THE APPLICATION OF ALGEBRA, &c. 159

1.

PROBLEM FITA 31T

Given the base of a right angled triangle 4 chains and the sum of the hypothenuse and perpendicular 8, chains. Required the area of the triangle.

In the right angled triangle D

A B C, let x=

the perpendngle

BC; then 8-x- the hypothe

nuse A C;

B

andog 2. ant. (Eu. 47, 1.) AB2+BCAC2, 1910 al old to Batent or 16+2=64–162fa Wiki by transposition, 16x=48; 3;sldong whence, the area of the triangle is 6 square chains, or 2r. 16p. did pot; te olavorder 503. udT is nomos sut vd OR PROBLEM II. #1 g of daizes adt The base of a right angled triangle is 4 chains, and hgion ou the difference between the hypothenuse and perpendi cular 2 chains. Required the hypothenuse and perpendicular, bookes na yem afoul borter doudw the perpendicular

In the above figure, let in BC then +2 the hypothenuse Acgo, ol way. (Ed. 47, 1.) AC2=AB+BC2, or x2+4x+4= natc964x2;

by transposition, 4x=12;.. x=3;

whence, t

the hypothenuse is 5, and perpendicular 3.

PROBLEM III aur vd tou yam Ji The diagonal of a rectangle is 50, and the sum of the four sides 140. Required its area, bedtim jedt viq In the above figure, let x = AB, the length; then 70-x=BC, the breadth;

.. (Eu. 47, 1.) AB2+BC2=AC2,

01 697) or x2+4900—140x+x2—2500;2 od stc18 by transposition, 22-140x2400, this equal tion solved gives r—40';

whence, its area is 1200.orɔɔnq asvig tod‡u A sďI

PROBLEM IV.

The area of a right angled triangle is 600, and the length and breadth of the inscribed rectangle 24 and 8. Required the sides of the triangle.

24x x

=AB;

then,600, or x2-50x400, which

ad or equation being solved, gives x=40.

Also, by similar triangles, FC DFBCVAB, or 32: 24: 401; 30;

whence, the three sides are 30, 40, and 50.

PROBLEM V.

The base of a plane triangle is 40, its area 400, and the area of the inscribed rectangle 150. Required the sides of the rectangle.

Let ABC be the triangle, and abcd the inscribed rectangle. Let x ab or cd; then by similar triangles, (Eu. vi. 4.) AB cd Sc: Cr;

then x. (20)=150;., x2-40x=-300;

PROBLEM VI.

The area of an isosceles triangle is 675, and the sum of its equal sides is equal to the sum of the base. and perpendicular. Required the sides and perpendicular of the triangle.

4 then ry=675 ; and √x2 + y2=AC or BC, one of

the equal sides; ·1 2√x2 + y2=2x+y;

... /

squaring each side, 4x2+4y2=4x2+4xy+y2

doud by transposition, &c. 3y=4x.

These equations being solved, is found to be 22.5 and y=30;

Whence the base is 45, the perpendicular 30, and each of the equal sides 37.5. acnltd somew

PROBLEM VII.

Given the segments of the base, made by the

per

pendicular from the verticle angle upon the base of a

plane triangle, 20 and 15; and the ratio of the sides as

5 to 3. Required the sides.

C

Let AD be the greater segment, his bodo and BD the less; put x=AC; no

A

Bra

.. (Eu. 47. 1) AC-AD-DC-BC2- BD2,

25

this equation being solved, x is found 23.0

22.0479276. Whenee AC-22.0479276 and BC-17.63834208.

24 ↑ PROBLEM VIÍI.

Given the base 20, the sum of the sides 26, and the line drawn from the verticle angle to the middle of the base 8.5 to determine the triangle,

Let AB represent the base, Ac and BC the two sides, and CD the line drawn from the vertical angle to the middle of the base. Put AC; then 26-x=BC.

C

4

A

B

D

Then, by Bonnycastle's Euclid or Geometry, Book

2. Prop. 13,

#AGM pen?" AC2+BC2=2CD2+2AD",

dor467652x+x=144:5+200;

by transposition, &c. x2-26x—— 165.75 this equation solved, x=14.80277564. and BC= 11.19722436.

PROBLEM IX.

5. al Given the sides and the line besecting the vertical angle, of any plane triangle, to find the base. to land N. B. The above figure being similar to the ones required for this question, the Author has conlo sidered it the real figure in the following solution..oura

J

"

Put the side Ac=a; BC=b; CD, (the line which besects the vertical angle ACB)=c, and the baseT AB=a; then, by (Euc. 3. vi.)

Also, by (Euc. 13. VI.) AD X DB + DC2=AC × BC,

then, abx2=(a+b)2 × (ab—c2); æft og han

A Carpenter's Apprentice applied to his Master to know the radius he must take to strike the are of a centre for the top of a circular door, its width being forty-eight inches and the arch to rise fifteen inches. The master not being able to inform him, he begs that the young Algebraist will give a general solution to the above, so as he may in future be enabled to find the radius of any circular centre for an arch.

A

In the annexed figure, let AB denote the width of the door, PC the height of the arch or centre, ACB the arch or centre, and Dc the diameter oo of the circle, of which the arch or circular door top ACB is a part..

C

B

Put AB 2a, PC b and DC. hi, ada ap¶ by the property of the circle, Euca

Then PD=x+b;

35. III.

Then substituting the values of a and b, in the 242+152 576+225

equation x=

53.4 the diameter of

the circle, half of which is 26.7➡oc, the radius required.

NOTE. This question is similar to question 42, page 74.

A