Page images
PDF
EPUB

,9180PROBLEM XI.YTOJ

Given the perimeter of a right angled triangle, and the radius of the inscribed circle; to find the sides of the triangle.

[blocks in formation]

[merged small][ocr errors][merged small][merged small][merged small][ocr errors]

Prepare the figure, for solution, by drawing the lines AO, BO, and co; then AE=AD, CE=CF, and Bo, and

BDBF.

kui 943 16 2pibus Then put the given perimeter=p, the radius of the inscribed circle=r, AE or its equal AD, and CE or CF=y.

By the Problem,

AD+DO=AB, or x+r=AB;

AE+EC AC, or x+y=ac;

081 A

ན ན ོག 1:|: :ས

and CF+FOBC, or y+r=BC.3472titedge2 .. x + r + x + y + y+rp, or 2x+2y+2r=p;

1

by transposition, &c. x+y=&p—r;

By Euc. 47. 1. (x+r)2+(y+r)2=(x+y)3,
or x2+2rx+r+y2+2yr+r2=x2+2xy + y2,
by transposition and division,
r(x+y)=xy

Substituting the value of r in this equation,

[ocr errors]

܂

completing the square,

y2 —(\pr) y+\({p_r)2=\(\px)2iprino extracting the square root, to eurban sit

[ocr errors]

Я

y—ž({p—r)=F√‡({R—r)

[ocr errors]

-pr

[ocr errors]

.. y=F√4(±p− r)2—{pr+\({p—r),
and by a similar operation,"

[ocr errors]
[ocr errors]
[ocr errors]

AB= ± √ ÷ ({p—r)2 — {pr+{({p—r)+r.

and BC({p—r)2—\pr+≥(žp−r)+r.

[ocr errors]
[ocr errors]
[merged small][ocr errors]

If the perimeter of the triangle be 120, and the radius of the inscribed circle 10.

[ocr errors]

then +10+y+10+a+y=120; chemi

[ocr errors]

x+y=50; and x=50-y.

Also, (x+10)+(y+10)?(x+y),

[ocr errors]

or x2+20x+100+ y2+20y+100=x+2xy+y2;
.. 20x+20y=2xy-200.2

Substituting the value of a in this equation,
q-1000-20y+20y=100y 2y200;

y2-50y=-600;

completing the square y2-50y+625-625-600-25 extracting the root y-25=5;

.. y=20, and r=30; 2

Whence AB=40, BC=30 and ▲c=50.

PROBLEM XII.

10

If a globe o of two inches diameter be suspended

against an upright wall, at what distance from the wall must a lighted candle be placed, so that the

enlightened part of the surface of the globe, may be equal to the area of the shadow on the wall; the light and centre of the globe being in the same hōrizontal line?

[blocks in formation]

ax-2a2

[ocr errors]

consequently the area of the enlightened part

of the globe EFD=8p.(a2a—20

Again CD2-2ax; then by similar triangles

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]

1. The base of a right angled triangle is 36, and the difference between the perpendicular and hypothenuse 18. Required the area of the triangle.

Ans. 486.

2. The difference between the base and hypothenuse of a right angled triangle is 9, and the difference

between the hypothenuse and perpendicular 18. Required the sides of the triangle.

[ocr errors]

Ans. Base 36, perpendicular 27, and hypothenuse 45.

3. The difference between the base and perpendicular of a right angled triangle is 9, and the perpendicular let fall from the right angle upon the hypothenuse 21.6. Required the sides of the triangle. Ans. 27. 36, and 45.

4. The base of a right angled triangle is 40, and the alternate segment of the hypothenuse, made by a perpendicular from the right angle upon the hypothenuse, 18. Required the sides of the triangle.

5. The perimeter of is 144, and its area 864.

Ans. 30, 40, and 50. a right angled triangle Required the sides

Ans. 36, 48, and 60.

6. The hypothenuse of a right angled triangle is 50, and the perpendicular let fall from the right angle upon the hypothenuse 24. Required the segments made by this perpendicular.

Ans. 18, and 32.

7. In a right angled triangle the sum of the greater segment, made by a perpendicular let fall from the right angle upon the hypothenuse, and the adjacent side is 72. Required the sides of the triangle.

Ans. 30, 40, and 50.

8. The base of a right angled triangle is 40, and the perpendicular 30. Required the length and breadth of that inscribed rectangle which takes up one-half of the area of the triangle.

Ans. 20, and 15.

9. The base of a plain triangle is 30, and the perpendicular 20. Required the area of the inscribed

square.

Ans. 144,

P

f

10. The diameter of a semicircle is 20, and the breadth of the inscribed rectangle 6. Required the area of the rectangle.

Ans. 96.

11. The diameter of a semicircle is 10; required the length and breadth of that inscribed rectangle, whose area is equal to one third of the area of the semicircle.

[ocr errors]

Ans. 9.6227 and 1.36035.

12. The diameter of a semicircle is 25, it is required to find a point in its arc, such, that if lines be drawn from the extremities of the diameter to that point, the triangle so formed shall be equal to one-half of the semicircle.

Ans. 11.2901 from the extremity of the diameter.

13. In an isosceles triangle two circles, whose diameters are 8 and 12, touching each other and the sides of the triangle. Required the two equal sides. Ans. 36.7423 each of the equal sides.

14. Six equal circles, each two inches in diameter, are inscribed in an equilateral triangle, touching each other and the sides of the triangle. Required each side of the triangle.

[ocr errors]

Ans. 7.4641016.

15. The three lines drawn from the angles of a plain triangle to the middle of the opposite sides are 18, 24, and 30. Required the sides of the triangle. Ans. 34.176; 28.844; and 20.

16. The difference of the sides of a plain triangle is 405, the difference of the segments of the base 495, and the difference between the base and greater side 165. Required the sides of the triangle.

Ans. Base 945, greater side 780, and less side 375. 17. The difference of the sides of a plain triangle is 405, the difference of the segments of the base 495,

« PreviousContinue »