to the product of the hypotenuse and the perpendicular drawn to the hypotenuse from the vertex of the right angle.

311. If ABC is a right triangle, C the vertex of the right angle, BD a line cutting AC in D, then BD2 + AC2 = AB2 + DC2.

312. Upon the sides of a right triangle as homologous sides three similar polygons are constructed. Prove that the polygon upon the hypotenuse is equivalent to the sum of the polygons upon the legs. 313. Two isosceles triangles are equivalent if their legs are equal each to each, and the altitude of one is equal to half the base of the other. 314. Three times the square of one side of an equilateral triangle is equal to four times the square of the altitude.

315. Two parallelograms are equivalent if two adjacent sides of the one are equal respectively to two adjacent sides of the other, and the included angles are supplementary.

316. Every straight line drawn through the centre of a parallelogram divides it into two equivalent parts.

317. If the middle points of two adjacent sides of a parallelogram are joined, a triangle is formed which is equivalent to one-eighth of the entire parallelogram.

318. If any point within a parallelogram is joined to the four vertices, the sum of either pair of triangles having parallel bases is equivalent to one-half the parallelogram.

319. The line which joins the middle points of the bases of a trapezoid divides the trapezoid into two equivalent parts.

320. The area of a trapezoid is equal to the product of one of the legs and the distance from this leg to the middle point of the other leg.

321. The lines joining the middle point of the diagonal of a quadrilateral to the opposite vertices divide the quadrilateral into two equivalent parts.

322. The figure whose vertices are the middle points of any quadrilateral is equivalent to one-half of the quadrilateral.

323. ABC is a triangle, M the middle point of AB, P any point in AB between A and M. If MD is drawn parallel to CP, and meeting BC at D, the triangle RPD is equivalent to one-half the triangle ABC.

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