393. To construct a parallelogram equivalent given square, and having the difference of its and altitude equal to a given line. Let R be the given square, and let the differe the base and altitude of the required parallelo be equal to the given line MN. To construct a☐ equivalent to R, with the difference base and altitude equal to MN. Construction. Upon the given line MN as a diameter, de a circle. From M draw MS, tangent to the O, and equal to of the given square R. Through the centre of the O draw SB intersecting th cumference at C and B. Then any □, as R', having SB for its base and SC f altitude, is equivalent to R. (if from a point without a O a secant and a tangent are drawn, the tan a mean proportional between the whole secant and the part without t Then SMSBX SC, and the difference between SB and SC is the diameter O, that is, MN. NOTE. This problem may be stated: To construct two straigh the difference and product of which are known 394. To construct a polygon similar to a given polygon P, and equivalent to a given polygon Q. 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 § 351 and let m and n respectively denote their sides. Find A'B', a fourth proportional to m, n, and AB. Upon A'B', homologous to AB, construct P' similar to P. Then P' is the polygon required. (similar polygons are to each other as the squares of their homologous sides) .. P' is equivalent to Q, and is similar to P by construction Ex. 307. To find the area of an equilateral triangle in terms side. Denote the side by a, the altitude by h, and the area by S. Ex. 308. To find the area of a triangle in terms of its sides. Ex. 309. To find the area of a triangle in terms of the radius circumscribing circle. If R denote the radius of the circumscribing circle, and h the a of the triangle, we have, by Ex. 222, NOTE. The radius of the circumscribing circle is equal to abc 4 S 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. The area of a circumscribed polygon is equal to half the product of its perimeter by the radius of the inscribed circle. 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 D the trionala RDD is canivelent to one half the triangle ARO 324. Find the area of a rhombus, if the sum of its diagonals is and their ratio is 3; 5. 325. Find the area of an isosceles right triangle if the hyp is 20 feet. 326. In a right triangle, the hypotenuse is 13 feet, one leg i Find the area. 8. 327. Find the area of an isosceles triangle if the base = b, and 328. Find the area of an equilateral triangle if one side 329. Find the area of an equilateral triangle if the altitude = 330. A house is 40 feet long, 30 feet wide, 25 feet high to t and 35 feet high to the ridge-pole. Find the number of square its entire exterior surface. 331. The sides of a right triangle are as 3:4:5. The altitud the hypotenuse is 12 feet. Find the area. 332. Find the area of a right triangle if one leg = a, and the upon the hypotenuse = h. 333. Find the area of a triangle if the lengths of the sides feet, 111 feet, and 175 feet. 334. The area of a trapezoid is 700 square feet. The bases are and 40 feet respectively. Find the distance between the bases. 335. ABCD is a trapezium; AB=87 feet, BC= 119 feet, feet, DA 169 feet, AC 200 feet. Find the area. 336. What is the area of a quadrilateral circumscribed about whose radius is 25 feet, if the perimeter of the quadrilateral is 4 What is the area of a hexagon having an equal perimeter and scribed about the same circle? 337. The base of a triangle is 15 feet, and its altitude is 8 feet the perimeter of an equivalent rhombus if the altitude is 6 feet. 338. Upon the diagonal of a rectangle 24 feet by 10 feet a equivalent to the rectangle is constructed. What is its altitude? 339. Find the side of a square equivalent to a trapezoid whos are 56 feet and 44 feet, and each leg is 10 feet. 340. Through a point P in the side AB of a triangle ABC, a drawn parallel to BC, and so as to divide the triangle into two lent parts. Find the value of APin terms of AR |